Research ArticleEquilibrium Customer Strategies in the Single-Server ConstantRetrial Queue with Breakdowns and Repairs

Zhengwu Zhang Jinting Wang and Feng Zhang

Department of Mathematics Beijing Jiaotong University Beijing 100044 China

Correspondence should be addressed to Jinting Wang jtwangbjtueducn

Received 23 September 2013 Accepted 28 November 2013 Published 2 January 2014

Academic Editor Carsten Proppe

Copyright copy 2014 Zhengwu Zhang et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We consider a single-server constant retrial queueing system with a Poisson arrival process and exponential service and retrialtimes in which the server may break down when it is workingThe lifetime of the server is assumed to be exponentially distributedand once the server breaks down it will be sent for repair immediately and the repair time is also exponentially distributed Thereis no waiting space in front of the server and arriving customers decide whether to enter the retrial orbit or to balk depending onthe available information they get upon arrival In the paper Nash equilibrium analysis for customersrsquo joining strategies as well asthe related social and profit maximization problems is investigated We consider separately the partially observable case where anarriving customer knows the state of the server but does not observe the exact number of customers waiting for service and the fullyobservable case where customer gets informed not only about the state of the server but also about the exact number of customersin the orbit Some numerical examples are presented to illustrate the effect of the information levels and several parameters on thecustomersrsquo equilibrium and optimal strategies

1 Introduction

Queueing systems in which arriving customers who findthe server occupied may retry for service after a periodof time are called retrial queues or queues with repeatedorders Among trials a customer is said to be in ldquoorbitrdquo Theretrial queueing systemhas been studied extensively due to itswide applications in telephone switching systems telecom-munication networks and computer networks The retrialqueueing literature is quite extensive For a recent accountthe readers are referred to the books of Falin and Templeton[1] and Artalejo and Gomez-Corral [2] in which they havesummarized the main models and methods thoroughly

In the retrial queueing literature the vast majorityof articles assume that each customer seeks for serviceindependently of other customers in orbit after a randomtime In such cases the total retrial rate of the system dependson the number of retrial customers in orbit Neverthelessin reality there are other types of queueing situations inwhich the repeated customers form a queue in orbit and only

the head customer of the orbit can request a service after arandom retrial time This type of retrial discipline is calledldquoconstant retrial policyrdquo and it arises from some applicationsin the computer and communication networks where theretrial rate is controlled by some automatic mechanismIt was introduced in Fayolle [3] who studied a telephoneexchange model as an MM1 retrial queue in which theorbiting customer who finds the server unavailable joins thetail of the queue Farahmand [4] called this discipline a retrialqueue with FCFS orbit Later on both retrial policies areincorporated by assuming the linear retrial policy introducedin Artalejo and Gomez-Corral [5]

Recently Economou and Kanta [6] studied the equi-librium customer strategies and the social and profit max-imization problems in the constant retrial system Theyinvestigated two information cases the unobservable casewhere the customers know only the state of the server (busyor idle) and the observable case where they also get informedabout the number of customers in the retrial orbit Thecustomersrsquo behavior is taken into account to identify theNash

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 379572 14 pageshttpdxdoiorg1011552014379572

2 Mathematical Problems in Engineering

equilibrium joining strategies It differs from equilibriumanalysis of retrial systems with the classical retrial policy seefor instance Kulkarni [7] Elcan [8] Hassin and Haviv [9]and Zhang et al [10] and Wang et al [11] among othersHowever although we have found several retrial models withservers subject to breakdowns that consider orbits as FCFSqueues for example in Atencia et al [12] and Li and Zhao[10] To the authorsrsquo knowledge no work has been done onsuch queueing systems from an economic view Thus in thispaper we propose to study such an MM1 retrial queuewith the constant retrial policy and serverrsquos breakdowns andrepairs as an extension of the paper considered by Economouand Kanta [6] who have also considered the observablesingle-server queue with breakdowns and repairs see [13]To this end the methodology will be based on the game-theoretic analysis and optimization

This paper is organized as follows In Section 2 we givethe model description and the natural reward-cost structureIn Sections 3 and 4 we determine Nash equilibrium socialand profit maximization strategies for joining the retrial orbitin the partially observable case and the fully observable caserespectively In Section 5 some numerical examples are givento illustrate the effect of the information levels and severalparameters on the customersrsquo strategies Finally in Section 6some conclusions are given

2 Model Description

We consider a single-server retrial queue without waiting linein which customers arrive according to a Poisson process atrate 120582 The service times are assumed to be exponentiallydistributed with parameter 120583 During the busy periods theserver may break down and once breakdown occurs theserver will be repaired immediately The customer beingserved will stay at the service area waiting for the serverrestoredThe lifetime of the server is assumed to be exponen-tially distributed with parameter 120585 and the repair time is alsoexponentially distributed with parameter 120578

Customers that find the server idle upon arrival willoccupy the server immediately and start being served Oncompletion of the service they leave the system Uponarrival if the server is busy the arriving customers will jointhe retrial orbit and then retry for service according to anFCFS discipline That is only the customer at the head ofthe orbiting queue can repeat his request for service Weassume that retrial times are exponentially distributed withparameter120572 Obviously the retrials can success onlywhen theserver is idle In addition newly arriving customers that findthe server broken will leave the system and never enter theretrial orbit Finally it is assumed that the interarrival timesservice times lifetime of the server repair times and retrialtimes are mutually independent

We denote the state of the system at time 119905 by a randomvector (119868(119905)119873(119905)) where 119868(119905) denotes the state of the server0 1 2 corresponding to the state of idle busy or brokenrespectively And 119873(119905) is the number of customers in theorbit Then (119868(119905)119873(119905)) is a continuous-time Markov chain

120582 120582 120582

120582 120583 120582 120583 120582 120583120572120572 120572

120585 120578

120582 120583

120585 120578 120585 120578 120585 120578

middot middot middot

middot middot middot

middot middot middot0 0 0 1

1 1

2 1

0 2

1 2

2 2

0 3

1 3

2 3

1 0

2 0

Figure 1 Transition rate diagram of the original model

with state space 119878 = 0 1 2 times 0 1 2 and the relatedtransition rates are given by

119902(0119895)(1119895) = 120582 119895 ge 0 (1)

119902(1119895)(0119895) = 120583 119895 ge 0 (2)

119902(1119895)(1119895+1) = 120582 119895 ge 0 (3)

119902(0119895)(1119895minus1) = 120572 119895 ge 1 (4)

119902(1119895)(2119895) = 120585 119895 ge 0 (5)

119902(2119895)(1119895) = 120578 119895 ge 0 (6)

The corresponding transition rate diagram is shown inFigure 1

We focus on studying the behavior of customers whenthey are allowed to decidewhether to join or balk based on theavailable information upon their arrival including the queuelength andor the state of the server In this paper we assumethat on completion of service every customer receives areward of 119877 units Besides we assume that there exists awaiting cost of 119862 units per time unit that is continuouslyaccumulated from the time that the customer arrives at thesystem till the time he leaves the system after being servedWe further assume that

119877 gt119862 (120585 + 120578)

120583120578 (7)

If this condition fails to hold even the customer that finds theserver idle will never enter the system because of the negativenet benefit So the above condition ensures that this queue isnot empty

We consider separately two information cases that is(1) partially observable case arriving customers just observethe state of the server (2) fully observable case arrivingcustomers get informed not only about the state of the serverbut also about the exact number of customers in the orbitWhen the server is idle customers will immediately enter thesystem but when the server is busy customers have to decidewhether to enter the orbit or leave the system We furtherassume that decisions are irrevocable retrials of balking andreneging of entering customer are not allowed

Mathematical Problems in Engineering 3

3 The Partially Observable Case

As we have mentioned a customer who finds the server idlewill enter the system because condition (7) ensures that hisreward exceeds his expected waiting cost In addition thedecision of the customer who finds the server idle will neverbe affected by the other customers So we just need to studythe behavior of the customers that find the server busy uponarrival We further assume that when an arriving customerfinds the server busy he will enter the orbit with probability119903 because the customer does not know the exact numberof customers in the orbit In order to identify the individualequilibrium social and profitmaximizing strategies we needto examine the stationary probabilities of the system and thenevaluate other key performances of the system Based on theabove assumptions the transition rate of the correspondingMarkov chain given in (3) should be substituted by

119902(1119895)(1119895+1) = 120582119903 119895 ge 0 (8)

Then we have the following propositions

Proposition 1 Consider the partially observable constantretrial queue with breakdowns and repairs in which customersenter the system with probability 119903 whenever the server isbusy with probability 1 whenever the server is idle and neverenter the system whenever the server is broken The stationaryprobabilities of the system are given by

1199010 (1) =120578 (120583 minus 120582119903)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

1199011 (1) =120582120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

1199012 (1) =120582120585

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

(9)

Proof Let 119901119894119899 (119894 = 0 1 2 119899 = 0 1 2 ) denote thestationary probabilities of the system at state (119894 119899) in which119894 denotes the state of the server and 119899 denotes the number ofcustomers in the orbit According to the transition principleof the stationary probabilities we get the following balanceequations

12058211990100 = 12058311990110 (10)

(120582 + 120572) 1199010119899 = 1205831199011119899 119899 ge 1 (11)

(120583 + 120585 + 120582119903) 11990110 = 12057211990101 + 12057811990120 + 12058211990100 (12)

(120583 + 120585 + 120582119903) 1199011119899 = 1205721199010119899+1 + 1205781199012119899 + 1205821199010119899 + 1205821199031199011119899minus1

119899 ge 1

(13)

1205781199012119899 = 1205851199011119899 119899 ge 0 (14)

The transition rate diagram is shown in Figure 2

120582

120582r 120582r

120583 120582 120583 120582 120583 120582 120583120572 120572 120572

120585 120578 120585 120578 120585 120578 120585 120578

middot middot middot

middot middot middot

middot middot middot

120582r

0 0 0 1

1 1

2 1

0 2

1 2

2 2

0 3

1 3

2 3

1 0

2 0

Figure 2 Transition rate diagram of the partially observable case

In order to solve the stationary probabilities of the systemwe define the following generating functions

119901119894 (119911) =

infin

sum

119899=0

119901119894119899119911119899 (15)

where 119894 = 0 1 2 Multiplying (10) by 1199110 and (11) by 119911119899 119899 ge 1

and summing over 119899 we get

(120582 + 120572) 1199010 (119911) = 1205831199011 (119911) + 12057211990100 (16)

Similarly we can obtain 1199011(119911) and 1199012(119911) from (12)ndash(14)

[minus1205821199031199112+ 119911 (120583 + 120585 + 120582119903)] 1199011 (119911)

= (120582119911 + 120572) 1199010 (119911) + 1205781199111199012 (119911) minus 12057211990100

1205781199012 (119911) = 1205851199011 (119911)

(17)

Using the above three equations we yield

1199010 (119911) =120572 (120583 minus 120582119903119911)

[120572120583 minus (120582 + 120572) 120582119903119911]11990100

1199011 (119911) =120582120572

[120572120583 minus (120582 + 120572) 120582119903119911]11990100

1199012 (119911) =120585120582120572

120578 [120572120583 minus (120582 + 120572) 120582119903119911]11990100

(18)

Using normalizing equation 1199010(1) + 1199011(1) + 1199012(1) = 1 weobtain

11990100 =120578

120572

120572120583 minus (120582 + 120572) 120582119903

120582 (120585 + 120578) + 120578 (120583 minus 120582119903) (19)

Therefore it is readily seen that the conclusion (9) can bederived based on the above results

Remark 2 We need to determine the stable condition of thismodel in fact from (10) (12) and (14) we obtain 12058211990311990110 =

12057211990101 In addition from (11) (13) and (14) we get 1205821199031199011119895 minus1205721199010119895+1 = 1205821199031199011119895minus1 minus1205721199010119895 (119895 ge 1) Proceeding to this recursiveprocess we get 1205821199031199011119895 = 1205721199010119895+1 (119895 ge 1) Using (11) again weobtain

1199010119895+1 =120582119903 (120582 + 120572)

1205721205831199010119895 119895 ge 1 (20)

4 Mathematical Problems in Engineering

So the system is stable if and only if

120582119903 (120582 + 120572)

120572120583lt 1 (21)

Proposition 3 In the partially observable case the expectedoverall queue length of this constant retrial queueing systemand the expected mean sojourn time of an arriving customerthat finds the server busy and decides to join the retrial orbitare given respectively by

ELoverall =1205822119903120583120578 + 120582120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(22)

ESsystem =(120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus (120582 + 120572) 120582119903]+120578 + 120585

120583120578 (23)

Proof According to the PASTA property we know that thetotal arrival rate in the orbit is given by

120582 = 1205821199031199011 (1) (24)

Then the expected mean queue length in the orbit ENorbit isgiven by

ENorbit =infin

sum

119899=1

119899 (1199010119899 + 1199011119899 + 1199012119899)

= 1199011015840

0 (1) + 1199011015840

1 (1) + 1199011015840

2 (1)

(25)

Differentiating (18) with respect to 119911 and taking 119911 = 1 yields

ENorbit =1205822119903 (120582 + 120572) (120585 + 120578) + 120582

2119903120583120578

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(26)

Then the expected sojourn time of an arriving customer thatfinds the server busy and decides to join in the retrial orbit isgiven by

ESsystem =ENorbit

120582+120585 + 120578

120583120578 (27)

In the above equation the first term ENorbit120582 equals to themean waiting time in the orbit and the second term (120585 +

120578)120583120578 means the generalized service time in a unreliablesystem (see [14]) Using (24) and (26) yields (23) Theexpected overall queue length in the system ELoverall satisfies

ELoverall

= 1199011015840

0 (1) + 1199011015840

1 (1) + 1199011015840

2 (1) + 1199011 (1) + 1199012 (1)

=1205822119903120583120578 + 120582120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(28)

In the following part of this paper we assume that

120582 (120582 + 120572)

120572120583lt 1 (29)

This assumption originally comes from (21) but it has beenmodified slightly This means the system is stable under anystrategy 119903 followed by the customers Under this conditionwe are going to study the customersrsquo equilibrium joiningstrategy We have the following theorem

Theorem 4 In the partially observable single-server constantretrial queue we can derive a uniquemixed equilibrium joiningstrategy ldquoenter the orbit with probability 119903119890 when finding theserver busy upon arrivalrdquo when condition (29) holds Theprobability 119903119890 is given by

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119905119871119890 lt

119877

119862lt 119905119880119890

1 if 119877119862ge 119905119880119890

(30)

where

119903lowast

119890 =120572120583

120582 (120582 + 120572)minus(120582 + 120572) (120585 + 120578) + 120583120578

120582120578 (120582 + 120572)(119877

119862minus120585 + 120578

120583120578)

minus1

(31)

119905119871119890 =(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (32)

119905119880119890 =(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120583120572 minus 120582 (120582 + 120572))+120585 + 120578

120583120578 (33)

Proof Suppose that all customers follow the same joiningstrategy 119903when the server is busy at their arrival instantThenwe can get the expected net benefit of a customer who decidesto enter the system

119878119890 (119903) = 119877 minus CESsystem

= 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120572120583 minus (120582 + 120572) 120582119903)+120578 + 120585

120583120578]

(34)

We observe that 119878119890(119903) is strictly decreasing whenever 119903 isin

[0 1] and it has a unique maximum

119878119890 (0) = 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120578 + 120585

120583120578] (35)

and a unique minimum

119878119890 (1) = 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120572120583 minus 120582 (120582 + 120572))+120578 + 120585

120583120578] (36)

Therefore we consider the following cases

(i) When 119877119862 isin ((120578 + 120585)120583120578 119905119871119890] 119878119890(119903) is nonpositivewhenever 119903 isin [0 1] so the best response is balkingand the equilibrium joining probability is 119903119890 = 0

Mathematical Problems in Engineering 5

(ii) When 119877119862 isin (119905119871119890 119905119880119890) there exists a unique solutionof the equation 119878119890(119903) = 0 which is given by (31)

(iii) Similarly when 119877119862 isin [119905119880119890infin) the net benefit isalways positive so we obtain the rest branch of thetheorem

Because of themonotonicity of the function 119878119890(119903) the bestresponse is 1 whenever the joining probability 119903 is smallerthan 119903119890 If 119903 gt 119903119890 the best response is 0 for the net benefitis negative If 119903 = 119903119890 any strategy is a best response andcustomers are indifferent between entering the system andleaving This shows that an individualrsquos best response is adecreasing function of the strategy selected by the othercustomers that is the higher the joining probability of othercustomers the lower the net benefit Therefore we have anldquoavoid-the-crowdrdquo (ATC) situation

We will proceed to determine the solutions of the socialand profitmaximization problems that is we need to find theoptimal joining probabilities 119903soc and 119903prof that maximize thesocial net benefit and the administratorrsquos profit per unit timeWefirstly consider the socialmaximization problemWehavethe following theorem

Theorem 5 In the partially observable single-sever constantretrial queue there exists a uniquemixed joining strategy ldquoenterthe retrial orbit with probability 119903soc when the sever is busyrdquothat maximizes the social net benefit per time unit in whichcondition (29) holds The probability 119903soc is given by

119903soc =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119905119871soc lt

119877

119862lt 119905119880soc

1 if 119877119862ge 119905119880soc

(37)

where

119903lowast

soc =1198610120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 1198610 minus 120582120578 (38)

1198610 =radic1205782(120582119877 + 119862)

119862120572 (120582 + 120572) (39)

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2minus 12058321205782120572

12058212058321205782120572 (40)

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2minus 1205782[120583120572 minus 120582 (120582 + 120572)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

(41)

Proof For a given joining strategy 119903 the system behavesstationarily when condition (29) holds Customers that findthe server busy will join the orbit with probability 119903 Whenall customers follow the same strategy 119903 the social net benefitper time unit is given by

119878soc (119903) = 120582lowast(119903) 119877 minus CELoverall (42)

where 120582lowast(119903) denotes the mean effective arrival rate of thesystem and ELoverall denotes the expectedmean queue lengthof the system (including the customer in the server) underthe strategy 119903 The mean effective arrival rate is given by

120582lowast(119903) = 1205821199010 (1) + 1205821199031199011 (1) =

120582120583120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903) (43)

Using (22) and (43) social net benefit is given by

119878soc (119903) =120582120583120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)119877 minus 119862120582

times120582120583120578119903 + 120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] [120583120572 minus 120582119903 (120582 + 120572)]

(44)

It can be written as

119878soc (119903) =120583120578 (120582119877 + 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus

119862120583120572

120583120572 minus 120582119903 (120582 + 120572) (45)

Differentiating the above equation with respect to 119903 we get

119889

119889119903119878soc (119903) =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2minus

119862120583120572120582 (120582 + 120572)

[120583120572 minus 120582119903 (120582 + 120572)]2

(46)

119889

119889119903119878soc (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(120582119877 + 119862)

119862120572 (120582 + 120572)

(47)

From (46) we obtain

119889

119889119903119878soc (119903) |119903=0 =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120583120578]2minus119862120583120572120582 (120582 + 120572)

(120583120572)2

119889

119889119903119878soc (119903) |119903=1 =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582)]2

minus119862120583120572120582 (120582 + 120572)

[120583120572 minus 120582 (120582 + 120572)]2

(48)

Letting (119889119889119903)119878soc(119903)|119903=0 = 0 and (119889119889119903)119878soc(119903)|119903=1 = 0 weobtain 119877119862 = 119905119871soc and 119877119862 = 119905119880soc respectively Then

119889

119889119903119878soc (119903) |119903=0 le 0 lArrrArr

119877

119862le 119905119871soc (49)

119889

119889119903119878soc (119903) |119903=1 ge 0 lArrrArr

119877

119862ge 119905119880soc (50)

Therefore we consider the following situations

(i) When 119877119862 le 119905119871soc function 119878soc(119903) is decreasingwhenever 119903 isin [0 1] so the best response is balkingthen 119903soc = 0

6 Mathematical Problems in Engineering

(ii) When 119905119871soc lt 119877119862 lt 119905119880soc there exists a positivenumber 1199030 so that function 119878soc(119903) increases when119903 isin (0 1199030] and decreases when 119903 isin [1199030 1) then 1199030 is thepoint that maximizes the function 119878soc(119903) By solving(47) we know that the unique maximum of 119878soc(119903) isattained at (1198610120583120572 minus 120582(120585 + 120578) minus 120583120578)(120582(120582 + 120572)1198610 minus 120582120578)Therefore 119903soc given by (38) is the social maximizingstrategy

(iii) When 119877119862 ge 119905119880soc function 119878soc(119903) is increasingwhenever 119903 isin [0 1] so the best response is 1

We now compare the optimal joining strategy of thesocial maximization problem and the individual equilibriumjoining strategy Their relation is given by the followingtheorem

Theorem 6 In our single-server constant retrial rate queuewith breakdowns and repairs the joining probabilities 119903119890 and119903soc are ordered as

119903soc le 119903119890 (51)

Proof When the stable condition (29) holds it is easy to showthat 119905119871119890 lt 119905119871soc lt 119905119880119890 lt 119905119880soc where 119905119871119890 119905119871soc 119905119880119890 119905119880socare defined in Theorems 4 and 5 We consider the followingcases

(i) If 119877119862 le 119905119871119890 according to (30) and (37) 119903soc = 119903119890 = 0

(ii) If 119905119871119890 lt 119877119862 le 119905119871soc then 119903119890 isin (0 1) and 119903soc = 0 So119903soc lt 119903119890

(iii) If 119905119871soc lt 119877119862 lt 119905119880119890 then 119903119890 isin (0 1) and 119903soc isin (0 1)But we can get that 119903soc lt 119903119890 after some algebra

(iv) If 119905119880119890 le 119877119862 lt 119905119880soc then 119903119890 = 1 and 119903soc isin (0 1)Therefore 119903soc lt 119903119890 is obviously satisfied

(v) Finally if 119877119862 ge 119905119880soc then 119903119890 = 119903soc = 1

From the above analysis we complete the proof

The above result can be explained by the perspective of aneconomic viewpoint In the process of individual equilibriumanalysis we notice that customers will enter the orbit as longas their net benefit is nonnegative Then individualrsquos rationalbehavior causes excessive use of the resource in equilibriumthat is to say individualrsquos optimization leads to the systemhavingmore congestion thanwhat is socially desirable whichis the appearance of the negative externality

Having considered the individual equilibrium strategyand social maximization problem we will further identifythe profit optimization joining strategy In our model wesuppose that every customer will be imposed an entrancefee 119901 by the administrator By imposing the entrance feecustomersrsquo reward decreases to 119877 minus 119901 then their strategy willdiffer We will determine the strategy 119903prof that maximizes theadministratorrsquos profit per time unit The result is given by thefollowing

Theorem 7 In the partially observable single-server constantretrial queue we can derive a unique mixed joining strategyldquoenter the orbit with probability 119903prof when the server is busyrdquothat maximizes the administratorrsquos net profit per time unitwhen condition (29) holds The probability 119903prof is given by

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119905119871prof lt

119877

119862lt 119905119880prof

1 if 119877119862ge 119905119880prof

(52)

where

119903lowast

prof =1198620120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572)1198620 minus 120582120578 (53)

1198620 =radic1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2 (54)

1198771015840= 119877 minus 119862

120585 + 120578

120583120578 (55)

119905119871prof

=(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2+ 120582120583120578120572

2(120585 + 120578) minus 120583

212057821205722

120582120583212057821205722

(56)

119905119880prof = (120583(120582 + 120572)2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

+ [120582120578 (120585 + 120578) minus 1205831205782] [120583120572 minus 120582 (120582 + 120572)]

2)

times (1205821205831205782[120583120572 minus 120582 (120582 + 120572)]

2)minus1

(57)

Proof For a given joining strategy 119903 we define the function119878prof(119903) to be the administratorrsquos net profit per time unit Let120582lowast(119903) be the mean effective arrival rate in the system we

further assume that the entrance fee that the administratorimposes on customers is 119901(119903) Then we have

119878prof (119903) = 120582lowast(119903) 119901 (119903) (58)

where 120582lowast(119903) is given by (43) In order to obtain 119878prof(119903) weneed to determine the entrance fee 119901(119903) Noticing (34) weyield

119877 minus 119901 (119903) = 119862((120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)]+120585 + 120578

120583120578) (59)

from which we obtain

119901 (119903) = 119877 minus 119862120585 + 120578

120583120578minus 119862

(120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)] (60)

Because if the entrance fee 119901 lt 119901(119903) the administrator canimprove their net benefit through increasing the entrancefee but if 119901 gt 119901(119903) the individualrsquos net benefit is negative

Mathematical Problems in Engineering 7

customers will balk So (59) always holds whenever 119903 isin [0 1]Therefore the net profit of the administrator can be calculatedas follows119878prof (119903) = 120582

lowast(119903) 119901 (119903)

=120583120578 (120582119877

1015840+ 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus 119862

120583 (120582 + 120572)

120583120572 minus 120582119903 (120582 + 120572)

(61)

where 1198771015840 is given by (55) Differentiating the equation withrespect to 119903 we get

119889

119889119903119878prof (119903) =

1205821205831205782(1205821198771015840+ 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2

minus119862120583120582(120582 + 120572)

2

[120583120572 minus 120582119903 (120582 + 120572)]2

119889

119889119903119878prof (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2

(62)

The rest of the proof can proceed along the same line with theproof of Theorem 5

Now we are going to compare the joining probabilitieswhen considering the social and profit maximization prob-lems The result is given by the following theorem

Theorem 8 In our model when condition (29) holds theoptimal joining probabilities 119903soc and 119903prof are ordered as

119903prof le 119903soc (63)

Proof Firstly we proceed to compare the relationsamong 119905119871soc 119905119880soc 119905119871prof 119905119880prof Rewrite the expressionsof 119905119871soc 119905119880soc 119905119871prof 119905119880prof given inTheorems 5 and 7

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2

12058212058321205782120572minus1

120582

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

minus1

120582

119905119871prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2

120582120583212057821205722+120585 + 120578

120583120578minus1

120582

119905119880prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

+120585 + 120578

120583120578minus1

120582

(64)

It is readily shown that 119905119871soc lt 119905119871prof and 119905119880soc lt 119905119880prof Butthe relation between 119905119880soc and 119905119871prof is undefined Two casesmay take place that is 119905119880soc lt 119905119871prof or 119905119880soc ge 119905119871prof Weconsider the situation when 119905119880soc ge 119905119871prof the other case canbe analyzed in the same wayThen we consider the followingcases

(i) If 119877119862 isin [(120585 + 120578)120583120578 119905119871soc) then 119903soc = 119903prof = 0(ii) If 119877119862 isin [119905119871soc 119905119871prof) then 119903soc isin (0 1) 119903prof = 0(iii) If 119877119862 isin [119905119871prof 119905119880soc) then 119903soc isin (0 1) and 119903prof isin

(0 1) We can take a close look at the expressions of119903soc 119903prof that are presented in Theorems 5 and 7 Wedefine a function

119892 (119909) =119909120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 119909 minus 120582120578 (65)

Differentiating the above function with respect to 119909 weobtain

1198921015840(119909) =

1205822((120582 + 120572) (120585 + 120578) + 120583120578)

(120582 (120582 + 120572) 119909 minus 120582120578)2

(66)

It means that function 119892(119909) increases with respect to 119909 Onthe other hand we can easily derive 1198610 gt 1198620 where 1198610 and1198620 are given inTheorems 5 and 7Then we obtain 119903prof lt 119903soc

(i) If 119877119862 isin [119905119880soc 119905119880prof) then 119903soc = 1 and 119903prof isin (0 1)(ii) If 119877119862 isin [119905119880socinfin) then 119903soc = 1 and 119903prof = 1

We obtain 119903soc ge 119903prof from the above analysis

Remark 9 Economou and Kanta [6] have studied the indi-vidual equilibrium strategy and social and profit maximiza-tion problems in the single-server constant retrial queue inthe partially observable case In our model if we let 120585 = 0we can derive the same results as given in [6] This is notaccidentally happened because when 120585 = 0 breakdownswill never occur then these two models are equivalent Inaddition notice that customersrsquo strategies 119903119890 119903soc 119903prof arefunctions of 120585120578 where other parameters are fixed It meansthat 119903119890 119903soc 119903prof only relate to 120585120578 whatever was the value 120585and 120578 we take in this model

Remark 10 Another case is that 120582(120582 + 120572)120572120583 ge 1 is left tobe considered In this situation customersrsquo joining strategieswill be less than 1 when considering the equilibrium socialand profit maximization problems otherwise the server willbehave unstably That is to say we have the following joiningstrategies

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119877

119862gt 119905119871119890

119903119904119900119888 =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119877

119862gt 119905119871soc

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119877

119862gt 119905119871prof

(67)

and the relation 119903119890 ge 119903soc ge 119903prof holds all the same

8 Mathematical Problems in Engineering

Till now we have completed the analysis of the customersrsquostrategies of the partially observable retrial queueing systemWe will turn to the fully observable case

4 The Fully Observable Case

We now focus on the fully observable case assuming cus-tomers observe not only the state of the server but also theexact number of customers in the retrial orbit Customerthat finds the server idle will enter the system and begin tobe served immediately so this information is valuable onlyfor the customer who finds the server busy We know thatafter every service completion and there is no new customerarriving the customer at the head of the orbit queue willsuccessfully retry for service That means customers in theorbit are served on an FCFS basis Therefore observing theposition in the orbit they can assess precisely whether toenter or balk

Consider a tagged customer that finds the server busyupon arrival This customerrsquos mean overall sojourn time inthe system is not affected by the joiningbalking behavior ofthe other customers that find the server busy because if thetagged customer joins the orbit queue at the 119895th positionthen the late customers who find the server busy will jointhe queue at the 119895 + 1th 119895 + 2th positions behind himin the orbit So his sojourn time does not depend on theirdecisions but it depends on the arrival rate because the newlyarriving customer who finds the server idle will be servedimmediately

To study the general threshold strategy adopted by allcustomers in the fully observable case we will firstly considerthe mean overall sojourn time of the customer who finds theserver idle busy or broken upon arrival Let 119879(1 0) be themean sojourn time when the server is idle and there is nocustomers in the orbit at his arrival instant and let 119879(119894 119895)be the mean sojourn time of a tagged customer at the 119895thposition in the orbit given that the server is at state 119894 Whenall customers follow the general strategy the mean overallsojourn times 119879(119894 119895) are given by

119879 (1 0) =120585 + 120578

120583120578 (68)

119879 (1 119895) =1

120583 + 120585+

120585

120583 + 120585119879 (2 119895) +

120583

120583 + 120585119879 (0 119895) 119895 ge 1

(69)

119879 (0 119895) =1

120582 + 120572+

120582

120582 + 120572119879 (1 119895) +

120572

120582 + 120572119879 (1 119895 minus 1)

119895 ge 1

(70)

119879 (2 119895) =1

120578+ 119879 (1 119895) 119895 ge 0 (71)

Let us consider a customer in the 119895th position of the orbitwhen the server is busy Then this customer has to wait foran exponentially distributed time with rate 120585 + 120583 for thenext event to occur in which the server breaks down or the

service is completed With probability 120585(120585 + 120583) breakdownoccurs then the mean overall sojourn time becomes 119879(2 119895)and the tagged customer remains in the 119895th position of theorbit On the other hand with probability 120583(120585 + 120583) theservice is completed then the mean overall sojourn timebecomes 119879(1 119895 minus 1) and the tagged customer moves to the119895 minus 1th position Through the above analysis we obtain (69)Considering the other cases in the same line we obtain therest equations

Solving the above equations the mean overall sojourntimes are given by

119879 (0 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578 119895 ge 1

119879 (1 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 119895 ge 0

119879 (2 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120583 + 120585 + 120578

120583120578

119895 ge 0

(72)

Then we need to find the general equilibrium strategyfollowed by the customers when the server is busy To ensurethat customer will enter the orbit when finding the serverbusy and the queue in the orbit is empty we further assumethat 119877 minus 119862119879(1 1) gt 0 The condition can be written as

119877

119862gt(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (73)

This means customersrsquo reward should be larger in the fullyobservable case than in the former case even if the other sys-tematic parameters are equal We will identify the customersrsquoequilibrium threshold strategy and we have the followingtheorem

Theorem 11 In the fully observable single-server constantretrial queue there exists a unique threshold joining strategyldquoenter the retrial orbit if there are at most 119899119890 customers in theorbit whenever finding the server busyrdquo in which condition (73)holds The threshold 119899119890 is given by

119899119890 = lfloor119909119890rfloor (74)

where

119909119890 =120572120583120578119877 minus 119862120572 (120585 + 120578)

119862 [(120582 + 120572) (120585 + 120578) + 120583120578]minus 1 (75)

This strategy is the unique equilibrium strategy among allpossible strategies

Proof Consider a tagged customer that finds the system atthe state (1 119895) upon arrival If he decides to join he goesdirectly to the orbit and occupies the 119895 + 1th position Thenhis expected net benefit is 119878119890(119895) = 119877 minus 119862119879(1 119895 + 1) where

119878119890 (119895) = 119877 minus 119862[(119895 + 1)(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578]

(76)

Mathematical Problems in Engineering 9

middot middot middot

middot middot middot

middot middot middot

120582

120582 120582 120582

120583 120572 120582 120583 120582 120583120572 120572

120585 120578 120585 120578 120585 120578

120582 120583

120585 120578

120582 120583

120585 120578

0 n0 0 0 1

1 1

2 1

0 2

1 2

2 2

1 0

2 0

1 n

2 n

0 n + 1

1 n + 1

2 n + 1

Figure 3 Transition rate diagram of the fully observable case

The customer prefers to join if 119878119890(119895) gt 0 and he is indifferentbetween joining and balking while 119878119890(119895) = 0 and prefers tobalk if 119878119890(119895) lt 0 A newly arriving customer will enter ifand only if 119895 le 119899119890 when finding the server busy By solvinginequality 119878119890(119895) ge 0 we obtain the result

We will elucidate the uniqueness of the equilibriumthreshold joining strategy followed by all customers Indeedif 119909119890 is an integer number we just need to let 119899119890 = 119909119890 thenwe know 119878119890(119899119890) = 0 and the mixed threshold strategy isthat prescribes to enter if 119899 lt 119899119890 to balk if 119899 gt 119899119890 and beindifferent if 119899 = 119899119890 When 119909119890 is any positive value we let119899119890 = lfloor119909119890rfloor Hence in this case the equilibrium strategy is alsounique

We now turn our attention to the social and profit maxi-mization problems and identify the thresholds 119899soc and 119899profthat maximize the social net benefit and the administratorrsquosprofit per time unit To consummate this goal we need todetermine the stationary behavior of the system when allcustomers follow the same threshold strategy

Suppose that all customers that find the server busy followthe same threshold strategies 119899soc and 119899prof respectively Thatis to say customers who find the sever busy will enter thesystem as long as the number of customers in the orbitis at most 119899soc when considering the social maximizationproblem and there is at most 119899prof customers in the orbitwhen considering the administratorrsquos optimization problemHence the maximum number of customers in the orbitwill never be larger than 119899 + 1 when customers follow thesame threshold-119899 strategyThen we obtain a continuous-timeMarkov chain (119868(119905)119873(119905)) with state space 119878 = 0 1 2 times

0 1 2 119899 + 1 and its transition diagram is given byFigure 3

The stationary distribution of the system (119901119894119895(119894 119895) isin 119878)is given by the following proposition

Proposition 12 Consider the fully observable single-serverconstant retrial queue with breakdowns and repairs in whichcustomers follow a threshold-119899 strategy then the stationarydistribution of the system is given by

11990100 = 119861 (119899)120583120572

1205822

1199010119895 = 119861 (119899) 120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 119861 (119899)120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 119861 (119899)120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(77)

where

120588 =120582 (120582 + 120572)

120572120583

119861 (119899) = 120583120572

1205822+ (1 + 120588 + sdot sdot sdot + 120588

119899) +

120572

120582(1 + 120588 + sdot sdot sdot + 120588

119899+1)

+120572120585

120582120578(1 + 120588 + sdot sdot sdot + 120588

119899+1)

minus1

(78)

Proof The stationary distribution is obtained from the fol-lowing balance equations

12058211990100 = 12058311990110 (79)

(120582 + 120572) 1199010119895 = 1205831199011119895 119895 = 1 2 119899 + 1 (80)

1205821199011119895 = 1205721199010119895+1 119895 = 0 1 2 119899 (81)

1205781199012119895 = 1205851199011119895 119895 = 0 1 119899 + 1 (82)

(120583 + 120585) 1199011119899+1 = 1205781199012119899+1 + 1205821199011119899 + 1205821199010119899+1 (83)

Combining (80) with (81) we can obtain the recursiverelation

1199011119895 = 11990110120588119895 119895 = 1 2 119899 minus 1 (84)

Then we derive the expressions of 1199012119895 from (82)

1199012119895 =120585

12057812058811989511990110 119895 = 0 1 119899 + 1 (85)

Using (80)ndash(82) we can get

11990100 = 11990110

120583120572

1205822

1199010119895 = 11990110120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 11990110120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 11990110120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(86)

Then with the help of the normalizing equation sum119899+1119895=0 1199010119895 +sum119899+1

119895=0 1199011119895 + sum119899+1

119895=0 1199012119895 = 1 we get the expression of 11990110

By considering the stationary distribution we examinethe social welfare maximization threshold strategy We havethe following result

10 Mathematical Problems in Engineering

Theorem 13 In our single-server constant retrial queue inwhich the condition (73) holds we define the following func-tion

119891 (119909) =(120583120572 + (120585120578) 120582120572) (119909 + 1)

120583120572 (1 minus 120588)

minus120582 (120582 + 120572120588 (1 + (120585120578))) (1 minus 120588

119909+1)

120583120572(1 minus 120588)2

minus 1

(87)

(88)

If 119891(0) gt 119909119890 then the customersrsquo best response is balkingOtherwise there exists a unique threshold-119899soc strategy ldquo enterthe retrial orbit if the number of customers in the orbit is atmost119899soc whenever customers find the server busyrdquo that maximizesthe social net benefit per time unit The threshold 119899soc is givenby

119899soc = lfloor119909119904119900119888rfloor (89)

where 119909soc is the unique nonnegative root of the equation119891(119909) = 119909119890

Proof In order to determine the optimal threshold 119899soc weneed to compute the effective arrival rate of the systemand the expected mean sojourn time of the system On theone hand when all customers follow the same threshold-119899strategy the effective arrival rate is given by

120582 (119899) = 120582(

119899+1

sum

119895=0

1199010119895 (119899) +

119899

sum

119895=0

1199011119895 (119899)) (90)

On the other hand by conditioning on the state found by acustomer in the system upon arrival we obtain the expectedmean sojourn time if entering the system with threshold-119899policy

119864 [119878 (119899)] = 119860 + 119861 (91)

where

119860 =

119899+1

sum

119895=0

1199010119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

120585 + 120578

120583120578

119861 =

119899

sum

119895=0

1199011119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

times (120585 + 120578

120583120578+ (119895 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(92)

The expected social net benefit is given by

119878soc (119899) = 120582 (119899) 119877 minus 119862120582 (119899) 119864 [119878 (119899)] (93)

Using (90) and (91) and after some algebra we obtain

119878soc (119899) = 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578

times (120582 (119899) (119909119890 + 1) minus 120582

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899))

(94)

Next we need to show that 119878soc(119899) is unimodal and hasonly one maximal point We firstly consider the increments119878soc(119899) minus 119878soc(119899 minus 1) It is not difficult to show that

119878soc (119899) minus 119878soc (119899 minus 1) ge 0 lArrrArr 119891 (119899) le 119909119890 (95)

where 119891(119899) is

119891 (119899) =1198911 (119899)

1198912 (119899)minus 1 (96)

1198911 (119899) = 120582(

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899) minus

119899minus1

sum

119895=0

(119895 + 1) 1199011119895 (119899 minus 1))

(97)

1198912 (119899) = 120582 (119899) minus 120582 (119899 minus 1) (98)

Using the stationary distribution we can simplify the expres-sions of (97) and (98) as follows

1198911 (119899) =120572119861 (119899) 119861 (119899 minus 1) 120588

119899

1205822(1 minus 120588)2

times 119863 (99)

1198912 (119899) =1205831205722120588119899

1205822119861 (119899) 119861 (119899 minus 1) (100)

where

119863 = (120583120572 + 120582120572120585

120578) (119899 + 1) (1 minus 120588) minus 120582

2

minus120582120572120588(1 +120585

120578) + 1205822120588119899+1

+ 120582120572(1 +120585

120578) 120588119899+2

(101)

Plugging (99) and (100) in (96) and replacing 119899 by 119909 weobtain (87) To prove the unimodality of 119878soc(119899) we need toshow that function119891(119909) is increasingWe define the functionℎ [0infin) rarr 119877 with

ℎ (119909) = 119891 (119909) minus 119909 (102)

We have

1198892

1198891199092ℎ (119909) =

[120582 + 120572120588 (1 + (120585120578))] 120588119909+2

(ln 120588)2

(120582 + 120572) (1 minus 120588)2

gt 0

119909 ge 0

(103)

Then function ℎ(119909) is a strictly convex function and thereis no flex point with respect to 120588 and 119909 Therefore ℎ(119909) isincreasing and so is 119891(119909) = ℎ(119909) + 119909 Therefore two casesmay take place

(i) 119891(0) gt 119909119890 then 119878soc(119899) is decreasing and customerswho find the server busy will balk

(ii) 119891(119899soc) le 119909119890 lt 119891(119899soc + 1) then we have 119878soc(119899) minus119878soc(119899minus1) ge 0 for 119899 le 119899soc and 119878soc(119899)minus119878soc(119899minus1) lt 0for 119899 gt 119899soc Then the maximum of 119878soc(119899) occurs in119899soc = lfloor119909socrfloor in which 119909soc is the unique solution ofthe equation 119891(119909) = 119909119890

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

equilibrium joining strategies It differs from equilibriumanalysis of retrial systems with the classical retrial policy seefor instance Kulkarni [7] Elcan [8] Hassin and Haviv [9]and Zhang et al [10] and Wang et al [11] among othersHowever although we have found several retrial models withservers subject to breakdowns that consider orbits as FCFSqueues for example in Atencia et al [12] and Li and Zhao[10] To the authorsrsquo knowledge no work has been done onsuch queueing systems from an economic view Thus in thispaper we propose to study such an MM1 retrial queuewith the constant retrial policy and serverrsquos breakdowns andrepairs as an extension of the paper considered by Economouand Kanta [6] who have also considered the observablesingle-server queue with breakdowns and repairs see [13]To this end the methodology will be based on the game-theoretic analysis and optimization

This paper is organized as follows In Section 2 we givethe model description and the natural reward-cost structureIn Sections 3 and 4 we determine Nash equilibrium socialand profit maximization strategies for joining the retrial orbitin the partially observable case and the fully observable caserespectively In Section 5 some numerical examples are givento illustrate the effect of the information levels and severalparameters on the customersrsquo strategies Finally in Section 6some conclusions are given

2 Model Description

We consider a single-server retrial queue without waiting linein which customers arrive according to a Poisson process atrate 120582 The service times are assumed to be exponentiallydistributed with parameter 120583 During the busy periods theserver may break down and once breakdown occurs theserver will be repaired immediately The customer beingserved will stay at the service area waiting for the serverrestoredThe lifetime of the server is assumed to be exponen-tially distributed with parameter 120585 and the repair time is alsoexponentially distributed with parameter 120578

Customers that find the server idle upon arrival willoccupy the server immediately and start being served Oncompletion of the service they leave the system Uponarrival if the server is busy the arriving customers will jointhe retrial orbit and then retry for service according to anFCFS discipline That is only the customer at the head ofthe orbiting queue can repeat his request for service Weassume that retrial times are exponentially distributed withparameter120572 Obviously the retrials can success onlywhen theserver is idle In addition newly arriving customers that findthe server broken will leave the system and never enter theretrial orbit Finally it is assumed that the interarrival timesservice times lifetime of the server repair times and retrialtimes are mutually independent

We denote the state of the system at time 119905 by a randomvector (119868(119905)119873(119905)) where 119868(119905) denotes the state of the server0 1 2 corresponding to the state of idle busy or brokenrespectively And 119873(119905) is the number of customers in theorbit Then (119868(119905)119873(119905)) is a continuous-time Markov chain

120582 120582 120582

120582 120583 120582 120583 120582 120583120572120572 120572

120585 120578

120582 120583

120585 120578 120585 120578 120585 120578

middot middot middot

middot middot middot

middot middot middot0 0 0 1

1 1

2 1

0 2

1 2

2 2

0 3

1 3

2 3

1 0

2 0

Figure 1 Transition rate diagram of the original model

with state space 119878 = 0 1 2 times 0 1 2 and the relatedtransition rates are given by

119902(0119895)(1119895) = 120582 119895 ge 0 (1)

119902(1119895)(0119895) = 120583 119895 ge 0 (2)

119902(1119895)(1119895+1) = 120582 119895 ge 0 (3)

119902(0119895)(1119895minus1) = 120572 119895 ge 1 (4)

119902(1119895)(2119895) = 120585 119895 ge 0 (5)

119902(2119895)(1119895) = 120578 119895 ge 0 (6)

The corresponding transition rate diagram is shown inFigure 1

We focus on studying the behavior of customers whenthey are allowed to decidewhether to join or balk based on theavailable information upon their arrival including the queuelength andor the state of the server In this paper we assumethat on completion of service every customer receives areward of 119877 units Besides we assume that there exists awaiting cost of 119862 units per time unit that is continuouslyaccumulated from the time that the customer arrives at thesystem till the time he leaves the system after being servedWe further assume that

119877 gt119862 (120585 + 120578)

120583120578 (7)

If this condition fails to hold even the customer that finds theserver idle will never enter the system because of the negativenet benefit So the above condition ensures that this queue isnot empty

We consider separately two information cases that is(1) partially observable case arriving customers just observethe state of the server (2) fully observable case arrivingcustomers get informed not only about the state of the serverbut also about the exact number of customers in the orbitWhen the server is idle customers will immediately enter thesystem but when the server is busy customers have to decidewhether to enter the orbit or leave the system We furtherassume that decisions are irrevocable retrials of balking andreneging of entering customer are not allowed

Mathematical Problems in Engineering 3

3 The Partially Observable Case

As we have mentioned a customer who finds the server idlewill enter the system because condition (7) ensures that hisreward exceeds his expected waiting cost In addition thedecision of the customer who finds the server idle will neverbe affected by the other customers So we just need to studythe behavior of the customers that find the server busy uponarrival We further assume that when an arriving customerfinds the server busy he will enter the orbit with probability119903 because the customer does not know the exact numberof customers in the orbit In order to identify the individualequilibrium social and profitmaximizing strategies we needto examine the stationary probabilities of the system and thenevaluate other key performances of the system Based on theabove assumptions the transition rate of the correspondingMarkov chain given in (3) should be substituted by

119902(1119895)(1119895+1) = 120582119903 119895 ge 0 (8)

Then we have the following propositions

Proposition 1 Consider the partially observable constantretrial queue with breakdowns and repairs in which customersenter the system with probability 119903 whenever the server isbusy with probability 1 whenever the server is idle and neverenter the system whenever the server is broken The stationaryprobabilities of the system are given by

1199010 (1) =120578 (120583 minus 120582119903)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

1199011 (1) =120582120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

1199012 (1) =120582120585

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

(9)

Proof Let 119901119894119899 (119894 = 0 1 2 119899 = 0 1 2 ) denote thestationary probabilities of the system at state (119894 119899) in which119894 denotes the state of the server and 119899 denotes the number ofcustomers in the orbit According to the transition principleof the stationary probabilities we get the following balanceequations

12058211990100 = 12058311990110 (10)

(120582 + 120572) 1199010119899 = 1205831199011119899 119899 ge 1 (11)

(120583 + 120585 + 120582119903) 11990110 = 12057211990101 + 12057811990120 + 12058211990100 (12)

(120583 + 120585 + 120582119903) 1199011119899 = 1205721199010119899+1 + 1205781199012119899 + 1205821199010119899 + 1205821199031199011119899minus1

119899 ge 1

(13)

1205781199012119899 = 1205851199011119899 119899 ge 0 (14)

The transition rate diagram is shown in Figure 2

120582

120582r 120582r

120583 120582 120583 120582 120583 120582 120583120572 120572 120572

120585 120578 120585 120578 120585 120578 120585 120578

middot middot middot

middot middot middot

middot middot middot

120582r

0 0 0 1

1 1

2 1

0 2

1 2

2 2

0 3

1 3

2 3

1 0

2 0

Figure 2 Transition rate diagram of the partially observable case

In order to solve the stationary probabilities of the systemwe define the following generating functions

119901119894 (119911) =

infin

sum

119899=0

119901119894119899119911119899 (15)

where 119894 = 0 1 2 Multiplying (10) by 1199110 and (11) by 119911119899 119899 ge 1

and summing over 119899 we get

(120582 + 120572) 1199010 (119911) = 1205831199011 (119911) + 12057211990100 (16)

Similarly we can obtain 1199011(119911) and 1199012(119911) from (12)ndash(14)

[minus1205821199031199112+ 119911 (120583 + 120585 + 120582119903)] 1199011 (119911)

= (120582119911 + 120572) 1199010 (119911) + 1205781199111199012 (119911) minus 12057211990100

1205781199012 (119911) = 1205851199011 (119911)

(17)

Using the above three equations we yield

1199010 (119911) =120572 (120583 minus 120582119903119911)

[120572120583 minus (120582 + 120572) 120582119903119911]11990100

1199011 (119911) =120582120572

[120572120583 minus (120582 + 120572) 120582119903119911]11990100

1199012 (119911) =120585120582120572

120578 [120572120583 minus (120582 + 120572) 120582119903119911]11990100

(18)

Using normalizing equation 1199010(1) + 1199011(1) + 1199012(1) = 1 weobtain

11990100 =120578

120572

120572120583 minus (120582 + 120572) 120582119903

120582 (120585 + 120578) + 120578 (120583 minus 120582119903) (19)

Therefore it is readily seen that the conclusion (9) can bederived based on the above results

Remark 2 We need to determine the stable condition of thismodel in fact from (10) (12) and (14) we obtain 12058211990311990110 =

12057211990101 In addition from (11) (13) and (14) we get 1205821199031199011119895 minus1205721199010119895+1 = 1205821199031199011119895minus1 minus1205721199010119895 (119895 ge 1) Proceeding to this recursiveprocess we get 1205821199031199011119895 = 1205721199010119895+1 (119895 ge 1) Using (11) again weobtain

1199010119895+1 =120582119903 (120582 + 120572)

1205721205831199010119895 119895 ge 1 (20)

4 Mathematical Problems in Engineering

So the system is stable if and only if

120582119903 (120582 + 120572)

120572120583lt 1 (21)

Proposition 3 In the partially observable case the expectedoverall queue length of this constant retrial queueing systemand the expected mean sojourn time of an arriving customerthat finds the server busy and decides to join the retrial orbitare given respectively by

ELoverall =1205822119903120583120578 + 120582120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(22)

ESsystem =(120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus (120582 + 120572) 120582119903]+120578 + 120585

120583120578 (23)

Proof According to the PASTA property we know that thetotal arrival rate in the orbit is given by

120582 = 1205821199031199011 (1) (24)

Then the expected mean queue length in the orbit ENorbit isgiven by

ENorbit =infin

sum

119899=1

119899 (1199010119899 + 1199011119899 + 1199012119899)

= 1199011015840

0 (1) + 1199011015840

1 (1) + 1199011015840

2 (1)

(25)

Differentiating (18) with respect to 119911 and taking 119911 = 1 yields

ENorbit =1205822119903 (120582 + 120572) (120585 + 120578) + 120582

2119903120583120578

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(26)

Then the expected sojourn time of an arriving customer thatfinds the server busy and decides to join in the retrial orbit isgiven by

ESsystem =ENorbit

120582+120585 + 120578

120583120578 (27)

In the above equation the first term ENorbit120582 equals to themean waiting time in the orbit and the second term (120585 +

120578)120583120578 means the generalized service time in a unreliablesystem (see [14]) Using (24) and (26) yields (23) Theexpected overall queue length in the system ELoverall satisfies

ELoverall

= 1199011015840

0 (1) + 1199011015840

1 (1) + 1199011015840

2 (1) + 1199011 (1) + 1199012 (1)

=1205822119903120583120578 + 120582120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(28)

In the following part of this paper we assume that

120582 (120582 + 120572)

120572120583lt 1 (29)

This assumption originally comes from (21) but it has beenmodified slightly This means the system is stable under anystrategy 119903 followed by the customers Under this conditionwe are going to study the customersrsquo equilibrium joiningstrategy We have the following theorem

Theorem 4 In the partially observable single-server constantretrial queue we can derive a uniquemixed equilibrium joiningstrategy ldquoenter the orbit with probability 119903119890 when finding theserver busy upon arrivalrdquo when condition (29) holds Theprobability 119903119890 is given by

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119905119871119890 lt

119877

119862lt 119905119880119890

1 if 119877119862ge 119905119880119890

(30)

where

119903lowast

119890 =120572120583

120582 (120582 + 120572)minus(120582 + 120572) (120585 + 120578) + 120583120578

120582120578 (120582 + 120572)(119877

119862minus120585 + 120578

120583120578)

minus1

(31)

119905119871119890 =(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (32)

119905119880119890 =(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120583120572 minus 120582 (120582 + 120572))+120585 + 120578

120583120578 (33)

Proof Suppose that all customers follow the same joiningstrategy 119903when the server is busy at their arrival instantThenwe can get the expected net benefit of a customer who decidesto enter the system

119878119890 (119903) = 119877 minus CESsystem

= 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120572120583 minus (120582 + 120572) 120582119903)+120578 + 120585

120583120578]

(34)

We observe that 119878119890(119903) is strictly decreasing whenever 119903 isin

[0 1] and it has a unique maximum

119878119890 (0) = 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120578 + 120585

120583120578] (35)

and a unique minimum

119878119890 (1) = 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120572120583 minus 120582 (120582 + 120572))+120578 + 120585

120583120578] (36)

Therefore we consider the following cases

(i) When 119877119862 isin ((120578 + 120585)120583120578 119905119871119890] 119878119890(119903) is nonpositivewhenever 119903 isin [0 1] so the best response is balkingand the equilibrium joining probability is 119903119890 = 0

Mathematical Problems in Engineering 5

(ii) When 119877119862 isin (119905119871119890 119905119880119890) there exists a unique solutionof the equation 119878119890(119903) = 0 which is given by (31)

(iii) Similarly when 119877119862 isin [119905119880119890infin) the net benefit isalways positive so we obtain the rest branch of thetheorem

Because of themonotonicity of the function 119878119890(119903) the bestresponse is 1 whenever the joining probability 119903 is smallerthan 119903119890 If 119903 gt 119903119890 the best response is 0 for the net benefitis negative If 119903 = 119903119890 any strategy is a best response andcustomers are indifferent between entering the system andleaving This shows that an individualrsquos best response is adecreasing function of the strategy selected by the othercustomers that is the higher the joining probability of othercustomers the lower the net benefit Therefore we have anldquoavoid-the-crowdrdquo (ATC) situation

We will proceed to determine the solutions of the socialand profitmaximization problems that is we need to find theoptimal joining probabilities 119903soc and 119903prof that maximize thesocial net benefit and the administratorrsquos profit per unit timeWefirstly consider the socialmaximization problemWehavethe following theorem

Theorem 5 In the partially observable single-sever constantretrial queue there exists a uniquemixed joining strategy ldquoenterthe retrial orbit with probability 119903soc when the sever is busyrdquothat maximizes the social net benefit per time unit in whichcondition (29) holds The probability 119903soc is given by

119903soc =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119905119871soc lt

119877

119862lt 119905119880soc

1 if 119877119862ge 119905119880soc

(37)

where

119903lowast

soc =1198610120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 1198610 minus 120582120578 (38)

1198610 =radic1205782(120582119877 + 119862)

119862120572 (120582 + 120572) (39)

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2minus 12058321205782120572

12058212058321205782120572 (40)

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2minus 1205782[120583120572 minus 120582 (120582 + 120572)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

(41)

Proof For a given joining strategy 119903 the system behavesstationarily when condition (29) holds Customers that findthe server busy will join the orbit with probability 119903 Whenall customers follow the same strategy 119903 the social net benefitper time unit is given by

119878soc (119903) = 120582lowast(119903) 119877 minus CELoverall (42)

where 120582lowast(119903) denotes the mean effective arrival rate of thesystem and ELoverall denotes the expectedmean queue lengthof the system (including the customer in the server) underthe strategy 119903 The mean effective arrival rate is given by

120582lowast(119903) = 1205821199010 (1) + 1205821199031199011 (1) =

120582120583120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903) (43)

Using (22) and (43) social net benefit is given by

119878soc (119903) =120582120583120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)119877 minus 119862120582

times120582120583120578119903 + 120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] [120583120572 minus 120582119903 (120582 + 120572)]

(44)

It can be written as

119878soc (119903) =120583120578 (120582119877 + 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus

119862120583120572

120583120572 minus 120582119903 (120582 + 120572) (45)

Differentiating the above equation with respect to 119903 we get

119889

119889119903119878soc (119903) =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2minus

119862120583120572120582 (120582 + 120572)

[120583120572 minus 120582119903 (120582 + 120572)]2

(46)

119889

119889119903119878soc (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(120582119877 + 119862)

119862120572 (120582 + 120572)

(47)

From (46) we obtain

119889

119889119903119878soc (119903) |119903=0 =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120583120578]2minus119862120583120572120582 (120582 + 120572)

(120583120572)2

119889

119889119903119878soc (119903) |119903=1 =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582)]2

minus119862120583120572120582 (120582 + 120572)

[120583120572 minus 120582 (120582 + 120572)]2

(48)

Letting (119889119889119903)119878soc(119903)|119903=0 = 0 and (119889119889119903)119878soc(119903)|119903=1 = 0 weobtain 119877119862 = 119905119871soc and 119877119862 = 119905119880soc respectively Then

119889

119889119903119878soc (119903) |119903=0 le 0 lArrrArr

119877

119862le 119905119871soc (49)

119889

119889119903119878soc (119903) |119903=1 ge 0 lArrrArr

119877

119862ge 119905119880soc (50)

Therefore we consider the following situations

(i) When 119877119862 le 119905119871soc function 119878soc(119903) is decreasingwhenever 119903 isin [0 1] so the best response is balkingthen 119903soc = 0

6 Mathematical Problems in Engineering

(ii) When 119905119871soc lt 119877119862 lt 119905119880soc there exists a positivenumber 1199030 so that function 119878soc(119903) increases when119903 isin (0 1199030] and decreases when 119903 isin [1199030 1) then 1199030 is thepoint that maximizes the function 119878soc(119903) By solving(47) we know that the unique maximum of 119878soc(119903) isattained at (1198610120583120572 minus 120582(120585 + 120578) minus 120583120578)(120582(120582 + 120572)1198610 minus 120582120578)Therefore 119903soc given by (38) is the social maximizingstrategy

(iii) When 119877119862 ge 119905119880soc function 119878soc(119903) is increasingwhenever 119903 isin [0 1] so the best response is 1

We now compare the optimal joining strategy of thesocial maximization problem and the individual equilibriumjoining strategy Their relation is given by the followingtheorem

Theorem 6 In our single-server constant retrial rate queuewith breakdowns and repairs the joining probabilities 119903119890 and119903soc are ordered as

119903soc le 119903119890 (51)

Proof When the stable condition (29) holds it is easy to showthat 119905119871119890 lt 119905119871soc lt 119905119880119890 lt 119905119880soc where 119905119871119890 119905119871soc 119905119880119890 119905119880socare defined in Theorems 4 and 5 We consider the followingcases

(i) If 119877119862 le 119905119871119890 according to (30) and (37) 119903soc = 119903119890 = 0

(ii) If 119905119871119890 lt 119877119862 le 119905119871soc then 119903119890 isin (0 1) and 119903soc = 0 So119903soc lt 119903119890

(iii) If 119905119871soc lt 119877119862 lt 119905119880119890 then 119903119890 isin (0 1) and 119903soc isin (0 1)But we can get that 119903soc lt 119903119890 after some algebra

(iv) If 119905119880119890 le 119877119862 lt 119905119880soc then 119903119890 = 1 and 119903soc isin (0 1)Therefore 119903soc lt 119903119890 is obviously satisfied

(v) Finally if 119877119862 ge 119905119880soc then 119903119890 = 119903soc = 1

From the above analysis we complete the proof

The above result can be explained by the perspective of aneconomic viewpoint In the process of individual equilibriumanalysis we notice that customers will enter the orbit as longas their net benefit is nonnegative Then individualrsquos rationalbehavior causes excessive use of the resource in equilibriumthat is to say individualrsquos optimization leads to the systemhavingmore congestion thanwhat is socially desirable whichis the appearance of the negative externality

Having considered the individual equilibrium strategyand social maximization problem we will further identifythe profit optimization joining strategy In our model wesuppose that every customer will be imposed an entrancefee 119901 by the administrator By imposing the entrance feecustomersrsquo reward decreases to 119877 minus 119901 then their strategy willdiffer We will determine the strategy 119903prof that maximizes theadministratorrsquos profit per time unit The result is given by thefollowing

Theorem 7 In the partially observable single-server constantretrial queue we can derive a unique mixed joining strategyldquoenter the orbit with probability 119903prof when the server is busyrdquothat maximizes the administratorrsquos net profit per time unitwhen condition (29) holds The probability 119903prof is given by

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119905119871prof lt

119877

119862lt 119905119880prof

1 if 119877119862ge 119905119880prof

(52)

where

119903lowast

prof =1198620120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572)1198620 minus 120582120578 (53)

1198620 =radic1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2 (54)

1198771015840= 119877 minus 119862

120585 + 120578

120583120578 (55)

119905119871prof

=(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2+ 120582120583120578120572

2(120585 + 120578) minus 120583

212057821205722

120582120583212057821205722

(56)

119905119880prof = (120583(120582 + 120572)2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

+ [120582120578 (120585 + 120578) minus 1205831205782] [120583120572 minus 120582 (120582 + 120572)]

2)

times (1205821205831205782[120583120572 minus 120582 (120582 + 120572)]

2)minus1

(57)

Proof For a given joining strategy 119903 we define the function119878prof(119903) to be the administratorrsquos net profit per time unit Let120582lowast(119903) be the mean effective arrival rate in the system we

further assume that the entrance fee that the administratorimposes on customers is 119901(119903) Then we have

119878prof (119903) = 120582lowast(119903) 119901 (119903) (58)

where 120582lowast(119903) is given by (43) In order to obtain 119878prof(119903) weneed to determine the entrance fee 119901(119903) Noticing (34) weyield

119877 minus 119901 (119903) = 119862((120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)]+120585 + 120578

120583120578) (59)

from which we obtain

119901 (119903) = 119877 minus 119862120585 + 120578

120583120578minus 119862

(120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)] (60)

Because if the entrance fee 119901 lt 119901(119903) the administrator canimprove their net benefit through increasing the entrancefee but if 119901 gt 119901(119903) the individualrsquos net benefit is negative

Mathematical Problems in Engineering 7

customers will balk So (59) always holds whenever 119903 isin [0 1]Therefore the net profit of the administrator can be calculatedas follows119878prof (119903) = 120582

lowast(119903) 119901 (119903)

=120583120578 (120582119877

1015840+ 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus 119862

120583 (120582 + 120572)

120583120572 minus 120582119903 (120582 + 120572)

(61)

where 1198771015840 is given by (55) Differentiating the equation withrespect to 119903 we get

119889

119889119903119878prof (119903) =

1205821205831205782(1205821198771015840+ 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2

minus119862120583120582(120582 + 120572)

2

[120583120572 minus 120582119903 (120582 + 120572)]2

119889

119889119903119878prof (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2

(62)

The rest of the proof can proceed along the same line with theproof of Theorem 5

Now we are going to compare the joining probabilitieswhen considering the social and profit maximization prob-lems The result is given by the following theorem

Theorem 8 In our model when condition (29) holds theoptimal joining probabilities 119903soc and 119903prof are ordered as

119903prof le 119903soc (63)

Proof Firstly we proceed to compare the relationsamong 119905119871soc 119905119880soc 119905119871prof 119905119880prof Rewrite the expressionsof 119905119871soc 119905119880soc 119905119871prof 119905119880prof given inTheorems 5 and 7

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2

12058212058321205782120572minus1

120582

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

minus1

120582

119905119871prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2

120582120583212057821205722+120585 + 120578

120583120578minus1

120582

119905119880prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

+120585 + 120578

120583120578minus1

120582

(64)

It is readily shown that 119905119871soc lt 119905119871prof and 119905119880soc lt 119905119880prof Butthe relation between 119905119880soc and 119905119871prof is undefined Two casesmay take place that is 119905119880soc lt 119905119871prof or 119905119880soc ge 119905119871prof Weconsider the situation when 119905119880soc ge 119905119871prof the other case canbe analyzed in the same wayThen we consider the followingcases

(i) If 119877119862 isin [(120585 + 120578)120583120578 119905119871soc) then 119903soc = 119903prof = 0(ii) If 119877119862 isin [119905119871soc 119905119871prof) then 119903soc isin (0 1) 119903prof = 0(iii) If 119877119862 isin [119905119871prof 119905119880soc) then 119903soc isin (0 1) and 119903prof isin

(0 1) We can take a close look at the expressions of119903soc 119903prof that are presented in Theorems 5 and 7 Wedefine a function

119892 (119909) =119909120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 119909 minus 120582120578 (65)

Differentiating the above function with respect to 119909 weobtain

1198921015840(119909) =

1205822((120582 + 120572) (120585 + 120578) + 120583120578)

(120582 (120582 + 120572) 119909 minus 120582120578)2

(66)

It means that function 119892(119909) increases with respect to 119909 Onthe other hand we can easily derive 1198610 gt 1198620 where 1198610 and1198620 are given inTheorems 5 and 7Then we obtain 119903prof lt 119903soc

(i) If 119877119862 isin [119905119880soc 119905119880prof) then 119903soc = 1 and 119903prof isin (0 1)(ii) If 119877119862 isin [119905119880socinfin) then 119903soc = 1 and 119903prof = 1

We obtain 119903soc ge 119903prof from the above analysis

Remark 9 Economou and Kanta [6] have studied the indi-vidual equilibrium strategy and social and profit maximiza-tion problems in the single-server constant retrial queue inthe partially observable case In our model if we let 120585 = 0we can derive the same results as given in [6] This is notaccidentally happened because when 120585 = 0 breakdownswill never occur then these two models are equivalent Inaddition notice that customersrsquo strategies 119903119890 119903soc 119903prof arefunctions of 120585120578 where other parameters are fixed It meansthat 119903119890 119903soc 119903prof only relate to 120585120578 whatever was the value 120585and 120578 we take in this model

Remark 10 Another case is that 120582(120582 + 120572)120572120583 ge 1 is left tobe considered In this situation customersrsquo joining strategieswill be less than 1 when considering the equilibrium socialand profit maximization problems otherwise the server willbehave unstably That is to say we have the following joiningstrategies

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119877

119862gt 119905119871119890

119903119904119900119888 =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119877

119862gt 119905119871soc

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119877

119862gt 119905119871prof

(67)

and the relation 119903119890 ge 119903soc ge 119903prof holds all the same

8 Mathematical Problems in Engineering

Till now we have completed the analysis of the customersrsquostrategies of the partially observable retrial queueing systemWe will turn to the fully observable case

4 The Fully Observable Case

We now focus on the fully observable case assuming cus-tomers observe not only the state of the server but also theexact number of customers in the retrial orbit Customerthat finds the server idle will enter the system and begin tobe served immediately so this information is valuable onlyfor the customer who finds the server busy We know thatafter every service completion and there is no new customerarriving the customer at the head of the orbit queue willsuccessfully retry for service That means customers in theorbit are served on an FCFS basis Therefore observing theposition in the orbit they can assess precisely whether toenter or balk

Consider a tagged customer that finds the server busyupon arrival This customerrsquos mean overall sojourn time inthe system is not affected by the joiningbalking behavior ofthe other customers that find the server busy because if thetagged customer joins the orbit queue at the 119895th positionthen the late customers who find the server busy will jointhe queue at the 119895 + 1th 119895 + 2th positions behind himin the orbit So his sojourn time does not depend on theirdecisions but it depends on the arrival rate because the newlyarriving customer who finds the server idle will be servedimmediately

To study the general threshold strategy adopted by allcustomers in the fully observable case we will firstly considerthe mean overall sojourn time of the customer who finds theserver idle busy or broken upon arrival Let 119879(1 0) be themean sojourn time when the server is idle and there is nocustomers in the orbit at his arrival instant and let 119879(119894 119895)be the mean sojourn time of a tagged customer at the 119895thposition in the orbit given that the server is at state 119894 Whenall customers follow the general strategy the mean overallsojourn times 119879(119894 119895) are given by

119879 (1 0) =120585 + 120578

120583120578 (68)

119879 (1 119895) =1

120583 + 120585+

120585

120583 + 120585119879 (2 119895) +

120583

120583 + 120585119879 (0 119895) 119895 ge 1

(69)

119879 (0 119895) =1

120582 + 120572+

120582

120582 + 120572119879 (1 119895) +

120572

120582 + 120572119879 (1 119895 minus 1)

119895 ge 1

(70)

119879 (2 119895) =1

120578+ 119879 (1 119895) 119895 ge 0 (71)

Let us consider a customer in the 119895th position of the orbitwhen the server is busy Then this customer has to wait foran exponentially distributed time with rate 120585 + 120583 for thenext event to occur in which the server breaks down or the

service is completed With probability 120585(120585 + 120583) breakdownoccurs then the mean overall sojourn time becomes 119879(2 119895)and the tagged customer remains in the 119895th position of theorbit On the other hand with probability 120583(120585 + 120583) theservice is completed then the mean overall sojourn timebecomes 119879(1 119895 minus 1) and the tagged customer moves to the119895 minus 1th position Through the above analysis we obtain (69)Considering the other cases in the same line we obtain therest equations

Solving the above equations the mean overall sojourntimes are given by

119879 (0 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578 119895 ge 1

119879 (1 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 119895 ge 0

119879 (2 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120583 + 120585 + 120578

120583120578

119895 ge 0

(72)

Then we need to find the general equilibrium strategyfollowed by the customers when the server is busy To ensurethat customer will enter the orbit when finding the serverbusy and the queue in the orbit is empty we further assumethat 119877 minus 119862119879(1 1) gt 0 The condition can be written as

119877

119862gt(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (73)

This means customersrsquo reward should be larger in the fullyobservable case than in the former case even if the other sys-tematic parameters are equal We will identify the customersrsquoequilibrium threshold strategy and we have the followingtheorem

Theorem 11 In the fully observable single-server constantretrial queue there exists a unique threshold joining strategyldquoenter the retrial orbit if there are at most 119899119890 customers in theorbit whenever finding the server busyrdquo in which condition (73)holds The threshold 119899119890 is given by

119899119890 = lfloor119909119890rfloor (74)

where

119909119890 =120572120583120578119877 minus 119862120572 (120585 + 120578)

119862 [(120582 + 120572) (120585 + 120578) + 120583120578]minus 1 (75)

This strategy is the unique equilibrium strategy among allpossible strategies

Proof Consider a tagged customer that finds the system atthe state (1 119895) upon arrival If he decides to join he goesdirectly to the orbit and occupies the 119895 + 1th position Thenhis expected net benefit is 119878119890(119895) = 119877 minus 119862119879(1 119895 + 1) where

119878119890 (119895) = 119877 minus 119862[(119895 + 1)(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578]

(76)

Mathematical Problems in Engineering 9

middot middot middot

middot middot middot

middot middot middot

120582

120582 120582 120582

120583 120572 120582 120583 120582 120583120572 120572

120585 120578 120585 120578 120585 120578

120582 120583

120585 120578

120582 120583

120585 120578

0 n0 0 0 1

1 1

2 1

0 2

1 2

2 2

1 0

2 0

1 n

2 n

0 n + 1

1 n + 1

2 n + 1

Figure 3 Transition rate diagram of the fully observable case

The customer prefers to join if 119878119890(119895) gt 0 and he is indifferentbetween joining and balking while 119878119890(119895) = 0 and prefers tobalk if 119878119890(119895) lt 0 A newly arriving customer will enter ifand only if 119895 le 119899119890 when finding the server busy By solvinginequality 119878119890(119895) ge 0 we obtain the result

We will elucidate the uniqueness of the equilibriumthreshold joining strategy followed by all customers Indeedif 119909119890 is an integer number we just need to let 119899119890 = 119909119890 thenwe know 119878119890(119899119890) = 0 and the mixed threshold strategy isthat prescribes to enter if 119899 lt 119899119890 to balk if 119899 gt 119899119890 and beindifferent if 119899 = 119899119890 When 119909119890 is any positive value we let119899119890 = lfloor119909119890rfloor Hence in this case the equilibrium strategy is alsounique

We now turn our attention to the social and profit maxi-mization problems and identify the thresholds 119899soc and 119899profthat maximize the social net benefit and the administratorrsquosprofit per time unit To consummate this goal we need todetermine the stationary behavior of the system when allcustomers follow the same threshold strategy

Suppose that all customers that find the server busy followthe same threshold strategies 119899soc and 119899prof respectively Thatis to say customers who find the sever busy will enter thesystem as long as the number of customers in the orbitis at most 119899soc when considering the social maximizationproblem and there is at most 119899prof customers in the orbitwhen considering the administratorrsquos optimization problemHence the maximum number of customers in the orbitwill never be larger than 119899 + 1 when customers follow thesame threshold-119899 strategyThen we obtain a continuous-timeMarkov chain (119868(119905)119873(119905)) with state space 119878 = 0 1 2 times

0 1 2 119899 + 1 and its transition diagram is given byFigure 3

The stationary distribution of the system (119901119894119895(119894 119895) isin 119878)is given by the following proposition

Proposition 12 Consider the fully observable single-serverconstant retrial queue with breakdowns and repairs in whichcustomers follow a threshold-119899 strategy then the stationarydistribution of the system is given by

11990100 = 119861 (119899)120583120572

1205822

1199010119895 = 119861 (119899) 120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 119861 (119899)120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 119861 (119899)120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(77)

where

120588 =120582 (120582 + 120572)

120572120583

119861 (119899) = 120583120572

1205822+ (1 + 120588 + sdot sdot sdot + 120588

119899) +

120572

120582(1 + 120588 + sdot sdot sdot + 120588

119899+1)

+120572120585

120582120578(1 + 120588 + sdot sdot sdot + 120588

119899+1)

minus1

(78)

Proof The stationary distribution is obtained from the fol-lowing balance equations

12058211990100 = 12058311990110 (79)

(120582 + 120572) 1199010119895 = 1205831199011119895 119895 = 1 2 119899 + 1 (80)

1205821199011119895 = 1205721199010119895+1 119895 = 0 1 2 119899 (81)

1205781199012119895 = 1205851199011119895 119895 = 0 1 119899 + 1 (82)

(120583 + 120585) 1199011119899+1 = 1205781199012119899+1 + 1205821199011119899 + 1205821199010119899+1 (83)

Combining (80) with (81) we can obtain the recursiverelation

1199011119895 = 11990110120588119895 119895 = 1 2 119899 minus 1 (84)

Then we derive the expressions of 1199012119895 from (82)

1199012119895 =120585

12057812058811989511990110 119895 = 0 1 119899 + 1 (85)

Using (80)ndash(82) we can get

11990100 = 11990110

120583120572

1205822

1199010119895 = 11990110120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 11990110120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 11990110120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(86)

Then with the help of the normalizing equation sum119899+1119895=0 1199010119895 +sum119899+1

119895=0 1199011119895 + sum119899+1

119895=0 1199012119895 = 1 we get the expression of 11990110

By considering the stationary distribution we examinethe social welfare maximization threshold strategy We havethe following result

10 Mathematical Problems in Engineering

Theorem 13 In our single-server constant retrial queue inwhich the condition (73) holds we define the following func-tion

119891 (119909) =(120583120572 + (120585120578) 120582120572) (119909 + 1)

120583120572 (1 minus 120588)

minus120582 (120582 + 120572120588 (1 + (120585120578))) (1 minus 120588

119909+1)

120583120572(1 minus 120588)2

minus 1

(87)

(88)

If 119891(0) gt 119909119890 then the customersrsquo best response is balkingOtherwise there exists a unique threshold-119899soc strategy ldquo enterthe retrial orbit if the number of customers in the orbit is atmost119899soc whenever customers find the server busyrdquo that maximizesthe social net benefit per time unit The threshold 119899soc is givenby

119899soc = lfloor119909119904119900119888rfloor (89)

where 119909soc is the unique nonnegative root of the equation119891(119909) = 119909119890

Proof In order to determine the optimal threshold 119899soc weneed to compute the effective arrival rate of the systemand the expected mean sojourn time of the system On theone hand when all customers follow the same threshold-119899strategy the effective arrival rate is given by

120582 (119899) = 120582(

119899+1

sum

119895=0

1199010119895 (119899) +

119899

sum

119895=0

1199011119895 (119899)) (90)

On the other hand by conditioning on the state found by acustomer in the system upon arrival we obtain the expectedmean sojourn time if entering the system with threshold-119899policy

119864 [119878 (119899)] = 119860 + 119861 (91)

where

119860 =

119899+1

sum

119895=0

1199010119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

120585 + 120578

120583120578

119861 =

119899

sum

119895=0

1199011119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

times (120585 + 120578

120583120578+ (119895 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(92)

The expected social net benefit is given by

119878soc (119899) = 120582 (119899) 119877 minus 119862120582 (119899) 119864 [119878 (119899)] (93)

Using (90) and (91) and after some algebra we obtain

119878soc (119899) = 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578

times (120582 (119899) (119909119890 + 1) minus 120582

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899))

(94)

Next we need to show that 119878soc(119899) is unimodal and hasonly one maximal point We firstly consider the increments119878soc(119899) minus 119878soc(119899 minus 1) It is not difficult to show that

119878soc (119899) minus 119878soc (119899 minus 1) ge 0 lArrrArr 119891 (119899) le 119909119890 (95)

where 119891(119899) is

119891 (119899) =1198911 (119899)

1198912 (119899)minus 1 (96)

1198911 (119899) = 120582(

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899) minus

119899minus1

sum

119895=0

(119895 + 1) 1199011119895 (119899 minus 1))

(97)

1198912 (119899) = 120582 (119899) minus 120582 (119899 minus 1) (98)

Using the stationary distribution we can simplify the expres-sions of (97) and (98) as follows

1198911 (119899) =120572119861 (119899) 119861 (119899 minus 1) 120588

119899

1205822(1 minus 120588)2

times 119863 (99)

1198912 (119899) =1205831205722120588119899

1205822119861 (119899) 119861 (119899 minus 1) (100)

where

119863 = (120583120572 + 120582120572120585

120578) (119899 + 1) (1 minus 120588) minus 120582

2

minus120582120572120588(1 +120585

120578) + 1205822120588119899+1

+ 120582120572(1 +120585

120578) 120588119899+2

(101)

Plugging (99) and (100) in (96) and replacing 119899 by 119909 weobtain (87) To prove the unimodality of 119878soc(119899) we need toshow that function119891(119909) is increasingWe define the functionℎ [0infin) rarr 119877 with

ℎ (119909) = 119891 (119909) minus 119909 (102)

We have

1198892

1198891199092ℎ (119909) =

[120582 + 120572120588 (1 + (120585120578))] 120588119909+2

(ln 120588)2

(120582 + 120572) (1 minus 120588)2

gt 0

119909 ge 0

(103)

Then function ℎ(119909) is a strictly convex function and thereis no flex point with respect to 120588 and 119909 Therefore ℎ(119909) isincreasing and so is 119891(119909) = ℎ(119909) + 119909 Therefore two casesmay take place

(i) 119891(0) gt 119909119890 then 119878soc(119899) is decreasing and customerswho find the server busy will balk

(ii) 119891(119899soc) le 119909119890 lt 119891(119899soc + 1) then we have 119878soc(119899) minus119878soc(119899minus1) ge 0 for 119899 le 119899soc and 119878soc(119899)minus119878soc(119899minus1) lt 0for 119899 gt 119899soc Then the maximum of 119878soc(119899) occurs in119899soc = lfloor119909socrfloor in which 119909soc is the unique solution ofthe equation 119891(119909) = 119909119890

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

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Mathematical Problems in Engineering

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Mathematical Problems in Engineering 3

3 The Partially Observable Case

As we have mentioned a customer who finds the server idlewill enter the system because condition (7) ensures that hisreward exceeds his expected waiting cost In addition thedecision of the customer who finds the server idle will neverbe affected by the other customers So we just need to studythe behavior of the customers that find the server busy uponarrival We further assume that when an arriving customerfinds the server busy he will enter the orbit with probability119903 because the customer does not know the exact numberof customers in the orbit In order to identify the individualequilibrium social and profitmaximizing strategies we needto examine the stationary probabilities of the system and thenevaluate other key performances of the system Based on theabove assumptions the transition rate of the correspondingMarkov chain given in (3) should be substituted by

119902(1119895)(1119895+1) = 120582119903 119895 ge 0 (8)

Then we have the following propositions

Proposition 1 Consider the partially observable constantretrial queue with breakdowns and repairs in which customersenter the system with probability 119903 whenever the server isbusy with probability 1 whenever the server is idle and neverenter the system whenever the server is broken The stationaryprobabilities of the system are given by

1199010 (1) =120578 (120583 minus 120582119903)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

1199011 (1) =120582120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

1199012 (1) =120582120585

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

(9)

Proof Let 119901119894119899 (119894 = 0 1 2 119899 = 0 1 2 ) denote thestationary probabilities of the system at state (119894 119899) in which119894 denotes the state of the server and 119899 denotes the number ofcustomers in the orbit According to the transition principleof the stationary probabilities we get the following balanceequations

12058211990100 = 12058311990110 (10)

(120582 + 120572) 1199010119899 = 1205831199011119899 119899 ge 1 (11)

(120583 + 120585 + 120582119903) 11990110 = 12057211990101 + 12057811990120 + 12058211990100 (12)

(120583 + 120585 + 120582119903) 1199011119899 = 1205721199010119899+1 + 1205781199012119899 + 1205821199010119899 + 1205821199031199011119899minus1

119899 ge 1

(13)

1205781199012119899 = 1205851199011119899 119899 ge 0 (14)

The transition rate diagram is shown in Figure 2

120582

120582r 120582r

120583 120582 120583 120582 120583 120582 120583120572 120572 120572

120585 120578 120585 120578 120585 120578 120585 120578

middot middot middot

middot middot middot

middot middot middot

120582r

0 0 0 1

1 1

2 1

0 2

1 2

2 2

0 3

1 3

2 3

1 0

2 0

Figure 2 Transition rate diagram of the partially observable case

In order to solve the stationary probabilities of the systemwe define the following generating functions

119901119894 (119911) =

infin

sum

119899=0

119901119894119899119911119899 (15)

where 119894 = 0 1 2 Multiplying (10) by 1199110 and (11) by 119911119899 119899 ge 1

and summing over 119899 we get

(120582 + 120572) 1199010 (119911) = 1205831199011 (119911) + 12057211990100 (16)

Similarly we can obtain 1199011(119911) and 1199012(119911) from (12)ndash(14)

[minus1205821199031199112+ 119911 (120583 + 120585 + 120582119903)] 1199011 (119911)

= (120582119911 + 120572) 1199010 (119911) + 1205781199111199012 (119911) minus 12057211990100

1205781199012 (119911) = 1205851199011 (119911)

(17)

Using the above three equations we yield

1199010 (119911) =120572 (120583 minus 120582119903119911)

[120572120583 minus (120582 + 120572) 120582119903119911]11990100

1199011 (119911) =120582120572

[120572120583 minus (120582 + 120572) 120582119903119911]11990100

1199012 (119911) =120585120582120572

120578 [120572120583 minus (120582 + 120572) 120582119903119911]11990100

(18)

Using normalizing equation 1199010(1) + 1199011(1) + 1199012(1) = 1 weobtain

11990100 =120578

120572

120572120583 minus (120582 + 120572) 120582119903

120582 (120585 + 120578) + 120578 (120583 minus 120582119903) (19)

Therefore it is readily seen that the conclusion (9) can bederived based on the above results

Remark 2 We need to determine the stable condition of thismodel in fact from (10) (12) and (14) we obtain 12058211990311990110 =

12057211990101 In addition from (11) (13) and (14) we get 1205821199031199011119895 minus1205721199010119895+1 = 1205821199031199011119895minus1 minus1205721199010119895 (119895 ge 1) Proceeding to this recursiveprocess we get 1205821199031199011119895 = 1205721199010119895+1 (119895 ge 1) Using (11) again weobtain

1199010119895+1 =120582119903 (120582 + 120572)

1205721205831199010119895 119895 ge 1 (20)

4 Mathematical Problems in Engineering

So the system is stable if and only if

120582119903 (120582 + 120572)

120572120583lt 1 (21)

Proposition 3 In the partially observable case the expectedoverall queue length of this constant retrial queueing systemand the expected mean sojourn time of an arriving customerthat finds the server busy and decides to join the retrial orbitare given respectively by

ELoverall =1205822119903120583120578 + 120582120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(22)

ESsystem =(120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus (120582 + 120572) 120582119903]+120578 + 120585

120583120578 (23)

Proof According to the PASTA property we know that thetotal arrival rate in the orbit is given by

120582 = 1205821199031199011 (1) (24)

Then the expected mean queue length in the orbit ENorbit isgiven by

ENorbit =infin

sum

119899=1

119899 (1199010119899 + 1199011119899 + 1199012119899)

= 1199011015840

0 (1) + 1199011015840

1 (1) + 1199011015840

2 (1)

(25)

Differentiating (18) with respect to 119911 and taking 119911 = 1 yields

ENorbit =1205822119903 (120582 + 120572) (120585 + 120578) + 120582

2119903120583120578

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(26)

Then the expected sojourn time of an arriving customer thatfinds the server busy and decides to join in the retrial orbit isgiven by

ESsystem =ENorbit

120582+120585 + 120578

120583120578 (27)

In the above equation the first term ENorbit120582 equals to themean waiting time in the orbit and the second term (120585 +

120578)120583120578 means the generalized service time in a unreliablesystem (see [14]) Using (24) and (26) yields (23) Theexpected overall queue length in the system ELoverall satisfies

ELoverall

= 1199011015840

0 (1) + 1199011015840

1 (1) + 1199011015840

2 (1) + 1199011 (1) + 1199012 (1)

=1205822119903120583120578 + 120582120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(28)

In the following part of this paper we assume that

120582 (120582 + 120572)

120572120583lt 1 (29)

This assumption originally comes from (21) but it has beenmodified slightly This means the system is stable under anystrategy 119903 followed by the customers Under this conditionwe are going to study the customersrsquo equilibrium joiningstrategy We have the following theorem

Theorem 4 In the partially observable single-server constantretrial queue we can derive a uniquemixed equilibrium joiningstrategy ldquoenter the orbit with probability 119903119890 when finding theserver busy upon arrivalrdquo when condition (29) holds Theprobability 119903119890 is given by

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119905119871119890 lt

119877

119862lt 119905119880119890

1 if 119877119862ge 119905119880119890

(30)

where

119903lowast

119890 =120572120583

120582 (120582 + 120572)minus(120582 + 120572) (120585 + 120578) + 120583120578

120582120578 (120582 + 120572)(119877

119862minus120585 + 120578

120583120578)

minus1

(31)

119905119871119890 =(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (32)

119905119880119890 =(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120583120572 minus 120582 (120582 + 120572))+120585 + 120578

120583120578 (33)

Proof Suppose that all customers follow the same joiningstrategy 119903when the server is busy at their arrival instantThenwe can get the expected net benefit of a customer who decidesto enter the system

119878119890 (119903) = 119877 minus CESsystem

= 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120572120583 minus (120582 + 120572) 120582119903)+120578 + 120585

120583120578]

(34)

We observe that 119878119890(119903) is strictly decreasing whenever 119903 isin

[0 1] and it has a unique maximum

119878119890 (0) = 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120578 + 120585

120583120578] (35)

and a unique minimum

119878119890 (1) = 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120572120583 minus 120582 (120582 + 120572))+120578 + 120585

120583120578] (36)

Therefore we consider the following cases

(i) When 119877119862 isin ((120578 + 120585)120583120578 119905119871119890] 119878119890(119903) is nonpositivewhenever 119903 isin [0 1] so the best response is balkingand the equilibrium joining probability is 119903119890 = 0

Mathematical Problems in Engineering 5

(ii) When 119877119862 isin (119905119871119890 119905119880119890) there exists a unique solutionof the equation 119878119890(119903) = 0 which is given by (31)

(iii) Similarly when 119877119862 isin [119905119880119890infin) the net benefit isalways positive so we obtain the rest branch of thetheorem

Because of themonotonicity of the function 119878119890(119903) the bestresponse is 1 whenever the joining probability 119903 is smallerthan 119903119890 If 119903 gt 119903119890 the best response is 0 for the net benefitis negative If 119903 = 119903119890 any strategy is a best response andcustomers are indifferent between entering the system andleaving This shows that an individualrsquos best response is adecreasing function of the strategy selected by the othercustomers that is the higher the joining probability of othercustomers the lower the net benefit Therefore we have anldquoavoid-the-crowdrdquo (ATC) situation

We will proceed to determine the solutions of the socialand profitmaximization problems that is we need to find theoptimal joining probabilities 119903soc and 119903prof that maximize thesocial net benefit and the administratorrsquos profit per unit timeWefirstly consider the socialmaximization problemWehavethe following theorem

Theorem 5 In the partially observable single-sever constantretrial queue there exists a uniquemixed joining strategy ldquoenterthe retrial orbit with probability 119903soc when the sever is busyrdquothat maximizes the social net benefit per time unit in whichcondition (29) holds The probability 119903soc is given by

119903soc =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119905119871soc lt

119877

119862lt 119905119880soc

1 if 119877119862ge 119905119880soc

(37)

where

119903lowast

soc =1198610120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 1198610 minus 120582120578 (38)

1198610 =radic1205782(120582119877 + 119862)

119862120572 (120582 + 120572) (39)

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2minus 12058321205782120572

12058212058321205782120572 (40)

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2minus 1205782[120583120572 minus 120582 (120582 + 120572)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

(41)

Proof For a given joining strategy 119903 the system behavesstationarily when condition (29) holds Customers that findthe server busy will join the orbit with probability 119903 Whenall customers follow the same strategy 119903 the social net benefitper time unit is given by

119878soc (119903) = 120582lowast(119903) 119877 minus CELoverall (42)

where 120582lowast(119903) denotes the mean effective arrival rate of thesystem and ELoverall denotes the expectedmean queue lengthof the system (including the customer in the server) underthe strategy 119903 The mean effective arrival rate is given by

120582lowast(119903) = 1205821199010 (1) + 1205821199031199011 (1) =

120582120583120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903) (43)

Using (22) and (43) social net benefit is given by

119878soc (119903) =120582120583120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)119877 minus 119862120582

times120582120583120578119903 + 120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] [120583120572 minus 120582119903 (120582 + 120572)]

(44)

It can be written as

119878soc (119903) =120583120578 (120582119877 + 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus

119862120583120572

120583120572 minus 120582119903 (120582 + 120572) (45)

Differentiating the above equation with respect to 119903 we get

119889

119889119903119878soc (119903) =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2minus

119862120583120572120582 (120582 + 120572)

[120583120572 minus 120582119903 (120582 + 120572)]2

(46)

119889

119889119903119878soc (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(120582119877 + 119862)

119862120572 (120582 + 120572)

(47)

From (46) we obtain

119889

119889119903119878soc (119903) |119903=0 =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120583120578]2minus119862120583120572120582 (120582 + 120572)

(120583120572)2

119889

119889119903119878soc (119903) |119903=1 =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582)]2

minus119862120583120572120582 (120582 + 120572)

[120583120572 minus 120582 (120582 + 120572)]2

(48)

Letting (119889119889119903)119878soc(119903)|119903=0 = 0 and (119889119889119903)119878soc(119903)|119903=1 = 0 weobtain 119877119862 = 119905119871soc and 119877119862 = 119905119880soc respectively Then

119889

119889119903119878soc (119903) |119903=0 le 0 lArrrArr

119877

119862le 119905119871soc (49)

119889

119889119903119878soc (119903) |119903=1 ge 0 lArrrArr

119877

119862ge 119905119880soc (50)

Therefore we consider the following situations

(i) When 119877119862 le 119905119871soc function 119878soc(119903) is decreasingwhenever 119903 isin [0 1] so the best response is balkingthen 119903soc = 0

6 Mathematical Problems in Engineering

(ii) When 119905119871soc lt 119877119862 lt 119905119880soc there exists a positivenumber 1199030 so that function 119878soc(119903) increases when119903 isin (0 1199030] and decreases when 119903 isin [1199030 1) then 1199030 is thepoint that maximizes the function 119878soc(119903) By solving(47) we know that the unique maximum of 119878soc(119903) isattained at (1198610120583120572 minus 120582(120585 + 120578) minus 120583120578)(120582(120582 + 120572)1198610 minus 120582120578)Therefore 119903soc given by (38) is the social maximizingstrategy

(iii) When 119877119862 ge 119905119880soc function 119878soc(119903) is increasingwhenever 119903 isin [0 1] so the best response is 1

We now compare the optimal joining strategy of thesocial maximization problem and the individual equilibriumjoining strategy Their relation is given by the followingtheorem

Theorem 6 In our single-server constant retrial rate queuewith breakdowns and repairs the joining probabilities 119903119890 and119903soc are ordered as

119903soc le 119903119890 (51)

Proof When the stable condition (29) holds it is easy to showthat 119905119871119890 lt 119905119871soc lt 119905119880119890 lt 119905119880soc where 119905119871119890 119905119871soc 119905119880119890 119905119880socare defined in Theorems 4 and 5 We consider the followingcases

(i) If 119877119862 le 119905119871119890 according to (30) and (37) 119903soc = 119903119890 = 0

(ii) If 119905119871119890 lt 119877119862 le 119905119871soc then 119903119890 isin (0 1) and 119903soc = 0 So119903soc lt 119903119890

(iii) If 119905119871soc lt 119877119862 lt 119905119880119890 then 119903119890 isin (0 1) and 119903soc isin (0 1)But we can get that 119903soc lt 119903119890 after some algebra

(iv) If 119905119880119890 le 119877119862 lt 119905119880soc then 119903119890 = 1 and 119903soc isin (0 1)Therefore 119903soc lt 119903119890 is obviously satisfied

(v) Finally if 119877119862 ge 119905119880soc then 119903119890 = 119903soc = 1

From the above analysis we complete the proof

The above result can be explained by the perspective of aneconomic viewpoint In the process of individual equilibriumanalysis we notice that customers will enter the orbit as longas their net benefit is nonnegative Then individualrsquos rationalbehavior causes excessive use of the resource in equilibriumthat is to say individualrsquos optimization leads to the systemhavingmore congestion thanwhat is socially desirable whichis the appearance of the negative externality

Having considered the individual equilibrium strategyand social maximization problem we will further identifythe profit optimization joining strategy In our model wesuppose that every customer will be imposed an entrancefee 119901 by the administrator By imposing the entrance feecustomersrsquo reward decreases to 119877 minus 119901 then their strategy willdiffer We will determine the strategy 119903prof that maximizes theadministratorrsquos profit per time unit The result is given by thefollowing

Theorem 7 In the partially observable single-server constantretrial queue we can derive a unique mixed joining strategyldquoenter the orbit with probability 119903prof when the server is busyrdquothat maximizes the administratorrsquos net profit per time unitwhen condition (29) holds The probability 119903prof is given by

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119905119871prof lt

119877

119862lt 119905119880prof

1 if 119877119862ge 119905119880prof

(52)

where

119903lowast

prof =1198620120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572)1198620 minus 120582120578 (53)

1198620 =radic1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2 (54)

1198771015840= 119877 minus 119862

120585 + 120578

120583120578 (55)

119905119871prof

=(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2+ 120582120583120578120572

2(120585 + 120578) minus 120583

212057821205722

120582120583212057821205722

(56)

119905119880prof = (120583(120582 + 120572)2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

+ [120582120578 (120585 + 120578) minus 1205831205782] [120583120572 minus 120582 (120582 + 120572)]

2)

times (1205821205831205782[120583120572 minus 120582 (120582 + 120572)]

2)minus1

(57)

Proof For a given joining strategy 119903 we define the function119878prof(119903) to be the administratorrsquos net profit per time unit Let120582lowast(119903) be the mean effective arrival rate in the system we

further assume that the entrance fee that the administratorimposes on customers is 119901(119903) Then we have

119878prof (119903) = 120582lowast(119903) 119901 (119903) (58)

where 120582lowast(119903) is given by (43) In order to obtain 119878prof(119903) weneed to determine the entrance fee 119901(119903) Noticing (34) weyield

119877 minus 119901 (119903) = 119862((120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)]+120585 + 120578

120583120578) (59)

from which we obtain

119901 (119903) = 119877 minus 119862120585 + 120578

120583120578minus 119862

(120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)] (60)

Because if the entrance fee 119901 lt 119901(119903) the administrator canimprove their net benefit through increasing the entrancefee but if 119901 gt 119901(119903) the individualrsquos net benefit is negative

Mathematical Problems in Engineering 7

customers will balk So (59) always holds whenever 119903 isin [0 1]Therefore the net profit of the administrator can be calculatedas follows119878prof (119903) = 120582

lowast(119903) 119901 (119903)

=120583120578 (120582119877

1015840+ 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus 119862

120583 (120582 + 120572)

120583120572 minus 120582119903 (120582 + 120572)

(61)

where 1198771015840 is given by (55) Differentiating the equation withrespect to 119903 we get

119889

119889119903119878prof (119903) =

1205821205831205782(1205821198771015840+ 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2

minus119862120583120582(120582 + 120572)

2

[120583120572 minus 120582119903 (120582 + 120572)]2

119889

119889119903119878prof (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2

(62)

The rest of the proof can proceed along the same line with theproof of Theorem 5

Now we are going to compare the joining probabilitieswhen considering the social and profit maximization prob-lems The result is given by the following theorem

Theorem 8 In our model when condition (29) holds theoptimal joining probabilities 119903soc and 119903prof are ordered as

119903prof le 119903soc (63)

Proof Firstly we proceed to compare the relationsamong 119905119871soc 119905119880soc 119905119871prof 119905119880prof Rewrite the expressionsof 119905119871soc 119905119880soc 119905119871prof 119905119880prof given inTheorems 5 and 7

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2

12058212058321205782120572minus1

120582

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

minus1

120582

119905119871prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2

120582120583212057821205722+120585 + 120578

120583120578minus1

120582

119905119880prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

+120585 + 120578

120583120578minus1

120582

(64)

It is readily shown that 119905119871soc lt 119905119871prof and 119905119880soc lt 119905119880prof Butthe relation between 119905119880soc and 119905119871prof is undefined Two casesmay take place that is 119905119880soc lt 119905119871prof or 119905119880soc ge 119905119871prof Weconsider the situation when 119905119880soc ge 119905119871prof the other case canbe analyzed in the same wayThen we consider the followingcases

(i) If 119877119862 isin [(120585 + 120578)120583120578 119905119871soc) then 119903soc = 119903prof = 0(ii) If 119877119862 isin [119905119871soc 119905119871prof) then 119903soc isin (0 1) 119903prof = 0(iii) If 119877119862 isin [119905119871prof 119905119880soc) then 119903soc isin (0 1) and 119903prof isin

(0 1) We can take a close look at the expressions of119903soc 119903prof that are presented in Theorems 5 and 7 Wedefine a function

119892 (119909) =119909120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 119909 minus 120582120578 (65)

Differentiating the above function with respect to 119909 weobtain

1198921015840(119909) =

1205822((120582 + 120572) (120585 + 120578) + 120583120578)

(120582 (120582 + 120572) 119909 minus 120582120578)2

(66)

It means that function 119892(119909) increases with respect to 119909 Onthe other hand we can easily derive 1198610 gt 1198620 where 1198610 and1198620 are given inTheorems 5 and 7Then we obtain 119903prof lt 119903soc

(i) If 119877119862 isin [119905119880soc 119905119880prof) then 119903soc = 1 and 119903prof isin (0 1)(ii) If 119877119862 isin [119905119880socinfin) then 119903soc = 1 and 119903prof = 1

We obtain 119903soc ge 119903prof from the above analysis

Remark 9 Economou and Kanta [6] have studied the indi-vidual equilibrium strategy and social and profit maximiza-tion problems in the single-server constant retrial queue inthe partially observable case In our model if we let 120585 = 0we can derive the same results as given in [6] This is notaccidentally happened because when 120585 = 0 breakdownswill never occur then these two models are equivalent Inaddition notice that customersrsquo strategies 119903119890 119903soc 119903prof arefunctions of 120585120578 where other parameters are fixed It meansthat 119903119890 119903soc 119903prof only relate to 120585120578 whatever was the value 120585and 120578 we take in this model

Remark 10 Another case is that 120582(120582 + 120572)120572120583 ge 1 is left tobe considered In this situation customersrsquo joining strategieswill be less than 1 when considering the equilibrium socialand profit maximization problems otherwise the server willbehave unstably That is to say we have the following joiningstrategies

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119877

119862gt 119905119871119890

119903119904119900119888 =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119877

119862gt 119905119871soc

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119877

119862gt 119905119871prof

(67)

and the relation 119903119890 ge 119903soc ge 119903prof holds all the same

8 Mathematical Problems in Engineering

Till now we have completed the analysis of the customersrsquostrategies of the partially observable retrial queueing systemWe will turn to the fully observable case

4 The Fully Observable Case

We now focus on the fully observable case assuming cus-tomers observe not only the state of the server but also theexact number of customers in the retrial orbit Customerthat finds the server idle will enter the system and begin tobe served immediately so this information is valuable onlyfor the customer who finds the server busy We know thatafter every service completion and there is no new customerarriving the customer at the head of the orbit queue willsuccessfully retry for service That means customers in theorbit are served on an FCFS basis Therefore observing theposition in the orbit they can assess precisely whether toenter or balk

Consider a tagged customer that finds the server busyupon arrival This customerrsquos mean overall sojourn time inthe system is not affected by the joiningbalking behavior ofthe other customers that find the server busy because if thetagged customer joins the orbit queue at the 119895th positionthen the late customers who find the server busy will jointhe queue at the 119895 + 1th 119895 + 2th positions behind himin the orbit So his sojourn time does not depend on theirdecisions but it depends on the arrival rate because the newlyarriving customer who finds the server idle will be servedimmediately

To study the general threshold strategy adopted by allcustomers in the fully observable case we will firstly considerthe mean overall sojourn time of the customer who finds theserver idle busy or broken upon arrival Let 119879(1 0) be themean sojourn time when the server is idle and there is nocustomers in the orbit at his arrival instant and let 119879(119894 119895)be the mean sojourn time of a tagged customer at the 119895thposition in the orbit given that the server is at state 119894 Whenall customers follow the general strategy the mean overallsojourn times 119879(119894 119895) are given by

119879 (1 0) =120585 + 120578

120583120578 (68)

119879 (1 119895) =1

120583 + 120585+

120585

120583 + 120585119879 (2 119895) +

120583

120583 + 120585119879 (0 119895) 119895 ge 1

(69)

119879 (0 119895) =1

120582 + 120572+

120582

120582 + 120572119879 (1 119895) +

120572

120582 + 120572119879 (1 119895 minus 1)

119895 ge 1

(70)

119879 (2 119895) =1

120578+ 119879 (1 119895) 119895 ge 0 (71)

Let us consider a customer in the 119895th position of the orbitwhen the server is busy Then this customer has to wait foran exponentially distributed time with rate 120585 + 120583 for thenext event to occur in which the server breaks down or the

service is completed With probability 120585(120585 + 120583) breakdownoccurs then the mean overall sojourn time becomes 119879(2 119895)and the tagged customer remains in the 119895th position of theorbit On the other hand with probability 120583(120585 + 120583) theservice is completed then the mean overall sojourn timebecomes 119879(1 119895 minus 1) and the tagged customer moves to the119895 minus 1th position Through the above analysis we obtain (69)Considering the other cases in the same line we obtain therest equations

Solving the above equations the mean overall sojourntimes are given by

119879 (0 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578 119895 ge 1

119879 (1 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 119895 ge 0

119879 (2 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120583 + 120585 + 120578

120583120578

119895 ge 0

(72)

Then we need to find the general equilibrium strategyfollowed by the customers when the server is busy To ensurethat customer will enter the orbit when finding the serverbusy and the queue in the orbit is empty we further assumethat 119877 minus 119862119879(1 1) gt 0 The condition can be written as

119877

119862gt(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (73)

This means customersrsquo reward should be larger in the fullyobservable case than in the former case even if the other sys-tematic parameters are equal We will identify the customersrsquoequilibrium threshold strategy and we have the followingtheorem

Theorem 11 In the fully observable single-server constantretrial queue there exists a unique threshold joining strategyldquoenter the retrial orbit if there are at most 119899119890 customers in theorbit whenever finding the server busyrdquo in which condition (73)holds The threshold 119899119890 is given by

119899119890 = lfloor119909119890rfloor (74)

where

119909119890 =120572120583120578119877 minus 119862120572 (120585 + 120578)

119862 [(120582 + 120572) (120585 + 120578) + 120583120578]minus 1 (75)

This strategy is the unique equilibrium strategy among allpossible strategies

Proof Consider a tagged customer that finds the system atthe state (1 119895) upon arrival If he decides to join he goesdirectly to the orbit and occupies the 119895 + 1th position Thenhis expected net benefit is 119878119890(119895) = 119877 minus 119862119879(1 119895 + 1) where

119878119890 (119895) = 119877 minus 119862[(119895 + 1)(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578]

(76)

Mathematical Problems in Engineering 9

middot middot middot

middot middot middot

middot middot middot

120582

120582 120582 120582

120583 120572 120582 120583 120582 120583120572 120572

120585 120578 120585 120578 120585 120578

120582 120583

120585 120578

120582 120583

120585 120578

0 n0 0 0 1

1 1

2 1

0 2

1 2

2 2

1 0

2 0

1 n

2 n

0 n + 1

1 n + 1

2 n + 1

Figure 3 Transition rate diagram of the fully observable case

The customer prefers to join if 119878119890(119895) gt 0 and he is indifferentbetween joining and balking while 119878119890(119895) = 0 and prefers tobalk if 119878119890(119895) lt 0 A newly arriving customer will enter ifand only if 119895 le 119899119890 when finding the server busy By solvinginequality 119878119890(119895) ge 0 we obtain the result

We will elucidate the uniqueness of the equilibriumthreshold joining strategy followed by all customers Indeedif 119909119890 is an integer number we just need to let 119899119890 = 119909119890 thenwe know 119878119890(119899119890) = 0 and the mixed threshold strategy isthat prescribes to enter if 119899 lt 119899119890 to balk if 119899 gt 119899119890 and beindifferent if 119899 = 119899119890 When 119909119890 is any positive value we let119899119890 = lfloor119909119890rfloor Hence in this case the equilibrium strategy is alsounique

We now turn our attention to the social and profit maxi-mization problems and identify the thresholds 119899soc and 119899profthat maximize the social net benefit and the administratorrsquosprofit per time unit To consummate this goal we need todetermine the stationary behavior of the system when allcustomers follow the same threshold strategy

Suppose that all customers that find the server busy followthe same threshold strategies 119899soc and 119899prof respectively Thatis to say customers who find the sever busy will enter thesystem as long as the number of customers in the orbitis at most 119899soc when considering the social maximizationproblem and there is at most 119899prof customers in the orbitwhen considering the administratorrsquos optimization problemHence the maximum number of customers in the orbitwill never be larger than 119899 + 1 when customers follow thesame threshold-119899 strategyThen we obtain a continuous-timeMarkov chain (119868(119905)119873(119905)) with state space 119878 = 0 1 2 times

0 1 2 119899 + 1 and its transition diagram is given byFigure 3

The stationary distribution of the system (119901119894119895(119894 119895) isin 119878)is given by the following proposition

Proposition 12 Consider the fully observable single-serverconstant retrial queue with breakdowns and repairs in whichcustomers follow a threshold-119899 strategy then the stationarydistribution of the system is given by

11990100 = 119861 (119899)120583120572

1205822

1199010119895 = 119861 (119899) 120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 119861 (119899)120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 119861 (119899)120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(77)

where

120588 =120582 (120582 + 120572)

120572120583

119861 (119899) = 120583120572

1205822+ (1 + 120588 + sdot sdot sdot + 120588

119899) +

120572

120582(1 + 120588 + sdot sdot sdot + 120588

119899+1)

+120572120585

120582120578(1 + 120588 + sdot sdot sdot + 120588

119899+1)

minus1

(78)

Proof The stationary distribution is obtained from the fol-lowing balance equations

12058211990100 = 12058311990110 (79)

(120582 + 120572) 1199010119895 = 1205831199011119895 119895 = 1 2 119899 + 1 (80)

1205821199011119895 = 1205721199010119895+1 119895 = 0 1 2 119899 (81)

1205781199012119895 = 1205851199011119895 119895 = 0 1 119899 + 1 (82)

(120583 + 120585) 1199011119899+1 = 1205781199012119899+1 + 1205821199011119899 + 1205821199010119899+1 (83)

Combining (80) with (81) we can obtain the recursiverelation

1199011119895 = 11990110120588119895 119895 = 1 2 119899 minus 1 (84)

Then we derive the expressions of 1199012119895 from (82)

1199012119895 =120585

12057812058811989511990110 119895 = 0 1 119899 + 1 (85)

Using (80)ndash(82) we can get

11990100 = 11990110

120583120572

1205822

1199010119895 = 11990110120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 11990110120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 11990110120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(86)

Then with the help of the normalizing equation sum119899+1119895=0 1199010119895 +sum119899+1

119895=0 1199011119895 + sum119899+1

119895=0 1199012119895 = 1 we get the expression of 11990110

By considering the stationary distribution we examinethe social welfare maximization threshold strategy We havethe following result

10 Mathematical Problems in Engineering

Theorem 13 In our single-server constant retrial queue inwhich the condition (73) holds we define the following func-tion

119891 (119909) =(120583120572 + (120585120578) 120582120572) (119909 + 1)

120583120572 (1 minus 120588)

minus120582 (120582 + 120572120588 (1 + (120585120578))) (1 minus 120588

119909+1)

120583120572(1 minus 120588)2

minus 1

(87)

(88)

If 119891(0) gt 119909119890 then the customersrsquo best response is balkingOtherwise there exists a unique threshold-119899soc strategy ldquo enterthe retrial orbit if the number of customers in the orbit is atmost119899soc whenever customers find the server busyrdquo that maximizesthe social net benefit per time unit The threshold 119899soc is givenby

119899soc = lfloor119909119904119900119888rfloor (89)

where 119909soc is the unique nonnegative root of the equation119891(119909) = 119909119890

Proof In order to determine the optimal threshold 119899soc weneed to compute the effective arrival rate of the systemand the expected mean sojourn time of the system On theone hand when all customers follow the same threshold-119899strategy the effective arrival rate is given by

120582 (119899) = 120582(

119899+1

sum

119895=0

1199010119895 (119899) +

119899

sum

119895=0

1199011119895 (119899)) (90)

On the other hand by conditioning on the state found by acustomer in the system upon arrival we obtain the expectedmean sojourn time if entering the system with threshold-119899policy

119864 [119878 (119899)] = 119860 + 119861 (91)

where

119860 =

119899+1

sum

119895=0

1199010119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

120585 + 120578

120583120578

119861 =

119899

sum

119895=0

1199011119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

times (120585 + 120578

120583120578+ (119895 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(92)

The expected social net benefit is given by

119878soc (119899) = 120582 (119899) 119877 minus 119862120582 (119899) 119864 [119878 (119899)] (93)

Using (90) and (91) and after some algebra we obtain

119878soc (119899) = 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578

times (120582 (119899) (119909119890 + 1) minus 120582

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899))

(94)

Next we need to show that 119878soc(119899) is unimodal and hasonly one maximal point We firstly consider the increments119878soc(119899) minus 119878soc(119899 minus 1) It is not difficult to show that

119878soc (119899) minus 119878soc (119899 minus 1) ge 0 lArrrArr 119891 (119899) le 119909119890 (95)

where 119891(119899) is

119891 (119899) =1198911 (119899)

1198912 (119899)minus 1 (96)

1198911 (119899) = 120582(

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899) minus

119899minus1

sum

119895=0

(119895 + 1) 1199011119895 (119899 minus 1))

(97)

1198912 (119899) = 120582 (119899) minus 120582 (119899 minus 1) (98)

Using the stationary distribution we can simplify the expres-sions of (97) and (98) as follows

1198911 (119899) =120572119861 (119899) 119861 (119899 minus 1) 120588

119899

1205822(1 minus 120588)2

times 119863 (99)

1198912 (119899) =1205831205722120588119899

1205822119861 (119899) 119861 (119899 minus 1) (100)

where

119863 = (120583120572 + 120582120572120585

120578) (119899 + 1) (1 minus 120588) minus 120582

2

minus120582120572120588(1 +120585

120578) + 1205822120588119899+1

+ 120582120572(1 +120585

120578) 120588119899+2

(101)

Plugging (99) and (100) in (96) and replacing 119899 by 119909 weobtain (87) To prove the unimodality of 119878soc(119899) we need toshow that function119891(119909) is increasingWe define the functionℎ [0infin) rarr 119877 with

ℎ (119909) = 119891 (119909) minus 119909 (102)

We have

1198892

1198891199092ℎ (119909) =

[120582 + 120572120588 (1 + (120585120578))] 120588119909+2

(ln 120588)2

(120582 + 120572) (1 minus 120588)2

gt 0

119909 ge 0

(103)

Then function ℎ(119909) is a strictly convex function and thereis no flex point with respect to 120588 and 119909 Therefore ℎ(119909) isincreasing and so is 119891(119909) = ℎ(119909) + 119909 Therefore two casesmay take place

(i) 119891(0) gt 119909119890 then 119878soc(119899) is decreasing and customerswho find the server busy will balk

(ii) 119891(119899soc) le 119909119890 lt 119891(119899soc + 1) then we have 119878soc(119899) minus119878soc(119899minus1) ge 0 for 119899 le 119899soc and 119878soc(119899)minus119878soc(119899minus1) lt 0for 119899 gt 119899soc Then the maximum of 119878soc(119899) occurs in119899soc = lfloor119909socrfloor in which 119909soc is the unique solution ofthe equation 119891(119909) = 119909119890

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

So the system is stable if and only if

120582119903 (120582 + 120572)

120572120583lt 1 (21)

Proposition 3 In the partially observable case the expectedoverall queue length of this constant retrial queueing systemand the expected mean sojourn time of an arriving customerthat finds the server busy and decides to join the retrial orbitare given respectively by

ELoverall =1205822119903120583120578 + 120582120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(22)

ESsystem =(120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus (120582 + 120572) 120582119903]+120578 + 120585

120583120578 (23)

Proof According to the PASTA property we know that thetotal arrival rate in the orbit is given by

120582 = 1205821199031199011 (1) (24)

Then the expected mean queue length in the orbit ENorbit isgiven by

ENorbit =infin

sum

119899=1

119899 (1199010119899 + 1199011119899 + 1199012119899)

= 1199011015840

0 (1) + 1199011015840

1 (1) + 1199011015840

2 (1)

(25)

Differentiating (18) with respect to 119911 and taking 119911 = 1 yields

ENorbit =1205822119903 (120582 + 120572) (120585 + 120578) + 120582

2119903120583120578

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(26)

Then the expected sojourn time of an arriving customer thatfinds the server busy and decides to join in the retrial orbit isgiven by

ESsystem =ENorbit

120582+120585 + 120578

120583120578 (27)

In the above equation the first term ENorbit120582 equals to themean waiting time in the orbit and the second term (120585 +

120578)120583120578 means the generalized service time in a unreliablesystem (see [14]) Using (24) and (26) yields (23) Theexpected overall queue length in the system ELoverall satisfies

ELoverall

= 1199011015840

0 (1) + 1199011015840

1 (1) + 1199011015840

2 (1) + 1199011 (1) + 1199012 (1)

=1205822119903120583120578 + 120582120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] times [120572120583 minus 120582119903 (120582 + 120572)]

(28)

In the following part of this paper we assume that

120582 (120582 + 120572)

120572120583lt 1 (29)

This assumption originally comes from (21) but it has beenmodified slightly This means the system is stable under anystrategy 119903 followed by the customers Under this conditionwe are going to study the customersrsquo equilibrium joiningstrategy We have the following theorem

Theorem 4 In the partially observable single-server constantretrial queue we can derive a uniquemixed equilibrium joiningstrategy ldquoenter the orbit with probability 119903119890 when finding theserver busy upon arrivalrdquo when condition (29) holds Theprobability 119903119890 is given by

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119905119871119890 lt

119877

119862lt 119905119880119890

1 if 119877119862ge 119905119880119890

(30)

where

119903lowast

119890 =120572120583

120582 (120582 + 120572)minus(120582 + 120572) (120585 + 120578) + 120583120578

120582120578 (120582 + 120572)(119877

119862minus120585 + 120578

120583120578)

minus1

(31)

119905119871119890 =(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (32)

119905119880119890 =(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120583120572 minus 120582 (120582 + 120572))+120585 + 120578

120583120578 (33)

Proof Suppose that all customers follow the same joiningstrategy 119903when the server is busy at their arrival instantThenwe can get the expected net benefit of a customer who decidesto enter the system

119878119890 (119903) = 119877 minus CESsystem

= 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120572120583 minus (120582 + 120572) 120582119903)+120578 + 120585

120583120578]

(34)

We observe that 119878119890(119903) is strictly decreasing whenever 119903 isin

[0 1] and it has a unique maximum

119878119890 (0) = 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120578 + 120585

120583120578] (35)

and a unique minimum

119878119890 (1) = 119877 minus 119862[(120582 + 120572) (120585 + 120578) + 120583120578

120578 (120572120583 minus 120582 (120582 + 120572))+120578 + 120585

120583120578] (36)

Therefore we consider the following cases

(i) When 119877119862 isin ((120578 + 120585)120583120578 119905119871119890] 119878119890(119903) is nonpositivewhenever 119903 isin [0 1] so the best response is balkingand the equilibrium joining probability is 119903119890 = 0

Mathematical Problems in Engineering 5

(ii) When 119877119862 isin (119905119871119890 119905119880119890) there exists a unique solutionof the equation 119878119890(119903) = 0 which is given by (31)

(iii) Similarly when 119877119862 isin [119905119880119890infin) the net benefit isalways positive so we obtain the rest branch of thetheorem

Because of themonotonicity of the function 119878119890(119903) the bestresponse is 1 whenever the joining probability 119903 is smallerthan 119903119890 If 119903 gt 119903119890 the best response is 0 for the net benefitis negative If 119903 = 119903119890 any strategy is a best response andcustomers are indifferent between entering the system andleaving This shows that an individualrsquos best response is adecreasing function of the strategy selected by the othercustomers that is the higher the joining probability of othercustomers the lower the net benefit Therefore we have anldquoavoid-the-crowdrdquo (ATC) situation

We will proceed to determine the solutions of the socialand profitmaximization problems that is we need to find theoptimal joining probabilities 119903soc and 119903prof that maximize thesocial net benefit and the administratorrsquos profit per unit timeWefirstly consider the socialmaximization problemWehavethe following theorem

Theorem 5 In the partially observable single-sever constantretrial queue there exists a uniquemixed joining strategy ldquoenterthe retrial orbit with probability 119903soc when the sever is busyrdquothat maximizes the social net benefit per time unit in whichcondition (29) holds The probability 119903soc is given by

119903soc =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119905119871soc lt

119877

119862lt 119905119880soc

1 if 119877119862ge 119905119880soc

(37)

where

119903lowast

soc =1198610120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 1198610 minus 120582120578 (38)

1198610 =radic1205782(120582119877 + 119862)

119862120572 (120582 + 120572) (39)

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2minus 12058321205782120572

12058212058321205782120572 (40)

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2minus 1205782[120583120572 minus 120582 (120582 + 120572)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

(41)

Proof For a given joining strategy 119903 the system behavesstationarily when condition (29) holds Customers that findthe server busy will join the orbit with probability 119903 Whenall customers follow the same strategy 119903 the social net benefitper time unit is given by

119878soc (119903) = 120582lowast(119903) 119877 minus CELoverall (42)

where 120582lowast(119903) denotes the mean effective arrival rate of thesystem and ELoverall denotes the expectedmean queue lengthof the system (including the customer in the server) underthe strategy 119903 The mean effective arrival rate is given by

120582lowast(119903) = 1205821199010 (1) + 1205821199031199011 (1) =

120582120583120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903) (43)

Using (22) and (43) social net benefit is given by

119878soc (119903) =120582120583120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)119877 minus 119862120582

times120582120583120578119903 + 120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] [120583120572 minus 120582119903 (120582 + 120572)]

(44)

It can be written as

119878soc (119903) =120583120578 (120582119877 + 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus

119862120583120572

120583120572 minus 120582119903 (120582 + 120572) (45)

Differentiating the above equation with respect to 119903 we get

119889

119889119903119878soc (119903) =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2minus

119862120583120572120582 (120582 + 120572)

[120583120572 minus 120582119903 (120582 + 120572)]2

(46)

119889

119889119903119878soc (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(120582119877 + 119862)

119862120572 (120582 + 120572)

(47)

From (46) we obtain

119889

119889119903119878soc (119903) |119903=0 =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120583120578]2minus119862120583120572120582 (120582 + 120572)

(120583120572)2

119889

119889119903119878soc (119903) |119903=1 =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582)]2

minus119862120583120572120582 (120582 + 120572)

[120583120572 minus 120582 (120582 + 120572)]2

(48)

Letting (119889119889119903)119878soc(119903)|119903=0 = 0 and (119889119889119903)119878soc(119903)|119903=1 = 0 weobtain 119877119862 = 119905119871soc and 119877119862 = 119905119880soc respectively Then

119889

119889119903119878soc (119903) |119903=0 le 0 lArrrArr

119877

119862le 119905119871soc (49)

119889

119889119903119878soc (119903) |119903=1 ge 0 lArrrArr

119877

119862ge 119905119880soc (50)

Therefore we consider the following situations

(i) When 119877119862 le 119905119871soc function 119878soc(119903) is decreasingwhenever 119903 isin [0 1] so the best response is balkingthen 119903soc = 0

6 Mathematical Problems in Engineering

(ii) When 119905119871soc lt 119877119862 lt 119905119880soc there exists a positivenumber 1199030 so that function 119878soc(119903) increases when119903 isin (0 1199030] and decreases when 119903 isin [1199030 1) then 1199030 is thepoint that maximizes the function 119878soc(119903) By solving(47) we know that the unique maximum of 119878soc(119903) isattained at (1198610120583120572 minus 120582(120585 + 120578) minus 120583120578)(120582(120582 + 120572)1198610 minus 120582120578)Therefore 119903soc given by (38) is the social maximizingstrategy

(iii) When 119877119862 ge 119905119880soc function 119878soc(119903) is increasingwhenever 119903 isin [0 1] so the best response is 1

We now compare the optimal joining strategy of thesocial maximization problem and the individual equilibriumjoining strategy Their relation is given by the followingtheorem

Theorem 6 In our single-server constant retrial rate queuewith breakdowns and repairs the joining probabilities 119903119890 and119903soc are ordered as

119903soc le 119903119890 (51)

Proof When the stable condition (29) holds it is easy to showthat 119905119871119890 lt 119905119871soc lt 119905119880119890 lt 119905119880soc where 119905119871119890 119905119871soc 119905119880119890 119905119880socare defined in Theorems 4 and 5 We consider the followingcases

(i) If 119877119862 le 119905119871119890 according to (30) and (37) 119903soc = 119903119890 = 0

(ii) If 119905119871119890 lt 119877119862 le 119905119871soc then 119903119890 isin (0 1) and 119903soc = 0 So119903soc lt 119903119890

(iii) If 119905119871soc lt 119877119862 lt 119905119880119890 then 119903119890 isin (0 1) and 119903soc isin (0 1)But we can get that 119903soc lt 119903119890 after some algebra

(iv) If 119905119880119890 le 119877119862 lt 119905119880soc then 119903119890 = 1 and 119903soc isin (0 1)Therefore 119903soc lt 119903119890 is obviously satisfied

(v) Finally if 119877119862 ge 119905119880soc then 119903119890 = 119903soc = 1

From the above analysis we complete the proof

The above result can be explained by the perspective of aneconomic viewpoint In the process of individual equilibriumanalysis we notice that customers will enter the orbit as longas their net benefit is nonnegative Then individualrsquos rationalbehavior causes excessive use of the resource in equilibriumthat is to say individualrsquos optimization leads to the systemhavingmore congestion thanwhat is socially desirable whichis the appearance of the negative externality

Having considered the individual equilibrium strategyand social maximization problem we will further identifythe profit optimization joining strategy In our model wesuppose that every customer will be imposed an entrancefee 119901 by the administrator By imposing the entrance feecustomersrsquo reward decreases to 119877 minus 119901 then their strategy willdiffer We will determine the strategy 119903prof that maximizes theadministratorrsquos profit per time unit The result is given by thefollowing

Theorem 7 In the partially observable single-server constantretrial queue we can derive a unique mixed joining strategyldquoenter the orbit with probability 119903prof when the server is busyrdquothat maximizes the administratorrsquos net profit per time unitwhen condition (29) holds The probability 119903prof is given by

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119905119871prof lt

119877

119862lt 119905119880prof

1 if 119877119862ge 119905119880prof

(52)

where

119903lowast

prof =1198620120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572)1198620 minus 120582120578 (53)

1198620 =radic1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2 (54)

1198771015840= 119877 minus 119862

120585 + 120578

120583120578 (55)

119905119871prof

=(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2+ 120582120583120578120572

2(120585 + 120578) minus 120583

212057821205722

120582120583212057821205722

(56)

119905119880prof = (120583(120582 + 120572)2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

+ [120582120578 (120585 + 120578) minus 1205831205782] [120583120572 minus 120582 (120582 + 120572)]

2)

times (1205821205831205782[120583120572 minus 120582 (120582 + 120572)]

2)minus1

(57)

Proof For a given joining strategy 119903 we define the function119878prof(119903) to be the administratorrsquos net profit per time unit Let120582lowast(119903) be the mean effective arrival rate in the system we

further assume that the entrance fee that the administratorimposes on customers is 119901(119903) Then we have

119878prof (119903) = 120582lowast(119903) 119901 (119903) (58)

where 120582lowast(119903) is given by (43) In order to obtain 119878prof(119903) weneed to determine the entrance fee 119901(119903) Noticing (34) weyield

119877 minus 119901 (119903) = 119862((120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)]+120585 + 120578

120583120578) (59)

from which we obtain

119901 (119903) = 119877 minus 119862120585 + 120578

120583120578minus 119862

(120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)] (60)

Because if the entrance fee 119901 lt 119901(119903) the administrator canimprove their net benefit through increasing the entrancefee but if 119901 gt 119901(119903) the individualrsquos net benefit is negative

Mathematical Problems in Engineering 7

customers will balk So (59) always holds whenever 119903 isin [0 1]Therefore the net profit of the administrator can be calculatedas follows119878prof (119903) = 120582

lowast(119903) 119901 (119903)

=120583120578 (120582119877

1015840+ 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus 119862

120583 (120582 + 120572)

120583120572 minus 120582119903 (120582 + 120572)

(61)

where 1198771015840 is given by (55) Differentiating the equation withrespect to 119903 we get

119889

119889119903119878prof (119903) =

1205821205831205782(1205821198771015840+ 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2

minus119862120583120582(120582 + 120572)

2

[120583120572 minus 120582119903 (120582 + 120572)]2

119889

119889119903119878prof (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2

(62)

The rest of the proof can proceed along the same line with theproof of Theorem 5

Now we are going to compare the joining probabilitieswhen considering the social and profit maximization prob-lems The result is given by the following theorem

Theorem 8 In our model when condition (29) holds theoptimal joining probabilities 119903soc and 119903prof are ordered as

119903prof le 119903soc (63)

Proof Firstly we proceed to compare the relationsamong 119905119871soc 119905119880soc 119905119871prof 119905119880prof Rewrite the expressionsof 119905119871soc 119905119880soc 119905119871prof 119905119880prof given inTheorems 5 and 7

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2

12058212058321205782120572minus1

120582

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

minus1

120582

119905119871prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2

120582120583212057821205722+120585 + 120578

120583120578minus1

120582

119905119880prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

+120585 + 120578

120583120578minus1

120582

(64)

It is readily shown that 119905119871soc lt 119905119871prof and 119905119880soc lt 119905119880prof Butthe relation between 119905119880soc and 119905119871prof is undefined Two casesmay take place that is 119905119880soc lt 119905119871prof or 119905119880soc ge 119905119871prof Weconsider the situation when 119905119880soc ge 119905119871prof the other case canbe analyzed in the same wayThen we consider the followingcases

(i) If 119877119862 isin [(120585 + 120578)120583120578 119905119871soc) then 119903soc = 119903prof = 0(ii) If 119877119862 isin [119905119871soc 119905119871prof) then 119903soc isin (0 1) 119903prof = 0(iii) If 119877119862 isin [119905119871prof 119905119880soc) then 119903soc isin (0 1) and 119903prof isin

(0 1) We can take a close look at the expressions of119903soc 119903prof that are presented in Theorems 5 and 7 Wedefine a function

119892 (119909) =119909120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 119909 minus 120582120578 (65)

Differentiating the above function with respect to 119909 weobtain

1198921015840(119909) =

1205822((120582 + 120572) (120585 + 120578) + 120583120578)

(120582 (120582 + 120572) 119909 minus 120582120578)2

(66)

It means that function 119892(119909) increases with respect to 119909 Onthe other hand we can easily derive 1198610 gt 1198620 where 1198610 and1198620 are given inTheorems 5 and 7Then we obtain 119903prof lt 119903soc

(i) If 119877119862 isin [119905119880soc 119905119880prof) then 119903soc = 1 and 119903prof isin (0 1)(ii) If 119877119862 isin [119905119880socinfin) then 119903soc = 1 and 119903prof = 1

We obtain 119903soc ge 119903prof from the above analysis

Remark 9 Economou and Kanta [6] have studied the indi-vidual equilibrium strategy and social and profit maximiza-tion problems in the single-server constant retrial queue inthe partially observable case In our model if we let 120585 = 0we can derive the same results as given in [6] This is notaccidentally happened because when 120585 = 0 breakdownswill never occur then these two models are equivalent Inaddition notice that customersrsquo strategies 119903119890 119903soc 119903prof arefunctions of 120585120578 where other parameters are fixed It meansthat 119903119890 119903soc 119903prof only relate to 120585120578 whatever was the value 120585and 120578 we take in this model

Remark 10 Another case is that 120582(120582 + 120572)120572120583 ge 1 is left tobe considered In this situation customersrsquo joining strategieswill be less than 1 when considering the equilibrium socialand profit maximization problems otherwise the server willbehave unstably That is to say we have the following joiningstrategies

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119877

119862gt 119905119871119890

119903119904119900119888 =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119877

119862gt 119905119871soc

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119877

119862gt 119905119871prof

(67)

and the relation 119903119890 ge 119903soc ge 119903prof holds all the same

8 Mathematical Problems in Engineering

Till now we have completed the analysis of the customersrsquostrategies of the partially observable retrial queueing systemWe will turn to the fully observable case

4 The Fully Observable Case

We now focus on the fully observable case assuming cus-tomers observe not only the state of the server but also theexact number of customers in the retrial orbit Customerthat finds the server idle will enter the system and begin tobe served immediately so this information is valuable onlyfor the customer who finds the server busy We know thatafter every service completion and there is no new customerarriving the customer at the head of the orbit queue willsuccessfully retry for service That means customers in theorbit are served on an FCFS basis Therefore observing theposition in the orbit they can assess precisely whether toenter or balk

Consider a tagged customer that finds the server busyupon arrival This customerrsquos mean overall sojourn time inthe system is not affected by the joiningbalking behavior ofthe other customers that find the server busy because if thetagged customer joins the orbit queue at the 119895th positionthen the late customers who find the server busy will jointhe queue at the 119895 + 1th 119895 + 2th positions behind himin the orbit So his sojourn time does not depend on theirdecisions but it depends on the arrival rate because the newlyarriving customer who finds the server idle will be servedimmediately

To study the general threshold strategy adopted by allcustomers in the fully observable case we will firstly considerthe mean overall sojourn time of the customer who finds theserver idle busy or broken upon arrival Let 119879(1 0) be themean sojourn time when the server is idle and there is nocustomers in the orbit at his arrival instant and let 119879(119894 119895)be the mean sojourn time of a tagged customer at the 119895thposition in the orbit given that the server is at state 119894 Whenall customers follow the general strategy the mean overallsojourn times 119879(119894 119895) are given by

119879 (1 0) =120585 + 120578

120583120578 (68)

119879 (1 119895) =1

120583 + 120585+

120585

120583 + 120585119879 (2 119895) +

120583

120583 + 120585119879 (0 119895) 119895 ge 1

(69)

119879 (0 119895) =1

120582 + 120572+

120582

120582 + 120572119879 (1 119895) +

120572

120582 + 120572119879 (1 119895 minus 1)

119895 ge 1

(70)

119879 (2 119895) =1

120578+ 119879 (1 119895) 119895 ge 0 (71)

Let us consider a customer in the 119895th position of the orbitwhen the server is busy Then this customer has to wait foran exponentially distributed time with rate 120585 + 120583 for thenext event to occur in which the server breaks down or the

service is completed With probability 120585(120585 + 120583) breakdownoccurs then the mean overall sojourn time becomes 119879(2 119895)and the tagged customer remains in the 119895th position of theorbit On the other hand with probability 120583(120585 + 120583) theservice is completed then the mean overall sojourn timebecomes 119879(1 119895 minus 1) and the tagged customer moves to the119895 minus 1th position Through the above analysis we obtain (69)Considering the other cases in the same line we obtain therest equations

Solving the above equations the mean overall sojourntimes are given by

119879 (0 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578 119895 ge 1

119879 (1 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 119895 ge 0

119879 (2 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120583 + 120585 + 120578

120583120578

119895 ge 0

(72)

Then we need to find the general equilibrium strategyfollowed by the customers when the server is busy To ensurethat customer will enter the orbit when finding the serverbusy and the queue in the orbit is empty we further assumethat 119877 minus 119862119879(1 1) gt 0 The condition can be written as

119877

119862gt(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (73)

This means customersrsquo reward should be larger in the fullyobservable case than in the former case even if the other sys-tematic parameters are equal We will identify the customersrsquoequilibrium threshold strategy and we have the followingtheorem

Theorem 11 In the fully observable single-server constantretrial queue there exists a unique threshold joining strategyldquoenter the retrial orbit if there are at most 119899119890 customers in theorbit whenever finding the server busyrdquo in which condition (73)holds The threshold 119899119890 is given by

119899119890 = lfloor119909119890rfloor (74)

where

119909119890 =120572120583120578119877 minus 119862120572 (120585 + 120578)

119862 [(120582 + 120572) (120585 + 120578) + 120583120578]minus 1 (75)

This strategy is the unique equilibrium strategy among allpossible strategies

Proof Consider a tagged customer that finds the system atthe state (1 119895) upon arrival If he decides to join he goesdirectly to the orbit and occupies the 119895 + 1th position Thenhis expected net benefit is 119878119890(119895) = 119877 minus 119862119879(1 119895 + 1) where

119878119890 (119895) = 119877 minus 119862[(119895 + 1)(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578]

(76)

Mathematical Problems in Engineering 9

middot middot middot

middot middot middot

middot middot middot

120582

120582 120582 120582

120583 120572 120582 120583 120582 120583120572 120572

120585 120578 120585 120578 120585 120578

120582 120583

120585 120578

120582 120583

120585 120578

0 n0 0 0 1

1 1

2 1

0 2

1 2

2 2

1 0

2 0

1 n

2 n

0 n + 1

1 n + 1

2 n + 1

Figure 3 Transition rate diagram of the fully observable case

The customer prefers to join if 119878119890(119895) gt 0 and he is indifferentbetween joining and balking while 119878119890(119895) = 0 and prefers tobalk if 119878119890(119895) lt 0 A newly arriving customer will enter ifand only if 119895 le 119899119890 when finding the server busy By solvinginequality 119878119890(119895) ge 0 we obtain the result

We will elucidate the uniqueness of the equilibriumthreshold joining strategy followed by all customers Indeedif 119909119890 is an integer number we just need to let 119899119890 = 119909119890 thenwe know 119878119890(119899119890) = 0 and the mixed threshold strategy isthat prescribes to enter if 119899 lt 119899119890 to balk if 119899 gt 119899119890 and beindifferent if 119899 = 119899119890 When 119909119890 is any positive value we let119899119890 = lfloor119909119890rfloor Hence in this case the equilibrium strategy is alsounique

We now turn our attention to the social and profit maxi-mization problems and identify the thresholds 119899soc and 119899profthat maximize the social net benefit and the administratorrsquosprofit per time unit To consummate this goal we need todetermine the stationary behavior of the system when allcustomers follow the same threshold strategy

Suppose that all customers that find the server busy followthe same threshold strategies 119899soc and 119899prof respectively Thatis to say customers who find the sever busy will enter thesystem as long as the number of customers in the orbitis at most 119899soc when considering the social maximizationproblem and there is at most 119899prof customers in the orbitwhen considering the administratorrsquos optimization problemHence the maximum number of customers in the orbitwill never be larger than 119899 + 1 when customers follow thesame threshold-119899 strategyThen we obtain a continuous-timeMarkov chain (119868(119905)119873(119905)) with state space 119878 = 0 1 2 times

0 1 2 119899 + 1 and its transition diagram is given byFigure 3

The stationary distribution of the system (119901119894119895(119894 119895) isin 119878)is given by the following proposition

Proposition 12 Consider the fully observable single-serverconstant retrial queue with breakdowns and repairs in whichcustomers follow a threshold-119899 strategy then the stationarydistribution of the system is given by

11990100 = 119861 (119899)120583120572

1205822

1199010119895 = 119861 (119899) 120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 119861 (119899)120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 119861 (119899)120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(77)

where

120588 =120582 (120582 + 120572)

120572120583

119861 (119899) = 120583120572

1205822+ (1 + 120588 + sdot sdot sdot + 120588

119899) +

120572

120582(1 + 120588 + sdot sdot sdot + 120588

119899+1)

+120572120585

120582120578(1 + 120588 + sdot sdot sdot + 120588

119899+1)

minus1

(78)

Proof The stationary distribution is obtained from the fol-lowing balance equations

12058211990100 = 12058311990110 (79)

(120582 + 120572) 1199010119895 = 1205831199011119895 119895 = 1 2 119899 + 1 (80)

1205821199011119895 = 1205721199010119895+1 119895 = 0 1 2 119899 (81)

1205781199012119895 = 1205851199011119895 119895 = 0 1 119899 + 1 (82)

(120583 + 120585) 1199011119899+1 = 1205781199012119899+1 + 1205821199011119899 + 1205821199010119899+1 (83)

Combining (80) with (81) we can obtain the recursiverelation

1199011119895 = 11990110120588119895 119895 = 1 2 119899 minus 1 (84)

Then we derive the expressions of 1199012119895 from (82)

1199012119895 =120585

12057812058811989511990110 119895 = 0 1 119899 + 1 (85)

Using (80)ndash(82) we can get

11990100 = 11990110

120583120572

1205822

1199010119895 = 11990110120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 11990110120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 11990110120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(86)

Then with the help of the normalizing equation sum119899+1119895=0 1199010119895 +sum119899+1

119895=0 1199011119895 + sum119899+1

119895=0 1199012119895 = 1 we get the expression of 11990110

By considering the stationary distribution we examinethe social welfare maximization threshold strategy We havethe following result

10 Mathematical Problems in Engineering

Theorem 13 In our single-server constant retrial queue inwhich the condition (73) holds we define the following func-tion

119891 (119909) =(120583120572 + (120585120578) 120582120572) (119909 + 1)

120583120572 (1 minus 120588)

minus120582 (120582 + 120572120588 (1 + (120585120578))) (1 minus 120588

119909+1)

120583120572(1 minus 120588)2

minus 1

(87)

(88)

If 119891(0) gt 119909119890 then the customersrsquo best response is balkingOtherwise there exists a unique threshold-119899soc strategy ldquo enterthe retrial orbit if the number of customers in the orbit is atmost119899soc whenever customers find the server busyrdquo that maximizesthe social net benefit per time unit The threshold 119899soc is givenby

119899soc = lfloor119909119904119900119888rfloor (89)

where 119909soc is the unique nonnegative root of the equation119891(119909) = 119909119890

Proof In order to determine the optimal threshold 119899soc weneed to compute the effective arrival rate of the systemand the expected mean sojourn time of the system On theone hand when all customers follow the same threshold-119899strategy the effective arrival rate is given by

120582 (119899) = 120582(

119899+1

sum

119895=0

1199010119895 (119899) +

119899

sum

119895=0

1199011119895 (119899)) (90)

On the other hand by conditioning on the state found by acustomer in the system upon arrival we obtain the expectedmean sojourn time if entering the system with threshold-119899policy

119864 [119878 (119899)] = 119860 + 119861 (91)

where

119860 =

119899+1

sum

119895=0

1199010119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

120585 + 120578

120583120578

119861 =

119899

sum

119895=0

1199011119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

times (120585 + 120578

120583120578+ (119895 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(92)

The expected social net benefit is given by

119878soc (119899) = 120582 (119899) 119877 minus 119862120582 (119899) 119864 [119878 (119899)] (93)

Using (90) and (91) and after some algebra we obtain

119878soc (119899) = 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578

times (120582 (119899) (119909119890 + 1) minus 120582

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899))

(94)

Next we need to show that 119878soc(119899) is unimodal and hasonly one maximal point We firstly consider the increments119878soc(119899) minus 119878soc(119899 minus 1) It is not difficult to show that

119878soc (119899) minus 119878soc (119899 minus 1) ge 0 lArrrArr 119891 (119899) le 119909119890 (95)

where 119891(119899) is

119891 (119899) =1198911 (119899)

1198912 (119899)minus 1 (96)

1198911 (119899) = 120582(

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899) minus

119899minus1

sum

119895=0

(119895 + 1) 1199011119895 (119899 minus 1))

(97)

1198912 (119899) = 120582 (119899) minus 120582 (119899 minus 1) (98)

Using the stationary distribution we can simplify the expres-sions of (97) and (98) as follows

1198911 (119899) =120572119861 (119899) 119861 (119899 minus 1) 120588

119899

1205822(1 minus 120588)2

times 119863 (99)

1198912 (119899) =1205831205722120588119899

1205822119861 (119899) 119861 (119899 minus 1) (100)

where

119863 = (120583120572 + 120582120572120585

120578) (119899 + 1) (1 minus 120588) minus 120582

2

minus120582120572120588(1 +120585

120578) + 1205822120588119899+1

+ 120582120572(1 +120585

120578) 120588119899+2

(101)

Plugging (99) and (100) in (96) and replacing 119899 by 119909 weobtain (87) To prove the unimodality of 119878soc(119899) we need toshow that function119891(119909) is increasingWe define the functionℎ [0infin) rarr 119877 with

ℎ (119909) = 119891 (119909) minus 119909 (102)

We have

1198892

1198891199092ℎ (119909) =

[120582 + 120572120588 (1 + (120585120578))] 120588119909+2

(ln 120588)2

(120582 + 120572) (1 minus 120588)2

gt 0

119909 ge 0

(103)

Then function ℎ(119909) is a strictly convex function and thereis no flex point with respect to 120588 and 119909 Therefore ℎ(119909) isincreasing and so is 119891(119909) = ℎ(119909) + 119909 Therefore two casesmay take place

(i) 119891(0) gt 119909119890 then 119878soc(119899) is decreasing and customerswho find the server busy will balk

(ii) 119891(119899soc) le 119909119890 lt 119891(119899soc + 1) then we have 119878soc(119899) minus119878soc(119899minus1) ge 0 for 119899 le 119899soc and 119878soc(119899)minus119878soc(119899minus1) lt 0for 119899 gt 119899soc Then the maximum of 119878soc(119899) occurs in119899soc = lfloor119909socrfloor in which 119909soc is the unique solution ofthe equation 119891(119909) = 119909119890

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

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Mathematical Problems in Engineering

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Mathematical Problems in Engineering 5

(ii) When 119877119862 isin (119905119871119890 119905119880119890) there exists a unique solutionof the equation 119878119890(119903) = 0 which is given by (31)

(iii) Similarly when 119877119862 isin [119905119880119890infin) the net benefit isalways positive so we obtain the rest branch of thetheorem

Because of themonotonicity of the function 119878119890(119903) the bestresponse is 1 whenever the joining probability 119903 is smallerthan 119903119890 If 119903 gt 119903119890 the best response is 0 for the net benefitis negative If 119903 = 119903119890 any strategy is a best response andcustomers are indifferent between entering the system andleaving This shows that an individualrsquos best response is adecreasing function of the strategy selected by the othercustomers that is the higher the joining probability of othercustomers the lower the net benefit Therefore we have anldquoavoid-the-crowdrdquo (ATC) situation

We will proceed to determine the solutions of the socialand profitmaximization problems that is we need to find theoptimal joining probabilities 119903soc and 119903prof that maximize thesocial net benefit and the administratorrsquos profit per unit timeWefirstly consider the socialmaximization problemWehavethe following theorem

Theorem 5 In the partially observable single-sever constantretrial queue there exists a uniquemixed joining strategy ldquoenterthe retrial orbit with probability 119903soc when the sever is busyrdquothat maximizes the social net benefit per time unit in whichcondition (29) holds The probability 119903soc is given by

119903soc =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119905119871soc lt

119877

119862lt 119905119880soc

1 if 119877119862ge 119905119880soc

(37)

where

119903lowast

soc =1198610120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 1198610 minus 120582120578 (38)

1198610 =radic1205782(120582119877 + 119862)

119862120572 (120582 + 120572) (39)

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2minus 12058321205782120572

12058212058321205782120572 (40)

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2minus 1205782[120583120572 minus 120582 (120582 + 120572)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

(41)

Proof For a given joining strategy 119903 the system behavesstationarily when condition (29) holds Customers that findthe server busy will join the orbit with probability 119903 Whenall customers follow the same strategy 119903 the social net benefitper time unit is given by

119878soc (119903) = 120582lowast(119903) 119877 minus CELoverall (42)

where 120582lowast(119903) denotes the mean effective arrival rate of thesystem and ELoverall denotes the expectedmean queue lengthof the system (including the customer in the server) underthe strategy 119903 The mean effective arrival rate is given by

120582lowast(119903) = 1205821199010 (1) + 1205821199031199011 (1) =

120582120583120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903) (43)

Using (22) and (43) social net benefit is given by

119878soc (119903) =120582120583120578

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)119877 minus 119862120582

times120582120583120578119903 + 120583120572 (120585 + 120578)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)] [120583120572 minus 120582119903 (120582 + 120572)]

(44)

It can be written as

119878soc (119903) =120583120578 (120582119877 + 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus

119862120583120572

120583120572 minus 120582119903 (120582 + 120572) (45)

Differentiating the above equation with respect to 119903 we get

119889

119889119903119878soc (119903) =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2minus

119862120583120572120582 (120582 + 120572)

[120583120572 minus 120582119903 (120582 + 120572)]2

(46)

119889

119889119903119878soc (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(120582119877 + 119862)

119862120572 (120582 + 120572)

(47)

From (46) we obtain

119889

119889119903119878soc (119903) |119903=0 =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120583120578]2minus119862120583120572120582 (120582 + 120572)

(120583120572)2

119889

119889119903119878soc (119903) |119903=1 =

1205821205831205782(120582119877 + 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582)]2

minus119862120583120572120582 (120582 + 120572)

[120583120572 minus 120582 (120582 + 120572)]2

(48)

Letting (119889119889119903)119878soc(119903)|119903=0 = 0 and (119889119889119903)119878soc(119903)|119903=1 = 0 weobtain 119877119862 = 119905119871soc and 119877119862 = 119905119880soc respectively Then

119889

119889119903119878soc (119903) |119903=0 le 0 lArrrArr

119877

119862le 119905119871soc (49)

119889

119889119903119878soc (119903) |119903=1 ge 0 lArrrArr

119877

119862ge 119905119880soc (50)

Therefore we consider the following situations

(i) When 119877119862 le 119905119871soc function 119878soc(119903) is decreasingwhenever 119903 isin [0 1] so the best response is balkingthen 119903soc = 0

6 Mathematical Problems in Engineering

(ii) When 119905119871soc lt 119877119862 lt 119905119880soc there exists a positivenumber 1199030 so that function 119878soc(119903) increases when119903 isin (0 1199030] and decreases when 119903 isin [1199030 1) then 1199030 is thepoint that maximizes the function 119878soc(119903) By solving(47) we know that the unique maximum of 119878soc(119903) isattained at (1198610120583120572 minus 120582(120585 + 120578) minus 120583120578)(120582(120582 + 120572)1198610 minus 120582120578)Therefore 119903soc given by (38) is the social maximizingstrategy

(iii) When 119877119862 ge 119905119880soc function 119878soc(119903) is increasingwhenever 119903 isin [0 1] so the best response is 1

We now compare the optimal joining strategy of thesocial maximization problem and the individual equilibriumjoining strategy Their relation is given by the followingtheorem

Theorem 6 In our single-server constant retrial rate queuewith breakdowns and repairs the joining probabilities 119903119890 and119903soc are ordered as

119903soc le 119903119890 (51)

Proof When the stable condition (29) holds it is easy to showthat 119905119871119890 lt 119905119871soc lt 119905119880119890 lt 119905119880soc where 119905119871119890 119905119871soc 119905119880119890 119905119880socare defined in Theorems 4 and 5 We consider the followingcases

(i) If 119877119862 le 119905119871119890 according to (30) and (37) 119903soc = 119903119890 = 0

(ii) If 119905119871119890 lt 119877119862 le 119905119871soc then 119903119890 isin (0 1) and 119903soc = 0 So119903soc lt 119903119890

(iii) If 119905119871soc lt 119877119862 lt 119905119880119890 then 119903119890 isin (0 1) and 119903soc isin (0 1)But we can get that 119903soc lt 119903119890 after some algebra

(iv) If 119905119880119890 le 119877119862 lt 119905119880soc then 119903119890 = 1 and 119903soc isin (0 1)Therefore 119903soc lt 119903119890 is obviously satisfied

(v) Finally if 119877119862 ge 119905119880soc then 119903119890 = 119903soc = 1

From the above analysis we complete the proof

The above result can be explained by the perspective of aneconomic viewpoint In the process of individual equilibriumanalysis we notice that customers will enter the orbit as longas their net benefit is nonnegative Then individualrsquos rationalbehavior causes excessive use of the resource in equilibriumthat is to say individualrsquos optimization leads to the systemhavingmore congestion thanwhat is socially desirable whichis the appearance of the negative externality

Having considered the individual equilibrium strategyand social maximization problem we will further identifythe profit optimization joining strategy In our model wesuppose that every customer will be imposed an entrancefee 119901 by the administrator By imposing the entrance feecustomersrsquo reward decreases to 119877 minus 119901 then their strategy willdiffer We will determine the strategy 119903prof that maximizes theadministratorrsquos profit per time unit The result is given by thefollowing

Theorem 7 In the partially observable single-server constantretrial queue we can derive a unique mixed joining strategyldquoenter the orbit with probability 119903prof when the server is busyrdquothat maximizes the administratorrsquos net profit per time unitwhen condition (29) holds The probability 119903prof is given by

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119905119871prof lt

119877

119862lt 119905119880prof

1 if 119877119862ge 119905119880prof

(52)

where

119903lowast

prof =1198620120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572)1198620 minus 120582120578 (53)

1198620 =radic1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2 (54)

1198771015840= 119877 minus 119862

120585 + 120578

120583120578 (55)

119905119871prof

=(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2+ 120582120583120578120572

2(120585 + 120578) minus 120583

212057821205722

120582120583212057821205722

(56)

119905119880prof = (120583(120582 + 120572)2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

+ [120582120578 (120585 + 120578) minus 1205831205782] [120583120572 minus 120582 (120582 + 120572)]

2)

times (1205821205831205782[120583120572 minus 120582 (120582 + 120572)]

2)minus1

(57)

Proof For a given joining strategy 119903 we define the function119878prof(119903) to be the administratorrsquos net profit per time unit Let120582lowast(119903) be the mean effective arrival rate in the system we

further assume that the entrance fee that the administratorimposes on customers is 119901(119903) Then we have

119878prof (119903) = 120582lowast(119903) 119901 (119903) (58)

where 120582lowast(119903) is given by (43) In order to obtain 119878prof(119903) weneed to determine the entrance fee 119901(119903) Noticing (34) weyield

119877 minus 119901 (119903) = 119862((120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)]+120585 + 120578

120583120578) (59)

from which we obtain

119901 (119903) = 119877 minus 119862120585 + 120578

120583120578minus 119862

(120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)] (60)

Because if the entrance fee 119901 lt 119901(119903) the administrator canimprove their net benefit through increasing the entrancefee but if 119901 gt 119901(119903) the individualrsquos net benefit is negative

Mathematical Problems in Engineering 7

customers will balk So (59) always holds whenever 119903 isin [0 1]Therefore the net profit of the administrator can be calculatedas follows119878prof (119903) = 120582

lowast(119903) 119901 (119903)

=120583120578 (120582119877

1015840+ 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus 119862

120583 (120582 + 120572)

120583120572 minus 120582119903 (120582 + 120572)

(61)

where 1198771015840 is given by (55) Differentiating the equation withrespect to 119903 we get

119889

119889119903119878prof (119903) =

1205821205831205782(1205821198771015840+ 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2

minus119862120583120582(120582 + 120572)

2

[120583120572 minus 120582119903 (120582 + 120572)]2

119889

119889119903119878prof (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2

(62)

The rest of the proof can proceed along the same line with theproof of Theorem 5

Now we are going to compare the joining probabilitieswhen considering the social and profit maximization prob-lems The result is given by the following theorem

Theorem 8 In our model when condition (29) holds theoptimal joining probabilities 119903soc and 119903prof are ordered as

119903prof le 119903soc (63)

Proof Firstly we proceed to compare the relationsamong 119905119871soc 119905119880soc 119905119871prof 119905119880prof Rewrite the expressionsof 119905119871soc 119905119880soc 119905119871prof 119905119880prof given inTheorems 5 and 7

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2

12058212058321205782120572minus1

120582

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

minus1

120582

119905119871prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2

120582120583212057821205722+120585 + 120578

120583120578minus1

120582

119905119880prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

+120585 + 120578

120583120578minus1

120582

(64)

It is readily shown that 119905119871soc lt 119905119871prof and 119905119880soc lt 119905119880prof Butthe relation between 119905119880soc and 119905119871prof is undefined Two casesmay take place that is 119905119880soc lt 119905119871prof or 119905119880soc ge 119905119871prof Weconsider the situation when 119905119880soc ge 119905119871prof the other case canbe analyzed in the same wayThen we consider the followingcases

(i) If 119877119862 isin [(120585 + 120578)120583120578 119905119871soc) then 119903soc = 119903prof = 0(ii) If 119877119862 isin [119905119871soc 119905119871prof) then 119903soc isin (0 1) 119903prof = 0(iii) If 119877119862 isin [119905119871prof 119905119880soc) then 119903soc isin (0 1) and 119903prof isin

(0 1) We can take a close look at the expressions of119903soc 119903prof that are presented in Theorems 5 and 7 Wedefine a function

119892 (119909) =119909120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 119909 minus 120582120578 (65)

Differentiating the above function with respect to 119909 weobtain

1198921015840(119909) =

1205822((120582 + 120572) (120585 + 120578) + 120583120578)

(120582 (120582 + 120572) 119909 minus 120582120578)2

(66)

It means that function 119892(119909) increases with respect to 119909 Onthe other hand we can easily derive 1198610 gt 1198620 where 1198610 and1198620 are given inTheorems 5 and 7Then we obtain 119903prof lt 119903soc

(i) If 119877119862 isin [119905119880soc 119905119880prof) then 119903soc = 1 and 119903prof isin (0 1)(ii) If 119877119862 isin [119905119880socinfin) then 119903soc = 1 and 119903prof = 1

We obtain 119903soc ge 119903prof from the above analysis

Remark 9 Economou and Kanta [6] have studied the indi-vidual equilibrium strategy and social and profit maximiza-tion problems in the single-server constant retrial queue inthe partially observable case In our model if we let 120585 = 0we can derive the same results as given in [6] This is notaccidentally happened because when 120585 = 0 breakdownswill never occur then these two models are equivalent Inaddition notice that customersrsquo strategies 119903119890 119903soc 119903prof arefunctions of 120585120578 where other parameters are fixed It meansthat 119903119890 119903soc 119903prof only relate to 120585120578 whatever was the value 120585and 120578 we take in this model

Remark 10 Another case is that 120582(120582 + 120572)120572120583 ge 1 is left tobe considered In this situation customersrsquo joining strategieswill be less than 1 when considering the equilibrium socialand profit maximization problems otherwise the server willbehave unstably That is to say we have the following joiningstrategies

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119877

119862gt 119905119871119890

119903119904119900119888 =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119877

119862gt 119905119871soc

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119877

119862gt 119905119871prof

(67)

and the relation 119903119890 ge 119903soc ge 119903prof holds all the same

8 Mathematical Problems in Engineering

Till now we have completed the analysis of the customersrsquostrategies of the partially observable retrial queueing systemWe will turn to the fully observable case

4 The Fully Observable Case

We now focus on the fully observable case assuming cus-tomers observe not only the state of the server but also theexact number of customers in the retrial orbit Customerthat finds the server idle will enter the system and begin tobe served immediately so this information is valuable onlyfor the customer who finds the server busy We know thatafter every service completion and there is no new customerarriving the customer at the head of the orbit queue willsuccessfully retry for service That means customers in theorbit are served on an FCFS basis Therefore observing theposition in the orbit they can assess precisely whether toenter or balk

Consider a tagged customer that finds the server busyupon arrival This customerrsquos mean overall sojourn time inthe system is not affected by the joiningbalking behavior ofthe other customers that find the server busy because if thetagged customer joins the orbit queue at the 119895th positionthen the late customers who find the server busy will jointhe queue at the 119895 + 1th 119895 + 2th positions behind himin the orbit So his sojourn time does not depend on theirdecisions but it depends on the arrival rate because the newlyarriving customer who finds the server idle will be servedimmediately

To study the general threshold strategy adopted by allcustomers in the fully observable case we will firstly considerthe mean overall sojourn time of the customer who finds theserver idle busy or broken upon arrival Let 119879(1 0) be themean sojourn time when the server is idle and there is nocustomers in the orbit at his arrival instant and let 119879(119894 119895)be the mean sojourn time of a tagged customer at the 119895thposition in the orbit given that the server is at state 119894 Whenall customers follow the general strategy the mean overallsojourn times 119879(119894 119895) are given by

119879 (1 0) =120585 + 120578

120583120578 (68)

119879 (1 119895) =1

120583 + 120585+

120585

120583 + 120585119879 (2 119895) +

120583

120583 + 120585119879 (0 119895) 119895 ge 1

(69)

119879 (0 119895) =1

120582 + 120572+

120582

120582 + 120572119879 (1 119895) +

120572

120582 + 120572119879 (1 119895 minus 1)

119895 ge 1

(70)

119879 (2 119895) =1

120578+ 119879 (1 119895) 119895 ge 0 (71)

Let us consider a customer in the 119895th position of the orbitwhen the server is busy Then this customer has to wait foran exponentially distributed time with rate 120585 + 120583 for thenext event to occur in which the server breaks down or the

service is completed With probability 120585(120585 + 120583) breakdownoccurs then the mean overall sojourn time becomes 119879(2 119895)and the tagged customer remains in the 119895th position of theorbit On the other hand with probability 120583(120585 + 120583) theservice is completed then the mean overall sojourn timebecomes 119879(1 119895 minus 1) and the tagged customer moves to the119895 minus 1th position Through the above analysis we obtain (69)Considering the other cases in the same line we obtain therest equations

Solving the above equations the mean overall sojourntimes are given by

119879 (0 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578 119895 ge 1

119879 (1 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 119895 ge 0

119879 (2 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120583 + 120585 + 120578

120583120578

119895 ge 0

(72)

Then we need to find the general equilibrium strategyfollowed by the customers when the server is busy To ensurethat customer will enter the orbit when finding the serverbusy and the queue in the orbit is empty we further assumethat 119877 minus 119862119879(1 1) gt 0 The condition can be written as

119877

119862gt(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (73)

This means customersrsquo reward should be larger in the fullyobservable case than in the former case even if the other sys-tematic parameters are equal We will identify the customersrsquoequilibrium threshold strategy and we have the followingtheorem

Theorem 11 In the fully observable single-server constantretrial queue there exists a unique threshold joining strategyldquoenter the retrial orbit if there are at most 119899119890 customers in theorbit whenever finding the server busyrdquo in which condition (73)holds The threshold 119899119890 is given by

119899119890 = lfloor119909119890rfloor (74)

where

119909119890 =120572120583120578119877 minus 119862120572 (120585 + 120578)

119862 [(120582 + 120572) (120585 + 120578) + 120583120578]minus 1 (75)

This strategy is the unique equilibrium strategy among allpossible strategies

Proof Consider a tagged customer that finds the system atthe state (1 119895) upon arrival If he decides to join he goesdirectly to the orbit and occupies the 119895 + 1th position Thenhis expected net benefit is 119878119890(119895) = 119877 minus 119862119879(1 119895 + 1) where

119878119890 (119895) = 119877 minus 119862[(119895 + 1)(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578]

(76)

Mathematical Problems in Engineering 9

middot middot middot

middot middot middot

middot middot middot

120582

120582 120582 120582

120583 120572 120582 120583 120582 120583120572 120572

120585 120578 120585 120578 120585 120578

120582 120583

120585 120578

120582 120583

120585 120578

0 n0 0 0 1

1 1

2 1

0 2

1 2

2 2

1 0

2 0

1 n

2 n

0 n + 1

1 n + 1

2 n + 1

Figure 3 Transition rate diagram of the fully observable case

The customer prefers to join if 119878119890(119895) gt 0 and he is indifferentbetween joining and balking while 119878119890(119895) = 0 and prefers tobalk if 119878119890(119895) lt 0 A newly arriving customer will enter ifand only if 119895 le 119899119890 when finding the server busy By solvinginequality 119878119890(119895) ge 0 we obtain the result

We will elucidate the uniqueness of the equilibriumthreshold joining strategy followed by all customers Indeedif 119909119890 is an integer number we just need to let 119899119890 = 119909119890 thenwe know 119878119890(119899119890) = 0 and the mixed threshold strategy isthat prescribes to enter if 119899 lt 119899119890 to balk if 119899 gt 119899119890 and beindifferent if 119899 = 119899119890 When 119909119890 is any positive value we let119899119890 = lfloor119909119890rfloor Hence in this case the equilibrium strategy is alsounique

We now turn our attention to the social and profit maxi-mization problems and identify the thresholds 119899soc and 119899profthat maximize the social net benefit and the administratorrsquosprofit per time unit To consummate this goal we need todetermine the stationary behavior of the system when allcustomers follow the same threshold strategy

Suppose that all customers that find the server busy followthe same threshold strategies 119899soc and 119899prof respectively Thatis to say customers who find the sever busy will enter thesystem as long as the number of customers in the orbitis at most 119899soc when considering the social maximizationproblem and there is at most 119899prof customers in the orbitwhen considering the administratorrsquos optimization problemHence the maximum number of customers in the orbitwill never be larger than 119899 + 1 when customers follow thesame threshold-119899 strategyThen we obtain a continuous-timeMarkov chain (119868(119905)119873(119905)) with state space 119878 = 0 1 2 times

0 1 2 119899 + 1 and its transition diagram is given byFigure 3

The stationary distribution of the system (119901119894119895(119894 119895) isin 119878)is given by the following proposition

Proposition 12 Consider the fully observable single-serverconstant retrial queue with breakdowns and repairs in whichcustomers follow a threshold-119899 strategy then the stationarydistribution of the system is given by

11990100 = 119861 (119899)120583120572

1205822

1199010119895 = 119861 (119899) 120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 119861 (119899)120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 119861 (119899)120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(77)

where

120588 =120582 (120582 + 120572)

120572120583

119861 (119899) = 120583120572

1205822+ (1 + 120588 + sdot sdot sdot + 120588

119899) +

120572

120582(1 + 120588 + sdot sdot sdot + 120588

119899+1)

+120572120585

120582120578(1 + 120588 + sdot sdot sdot + 120588

119899+1)

minus1

(78)

Proof The stationary distribution is obtained from the fol-lowing balance equations

12058211990100 = 12058311990110 (79)

(120582 + 120572) 1199010119895 = 1205831199011119895 119895 = 1 2 119899 + 1 (80)

1205821199011119895 = 1205721199010119895+1 119895 = 0 1 2 119899 (81)

1205781199012119895 = 1205851199011119895 119895 = 0 1 119899 + 1 (82)

(120583 + 120585) 1199011119899+1 = 1205781199012119899+1 + 1205821199011119899 + 1205821199010119899+1 (83)

Combining (80) with (81) we can obtain the recursiverelation

1199011119895 = 11990110120588119895 119895 = 1 2 119899 minus 1 (84)

Then we derive the expressions of 1199012119895 from (82)

1199012119895 =120585

12057812058811989511990110 119895 = 0 1 119899 + 1 (85)

Using (80)ndash(82) we can get

11990100 = 11990110

120583120572

1205822

1199010119895 = 11990110120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 11990110120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 11990110120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(86)

Then with the help of the normalizing equation sum119899+1119895=0 1199010119895 +sum119899+1

119895=0 1199011119895 + sum119899+1

119895=0 1199012119895 = 1 we get the expression of 11990110

By considering the stationary distribution we examinethe social welfare maximization threshold strategy We havethe following result

10 Mathematical Problems in Engineering

Theorem 13 In our single-server constant retrial queue inwhich the condition (73) holds we define the following func-tion

119891 (119909) =(120583120572 + (120585120578) 120582120572) (119909 + 1)

120583120572 (1 minus 120588)

minus120582 (120582 + 120572120588 (1 + (120585120578))) (1 minus 120588

119909+1)

120583120572(1 minus 120588)2

minus 1

(87)

(88)

If 119891(0) gt 119909119890 then the customersrsquo best response is balkingOtherwise there exists a unique threshold-119899soc strategy ldquo enterthe retrial orbit if the number of customers in the orbit is atmost119899soc whenever customers find the server busyrdquo that maximizesthe social net benefit per time unit The threshold 119899soc is givenby

119899soc = lfloor119909119904119900119888rfloor (89)

where 119909soc is the unique nonnegative root of the equation119891(119909) = 119909119890

Proof In order to determine the optimal threshold 119899soc weneed to compute the effective arrival rate of the systemand the expected mean sojourn time of the system On theone hand when all customers follow the same threshold-119899strategy the effective arrival rate is given by

120582 (119899) = 120582(

119899+1

sum

119895=0

1199010119895 (119899) +

119899

sum

119895=0

1199011119895 (119899)) (90)

On the other hand by conditioning on the state found by acustomer in the system upon arrival we obtain the expectedmean sojourn time if entering the system with threshold-119899policy

119864 [119878 (119899)] = 119860 + 119861 (91)

where

119860 =

119899+1

sum

119895=0

1199010119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

120585 + 120578

120583120578

119861 =

119899

sum

119895=0

1199011119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

times (120585 + 120578

120583120578+ (119895 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(92)

The expected social net benefit is given by

119878soc (119899) = 120582 (119899) 119877 minus 119862120582 (119899) 119864 [119878 (119899)] (93)

Using (90) and (91) and after some algebra we obtain

119878soc (119899) = 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578

times (120582 (119899) (119909119890 + 1) minus 120582

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899))

(94)

Next we need to show that 119878soc(119899) is unimodal and hasonly one maximal point We firstly consider the increments119878soc(119899) minus 119878soc(119899 minus 1) It is not difficult to show that

119878soc (119899) minus 119878soc (119899 minus 1) ge 0 lArrrArr 119891 (119899) le 119909119890 (95)

where 119891(119899) is

119891 (119899) =1198911 (119899)

1198912 (119899)minus 1 (96)

1198911 (119899) = 120582(

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899) minus

119899minus1

sum

119895=0

(119895 + 1) 1199011119895 (119899 minus 1))

(97)

1198912 (119899) = 120582 (119899) minus 120582 (119899 minus 1) (98)

Using the stationary distribution we can simplify the expres-sions of (97) and (98) as follows

1198911 (119899) =120572119861 (119899) 119861 (119899 minus 1) 120588

119899

1205822(1 minus 120588)2

times 119863 (99)

1198912 (119899) =1205831205722120588119899

1205822119861 (119899) 119861 (119899 minus 1) (100)

where

119863 = (120583120572 + 120582120572120585

120578) (119899 + 1) (1 minus 120588) minus 120582

2

minus120582120572120588(1 +120585

120578) + 1205822120588119899+1

+ 120582120572(1 +120585

120578) 120588119899+2

(101)

Plugging (99) and (100) in (96) and replacing 119899 by 119909 weobtain (87) To prove the unimodality of 119878soc(119899) we need toshow that function119891(119909) is increasingWe define the functionℎ [0infin) rarr 119877 with

ℎ (119909) = 119891 (119909) minus 119909 (102)

We have

1198892

1198891199092ℎ (119909) =

[120582 + 120572120588 (1 + (120585120578))] 120588119909+2

(ln 120588)2

(120582 + 120572) (1 minus 120588)2

gt 0

119909 ge 0

(103)

Then function ℎ(119909) is a strictly convex function and thereis no flex point with respect to 120588 and 119909 Therefore ℎ(119909) isincreasing and so is 119891(119909) = ℎ(119909) + 119909 Therefore two casesmay take place

(i) 119891(0) gt 119909119890 then 119878soc(119899) is decreasing and customerswho find the server busy will balk

(ii) 119891(119899soc) le 119909119890 lt 119891(119899soc + 1) then we have 119878soc(119899) minus119878soc(119899minus1) ge 0 for 119899 le 119899soc and 119878soc(119899)minus119878soc(119899minus1) lt 0for 119899 gt 119899soc Then the maximum of 119878soc(119899) occurs in119899soc = lfloor119909socrfloor in which 119909soc is the unique solution ofthe equation 119891(119909) = 119909119890

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

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Mathematical Problems in Engineering

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6 Mathematical Problems in Engineering

(ii) When 119905119871soc lt 119877119862 lt 119905119880soc there exists a positivenumber 1199030 so that function 119878soc(119903) increases when119903 isin (0 1199030] and decreases when 119903 isin [1199030 1) then 1199030 is thepoint that maximizes the function 119878soc(119903) By solving(47) we know that the unique maximum of 119878soc(119903) isattained at (1198610120583120572 minus 120582(120585 + 120578) minus 120583120578)(120582(120582 + 120572)1198610 minus 120582120578)Therefore 119903soc given by (38) is the social maximizingstrategy

(iii) When 119877119862 ge 119905119880soc function 119878soc(119903) is increasingwhenever 119903 isin [0 1] so the best response is 1

We now compare the optimal joining strategy of thesocial maximization problem and the individual equilibriumjoining strategy Their relation is given by the followingtheorem

Theorem 6 In our single-server constant retrial rate queuewith breakdowns and repairs the joining probabilities 119903119890 and119903soc are ordered as

119903soc le 119903119890 (51)

Proof When the stable condition (29) holds it is easy to showthat 119905119871119890 lt 119905119871soc lt 119905119880119890 lt 119905119880soc where 119905119871119890 119905119871soc 119905119880119890 119905119880socare defined in Theorems 4 and 5 We consider the followingcases

(i) If 119877119862 le 119905119871119890 according to (30) and (37) 119903soc = 119903119890 = 0

(ii) If 119905119871119890 lt 119877119862 le 119905119871soc then 119903119890 isin (0 1) and 119903soc = 0 So119903soc lt 119903119890

(iii) If 119905119871soc lt 119877119862 lt 119905119880119890 then 119903119890 isin (0 1) and 119903soc isin (0 1)But we can get that 119903soc lt 119903119890 after some algebra

(iv) If 119905119880119890 le 119877119862 lt 119905119880soc then 119903119890 = 1 and 119903soc isin (0 1)Therefore 119903soc lt 119903119890 is obviously satisfied

(v) Finally if 119877119862 ge 119905119880soc then 119903119890 = 119903soc = 1

From the above analysis we complete the proof

The above result can be explained by the perspective of aneconomic viewpoint In the process of individual equilibriumanalysis we notice that customers will enter the orbit as longas their net benefit is nonnegative Then individualrsquos rationalbehavior causes excessive use of the resource in equilibriumthat is to say individualrsquos optimization leads to the systemhavingmore congestion thanwhat is socially desirable whichis the appearance of the negative externality

Having considered the individual equilibrium strategyand social maximization problem we will further identifythe profit optimization joining strategy In our model wesuppose that every customer will be imposed an entrancefee 119901 by the administrator By imposing the entrance feecustomersrsquo reward decreases to 119877 minus 119901 then their strategy willdiffer We will determine the strategy 119903prof that maximizes theadministratorrsquos profit per time unit The result is given by thefollowing

Theorem 7 In the partially observable single-server constantretrial queue we can derive a unique mixed joining strategyldquoenter the orbit with probability 119903prof when the server is busyrdquothat maximizes the administratorrsquos net profit per time unitwhen condition (29) holds The probability 119903prof is given by

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119905119871prof lt

119877

119862lt 119905119880prof

1 if 119877119862ge 119905119880prof

(52)

where

119903lowast

prof =1198620120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572)1198620 minus 120582120578 (53)

1198620 =radic1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2 (54)

1198771015840= 119877 minus 119862

120585 + 120578

120583120578 (55)

119905119871prof

=(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2+ 120582120583120578120572

2(120585 + 120578) minus 120583

212057821205722

120582120583212057821205722

(56)

119905119880prof = (120583(120582 + 120572)2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

+ [120582120578 (120585 + 120578) minus 1205831205782] [120583120572 minus 120582 (120582 + 120572)]

2)

times (1205821205831205782[120583120572 minus 120582 (120582 + 120572)]

2)minus1

(57)

Proof For a given joining strategy 119903 we define the function119878prof(119903) to be the administratorrsquos net profit per time unit Let120582lowast(119903) be the mean effective arrival rate in the system we

further assume that the entrance fee that the administratorimposes on customers is 119901(119903) Then we have

119878prof (119903) = 120582lowast(119903) 119901 (119903) (58)

where 120582lowast(119903) is given by (43) In order to obtain 119878prof(119903) weneed to determine the entrance fee 119901(119903) Noticing (34) weyield

119877 minus 119901 (119903) = 119862((120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)]+120585 + 120578

120583120578) (59)

from which we obtain

119901 (119903) = 119877 minus 119862120585 + 120578

120583120578minus 119862

(120582 + 120572) (120585 + 120578) + 120583120578

120578 [120572120583 minus 120582119903 (120582 + 120572)] (60)

Because if the entrance fee 119901 lt 119901(119903) the administrator canimprove their net benefit through increasing the entrancefee but if 119901 gt 119901(119903) the individualrsquos net benefit is negative

Mathematical Problems in Engineering 7

customers will balk So (59) always holds whenever 119903 isin [0 1]Therefore the net profit of the administrator can be calculatedas follows119878prof (119903) = 120582

lowast(119903) 119901 (119903)

=120583120578 (120582119877

1015840+ 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus 119862

120583 (120582 + 120572)

120583120572 minus 120582119903 (120582 + 120572)

(61)

where 1198771015840 is given by (55) Differentiating the equation withrespect to 119903 we get

119889

119889119903119878prof (119903) =

1205821205831205782(1205821198771015840+ 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2

minus119862120583120582(120582 + 120572)

2

[120583120572 minus 120582119903 (120582 + 120572)]2

119889

119889119903119878prof (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2

(62)

The rest of the proof can proceed along the same line with theproof of Theorem 5

Now we are going to compare the joining probabilitieswhen considering the social and profit maximization prob-lems The result is given by the following theorem

Theorem 8 In our model when condition (29) holds theoptimal joining probabilities 119903soc and 119903prof are ordered as

119903prof le 119903soc (63)

Proof Firstly we proceed to compare the relationsamong 119905119871soc 119905119880soc 119905119871prof 119905119880prof Rewrite the expressionsof 119905119871soc 119905119880soc 119905119871prof 119905119880prof given inTheorems 5 and 7

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2

12058212058321205782120572minus1

120582

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

minus1

120582

119905119871prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2

120582120583212057821205722+120585 + 120578

120583120578minus1

120582

119905119880prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

+120585 + 120578

120583120578minus1

120582

(64)

It is readily shown that 119905119871soc lt 119905119871prof and 119905119880soc lt 119905119880prof Butthe relation between 119905119880soc and 119905119871prof is undefined Two casesmay take place that is 119905119880soc lt 119905119871prof or 119905119880soc ge 119905119871prof Weconsider the situation when 119905119880soc ge 119905119871prof the other case canbe analyzed in the same wayThen we consider the followingcases

(i) If 119877119862 isin [(120585 + 120578)120583120578 119905119871soc) then 119903soc = 119903prof = 0(ii) If 119877119862 isin [119905119871soc 119905119871prof) then 119903soc isin (0 1) 119903prof = 0(iii) If 119877119862 isin [119905119871prof 119905119880soc) then 119903soc isin (0 1) and 119903prof isin

(0 1) We can take a close look at the expressions of119903soc 119903prof that are presented in Theorems 5 and 7 Wedefine a function

119892 (119909) =119909120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 119909 minus 120582120578 (65)

Differentiating the above function with respect to 119909 weobtain

1198921015840(119909) =

1205822((120582 + 120572) (120585 + 120578) + 120583120578)

(120582 (120582 + 120572) 119909 minus 120582120578)2

(66)

It means that function 119892(119909) increases with respect to 119909 Onthe other hand we can easily derive 1198610 gt 1198620 where 1198610 and1198620 are given inTheorems 5 and 7Then we obtain 119903prof lt 119903soc

(i) If 119877119862 isin [119905119880soc 119905119880prof) then 119903soc = 1 and 119903prof isin (0 1)(ii) If 119877119862 isin [119905119880socinfin) then 119903soc = 1 and 119903prof = 1

We obtain 119903soc ge 119903prof from the above analysis

Remark 9 Economou and Kanta [6] have studied the indi-vidual equilibrium strategy and social and profit maximiza-tion problems in the single-server constant retrial queue inthe partially observable case In our model if we let 120585 = 0we can derive the same results as given in [6] This is notaccidentally happened because when 120585 = 0 breakdownswill never occur then these two models are equivalent Inaddition notice that customersrsquo strategies 119903119890 119903soc 119903prof arefunctions of 120585120578 where other parameters are fixed It meansthat 119903119890 119903soc 119903prof only relate to 120585120578 whatever was the value 120585and 120578 we take in this model

Remark 10 Another case is that 120582(120582 + 120572)120572120583 ge 1 is left tobe considered In this situation customersrsquo joining strategieswill be less than 1 when considering the equilibrium socialand profit maximization problems otherwise the server willbehave unstably That is to say we have the following joiningstrategies

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119877

119862gt 119905119871119890

119903119904119900119888 =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119877

119862gt 119905119871soc

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119877

119862gt 119905119871prof

(67)

and the relation 119903119890 ge 119903soc ge 119903prof holds all the same

8 Mathematical Problems in Engineering

Till now we have completed the analysis of the customersrsquostrategies of the partially observable retrial queueing systemWe will turn to the fully observable case

4 The Fully Observable Case

We now focus on the fully observable case assuming cus-tomers observe not only the state of the server but also theexact number of customers in the retrial orbit Customerthat finds the server idle will enter the system and begin tobe served immediately so this information is valuable onlyfor the customer who finds the server busy We know thatafter every service completion and there is no new customerarriving the customer at the head of the orbit queue willsuccessfully retry for service That means customers in theorbit are served on an FCFS basis Therefore observing theposition in the orbit they can assess precisely whether toenter or balk

Consider a tagged customer that finds the server busyupon arrival This customerrsquos mean overall sojourn time inthe system is not affected by the joiningbalking behavior ofthe other customers that find the server busy because if thetagged customer joins the orbit queue at the 119895th positionthen the late customers who find the server busy will jointhe queue at the 119895 + 1th 119895 + 2th positions behind himin the orbit So his sojourn time does not depend on theirdecisions but it depends on the arrival rate because the newlyarriving customer who finds the server idle will be servedimmediately

To study the general threshold strategy adopted by allcustomers in the fully observable case we will firstly considerthe mean overall sojourn time of the customer who finds theserver idle busy or broken upon arrival Let 119879(1 0) be themean sojourn time when the server is idle and there is nocustomers in the orbit at his arrival instant and let 119879(119894 119895)be the mean sojourn time of a tagged customer at the 119895thposition in the orbit given that the server is at state 119894 Whenall customers follow the general strategy the mean overallsojourn times 119879(119894 119895) are given by

119879 (1 0) =120585 + 120578

120583120578 (68)

119879 (1 119895) =1

120583 + 120585+

120585

120583 + 120585119879 (2 119895) +

120583

120583 + 120585119879 (0 119895) 119895 ge 1

(69)

119879 (0 119895) =1

120582 + 120572+

120582

120582 + 120572119879 (1 119895) +

120572

120582 + 120572119879 (1 119895 minus 1)

119895 ge 1

(70)

119879 (2 119895) =1

120578+ 119879 (1 119895) 119895 ge 0 (71)

Let us consider a customer in the 119895th position of the orbitwhen the server is busy Then this customer has to wait foran exponentially distributed time with rate 120585 + 120583 for thenext event to occur in which the server breaks down or the

service is completed With probability 120585(120585 + 120583) breakdownoccurs then the mean overall sojourn time becomes 119879(2 119895)and the tagged customer remains in the 119895th position of theorbit On the other hand with probability 120583(120585 + 120583) theservice is completed then the mean overall sojourn timebecomes 119879(1 119895 minus 1) and the tagged customer moves to the119895 minus 1th position Through the above analysis we obtain (69)Considering the other cases in the same line we obtain therest equations

Solving the above equations the mean overall sojourntimes are given by

119879 (0 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578 119895 ge 1

119879 (1 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 119895 ge 0

119879 (2 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120583 + 120585 + 120578

120583120578

119895 ge 0

(72)

Then we need to find the general equilibrium strategyfollowed by the customers when the server is busy To ensurethat customer will enter the orbit when finding the serverbusy and the queue in the orbit is empty we further assumethat 119877 minus 119862119879(1 1) gt 0 The condition can be written as

119877

119862gt(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (73)

This means customersrsquo reward should be larger in the fullyobservable case than in the former case even if the other sys-tematic parameters are equal We will identify the customersrsquoequilibrium threshold strategy and we have the followingtheorem

Theorem 11 In the fully observable single-server constantretrial queue there exists a unique threshold joining strategyldquoenter the retrial orbit if there are at most 119899119890 customers in theorbit whenever finding the server busyrdquo in which condition (73)holds The threshold 119899119890 is given by

119899119890 = lfloor119909119890rfloor (74)

where

119909119890 =120572120583120578119877 minus 119862120572 (120585 + 120578)

119862 [(120582 + 120572) (120585 + 120578) + 120583120578]minus 1 (75)

This strategy is the unique equilibrium strategy among allpossible strategies

Proof Consider a tagged customer that finds the system atthe state (1 119895) upon arrival If he decides to join he goesdirectly to the orbit and occupies the 119895 + 1th position Thenhis expected net benefit is 119878119890(119895) = 119877 minus 119862119879(1 119895 + 1) where

119878119890 (119895) = 119877 minus 119862[(119895 + 1)(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578]

(76)

Mathematical Problems in Engineering 9

middot middot middot

middot middot middot

middot middot middot

120582

120582 120582 120582

120583 120572 120582 120583 120582 120583120572 120572

120585 120578 120585 120578 120585 120578

120582 120583

120585 120578

120582 120583

120585 120578

0 n0 0 0 1

1 1

2 1

0 2

1 2

2 2

1 0

2 0

1 n

2 n

0 n + 1

1 n + 1

2 n + 1

Figure 3 Transition rate diagram of the fully observable case

The customer prefers to join if 119878119890(119895) gt 0 and he is indifferentbetween joining and balking while 119878119890(119895) = 0 and prefers tobalk if 119878119890(119895) lt 0 A newly arriving customer will enter ifand only if 119895 le 119899119890 when finding the server busy By solvinginequality 119878119890(119895) ge 0 we obtain the result

We will elucidate the uniqueness of the equilibriumthreshold joining strategy followed by all customers Indeedif 119909119890 is an integer number we just need to let 119899119890 = 119909119890 thenwe know 119878119890(119899119890) = 0 and the mixed threshold strategy isthat prescribes to enter if 119899 lt 119899119890 to balk if 119899 gt 119899119890 and beindifferent if 119899 = 119899119890 When 119909119890 is any positive value we let119899119890 = lfloor119909119890rfloor Hence in this case the equilibrium strategy is alsounique

We now turn our attention to the social and profit maxi-mization problems and identify the thresholds 119899soc and 119899profthat maximize the social net benefit and the administratorrsquosprofit per time unit To consummate this goal we need todetermine the stationary behavior of the system when allcustomers follow the same threshold strategy

Suppose that all customers that find the server busy followthe same threshold strategies 119899soc and 119899prof respectively Thatis to say customers who find the sever busy will enter thesystem as long as the number of customers in the orbitis at most 119899soc when considering the social maximizationproblem and there is at most 119899prof customers in the orbitwhen considering the administratorrsquos optimization problemHence the maximum number of customers in the orbitwill never be larger than 119899 + 1 when customers follow thesame threshold-119899 strategyThen we obtain a continuous-timeMarkov chain (119868(119905)119873(119905)) with state space 119878 = 0 1 2 times

0 1 2 119899 + 1 and its transition diagram is given byFigure 3

The stationary distribution of the system (119901119894119895(119894 119895) isin 119878)is given by the following proposition

Proposition 12 Consider the fully observable single-serverconstant retrial queue with breakdowns and repairs in whichcustomers follow a threshold-119899 strategy then the stationarydistribution of the system is given by

11990100 = 119861 (119899)120583120572

1205822

1199010119895 = 119861 (119899) 120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 119861 (119899)120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 119861 (119899)120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(77)

where

120588 =120582 (120582 + 120572)

120572120583

119861 (119899) = 120583120572

1205822+ (1 + 120588 + sdot sdot sdot + 120588

119899) +

120572

120582(1 + 120588 + sdot sdot sdot + 120588

119899+1)

+120572120585

120582120578(1 + 120588 + sdot sdot sdot + 120588

119899+1)

minus1

(78)

Proof The stationary distribution is obtained from the fol-lowing balance equations

12058211990100 = 12058311990110 (79)

(120582 + 120572) 1199010119895 = 1205831199011119895 119895 = 1 2 119899 + 1 (80)

1205821199011119895 = 1205721199010119895+1 119895 = 0 1 2 119899 (81)

1205781199012119895 = 1205851199011119895 119895 = 0 1 119899 + 1 (82)

(120583 + 120585) 1199011119899+1 = 1205781199012119899+1 + 1205821199011119899 + 1205821199010119899+1 (83)

Combining (80) with (81) we can obtain the recursiverelation

1199011119895 = 11990110120588119895 119895 = 1 2 119899 minus 1 (84)

Then we derive the expressions of 1199012119895 from (82)

1199012119895 =120585

12057812058811989511990110 119895 = 0 1 119899 + 1 (85)

Using (80)ndash(82) we can get

11990100 = 11990110

120583120572

1205822

1199010119895 = 11990110120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 11990110120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 11990110120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(86)

Then with the help of the normalizing equation sum119899+1119895=0 1199010119895 +sum119899+1

119895=0 1199011119895 + sum119899+1

119895=0 1199012119895 = 1 we get the expression of 11990110

By considering the stationary distribution we examinethe social welfare maximization threshold strategy We havethe following result

10 Mathematical Problems in Engineering

Theorem 13 In our single-server constant retrial queue inwhich the condition (73) holds we define the following func-tion

119891 (119909) =(120583120572 + (120585120578) 120582120572) (119909 + 1)

120583120572 (1 minus 120588)

minus120582 (120582 + 120572120588 (1 + (120585120578))) (1 minus 120588

119909+1)

120583120572(1 minus 120588)2

minus 1

(87)

(88)

If 119891(0) gt 119909119890 then the customersrsquo best response is balkingOtherwise there exists a unique threshold-119899soc strategy ldquo enterthe retrial orbit if the number of customers in the orbit is atmost119899soc whenever customers find the server busyrdquo that maximizesthe social net benefit per time unit The threshold 119899soc is givenby

119899soc = lfloor119909119904119900119888rfloor (89)

where 119909soc is the unique nonnegative root of the equation119891(119909) = 119909119890

Proof In order to determine the optimal threshold 119899soc weneed to compute the effective arrival rate of the systemand the expected mean sojourn time of the system On theone hand when all customers follow the same threshold-119899strategy the effective arrival rate is given by

120582 (119899) = 120582(

119899+1

sum

119895=0

1199010119895 (119899) +

119899

sum

119895=0

1199011119895 (119899)) (90)

On the other hand by conditioning on the state found by acustomer in the system upon arrival we obtain the expectedmean sojourn time if entering the system with threshold-119899policy

119864 [119878 (119899)] = 119860 + 119861 (91)

where

119860 =

119899+1

sum

119895=0

1199010119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

120585 + 120578

120583120578

119861 =

119899

sum

119895=0

1199011119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

times (120585 + 120578

120583120578+ (119895 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(92)

The expected social net benefit is given by

119878soc (119899) = 120582 (119899) 119877 minus 119862120582 (119899) 119864 [119878 (119899)] (93)

Using (90) and (91) and after some algebra we obtain

119878soc (119899) = 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578

times (120582 (119899) (119909119890 + 1) minus 120582

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899))

(94)

Next we need to show that 119878soc(119899) is unimodal and hasonly one maximal point We firstly consider the increments119878soc(119899) minus 119878soc(119899 minus 1) It is not difficult to show that

119878soc (119899) minus 119878soc (119899 minus 1) ge 0 lArrrArr 119891 (119899) le 119909119890 (95)

where 119891(119899) is

119891 (119899) =1198911 (119899)

1198912 (119899)minus 1 (96)

1198911 (119899) = 120582(

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899) minus

119899minus1

sum

119895=0

(119895 + 1) 1199011119895 (119899 minus 1))

(97)

1198912 (119899) = 120582 (119899) minus 120582 (119899 minus 1) (98)

Using the stationary distribution we can simplify the expres-sions of (97) and (98) as follows

1198911 (119899) =120572119861 (119899) 119861 (119899 minus 1) 120588

119899

1205822(1 minus 120588)2

times 119863 (99)

1198912 (119899) =1205831205722120588119899

1205822119861 (119899) 119861 (119899 minus 1) (100)

where

119863 = (120583120572 + 120582120572120585

120578) (119899 + 1) (1 minus 120588) minus 120582

2

minus120582120572120588(1 +120585

120578) + 1205822120588119899+1

+ 120582120572(1 +120585

120578) 120588119899+2

(101)

Plugging (99) and (100) in (96) and replacing 119899 by 119909 weobtain (87) To prove the unimodality of 119878soc(119899) we need toshow that function119891(119909) is increasingWe define the functionℎ [0infin) rarr 119877 with

ℎ (119909) = 119891 (119909) minus 119909 (102)

We have

1198892

1198891199092ℎ (119909) =

[120582 + 120572120588 (1 + (120585120578))] 120588119909+2

(ln 120588)2

(120582 + 120572) (1 minus 120588)2

gt 0

119909 ge 0

(103)

Then function ℎ(119909) is a strictly convex function and thereis no flex point with respect to 120588 and 119909 Therefore ℎ(119909) isincreasing and so is 119891(119909) = ℎ(119909) + 119909 Therefore two casesmay take place

(i) 119891(0) gt 119909119890 then 119878soc(119899) is decreasing and customerswho find the server busy will balk

(ii) 119891(119899soc) le 119909119890 lt 119891(119899soc + 1) then we have 119878soc(119899) minus119878soc(119899minus1) ge 0 for 119899 le 119899soc and 119878soc(119899)minus119878soc(119899minus1) lt 0for 119899 gt 119899soc Then the maximum of 119878soc(119899) occurs in119899soc = lfloor119909socrfloor in which 119909soc is the unique solution ofthe equation 119891(119909) = 119909119890

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

customers will balk So (59) always holds whenever 119903 isin [0 1]Therefore the net profit of the administrator can be calculatedas follows119878prof (119903) = 120582

lowast(119903) 119901 (119903)

=120583120578 (120582119877

1015840+ 119862)

120582 (120585 + 120578) + 120578 (120583 minus 120582119903)minus 119862

120583 (120582 + 120572)

120583120572 minus 120582119903 (120582 + 120572)

(61)

where 1198771015840 is given by (55) Differentiating the equation withrespect to 119903 we get

119889

119889119903119878prof (119903) =

1205821205831205782(1205821198771015840+ 119862)

[120582 (120585 + 120578) + 120578 (120583 minus 120582119903)]2

minus119862120583120582(120582 + 120572)

2

[120583120572 minus 120582119903 (120582 + 120572)]2

119889

119889119903119878prof (119903) = 0

lArrrArr120582 (120585 + 120578) + 120578 (120583 minus 120582119903)

120583120572 minus 120582119903 (120582 + 120572)= radic

1205782(1205821198771015840+ 119862)

119862(120582 + 120572)2

(62)

The rest of the proof can proceed along the same line with theproof of Theorem 5

Now we are going to compare the joining probabilitieswhen considering the social and profit maximization prob-lems The result is given by the following theorem

Theorem 8 In our model when condition (29) holds theoptimal joining probabilities 119903soc and 119903prof are ordered as

119903prof le 119903soc (63)

Proof Firstly we proceed to compare the relationsamong 119905119871soc 119905119880soc 119905119871prof 119905119880prof Rewrite the expressionsof 119905119871soc 119905119880soc 119905119871prof 119905119880prof given inTheorems 5 and 7

119905119871soc =(120582 + 120572) [120582 (120585 + 120578) + 120583120578]

2

12058212058321205782120572minus1

120582

119905119880soc =120572 (120582 + 120572) [120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

minus1

120582

119905119871prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120583120578]

2

120582120583212057821205722+120585 + 120578

120583120578minus1

120582

119905119880prof =(120582 + 120572)

2[120582 (120585 + 120578) + 120578 (120583 minus 120582)]

2

1205821205782[120583120572 minus 120582 (120582 + 120572)]2

+120585 + 120578

120583120578minus1

120582

(64)

It is readily shown that 119905119871soc lt 119905119871prof and 119905119880soc lt 119905119880prof Butthe relation between 119905119880soc and 119905119871prof is undefined Two casesmay take place that is 119905119880soc lt 119905119871prof or 119905119880soc ge 119905119871prof Weconsider the situation when 119905119880soc ge 119905119871prof the other case canbe analyzed in the same wayThen we consider the followingcases

(i) If 119877119862 isin [(120585 + 120578)120583120578 119905119871soc) then 119903soc = 119903prof = 0(ii) If 119877119862 isin [119905119871soc 119905119871prof) then 119903soc isin (0 1) 119903prof = 0(iii) If 119877119862 isin [119905119871prof 119905119880soc) then 119903soc isin (0 1) and 119903prof isin

(0 1) We can take a close look at the expressions of119903soc 119903prof that are presented in Theorems 5 and 7 Wedefine a function

119892 (119909) =119909120583120572 minus 120582 (120585 + 120578) minus 120583120578

120582 (120582 + 120572) 119909 minus 120582120578 (65)

Differentiating the above function with respect to 119909 weobtain

1198921015840(119909) =

1205822((120582 + 120572) (120585 + 120578) + 120583120578)

(120582 (120582 + 120572) 119909 minus 120582120578)2

(66)

It means that function 119892(119909) increases with respect to 119909 Onthe other hand we can easily derive 1198610 gt 1198620 where 1198610 and1198620 are given inTheorems 5 and 7Then we obtain 119903prof lt 119903soc

(i) If 119877119862 isin [119905119880soc 119905119880prof) then 119903soc = 1 and 119903prof isin (0 1)(ii) If 119877119862 isin [119905119880socinfin) then 119903soc = 1 and 119903prof = 1

We obtain 119903soc ge 119903prof from the above analysis

Remark 9 Economou and Kanta [6] have studied the indi-vidual equilibrium strategy and social and profit maximiza-tion problems in the single-server constant retrial queue inthe partially observable case In our model if we let 120585 = 0we can derive the same results as given in [6] This is notaccidentally happened because when 120585 = 0 breakdownswill never occur then these two models are equivalent Inaddition notice that customersrsquo strategies 119903119890 119903soc 119903prof arefunctions of 120585120578 where other parameters are fixed It meansthat 119903119890 119903soc 119903prof only relate to 120585120578 whatever was the value 120585and 120578 we take in this model

Remark 10 Another case is that 120582(120582 + 120572)120572120583 ge 1 is left tobe considered In this situation customersrsquo joining strategieswill be less than 1 when considering the equilibrium socialand profit maximization problems otherwise the server willbehave unstably That is to say we have the following joiningstrategies

119903119890 =

0 if 119877119862le 119905119871119890

119903lowast119890 if 119877

119862gt 119905119871119890

119903119904119900119888 =

0 if 119877119862le 119905119871soc

119903lowastsoc if 119877

119862gt 119905119871soc

119903prof =

0 if 119877119862le 119905119871prof

119903lowastprof if 119877

119862gt 119905119871prof

(67)

and the relation 119903119890 ge 119903soc ge 119903prof holds all the same

8 Mathematical Problems in Engineering

Till now we have completed the analysis of the customersrsquostrategies of the partially observable retrial queueing systemWe will turn to the fully observable case

4 The Fully Observable Case

We now focus on the fully observable case assuming cus-tomers observe not only the state of the server but also theexact number of customers in the retrial orbit Customerthat finds the server idle will enter the system and begin tobe served immediately so this information is valuable onlyfor the customer who finds the server busy We know thatafter every service completion and there is no new customerarriving the customer at the head of the orbit queue willsuccessfully retry for service That means customers in theorbit are served on an FCFS basis Therefore observing theposition in the orbit they can assess precisely whether toenter or balk

Consider a tagged customer that finds the server busyupon arrival This customerrsquos mean overall sojourn time inthe system is not affected by the joiningbalking behavior ofthe other customers that find the server busy because if thetagged customer joins the orbit queue at the 119895th positionthen the late customers who find the server busy will jointhe queue at the 119895 + 1th 119895 + 2th positions behind himin the orbit So his sojourn time does not depend on theirdecisions but it depends on the arrival rate because the newlyarriving customer who finds the server idle will be servedimmediately

To study the general threshold strategy adopted by allcustomers in the fully observable case we will firstly considerthe mean overall sojourn time of the customer who finds theserver idle busy or broken upon arrival Let 119879(1 0) be themean sojourn time when the server is idle and there is nocustomers in the orbit at his arrival instant and let 119879(119894 119895)be the mean sojourn time of a tagged customer at the 119895thposition in the orbit given that the server is at state 119894 Whenall customers follow the general strategy the mean overallsojourn times 119879(119894 119895) are given by

119879 (1 0) =120585 + 120578

120583120578 (68)

119879 (1 119895) =1

120583 + 120585+

120585

120583 + 120585119879 (2 119895) +

120583

120583 + 120585119879 (0 119895) 119895 ge 1

(69)

119879 (0 119895) =1

120582 + 120572+

120582

120582 + 120572119879 (1 119895) +

120572

120582 + 120572119879 (1 119895 minus 1)

119895 ge 1

(70)

119879 (2 119895) =1

120578+ 119879 (1 119895) 119895 ge 0 (71)

Let us consider a customer in the 119895th position of the orbitwhen the server is busy Then this customer has to wait foran exponentially distributed time with rate 120585 + 120583 for thenext event to occur in which the server breaks down or the

service is completed With probability 120585(120585 + 120583) breakdownoccurs then the mean overall sojourn time becomes 119879(2 119895)and the tagged customer remains in the 119895th position of theorbit On the other hand with probability 120583(120585 + 120583) theservice is completed then the mean overall sojourn timebecomes 119879(1 119895 minus 1) and the tagged customer moves to the119895 minus 1th position Through the above analysis we obtain (69)Considering the other cases in the same line we obtain therest equations

Solving the above equations the mean overall sojourntimes are given by

119879 (0 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578 119895 ge 1

119879 (1 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 119895 ge 0

119879 (2 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120583 + 120585 + 120578

120583120578

119895 ge 0

(72)

Then we need to find the general equilibrium strategyfollowed by the customers when the server is busy To ensurethat customer will enter the orbit when finding the serverbusy and the queue in the orbit is empty we further assumethat 119877 minus 119862119879(1 1) gt 0 The condition can be written as

119877

119862gt(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (73)

This means customersrsquo reward should be larger in the fullyobservable case than in the former case even if the other sys-tematic parameters are equal We will identify the customersrsquoequilibrium threshold strategy and we have the followingtheorem

Theorem 11 In the fully observable single-server constantretrial queue there exists a unique threshold joining strategyldquoenter the retrial orbit if there are at most 119899119890 customers in theorbit whenever finding the server busyrdquo in which condition (73)holds The threshold 119899119890 is given by

119899119890 = lfloor119909119890rfloor (74)

where

119909119890 =120572120583120578119877 minus 119862120572 (120585 + 120578)

119862 [(120582 + 120572) (120585 + 120578) + 120583120578]minus 1 (75)

This strategy is the unique equilibrium strategy among allpossible strategies

Proof Consider a tagged customer that finds the system atthe state (1 119895) upon arrival If he decides to join he goesdirectly to the orbit and occupies the 119895 + 1th position Thenhis expected net benefit is 119878119890(119895) = 119877 minus 119862119879(1 119895 + 1) where

119878119890 (119895) = 119877 minus 119862[(119895 + 1)(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578]

(76)

Mathematical Problems in Engineering 9

middot middot middot

middot middot middot

middot middot middot

120582

120582 120582 120582

120583 120572 120582 120583 120582 120583120572 120572

120585 120578 120585 120578 120585 120578

120582 120583

120585 120578

120582 120583

120585 120578

0 n0 0 0 1

1 1

2 1

0 2

1 2

2 2

1 0

2 0

1 n

2 n

0 n + 1

1 n + 1

2 n + 1

Figure 3 Transition rate diagram of the fully observable case

The customer prefers to join if 119878119890(119895) gt 0 and he is indifferentbetween joining and balking while 119878119890(119895) = 0 and prefers tobalk if 119878119890(119895) lt 0 A newly arriving customer will enter ifand only if 119895 le 119899119890 when finding the server busy By solvinginequality 119878119890(119895) ge 0 we obtain the result

We will elucidate the uniqueness of the equilibriumthreshold joining strategy followed by all customers Indeedif 119909119890 is an integer number we just need to let 119899119890 = 119909119890 thenwe know 119878119890(119899119890) = 0 and the mixed threshold strategy isthat prescribes to enter if 119899 lt 119899119890 to balk if 119899 gt 119899119890 and beindifferent if 119899 = 119899119890 When 119909119890 is any positive value we let119899119890 = lfloor119909119890rfloor Hence in this case the equilibrium strategy is alsounique

We now turn our attention to the social and profit maxi-mization problems and identify the thresholds 119899soc and 119899profthat maximize the social net benefit and the administratorrsquosprofit per time unit To consummate this goal we need todetermine the stationary behavior of the system when allcustomers follow the same threshold strategy

Suppose that all customers that find the server busy followthe same threshold strategies 119899soc and 119899prof respectively Thatis to say customers who find the sever busy will enter thesystem as long as the number of customers in the orbitis at most 119899soc when considering the social maximizationproblem and there is at most 119899prof customers in the orbitwhen considering the administratorrsquos optimization problemHence the maximum number of customers in the orbitwill never be larger than 119899 + 1 when customers follow thesame threshold-119899 strategyThen we obtain a continuous-timeMarkov chain (119868(119905)119873(119905)) with state space 119878 = 0 1 2 times

0 1 2 119899 + 1 and its transition diagram is given byFigure 3

The stationary distribution of the system (119901119894119895(119894 119895) isin 119878)is given by the following proposition

Proposition 12 Consider the fully observable single-serverconstant retrial queue with breakdowns and repairs in whichcustomers follow a threshold-119899 strategy then the stationarydistribution of the system is given by

11990100 = 119861 (119899)120583120572

1205822

1199010119895 = 119861 (119899) 120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 119861 (119899)120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 119861 (119899)120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(77)

where

120588 =120582 (120582 + 120572)

120572120583

119861 (119899) = 120583120572

1205822+ (1 + 120588 + sdot sdot sdot + 120588

119899) +

120572

120582(1 + 120588 + sdot sdot sdot + 120588

119899+1)

+120572120585

120582120578(1 + 120588 + sdot sdot sdot + 120588

119899+1)

minus1

(78)

Proof The stationary distribution is obtained from the fol-lowing balance equations

12058211990100 = 12058311990110 (79)

(120582 + 120572) 1199010119895 = 1205831199011119895 119895 = 1 2 119899 + 1 (80)

1205821199011119895 = 1205721199010119895+1 119895 = 0 1 2 119899 (81)

1205781199012119895 = 1205851199011119895 119895 = 0 1 119899 + 1 (82)

(120583 + 120585) 1199011119899+1 = 1205781199012119899+1 + 1205821199011119899 + 1205821199010119899+1 (83)

Combining (80) with (81) we can obtain the recursiverelation

1199011119895 = 11990110120588119895 119895 = 1 2 119899 minus 1 (84)

Then we derive the expressions of 1199012119895 from (82)

1199012119895 =120585

12057812058811989511990110 119895 = 0 1 119899 + 1 (85)

Using (80)ndash(82) we can get

11990100 = 11990110

120583120572

1205822

1199010119895 = 11990110120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 11990110120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 11990110120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(86)

Then with the help of the normalizing equation sum119899+1119895=0 1199010119895 +sum119899+1

119895=0 1199011119895 + sum119899+1

119895=0 1199012119895 = 1 we get the expression of 11990110

By considering the stationary distribution we examinethe social welfare maximization threshold strategy We havethe following result

10 Mathematical Problems in Engineering

Theorem 13 In our single-server constant retrial queue inwhich the condition (73) holds we define the following func-tion

119891 (119909) =(120583120572 + (120585120578) 120582120572) (119909 + 1)

120583120572 (1 minus 120588)

minus120582 (120582 + 120572120588 (1 + (120585120578))) (1 minus 120588

119909+1)

120583120572(1 minus 120588)2

minus 1

(87)

(88)

If 119891(0) gt 119909119890 then the customersrsquo best response is balkingOtherwise there exists a unique threshold-119899soc strategy ldquo enterthe retrial orbit if the number of customers in the orbit is atmost119899soc whenever customers find the server busyrdquo that maximizesthe social net benefit per time unit The threshold 119899soc is givenby

119899soc = lfloor119909119904119900119888rfloor (89)

where 119909soc is the unique nonnegative root of the equation119891(119909) = 119909119890

Proof In order to determine the optimal threshold 119899soc weneed to compute the effective arrival rate of the systemand the expected mean sojourn time of the system On theone hand when all customers follow the same threshold-119899strategy the effective arrival rate is given by

120582 (119899) = 120582(

119899+1

sum

119895=0

1199010119895 (119899) +

119899

sum

119895=0

1199011119895 (119899)) (90)

On the other hand by conditioning on the state found by acustomer in the system upon arrival we obtain the expectedmean sojourn time if entering the system with threshold-119899policy

119864 [119878 (119899)] = 119860 + 119861 (91)

where

119860 =

119899+1

sum

119895=0

1199010119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

120585 + 120578

120583120578

119861 =

119899

sum

119895=0

1199011119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

times (120585 + 120578

120583120578+ (119895 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(92)

The expected social net benefit is given by

119878soc (119899) = 120582 (119899) 119877 minus 119862120582 (119899) 119864 [119878 (119899)] (93)

Using (90) and (91) and after some algebra we obtain

119878soc (119899) = 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578

times (120582 (119899) (119909119890 + 1) minus 120582

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899))

(94)

Next we need to show that 119878soc(119899) is unimodal and hasonly one maximal point We firstly consider the increments119878soc(119899) minus 119878soc(119899 minus 1) It is not difficult to show that

119878soc (119899) minus 119878soc (119899 minus 1) ge 0 lArrrArr 119891 (119899) le 119909119890 (95)

where 119891(119899) is

119891 (119899) =1198911 (119899)

1198912 (119899)minus 1 (96)

1198911 (119899) = 120582(

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899) minus

119899minus1

sum

119895=0

(119895 + 1) 1199011119895 (119899 minus 1))

(97)

1198912 (119899) = 120582 (119899) minus 120582 (119899 minus 1) (98)

Using the stationary distribution we can simplify the expres-sions of (97) and (98) as follows

1198911 (119899) =120572119861 (119899) 119861 (119899 minus 1) 120588

119899

1205822(1 minus 120588)2

times 119863 (99)

1198912 (119899) =1205831205722120588119899

1205822119861 (119899) 119861 (119899 minus 1) (100)

where

119863 = (120583120572 + 120582120572120585

120578) (119899 + 1) (1 minus 120588) minus 120582

2

minus120582120572120588(1 +120585

120578) + 1205822120588119899+1

+ 120582120572(1 +120585

120578) 120588119899+2

(101)

Plugging (99) and (100) in (96) and replacing 119899 by 119909 weobtain (87) To prove the unimodality of 119878soc(119899) we need toshow that function119891(119909) is increasingWe define the functionℎ [0infin) rarr 119877 with

ℎ (119909) = 119891 (119909) minus 119909 (102)

We have

1198892

1198891199092ℎ (119909) =

[120582 + 120572120588 (1 + (120585120578))] 120588119909+2

(ln 120588)2

(120582 + 120572) (1 minus 120588)2

gt 0

119909 ge 0

(103)

Then function ℎ(119909) is a strictly convex function and thereis no flex point with respect to 120588 and 119909 Therefore ℎ(119909) isincreasing and so is 119891(119909) = ℎ(119909) + 119909 Therefore two casesmay take place

(i) 119891(0) gt 119909119890 then 119878soc(119899) is decreasing and customerswho find the server busy will balk

(ii) 119891(119899soc) le 119909119890 lt 119891(119899soc + 1) then we have 119878soc(119899) minus119878soc(119899minus1) ge 0 for 119899 le 119899soc and 119878soc(119899)minus119878soc(119899minus1) lt 0for 119899 gt 119899soc Then the maximum of 119878soc(119899) occurs in119899soc = lfloor119909socrfloor in which 119909soc is the unique solution ofthe equation 119891(119909) = 119909119890

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

Till now we have completed the analysis of the customersrsquostrategies of the partially observable retrial queueing systemWe will turn to the fully observable case

4 The Fully Observable Case

We now focus on the fully observable case assuming cus-tomers observe not only the state of the server but also theexact number of customers in the retrial orbit Customerthat finds the server idle will enter the system and begin tobe served immediately so this information is valuable onlyfor the customer who finds the server busy We know thatafter every service completion and there is no new customerarriving the customer at the head of the orbit queue willsuccessfully retry for service That means customers in theorbit are served on an FCFS basis Therefore observing theposition in the orbit they can assess precisely whether toenter or balk

Consider a tagged customer that finds the server busyupon arrival This customerrsquos mean overall sojourn time inthe system is not affected by the joiningbalking behavior ofthe other customers that find the server busy because if thetagged customer joins the orbit queue at the 119895th positionthen the late customers who find the server busy will jointhe queue at the 119895 + 1th 119895 + 2th positions behind himin the orbit So his sojourn time does not depend on theirdecisions but it depends on the arrival rate because the newlyarriving customer who finds the server idle will be servedimmediately

To study the general threshold strategy adopted by allcustomers in the fully observable case we will firstly considerthe mean overall sojourn time of the customer who finds theserver idle busy or broken upon arrival Let 119879(1 0) be themean sojourn time when the server is idle and there is nocustomers in the orbit at his arrival instant and let 119879(119894 119895)be the mean sojourn time of a tagged customer at the 119895thposition in the orbit given that the server is at state 119894 Whenall customers follow the general strategy the mean overallsojourn times 119879(119894 119895) are given by

119879 (1 0) =120585 + 120578

120583120578 (68)

119879 (1 119895) =1

120583 + 120585+

120585

120583 + 120585119879 (2 119895) +

120583

120583 + 120585119879 (0 119895) 119895 ge 1

(69)

119879 (0 119895) =1

120582 + 120572+

120582

120582 + 120572119879 (1 119895) +

120572

120582 + 120572119879 (1 119895 minus 1)

119895 ge 1

(70)

119879 (2 119895) =1

120578+ 119879 (1 119895) 119895 ge 0 (71)

Let us consider a customer in the 119895th position of the orbitwhen the server is busy Then this customer has to wait foran exponentially distributed time with rate 120585 + 120583 for thenext event to occur in which the server breaks down or the

service is completed With probability 120585(120585 + 120583) breakdownoccurs then the mean overall sojourn time becomes 119879(2 119895)and the tagged customer remains in the 119895th position of theorbit On the other hand with probability 120583(120585 + 120583) theservice is completed then the mean overall sojourn timebecomes 119879(1 119895 minus 1) and the tagged customer moves to the119895 minus 1th position Through the above analysis we obtain (69)Considering the other cases in the same line we obtain therest equations

Solving the above equations the mean overall sojourntimes are given by

119879 (0 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578 119895 ge 1

119879 (1 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 119895 ge 0

119879 (2 119895) = 119895(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120583 + 120585 + 120578

120583120578

119895 ge 0

(72)

Then we need to find the general equilibrium strategyfollowed by the customers when the server is busy To ensurethat customer will enter the orbit when finding the serverbusy and the queue in the orbit is empty we further assumethat 119877 minus 119862119879(1 1) gt 0 The condition can be written as

119877

119862gt(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578 (73)

This means customersrsquo reward should be larger in the fullyobservable case than in the former case even if the other sys-tematic parameters are equal We will identify the customersrsquoequilibrium threshold strategy and we have the followingtheorem

Theorem 11 In the fully observable single-server constantretrial queue there exists a unique threshold joining strategyldquoenter the retrial orbit if there are at most 119899119890 customers in theorbit whenever finding the server busyrdquo in which condition (73)holds The threshold 119899119890 is given by

119899119890 = lfloor119909119890rfloor (74)

where

119909119890 =120572120583120578119877 minus 119862120572 (120585 + 120578)

119862 [(120582 + 120572) (120585 + 120578) + 120583120578]minus 1 (75)

This strategy is the unique equilibrium strategy among allpossible strategies

Proof Consider a tagged customer that finds the system atthe state (1 119895) upon arrival If he decides to join he goesdirectly to the orbit and occupies the 119895 + 1th position Thenhis expected net benefit is 119878119890(119895) = 119877 minus 119862119879(1 119895 + 1) where

119878119890 (119895) = 119877 minus 119862[(119895 + 1)(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578+120585 + 120578

120583120578]

(76)

Mathematical Problems in Engineering 9

middot middot middot

middot middot middot

middot middot middot

120582

120582 120582 120582

120583 120572 120582 120583 120582 120583120572 120572

120585 120578 120585 120578 120585 120578

120582 120583

120585 120578

120582 120583

120585 120578

0 n0 0 0 1

1 1

2 1

0 2

1 2

2 2

1 0

2 0

1 n

2 n

0 n + 1

1 n + 1

2 n + 1

Figure 3 Transition rate diagram of the fully observable case

The customer prefers to join if 119878119890(119895) gt 0 and he is indifferentbetween joining and balking while 119878119890(119895) = 0 and prefers tobalk if 119878119890(119895) lt 0 A newly arriving customer will enter ifand only if 119895 le 119899119890 when finding the server busy By solvinginequality 119878119890(119895) ge 0 we obtain the result

We will elucidate the uniqueness of the equilibriumthreshold joining strategy followed by all customers Indeedif 119909119890 is an integer number we just need to let 119899119890 = 119909119890 thenwe know 119878119890(119899119890) = 0 and the mixed threshold strategy isthat prescribes to enter if 119899 lt 119899119890 to balk if 119899 gt 119899119890 and beindifferent if 119899 = 119899119890 When 119909119890 is any positive value we let119899119890 = lfloor119909119890rfloor Hence in this case the equilibrium strategy is alsounique

We now turn our attention to the social and profit maxi-mization problems and identify the thresholds 119899soc and 119899profthat maximize the social net benefit and the administratorrsquosprofit per time unit To consummate this goal we need todetermine the stationary behavior of the system when allcustomers follow the same threshold strategy

Suppose that all customers that find the server busy followthe same threshold strategies 119899soc and 119899prof respectively Thatis to say customers who find the sever busy will enter thesystem as long as the number of customers in the orbitis at most 119899soc when considering the social maximizationproblem and there is at most 119899prof customers in the orbitwhen considering the administratorrsquos optimization problemHence the maximum number of customers in the orbitwill never be larger than 119899 + 1 when customers follow thesame threshold-119899 strategyThen we obtain a continuous-timeMarkov chain (119868(119905)119873(119905)) with state space 119878 = 0 1 2 times

0 1 2 119899 + 1 and its transition diagram is given byFigure 3

The stationary distribution of the system (119901119894119895(119894 119895) isin 119878)is given by the following proposition

Proposition 12 Consider the fully observable single-serverconstant retrial queue with breakdowns and repairs in whichcustomers follow a threshold-119899 strategy then the stationarydistribution of the system is given by

11990100 = 119861 (119899)120583120572

1205822

1199010119895 = 119861 (119899) 120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 119861 (119899)120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 119861 (119899)120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(77)

where

120588 =120582 (120582 + 120572)

120572120583

119861 (119899) = 120583120572

1205822+ (1 + 120588 + sdot sdot sdot + 120588

119899) +

120572

120582(1 + 120588 + sdot sdot sdot + 120588

119899+1)

+120572120585

120582120578(1 + 120588 + sdot sdot sdot + 120588

119899+1)

minus1

(78)

Proof The stationary distribution is obtained from the fol-lowing balance equations

12058211990100 = 12058311990110 (79)

(120582 + 120572) 1199010119895 = 1205831199011119895 119895 = 1 2 119899 + 1 (80)

1205821199011119895 = 1205721199010119895+1 119895 = 0 1 2 119899 (81)

1205781199012119895 = 1205851199011119895 119895 = 0 1 119899 + 1 (82)

(120583 + 120585) 1199011119899+1 = 1205781199012119899+1 + 1205821199011119899 + 1205821199010119899+1 (83)

Combining (80) with (81) we can obtain the recursiverelation

1199011119895 = 11990110120588119895 119895 = 1 2 119899 minus 1 (84)

Then we derive the expressions of 1199012119895 from (82)

1199012119895 =120585

12057812058811989511990110 119895 = 0 1 119899 + 1 (85)

Using (80)ndash(82) we can get

11990100 = 11990110

120583120572

1205822

1199010119895 = 11990110120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 11990110120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 11990110120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(86)

Then with the help of the normalizing equation sum119899+1119895=0 1199010119895 +sum119899+1

119895=0 1199011119895 + sum119899+1

119895=0 1199012119895 = 1 we get the expression of 11990110

By considering the stationary distribution we examinethe social welfare maximization threshold strategy We havethe following result

10 Mathematical Problems in Engineering

Theorem 13 In our single-server constant retrial queue inwhich the condition (73) holds we define the following func-tion

119891 (119909) =(120583120572 + (120585120578) 120582120572) (119909 + 1)

120583120572 (1 minus 120588)

minus120582 (120582 + 120572120588 (1 + (120585120578))) (1 minus 120588

119909+1)

120583120572(1 minus 120588)2

minus 1

(87)

(88)

If 119891(0) gt 119909119890 then the customersrsquo best response is balkingOtherwise there exists a unique threshold-119899soc strategy ldquo enterthe retrial orbit if the number of customers in the orbit is atmost119899soc whenever customers find the server busyrdquo that maximizesthe social net benefit per time unit The threshold 119899soc is givenby

119899soc = lfloor119909119904119900119888rfloor (89)

where 119909soc is the unique nonnegative root of the equation119891(119909) = 119909119890

Proof In order to determine the optimal threshold 119899soc weneed to compute the effective arrival rate of the systemand the expected mean sojourn time of the system On theone hand when all customers follow the same threshold-119899strategy the effective arrival rate is given by

120582 (119899) = 120582(

119899+1

sum

119895=0

1199010119895 (119899) +

119899

sum

119895=0

1199011119895 (119899)) (90)

On the other hand by conditioning on the state found by acustomer in the system upon arrival we obtain the expectedmean sojourn time if entering the system with threshold-119899policy

119864 [119878 (119899)] = 119860 + 119861 (91)

where

119860 =

119899+1

sum

119895=0

1199010119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

120585 + 120578

120583120578

119861 =

119899

sum

119895=0

1199011119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

times (120585 + 120578

120583120578+ (119895 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(92)

The expected social net benefit is given by

119878soc (119899) = 120582 (119899) 119877 minus 119862120582 (119899) 119864 [119878 (119899)] (93)

Using (90) and (91) and after some algebra we obtain

119878soc (119899) = 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578

times (120582 (119899) (119909119890 + 1) minus 120582

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899))

(94)

Next we need to show that 119878soc(119899) is unimodal and hasonly one maximal point We firstly consider the increments119878soc(119899) minus 119878soc(119899 minus 1) It is not difficult to show that

119878soc (119899) minus 119878soc (119899 minus 1) ge 0 lArrrArr 119891 (119899) le 119909119890 (95)

where 119891(119899) is

119891 (119899) =1198911 (119899)

1198912 (119899)minus 1 (96)

1198911 (119899) = 120582(

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899) minus

119899minus1

sum

119895=0

(119895 + 1) 1199011119895 (119899 minus 1))

(97)

1198912 (119899) = 120582 (119899) minus 120582 (119899 minus 1) (98)

Using the stationary distribution we can simplify the expres-sions of (97) and (98) as follows

1198911 (119899) =120572119861 (119899) 119861 (119899 minus 1) 120588

119899

1205822(1 minus 120588)2

times 119863 (99)

1198912 (119899) =1205831205722120588119899

1205822119861 (119899) 119861 (119899 minus 1) (100)

where

119863 = (120583120572 + 120582120572120585

120578) (119899 + 1) (1 minus 120588) minus 120582

2

minus120582120572120588(1 +120585

120578) + 1205822120588119899+1

+ 120582120572(1 +120585

120578) 120588119899+2

(101)

Plugging (99) and (100) in (96) and replacing 119899 by 119909 weobtain (87) To prove the unimodality of 119878soc(119899) we need toshow that function119891(119909) is increasingWe define the functionℎ [0infin) rarr 119877 with

ℎ (119909) = 119891 (119909) minus 119909 (102)

We have

1198892

1198891199092ℎ (119909) =

[120582 + 120572120588 (1 + (120585120578))] 120588119909+2

(ln 120588)2

(120582 + 120572) (1 minus 120588)2

gt 0

119909 ge 0

(103)

Then function ℎ(119909) is a strictly convex function and thereis no flex point with respect to 120588 and 119909 Therefore ℎ(119909) isincreasing and so is 119891(119909) = ℎ(119909) + 119909 Therefore two casesmay take place

(i) 119891(0) gt 119909119890 then 119878soc(119899) is decreasing and customerswho find the server busy will balk

(ii) 119891(119899soc) le 119909119890 lt 119891(119899soc + 1) then we have 119878soc(119899) minus119878soc(119899minus1) ge 0 for 119899 le 119899soc and 119878soc(119899)minus119878soc(119899minus1) lt 0for 119899 gt 119899soc Then the maximum of 119878soc(119899) occurs in119899soc = lfloor119909socrfloor in which 119909soc is the unique solution ofthe equation 119891(119909) = 119909119890

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

middot middot middot

middot middot middot

middot middot middot

120582

120582 120582 120582

120583 120572 120582 120583 120582 120583120572 120572

120585 120578 120585 120578 120585 120578

120582 120583

120585 120578

120582 120583

120585 120578

0 n0 0 0 1

1 1

2 1

0 2

1 2

2 2

1 0

2 0

1 n

2 n

0 n + 1

1 n + 1

2 n + 1

Figure 3 Transition rate diagram of the fully observable case

The customer prefers to join if 119878119890(119895) gt 0 and he is indifferentbetween joining and balking while 119878119890(119895) = 0 and prefers tobalk if 119878119890(119895) lt 0 A newly arriving customer will enter ifand only if 119895 le 119899119890 when finding the server busy By solvinginequality 119878119890(119895) ge 0 we obtain the result

We will elucidate the uniqueness of the equilibriumthreshold joining strategy followed by all customers Indeedif 119909119890 is an integer number we just need to let 119899119890 = 119909119890 thenwe know 119878119890(119899119890) = 0 and the mixed threshold strategy isthat prescribes to enter if 119899 lt 119899119890 to balk if 119899 gt 119899119890 and beindifferent if 119899 = 119899119890 When 119909119890 is any positive value we let119899119890 = lfloor119909119890rfloor Hence in this case the equilibrium strategy is alsounique

We now turn our attention to the social and profit maxi-mization problems and identify the thresholds 119899soc and 119899profthat maximize the social net benefit and the administratorrsquosprofit per time unit To consummate this goal we need todetermine the stationary behavior of the system when allcustomers follow the same threshold strategy

Suppose that all customers that find the server busy followthe same threshold strategies 119899soc and 119899prof respectively Thatis to say customers who find the sever busy will enter thesystem as long as the number of customers in the orbitis at most 119899soc when considering the social maximizationproblem and there is at most 119899prof customers in the orbitwhen considering the administratorrsquos optimization problemHence the maximum number of customers in the orbitwill never be larger than 119899 + 1 when customers follow thesame threshold-119899 strategyThen we obtain a continuous-timeMarkov chain (119868(119905)119873(119905)) with state space 119878 = 0 1 2 times

0 1 2 119899 + 1 and its transition diagram is given byFigure 3

The stationary distribution of the system (119901119894119895(119894 119895) isin 119878)is given by the following proposition

Proposition 12 Consider the fully observable single-serverconstant retrial queue with breakdowns and repairs in whichcustomers follow a threshold-119899 strategy then the stationarydistribution of the system is given by

11990100 = 119861 (119899)120583120572

1205822

1199010119895 = 119861 (119899) 120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 119861 (119899)120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 119861 (119899)120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(77)

where

120588 =120582 (120582 + 120572)

120572120583

119861 (119899) = 120583120572

1205822+ (1 + 120588 + sdot sdot sdot + 120588

119899) +

120572

120582(1 + 120588 + sdot sdot sdot + 120588

119899+1)

+120572120585

120582120578(1 + 120588 + sdot sdot sdot + 120588

119899+1)

minus1

(78)

Proof The stationary distribution is obtained from the fol-lowing balance equations

12058211990100 = 12058311990110 (79)

(120582 + 120572) 1199010119895 = 1205831199011119895 119895 = 1 2 119899 + 1 (80)

1205821199011119895 = 1205721199010119895+1 119895 = 0 1 2 119899 (81)

1205781199012119895 = 1205851199011119895 119895 = 0 1 119899 + 1 (82)

(120583 + 120585) 1199011119899+1 = 1205781199012119899+1 + 1205821199011119899 + 1205821199010119899+1 (83)

Combining (80) with (81) we can obtain the recursiverelation

1199011119895 = 11990110120588119895 119895 = 1 2 119899 minus 1 (84)

Then we derive the expressions of 1199012119895 from (82)

1199012119895 =120585

12057812058811989511990110 119895 = 0 1 119899 + 1 (85)

Using (80)ndash(82) we can get

11990100 = 11990110

120583120572

1205822

1199010119895 = 11990110120588119895minus1 119895 = 1 2 119899 + 1

1199011119895 = 11990110120572

120582120588119895 119895 = 0 1 2 119899 + 1

1199012119895 = 11990110120572

120582

120585

120578120588119895 119895 = 0 1 2 119899 + 1

(86)

Then with the help of the normalizing equation sum119899+1119895=0 1199010119895 +sum119899+1

119895=0 1199011119895 + sum119899+1

119895=0 1199012119895 = 1 we get the expression of 11990110

By considering the stationary distribution we examinethe social welfare maximization threshold strategy We havethe following result

10 Mathematical Problems in Engineering

Theorem 13 In our single-server constant retrial queue inwhich the condition (73) holds we define the following func-tion

119891 (119909) =(120583120572 + (120585120578) 120582120572) (119909 + 1)

120583120572 (1 minus 120588)

minus120582 (120582 + 120572120588 (1 + (120585120578))) (1 minus 120588

119909+1)

120583120572(1 minus 120588)2

minus 1

(87)

(88)

If 119891(0) gt 119909119890 then the customersrsquo best response is balkingOtherwise there exists a unique threshold-119899soc strategy ldquo enterthe retrial orbit if the number of customers in the orbit is atmost119899soc whenever customers find the server busyrdquo that maximizesthe social net benefit per time unit The threshold 119899soc is givenby

119899soc = lfloor119909119904119900119888rfloor (89)

where 119909soc is the unique nonnegative root of the equation119891(119909) = 119909119890

Proof In order to determine the optimal threshold 119899soc weneed to compute the effective arrival rate of the systemand the expected mean sojourn time of the system On theone hand when all customers follow the same threshold-119899strategy the effective arrival rate is given by

120582 (119899) = 120582(

119899+1

sum

119895=0

1199010119895 (119899) +

119899

sum

119895=0

1199011119895 (119899)) (90)

On the other hand by conditioning on the state found by acustomer in the system upon arrival we obtain the expectedmean sojourn time if entering the system with threshold-119899policy

119864 [119878 (119899)] = 119860 + 119861 (91)

where

119860 =

119899+1

sum

119895=0

1199010119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

120585 + 120578

120583120578

119861 =

119899

sum

119895=0

1199011119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

times (120585 + 120578

120583120578+ (119895 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(92)

The expected social net benefit is given by

119878soc (119899) = 120582 (119899) 119877 minus 119862120582 (119899) 119864 [119878 (119899)] (93)

Using (90) and (91) and after some algebra we obtain

119878soc (119899) = 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578

times (120582 (119899) (119909119890 + 1) minus 120582

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899))

(94)

Next we need to show that 119878soc(119899) is unimodal and hasonly one maximal point We firstly consider the increments119878soc(119899) minus 119878soc(119899 minus 1) It is not difficult to show that

119878soc (119899) minus 119878soc (119899 minus 1) ge 0 lArrrArr 119891 (119899) le 119909119890 (95)

where 119891(119899) is

119891 (119899) =1198911 (119899)

1198912 (119899)minus 1 (96)

1198911 (119899) = 120582(

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899) minus

119899minus1

sum

119895=0

(119895 + 1) 1199011119895 (119899 minus 1))

(97)

1198912 (119899) = 120582 (119899) minus 120582 (119899 minus 1) (98)

Using the stationary distribution we can simplify the expres-sions of (97) and (98) as follows

1198911 (119899) =120572119861 (119899) 119861 (119899 minus 1) 120588

119899

1205822(1 minus 120588)2

times 119863 (99)

1198912 (119899) =1205831205722120588119899

1205822119861 (119899) 119861 (119899 minus 1) (100)

where

119863 = (120583120572 + 120582120572120585

120578) (119899 + 1) (1 minus 120588) minus 120582

2

minus120582120572120588(1 +120585

120578) + 1205822120588119899+1

+ 120582120572(1 +120585

120578) 120588119899+2

(101)

Plugging (99) and (100) in (96) and replacing 119899 by 119909 weobtain (87) To prove the unimodality of 119878soc(119899) we need toshow that function119891(119909) is increasingWe define the functionℎ [0infin) rarr 119877 with

ℎ (119909) = 119891 (119909) minus 119909 (102)

We have

1198892

1198891199092ℎ (119909) =

[120582 + 120572120588 (1 + (120585120578))] 120588119909+2

(ln 120588)2

(120582 + 120572) (1 minus 120588)2

gt 0

119909 ge 0

(103)

Then function ℎ(119909) is a strictly convex function and thereis no flex point with respect to 120588 and 119909 Therefore ℎ(119909) isincreasing and so is 119891(119909) = ℎ(119909) + 119909 Therefore two casesmay take place

(i) 119891(0) gt 119909119890 then 119878soc(119899) is decreasing and customerswho find the server busy will balk

(ii) 119891(119899soc) le 119909119890 lt 119891(119899soc + 1) then we have 119878soc(119899) minus119878soc(119899minus1) ge 0 for 119899 le 119899soc and 119878soc(119899)minus119878soc(119899minus1) lt 0for 119899 gt 119899soc Then the maximum of 119878soc(119899) occurs in119899soc = lfloor119909socrfloor in which 119909soc is the unique solution ofthe equation 119891(119909) = 119909119890

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

Theorem 13 In our single-server constant retrial queue inwhich the condition (73) holds we define the following func-tion

119891 (119909) =(120583120572 + (120585120578) 120582120572) (119909 + 1)

120583120572 (1 minus 120588)

minus120582 (120582 + 120572120588 (1 + (120585120578))) (1 minus 120588

119909+1)

120583120572(1 minus 120588)2

minus 1

(87)

(88)

If 119891(0) gt 119909119890 then the customersrsquo best response is balkingOtherwise there exists a unique threshold-119899soc strategy ldquo enterthe retrial orbit if the number of customers in the orbit is atmost119899soc whenever customers find the server busyrdquo that maximizesthe social net benefit per time unit The threshold 119899soc is givenby

119899soc = lfloor119909119904119900119888rfloor (89)

where 119909soc is the unique nonnegative root of the equation119891(119909) = 119909119890

Proof In order to determine the optimal threshold 119899soc weneed to compute the effective arrival rate of the systemand the expected mean sojourn time of the system On theone hand when all customers follow the same threshold-119899strategy the effective arrival rate is given by

120582 (119899) = 120582(

119899+1

sum

119895=0

1199010119895 (119899) +

119899

sum

119895=0

1199011119895 (119899)) (90)

On the other hand by conditioning on the state found by acustomer in the system upon arrival we obtain the expectedmean sojourn time if entering the system with threshold-119899policy

119864 [119878 (119899)] = 119860 + 119861 (91)

where

119860 =

119899+1

sum

119895=0

1199010119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

120585 + 120578

120583120578

119861 =

119899

sum

119895=0

1199011119895 (119899)

sum119899+1

119896=0 1199010119896 (119899) + sum119899

119896=0 1199011119896 (119899)

times (120585 + 120578

120583120578+ (119895 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(92)

The expected social net benefit is given by

119878soc (119899) = 120582 (119899) 119877 minus 119862120582 (119899) 119864 [119878 (119899)] (93)

Using (90) and (91) and after some algebra we obtain

119878soc (119899) = 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578

times (120582 (119899) (119909119890 + 1) minus 120582

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899))

(94)

Next we need to show that 119878soc(119899) is unimodal and hasonly one maximal point We firstly consider the increments119878soc(119899) minus 119878soc(119899 minus 1) It is not difficult to show that

119878soc (119899) minus 119878soc (119899 minus 1) ge 0 lArrrArr 119891 (119899) le 119909119890 (95)

where 119891(119899) is

119891 (119899) =1198911 (119899)

1198912 (119899)minus 1 (96)

1198911 (119899) = 120582(

119899

sum

119895=0

(119895 + 1) 1199011119895 (119899) minus

119899minus1

sum

119895=0

(119895 + 1) 1199011119895 (119899 minus 1))

(97)

1198912 (119899) = 120582 (119899) minus 120582 (119899 minus 1) (98)

Using the stationary distribution we can simplify the expres-sions of (97) and (98) as follows

1198911 (119899) =120572119861 (119899) 119861 (119899 minus 1) 120588

119899

1205822(1 minus 120588)2

times 119863 (99)

1198912 (119899) =1205831205722120588119899

1205822119861 (119899) 119861 (119899 minus 1) (100)

where

119863 = (120583120572 + 120582120572120585

120578) (119899 + 1) (1 minus 120588) minus 120582

2

minus120582120572120588(1 +120585

120578) + 1205822120588119899+1

+ 120582120572(1 +120585

120578) 120588119899+2

(101)

Plugging (99) and (100) in (96) and replacing 119899 by 119909 weobtain (87) To prove the unimodality of 119878soc(119899) we need toshow that function119891(119909) is increasingWe define the functionℎ [0infin) rarr 119877 with

ℎ (119909) = 119891 (119909) minus 119909 (102)

We have

1198892

1198891199092ℎ (119909) =

[120582 + 120572120588 (1 + (120585120578))] 120588119909+2

(ln 120588)2

(120582 + 120572) (1 minus 120588)2

gt 0

119909 ge 0

(103)

Then function ℎ(119909) is a strictly convex function and thereis no flex point with respect to 120588 and 119909 Therefore ℎ(119909) isincreasing and so is 119891(119909) = ℎ(119909) + 119909 Therefore two casesmay take place

(i) 119891(0) gt 119909119890 then 119878soc(119899) is decreasing and customerswho find the server busy will balk

(ii) 119891(119899soc) le 119909119890 lt 119891(119899soc + 1) then we have 119878soc(119899) minus119878soc(119899minus1) ge 0 for 119899 le 119899soc and 119878soc(119899)minus119878soc(119899minus1) lt 0for 119899 gt 119899soc Then the maximum of 119878soc(119899) occurs in119899soc = lfloor119909socrfloor in which 119909soc is the unique solution ofthe equation 119891(119909) = 119909119890

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 11

The last purpose of this paper is to solve the profit max-imization problem By imposing an appropriate nonnegativeadmission fee 119901 on the customers the administrator canmaximize his net profit per time unit We have the followingtheorem

Theorem 14 In the fully observable single-server constantretrial queue with breakdowns and repairs in which condition(73) holds let

119892 (119909) = 119909 + 120583 (1 minus 120588119909+1

)

times(1+(120582120585120583120578)minus((120582+120572120588 (1+120585120578)) (120582+120572)) 120588119909+2

)

times (120582120588119909(1 minus 120588)

2)minus1

(104)

if119892(0) gt 119909119890 then customersrsquo best strategy is balking Otherwisethere exists a unique threshold joining strategy ldquo enter the retrialorbit when the number of customers in the orbit is at most119899prof whenever finding the server busyrdquo that maximizes theadministratorrsquos net profit per time unit The threshold 119899prof isgiven by

119899prof = lfloor119909profrfloor (105)

where 119909prof is the unique nonnegative root of the equation119892(119909) = 119909119890

Proof We first need to determine the appropriate entrancefee 119901 that the administrator imposes on the customers in thesystem As we have analyzed in Theorem 7 the entrance fee119901 also satisfies

119877 minus 119901 (119899) = 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(106)

From the above equation we obtain

119901 (119899) = 119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578)

(107)

The administratorrsquos profit per time unit is

119878prof (119899) = 120582 (119899) 119901 (119899)

= 120582 (119899) (119877 minus 119862(120585 + 120578

120583120578+ (119899 + 1)

times(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578))

= 119862(120582 + 120572) (120585 + 120578) + 120583120578

120572120583120578120582 (119899) (119909119890 minus 119899)

(108)

We will proceed to prove the unimodality of the function119892(119909) From (108) we get

119878prof (119899)

119878prof (119899 minus 1)ge 1 lArrrArr 119892 (119899) le 119909119890 (109)

where 119892(119899) = 119899 + 120582(119899 minus 1)(120582(119899) minus 120582(119899 minus 1)) Simplifying(90) and using (98) and (100) we obtain the expression of119892(119899) substituting 119899 by 119909 we obtain (100)We can examine thefunction 119892(119909) is increasing Then we consider the followingtwo cases

(i) 119892(0) gt 119909119890 then function 119878prof(119899) is decreasing and thecustomersrsquo best response is balking

(ii) 119892(119899prof) le 119909119890 lt 119892(119899prof + 1) then 119878prof(119899) minus 119878prof(119899 minus1) ge 0 for 119899 le 119899prof and 119878prof(119899) minus 119878prof(119899 minus 1) lt 0 for119899 gt 119899prof Then the maximum of 119878prof(119899) is attained at119899prof = lfloor119909profrfloor where 119909prof is the unique solution of119892(119909) = 119909119890

Remark 15 In this case if we take a close look at theexpressions of thresholds 119899119890 119899soc and 119899prof we will findthat these thresholds are also functions of 120585120578 when otherparameters are fixed Furthermore we let 120585 equal to zero thenthis case is equivalent to themodel studied in [6] andwe havethe same results when counterpart problems are consideredSo it is also a special case of this model

Now we finish the analysis of the single-server retrialqueueing systemwith constant retrial rate in the partially andfully observable casesWe obtain the entrance probabilities inthe partially observable case and the thresholds in the lattercase In the following section wewill present some numericalexamples to support our results

5 Numerical Examples

In this section wewill investigate the effects of the parameterson the customersrsquo mixed strategies in this constant retrialqueueing system Figures 4 5 and 6 present the influenceof the parameters on the equilibrium joining probability119903119890 and optimal entrance probabilities 119903soc and 119903prof whereparameters 120582 120572 and 120585120578 vary respectively In these figureswe can clearly see that 119903119890 ge 119903soc ge 119903prof which has been provedinTheorems 6 and 8

Now we pay our attention to the fully observable caseFigures 7ndash10 depict the equilibrium and optimal joiningthresholds vary with respect to parameters 120582 120583 120572 and 120585120578From these four figures we can see 119899prof le 119899soc le 119899119890 Indeedthe relation 119899soc le 119899119890 is obvious from Theorems 11 and 13On the one hand we can obtain that ℎ(0) = 119891(0) gt 0 andbecause function ℎ(119909) is increasing and ℎ(119909soc) ge 0 then weget 119891(119909soc) minus 119909soc ge 0 But on the other hand 119891(119909soc) = 119909119890which means 119909soc le 119909119890 Taking floors we obtain the result119899soc le 119899119890 The relation 119899prof le 119899soc is not proved theoreticallybut it can be traced numerically

Figure 7 shows the thresholds when we consider the indi-vidual equilibrium social and profitmaximization threshold

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Mathematical Problems in Engineering

0

Entr

ance

pro

babi

litie

s

02

03

04

05

06

07

0201 03 04 05 06 07

08

09

1

11

rersocrprof

120582

Figure 4 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120582 for 120583 = 1 120578 = 1 120585 = 002120572 = 2 119877 = 10 and 119862 = 1

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120572

Figure 5 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120572 120582 = 05 for 120583 = 1 120585 = 002119877 = 10 and 119862 = 1

strategies with respect to 120582 We can see that the thresholdsfinally converge to 0 On the one hand it is true that when 120582becomes larger there are more customers arriving per timeunit So customers in the orbit will wait longer to successfullyretry for service On the other hand from (74) we can see that119899119890 is decreasingwith respect to120582 And because of the function119878soc(119899) and 119878prof(119899) we can see that thresholds 119899soc and 119899prof

0 1 2 3 4 5 6

Entr

ance

pro

babi

litie

s

0

02

04

06

08

1

rersocrprof

120585120578

Figure 6 Individual equilibrium strategy and optimal entranceprobabilities vary with respect to 120585120578 for 120583 = 1 120582 = 05 120572 = 2119877 = 10 and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

120582

nensocnprof

Figure 7 Individual equilibrium threshold and optimal thresholdsvary with respect to 120582 for 120585 = 002 120583 = 1 120578 = 1 120572 = 2 119877 = 40 and119862 = 1

finally become zero for large value of 120582 We have 119899prof le 119899socas 120582 varies When considering the variation of parameters 120583and120572 in Figures 8 and 9 takingmean service and retrial timesof the system into consideration we can easily explain theoutcomes and we also have 119899prof le 119899soc

Figure 10 presents the thresholds vary with respect to 120585120578On the one hand from the expression of 119899119890 and functions

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 13

05

10152025303540

Opt

imal

thre

shol

ds

0 05 1 15 2 25 3

nensocnprof

120583

Figure 8 Individual equilibrium threshold and optimal thresholdsvary with respect to 120583 for 120585 = 002 120572 = 2 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

0 1 2 3 4 5 60

5

10

15

20

25

30

Opt

imal

thre

shol

ds

nensocnprof

120572

Figure 9 Individual equilibrium threshold and optimal thresholdsvary with respect to 120572 for 120585 = 002 120583 = 1 120578 = 1 120582 = 05 119877 = 40and 119862 = 1

119878soc(119899) and 119878prof(119899) we know that thresholds decrease to 0when 120585120578 exceeds a positive number On the other handwhen 120585120578 increases it means that the lifetime of the systembecomes shorter then the customers in the orbit need to waitlonger for the server restored So the thresholds decreasewith respect to 120585120578 When 120585120578 varies we can clearly see that119899prof le 119899soc

6 Conclusions

We have analyzed the equilibrium and optimal entranceprobabilities in the partially observable case and the coun-terpart thresholds in the fully observable case We have an

0 1 2 3 4 5 60

5

10

15

20

25

Opt

imal

thre

shol

ds

nensocnprof

120585120578

Figure 10 Individual equilibrium threshold and optimal thresholdsvary with respect to 120585120578 for 120582 = 1 120583 = 1 120572 = 2 119877 = 40 and 119862 = 1

ldquoavoid-the-crowdrdquo (ATC) situation in the former case Itwas observed that individualrsquos net benefit is decreasing withrespect to the entrance probability 119903 and the individualrsquos bestresponse is a decreasing function of the strategy selected bythe other customers In the fully observable case we obtainedthat the individualrsquos net benefit decreases with respect to thenumber of customers in the orbit but the social welfare andthe administratorrsquos net benefit are concave with respect to thenumber of customers in the orbit

Both in the partially observable case and the fullyobservable case the solutions of the social maximizationproblem are smaller than the solutions of the individualequilibrium problem This is aroused by the fact that theformer customers imposed negative externality on the latearriving customers Rational individuals in an economicsystem prefer to maximize their own benefit but if customerswant to maximize their social net benefit they need tocooperate rather than ignore the negative externality that isimposed on the other customers We observed that 119903119890 ge

119903soc ge 119903prof in the partially observable case and 119899prof le

119899soc le 119899119890 in the fully observable case which mean that whenconsidering optimization problems the effective service ratecan not satisfy the customersrsquo desirable service demand Thisfurther implicates that there are some customers who can notobtain service in some cases even if they are identical

The effect of the server unreliability on the systemperformance cannot be ignored It was readily seen that cus-tomersrsquo equilibrium entrance probability optimal entranceprobability and the related threshold strategies all decreaseas 120585120578 increases This is very important in managementprocess When 120585120578 increases some customersrsquo interest willbe undermined and decreases the social welfare and theadministratorrsquos net profit at the same time So it is very crucialto improve the reliability of the sever to achieve the goal ofoptimal management

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

14 Mathematical Problems in Engineering

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research was supported by the National Natural ScienceFoundation of China (Grants nos 11171019 and 713990334)and the Program for New Century Excellent Talents inUniversity (no NCET-11-0568)

References

[1] G I Falin and J G C Templeton Retrial Queues Chapman ampHall London UK 1997

[2] J R Artalejo andAGomez-CorralRetrial Queueing Systems AComputational Approach SpringerHeidelbergGermany 2008

[3] G Fayolle ldquoA simple telephone exchange with delayed feed-backsrdquo in Proceedings of the International Seminar on TeletrafficAnalysis and Computer Performance Evalution pp 245ndash253Amsterdam The Netherlands 1986

[4] K Farahmand ldquoSingle line queue with repeated demandsrdquoQueueing Systems vol 6 no 1 pp 223ndash228 1990

[5] J R Artalejo and A Gomez-Corral ldquoSteady state solution ofa single-server queue with linear repeated requestsrdquo Journal ofApplied Probability vol 34 no 1 pp 223ndash233 1997

[6] A Economou and S Kanta ldquoEquilibrium customer strategiesand social-profit maximization in the single-server constantretrial queuerdquo Naval Research Logistics vol 58 no 2 pp 107ndash122 2011

[7] V G Kulkarni ldquoA game theoretic model for two types ofcustomers competing for servicerdquo Operations Research Lettersvol 2 no 3 pp 119ndash122 1983

[8] A Elcan ldquoOptimal customer return rate for anMM1 queueingsystemwith retrialsrdquoProbability in the Engineering and Informa-tional Sciences vol 8 no 4 pp 521ndash539 1994

[9] R Hassin and M Haviv ldquoOn optimal and equilibrium retrialrates in a queueing systemrdquo Probability in the Engineering andInformational Sciences vol 10 no 2 pp 223ndash227 1996

[10] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquoJournal of Industrial and Management Optimization vol 8 no4 pp 861ndash875 2012

[11] J Wang and F Zhang ldquoStrategic joining in MM1 retrialqueuesrdquo European Journal of Operational Research vol 230 no1 pp 76ndash87 2013

[12] H Li and Y Q Zhao ldquoA retrial queue with a constant retrialrate server break downs and impatient customersrdquo StochasticModels vol 21 no 2-3 pp 531ndash550 2005

[13] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[14] J H Cao andK Cheng ldquoAnalysis of an1198721198661 queueing systemwith repairable service stationrdquo Acta Mathematicae ApplicataeSinica vol 5 no 2 pp 113ndash127 1982

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of