1502 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 34, NO. 6 NOVEMBER 1988

Retrials and Balks EDGAR N. GILBERT, FELLOW, IEEE

Abstract -An overloaded service system may reject new customers if it has no queue to store them. In practice, rejected customers return later to make retrials and may not leave pemanently (balk) until several retrials fail. A single-server system with Poisson arrivals is examined in which rejected customers balk or make retrials according to a simple probabilistic model. Customer service times are independent random variables, all with the same given distribution function b( t ) . The stationary probability distribution for the number of customers waiting to make retrials satisfies a complicated functional equation. The solution is elusive in general but can be obtained for special b ( t ) (exponential) or special values of model parameters. When the solution cannot be found, bounds on the fraction of customers served can be obtained.

DEDICATION

Y DEFT and imaginative twists of mathematics S. 0. B Rice could somehow wring the essence from the most unyielding problem. His specialty, noise theory, was just one of many interests. Working at a telephone laboratory, Rice was naturally attracted to queueing problems (e.g., see [15], [16]). One (see Riordan [17, p. 951) concerned a many-server system with retrials, a subject close to this paper. The pity is that Rice never published it, but that was typical; he had more good ideas than time to write about them. A true gentleman, he gladly gave his time and expert help to us all.

I. INTRODUCTION

A service system, unable to serve new customers, can dispose of them in two ways. A delay system puts them in a queue to wait for service later. A loss system rejects them. A telephone exchange is a loss system, using a busy signal to reject calls to busy lines. In real life, rejected customers do not simply leave. Many return later to make another attempt, a retrial. After one or more failed at- tempts, the customer may truly leave, i.e., balk. While common in practice, retrials and balks are usually ignored in theory.

This paper examines the effect of retrials and balks on a single-server system when new customers arrive in a Poisson stream. Service times for different customers are assumed independent and identically distributed with a general common service-time distribution b( t). At each rejection, a customer balks with probability /3 or waits for a retrial with probability 1-j?. The waiting time to the retrial is assumed random with an exponential distribu- tion. This model of retrials and balks agrees well with

Manuscript received October 9, 1987; revised February 22, 1988. The author is with AT&T Bell Laboratories, Murray Hill, NJ 07974. IEEE Log Number 8824879.

experimental observations on the habits of telephone dialers (see Section 11).

The general service-time distribution b( t ) is a departure from previous treatments that use the exponential distribu- tion

b ( t ) =1- e-p'. (1)

The exponential holding time law (1) has long been known to describe calls by telephone subscribers very well. Molina [13] and Wilkinson [19] give data on 7387 calls made in Newark in 1918 and Erlang [6] also remarks that his early measurements in Copenhagen support (1). However, su- pervisory signaling and other services may have other distributions. To offset the complication of a general dis- tribution b( t ) , the analysis here will consider only single- server systems.

Several authors have studied many-server queues (without retrials) in which customers, with exponential service times, either balk on arrival if they observe a long queue or defect after waiting a long time without being served (see Palm [14], Wilkinson [20], Barrer [l], [2], Haight [8], Finch [7]). Kosten [ l l ] added retrials to the problem and obtained an approximate solution. The exact analysis by Cohen [4] is very complicated. Syski [18] and Riordan [17] survey these results. For an analysis of retrials, but with no balking, in systems with one or two servers, see Keilson et al. [9].

11. WAITS FOR RETRIAL

Clos [3] reported an early experimental measurement of retrials. In 1944, the New York Telephone Company mon- itored 1107 subscribers who dialed local numbers and received busy signals. The study recorded the times they waited before dialing again, and also the times for subse- quent retrials if the first failed. One important observation was that the times waited for a second retrial were dis- tributed very nearly like the times for the first. That is consistent with the assumption that retrial times are ran- dom with a common distribution, the same for first, sec- ond, third, etc., retrial. Graphs in Clos' paper give the function f ( t ) = Pr {retrial time 2 t}. These curves, which can be read to two-digit accuracy, supplied the distribution shown here in Table I. Wilkinson [20] approximated these data by an exponential distribution f ( t ) = which also appears in Table I. The fit is good only for retrial times shorter than 4 min, i.e., for 60 percent of the trials. Clos tried to extrapolate his curves, by eye, to times beyond the 10-min maximum where monitoring ceased. He

0018-9448/88/1100-1502$01.00 01988 IEEE

GILBERT: RETRIALS AND BALKS 1503

concluded that ten percent of the retrials waited more than 100 min and interpreted that to mean that ten percent of the customers balked. Of course, that interpretation is hard to defend because the extrapolation would have given a quite different probability of balking if he had used a different wait, say 10 min or 1000 min, instead of 100 min.

For a different mathematical description of balking, suppose a customer balks with probability p each time he is denied service. Suppose also that a nonbalking customer makes a retrial after a waiting time exponentially dis- tributed with mean l / v . The retrial times are then de- scribed by

f( t ) = p + (1 - p ) e-" (2) with balks counted as retrials with t = 60. In fact, Table I shows that (2) fits Clos' data within a few percent for the parameter values p = 0.24, v = 0.36. The largest errors occur at small values of t , perhaps because (2) ignores the dialing time that is part of each retrial. The shortest retrial times in the data seem to be about 0.2 min. Allowing time for dialing, one might replace (2) by

f ( t ) = p + (1 - p ) e - Y ( t - 2 ) . (3) As Table I shows, (3) with /3 = 0.255 and v = 0.41 fits the data even better than (2). Fitting by (2) or (3) gives much higher estimates of the balking probability, about 1/4 as compared with Clos' 1/10. Also the mean time to retrial is estimated at only 2.78 min in (2) or 2.44 min in (3), as compared with Wilkinson's 4.17 min. Distributions (2) or (3) would hold if customers, after a wait of any duration t , have probability vdt of trying again in the next dt min.

TABLE I PROBABILITY f( t ) THAT DIALERS, RECEIVING A BUSY SIGNAL, WAIT

LONGER THAN t MINUTES BEFORE DIALING AGAIN

t (min)

0.2 0.6 1 2 3 4 5 6 7 8 9

10

Clos Data

Wilkinson e-0241

1 .oo 0.90 0.78 0.58 0.47 0.40 0.36 0.33 0.31 0.29 0.27 0.26

0.953 0.866 0.787 0.619 0.487 0.383 0.301 0.237 0.186 0.147 0.115 0.091

(2) (3)

0.947 1 .OOo 0.852 0.887 0.770 0.792 0.610 0.611 0.498 0.491 0.420 0.412 0.366 0.359 0.328 0.324 0.301 0.301 0.283 0.285 0.270 0.275 0.261 0.268

There appears to be no a priori reason to predict that telephone dialers behave this way. However, the good fit of (2) to the data lends credibility to the simple retrial model that the following sections use.

Cohen [4], used a slightly different model. His balking does not take place at retrials but occurs at constant rate y during the waiting period. That is, a waiting customer is assumed to have probabilities ydt of balking and v'dt of making a retrial during the next dt minutes. Write h ( t ) for the probability that the customer has neither balked nor

made a retrial during time t and g ( t ) for the probability that he has balked before time t . Then h ( t ) and g( t ) satisfy differential equations

-= dh - ( v ' + y ) h , h(0 ) = l , - = y h , dg dt dt

g(O)=O, f ( t ) = g ( t ) + h ( t ) .

The solution is again (2), now with v = v'+ y and p = y / v . Although both balking models predict (2), balking cus- tomers remain in the system longer in Cohen's model.

In telephony, retrials and balks can occur for several reasons. The called line may be busy, may not answer, or may be blocked by overload. The caller receives different indications of these failure conditions and reacts differ- ently to them, i.e., each kind of failure has its own retrial distribution. The overall distribution of telephone retrial times is then an average of distributions for the different failure conditions and cannot be expected to fit (2). Wilkinson and Radnik [21], Duffy and Mercer [5], and Liu [12] discuss recent measurements. For each failure condi- tion, Liu fits the measured distribution by three empirical formulas (Wilkinson's exponential, the log-normal, and a distribution with two exponential terms).

111. FRACTION SERVED

Fig. 1 shows a service system with retrials and balks. New customers appear in a Poisson stream with rate X customers per minute. On arrival or when making a retrial, if a customer finds the server idle, his service begins immediately and lasts a random time with a prescribed proability distribution Pr {service time 5 t } = b( t ) . If the customer finds the server busy, he may balk immediately with probability p, or wait for a retrial with probability 1 - p. The waiting time for a retrial will be exponentially distributed with mean l / v . Thus, counting a balk as an infinite retrial time, the probability f ( r ) of retrial times exceeding t is given by (2).

WAITING CUSTOMERS

I I I L _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ J

Fig. 1. Service system with retrials and balks

Note that customers balk with a probability ,8 that is independent of other customers' behavior. That is a rea- sonable assumption in telephony because customers do not observe one another. If customers knew how many others

1504 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 34, NO. 6 NOVEMBER 1988

were waiting for retrials, a more reasonable model might make the balking probability depend on the number wait- ing.

Perhaps the most important property of t h s service system is the fraction f of new customers that manage to receive service. This fraction, which will be examined in Section VIII, is simply related to other rates and moments that describe retrials and balks. In equilibrium customers must leave the system, shown as a dashed box in Fig. 1, at the arrival rate A. Served customers leave at rate f A ; then ( 1 - f ) A is the rate at which customers leave by balking. However, in Fig. 1 it is clear that customers return for retrials at a rate (l-P)/P times the balking rate. Then retrials occur at a rate R = (1 - P)(1- f)A/P. A customer makes a mean number R/A of retrials and so waits for a mean time R / ( A v ) before balking or beginning service. According to Little's theorem (see Kleinrock [lo]) the mean number of customers still waiting is A times the mean waiting time;

E(number waiting) = ( l - P ) ( l - f ) A / ( @ v ) . A service completion is an instant when the server fin-

ishes serving a customer. Consecutive service completions divide time into periods, each starting with an idlephase in which the server waits for a new arrival or retrial attempt and ending with a service phase in which one customer is served. The duration of a service phase is one service time t , with distribution function b ( t ) and mean T = E ( t ) . The duration of an idle phase has a more complicated distribu- tion, related to the number w of customers left waiting at the preceding service completion.

In the limiting case of small v the idle phases simplify enough to allow an easy derivation of f. With small U,

customers wait so long for retrials that retrial attempts and new arrivals may be considered independent Poisson pro- cesses with rates R and A. Then idle phases have mean duration 1 / ( A + R ) and periods have mean duration T + 1/( A + R ) . Customers are served, one per period, at mean rate

A + R f A =

1 + ( A + R ) T customers per minute. However, a formula

R = ( 1 - P ) w - f ) A / P was derived earlier, and so R can be eliminated from the above formula to obtain a quadratic equation for f

(1 - P ) A T f 2 - (1 + A T ) f +1= 0. (4) Because the quadratic in (4) has values 1, - PAT, and 60

at f = 0,1, and 60, only one root lies in 0 < f < 1. The other root, in 1 < f < 00, must be ignored.

When v is not small, the argument for (4) fails because the new arrivals and retrial attempts no longer are inde- pendent Poisson processes. Indeed, when v is very large, customers make retrials almost immediately and ultimately balk before the present service ends. Then idle periods begin with no customers waiting and last a mean time 1 / A . The service rate is then f A = A / ( l + AT).

IV. TRANSITION PROBABILITIES

An exact analysis for intermediate values of v will focus on the number w of customers waiting to make a retrial. At service completions w completely describes the state of Fig. 1, i.e., w at service completions is a discrete Markov process. The transition probabilities will now be derived.

Consider a typical period between service completions. In the idle phase there is probability A dt or wv dt of a new arrival or a retrial in the next interval dt. Thus there is probability A / ( A + wv) or w v / ( A + wv) that the idle phase ends with a new arrival or a retrial. These two events entail transitions w + m with m = w or m = w - 1 at the end of the idle phase.

Changes in w during t min of the service phase are of two kinds. Original waiting customers may leave by balk- ing at a retrial, and new customers may arrive and remain without balking until time t. Waiting customers each balk at a rate N = P v . Then each of m customers, originally waiting at the start of service, has probability eCNf of remaining for time t. The number at time t has a binomial probability distribution with generating function q", where

q = 1 - (1 - z ) e - N r . ( 5 ) A Poisson-distributed number of new customers arrive

during service, but only those that do not balk immedi- ately or at a retrial before time t contribute to w. For a given number of arrivals, each amval time s is uniformly distributed over 0 I s I t . A new customer has probability

p = ( I - B ) J d e N ( r - s ) d s / t = (I-P)(I- e - N ' ) / ( N t )

( 6 )

of not balking immediately or at a retrial. Because the number of new customers is Poisson-distributed and be- cause customers balk independently of one another in this model, the number remaining at time t is also Poisson-dis- tributed, with mean pAt.

Combining the results for the new arrivals and the m customers that were present when service began, one finds the probability u,( t ) of w waiting after t min of service. The generating function is

u,(z, t ) = C u , ( t ) z w = qme-phr(l-') W

(7 ) = q m e ( A / W ( z - d

where A = (1 - P)A is the arrival rate of customers that do not balk immediately.

The solution (7) may be checked in simple cases. The initial condition Um(z,O) = z m is satisfied. Also Um(l, t ) =1, as expected because h , ( t ) =l. When the time t becomes large, U,( z, 60) = e - (h /N) ( l - z ) , i.e., the number waiting becomes Poisson-distributed with mean A / N . Indeed, the waiters are then all new amvals, which were shown to be Poisson-distributed with mean pAt + A / N . Finally, (7) contains the result for systems without balking; as P + 0, N --* 0, q + z, ( z - q ) / N -+ z - 1, and Um(z, t ) +

, the exponential factor in (7) now counting all new arrivals.

me - h(l - z ) f

GILBERT: RETRIALS AND BALKS 1505

The generating function Um(z, t ) relates only to changes in the number of waiting customers during service. The transition probabilities sought must also account for changes in the number of waiting customers during the idle phase. If the initial service completion occurred with k customers waiting, the idle phase ends with m = k or m = k - 1, as explained earlier. The probability of w wait- ing after time t of service, given k waiting at the previous service completion, is then generated by

Equation ( 8 ) holds for k = 1,2; . ., and also for k = 0 with the convention U - 1( z , t ) = 0. At the service completion, t has the service time distribution b ( t ) and the transition probabilities p ( w l k ) (for k + w ) between service comple- tions are generated by

C p ( w ( k ) z w = / m V k ( z , t ) d b ( t ) . 0

V. STATIONARY PROBABILITIES AT SERVICE COMPLETION

Let pw denote the stationary probability of w waiting at a service completion, and call its generating function

P( z ) = C p w z w . ( 9 ) W

Then p , are to be obtained from the transition equations, i.e., from the generating function identity

‘(’) = i m F p k v k ( z , r ) d b ( t ) . (10)

Now (11) changes (10) into a functional equation

P ( z ) +-1( P ( q ) - ( A / v ) ( l - q ) q - x / ” 0

x i 9 1 ( k / / ’ ) - 1 P ( l ) d l } e ( A / N ) ( z - q ) d b ( t ) , (12)

with q depending on z and t as in (5) . Although a general solution of (12) has not been found,

certain special cases provide checks. First, suppose U + 00,

i.e., the times between retrials become very short. If @ # 0, N -+ 00, q + 1, and e(A/N)(z-q) + 1. Then (12) becomes

P ( z ) = p l ) d b ( t ) =l. 0

That result was predictable; waiting customers make retri- als so rapidly that they balk almost immediately and then p0 = 1. Likewise, for any U , customers balk immediately if @ = 1. Then A = 0 and again P ( z ) = 1 is easily verified to be a solution of (12).

For mathematical purposes it would be desirable to know that a solution to (12), with positive coefficients and P(1) =1, exists and is unique. That has not been proved, but it seems clear physically that probabilities p , exist; their generating function will be a solution.

VI. EXPONENTIAL SERVICE

If service times are exponentially distributed with mean l/p, the variable of integration in (12) may be changed from t to q using

b ( r ) =I - e-Pt =I - ((1 - q ) / ( 1 - z)lPlN. A functional equation for P ( z ) may be derived as fol-

lows. In (lo), xpkvk(z , t ) is the generating function for the probability of the number w waiting given that service started t minutes ago and is not yet finished. This proba- bility may, with the help of (7) and (8), be written

The result is an integral equation

(N/p)e-*’/N(l- Z ) ~ / ~ P ( Z )

With k v / ( A + k v ) rewritten as 1 - h / ( A + kv) , the sum from one to cc may be identified as

Two differentiations would remove the integrals, leaving a simple second-order differential equation. However, a sin- gle differentiation gives a more useful equation

( p - A z ) P ( z ) + N z P ’ ( z )

Then

c I ? k v k ( z , ‘) This integrodifferential equation is equivalent to a recur- rence system

P P , - A P w - 1 + NWP, = P P , / O + (./A 1 w 1

1506 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 34, NO. 6 NOVEMBER 1988

holding for w = 1,2, . . The solution is as z + 1 and so the solution with P(1) = 1 is

w h + k v (1- X T ) ( l - z ) A ( z ) (13) P ( z ) =

Po A ( . ) - .

with p0 determined by the condition C p , = 1 .

VII. No BALKING

When customers can balk, they do not remain in the system long enough to cause instability. By contrast, a nonbalking system (/3 = 0) will be stable only when X is sufficiently small. The non-balking system is also interest- ing theoretically because its service completions are the only times at which customers depart; it then follows that the stationary distribution for the number of customers in the system at a random time is the same as { p,}, the distribution for the number waiting at a service comple- tion.

Equation (12) can be solved in the limiting case p + 0. Then q + z, and ( A / N ) ( z - q ) + - h(1- z ) t . In (12) the integration over t introduces the function

the characteristic function of service times evaluated at X(1- z ) , and (12) becomes

Since the left side is just the derivative of the integral on the right, this equation is a simple differential equation for the integral. The general solution is

with C a constant of integration. Differentiation converts this equation into

hCz-(”/’) ( l - Z ) A ( Z ) v [ A ( z ) - z I

P ( z ) =

The integrand can be written as

Differentiating (15) produces a simple formula for the mean number waiting after a service completion

AT h Xu2

, - A T { v 7 ) E ( w ) = P’(1) = - l + - + - , /3=0 (16)

where u2 is the variance of the service times. The limiting case v + m provides a simple check on

(1 5). The exponential term then approaches 1, leaving P ( z ) = (1 - hT)(l- z ) A ( z ) / [ A ( z ) - z ] . This may be rec- ognized as the Pollaczek-Khinchine generating function for stationary probabilities of an M/G/1 queue (see Kleinrock [lo, eq. 5.861, Riordan [17, ch. 4, eq. (16))). Indeed, with v + bo and /3 = 0, the server wastes no time waiting for some customer to make a retrial. The system behaves as though waiting customers formed a queue, served in random order. Thus eliminating balking makes a qualitative difference for large v (recall the limiting solu- tion P ( z ) =1 when f i # 0).

When customers cannot leave by balking the pool of waiting customers can grow indefinitely. To determine the condition for this instability one may examine the proba- bility of no customer waiting, which (15) gives as

0 A ( 0 - S po = P (0) = (1 - AT) exp [ - (A/.) \

Since the exponential is positive, po > 0 requires AT < 1 . This inequality is the same stability requirement as for the limiting M / G / 1 queue, but now holding for all values of U. Evidently, when the system is nearly unstable, there are so many customers waiting that a retrial occurs after a negligible time, i.e., the system resembles a queue.

If service times are exponentially distributed with mean T = 1 / p , A ( z ) = p/{ p + X(1- z ) } and (15) reduces to

l + ( A / v )

p - X Z

or

As a check, (17) is (13) with /3 = 0.

VIII. MEANS

to obtain the simpler form

To derive f, the fraction of customers ultimately served, (I4) consider a service completion with w waiting. The next

period begins with an idle phase of mean duration E{1/ (A + w v ) } and ends with a service of mean duration T. Because just one customer is served per period, the system serves at a rate [T + E{l/(X + wv)} ] - ’ customers per

XC(1- Z) A ( Z ) P ( 2 ) =

v [ A ( z ) - z I

The condition P(1) = p0 + p1 + p 2 + * * = 1 will deter- mine C. Although (14) is indeterminate if z =1, A ( z ) = 1 - XT(1- z ) + O((1- z ) * ) where T is the mean service time. The right side of (14) approaches ( A / v ) C / ( l - AT)

GILBERT: RETRIALS AND BALKS 1507

minute. However, this rate is also f A; then 1

= A ( T + E{l/(A + w v ) } ) ’ (18)

To relate f to E ( w ) one may examine balks that occur during a period. Balks occur only during the serving phase. The number m waiting as service begins is m = w or m = w - 1, with probabilities A/ ( A + w v ) or w v / ( X + w v ) as explained at the start of Section IV. If service lasts for time t , each of the m waiters has probability 1 - e-8”‘ of ballung. All m together contribute a number of balks that is binomially distributed with mean m(1- e-8”) . With m unspecified, the mean number of balks by the original waiters is

.(l- e - 8 ” ‘ ) . (19) New arrivals during the service contribute additional balks, which may be treated by the argument for (6). The number of these balks is Poissson-distributed with mean

A{ t - (1 - P)(I - e - P y ‘ ) / ( P v ) } . (20) With t given, the expected total number of balks is the

sum of (19) and (20). With t random, taking further expectations over t introduces E ( t ) = T and a new symbol B = E ( e - 8 ” r ) , i.e., B is the Laplace-Stieltjes transform of b( t ) evaluated at P v . The expected number of balks in the period is then

E(ba1ks) = (1-B)

The expected number of customers leaving the system during a period is 1 + E(ba1ks) (the additional 1 counts the served customer). Since the period has mean duration T + E { 1/( A + w v ) } , customers depart at a mean rate

{ E (balks) + 1 } / { T + E [l/( A + w v ) ] } . In equilibrium this departure rate equals the arrival rate A. That condition, together with (21), supplies identities relat- ing E{l/(A + w v ) } , and hence f in (18), to E(w)

(1- B ) E ( w ) X(l-P) ( l -B) -

1

B VPB

7 f = (1- B ) E ( w ) A(l-P)( l - B ) ‘ (23)

In (22) and (23) the term 1 - B is positive because B =

E ( e - 8 ” ‘ ) < 1. If the probabilities p , are known, E{l/(A + w v ) } or

E ( w ) can be calculated to find f . One might also hope to

tion identity (12). However, differentiating (12) only leads, after disagreeable analysis, to another proof of (21) with no new insight. Without knowing p , one can still find simple bounds on f.

Since 0 I w < 00, (18) implies 1 1

+ AT sf 2 - AT’

Then bounds on E(w) follow from (23)

The lower bound (24) is achieved when P = 1 (and so w = 0). It is also achieved in the limit of long service times ( T + CO and B + 0), when almost all customers eventually balk, or when v + CO (see the end of Section 111). The upper bound is useless when AT is small but can be improved as follows.

Because l / ( X + w v ) is a concave function of w,

1 1 A + v E ( w )

Then (22) gives a bound E I E(w) where E is the positive root of the quadratic equation

+ ( l - B ) - . (26) ( :I2 Replacing E ( w ) by E converts (23) to an upper bound

on f. This bound appears in Table I1 as a function of V T and AT, the mean number of retrial attempts per customer and the mean number of new arrivals during a service. In Table I1 service times t are taken to be constant; then

TABLE I1 BOUNDS ON THE FRACTION f SERVED

Lower (24) Upper vT AT p =0.10 0.25 0.50 o r p = 1

0 0.05 0.2 1 5

20 0.2 0.05

0.2 1 5

20 1 0.05

0.2 1 5

20 5 0.05

0.2 1 5

20

0.995 0.976 0.760 0.195 0.050 0.995: 0.978 0.762 0.195 0.050 0.995* 0.980* 0.770 0.195 0.050 0.995* 0.980: 0.806 0.196 0.050

0.987 0.945 0.667 0.189 0.049 0.988* 0.948 0.669 0.189 0.049 0.988* 0.952: 0.681 0.189 0.049 0.988* 0.952* 0.732 0.189 0.049

0.975 0.952 0.901 0.833 0.586 0.500 0.180 0.167 0.049 0.048 0.976: 0.952 0.904 0.833 0.588 0.500 0.180 0.167 0.049 0.048 0.976; 0.952 0.909* 0.833 0.599 0.500 0.181 0.167 0.049 0.048 0.976: 0.952 0.909* 0.833 0.643 0.500 0.181 0.167 0.049 0.048

evaluate E(w) = f”(1) directly from the generating func- +Cases where bound from (28) is better than that from (26).

~

1508

B = e-flVT. Although E + bo in the limit of small vT, vE in (26) remains finite and the upper-bound on f becomes the exact value, given by (4). For larger values of vT the upper bound exceeds 1 for certain values of AT and p. The tabulated values marked with asterisks come from another bound (28) that follows.

Because customers are served at rate f A and service lasts a mean time T, the server is busy a fraction f AT of the time. Then, because customers arrive in a Poisson stream, a newly arrived customer finds the server busy with probability fAT. A fraction PfAT of new arrivals balk immediately, a fraction ( 1 - P ) f A T make one or more retrials, and a fraction

Pr (immediate service) = 1 - f AT (27) are served without waiting. Because a total fraction f is served, (27) shows that f (1 + AT) - 1 is the fraction served after one or more retrials. Then the conditional probability that a customer is served later, given that she was not served immediately but did not balk, is

f ( l + A T ) - 1 Pr (servedidid not balk) =

( 1 - P ) A T f *

This probability lies between 0 and 1, and so f has bounds 1 1

1+AT If<- 1+/3AT’ (28)

The upper bound (28) is sometimes better than the one obtained from (26) (in Table I1 these cases are marked with asterisks) and has the advantage of never exceeding 1.

IX. RANDOMTIME

The probability of finding w waiting at a random time, say at the arrival time of a customer, is not p , but can be derived from it. Consider first P( wlidle), the probability of w waiting if the random time is chosen during an idle phase. Although w is the number left waiting at the preceding service completion, P( wlidle) is not just p,. For, choosing an idle time at random favors long idle inter- vals and hence small numbers of waiting customers. Idle intervals with w waiting have mean duration l/(A+wv). Then

P , / ( A + W.1

E { l / ( A + W V ) } P( wlidle) =

To find P(wlbusy), the probability of w waiting at a random time when the server is busy, let t be the duration of the service phase in which the random time lies. Again, t does not just have distribution function b ( t ) because choosing a random time favors long service intervals. The

IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 34, NO. 6 NOVEMBER 1988

probability of picking an interval of duration t to t + dt is t d b ( t ) / T . With t given, the random time s is uniformly distributed over 0 I s I t. However, the number waiting at a time s after a service start has generating function Cp,V,(z, s ) , given by (11). Then P(w1busy) has generating function

U‘( w 1 busy) z = Jm JrZp,V, ( z , s ) h d b ( t ) / T .

According to (27), idle times and busy times have proba- bilities 1 - f AT and f AT. Then the probability of w wait- ing at a random time, with no condition on the server, is

P ( w ) = (1- fAT)P(wJidle)+ fATP(w1busy).

0 0

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