A Game-Theoretic Queueing Exercise From Seinfeld

7/30/2019 A Game-Theoretic Queueing Exercise From Seinfeld

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A Game-theoretic Queueing Exercise from SeinfeldBismark Singh

AbstractNo abstract for you. Next1!

1. IntroductionIn Season 7, episode 6 of the sitcom Seinfeld, people line up to be served by the Soup

Nazi and must adhere to his strict disciplinary demands. Failure to comply with

these makes a customer abandon the line. Due to the soups delicious taste, which

makes your knees buckle, customers nonetheless brave the hardships and seek to be

served. A customer standing outside the store, facing a decision of whether to enter

or not, wants to know how likely he is to be served and in what time. We quantify some

of these parameters for three standard queueing disciplines. We also analyze two

psychological queueing strategies, from a hawk and dove game perspective, which a

customer may adopt while waiting to attempt to decrease his queue time. Finally we

attempt to see if an evolutionary stable strategy amongst these two is possible or

not.

2. Modeling:We observed from the episode that customers are served in a First-In-First-Out fashion,

and we follow that throughout to highly simplify things. Further, we define thefollowing rewards and costs associated with the queue from the customers

perspective:

Cost to purchase the soup, if served Happiness rewardat achieving the soup (Jambalaya!) if served Discomfort cost arising from standing in a regimented line: $d/min. No soup for you! Dissatisfaction and insult cost of being kicked out of the

restaurant and from the line: $c if not served, $0 if served.

The first two costs can be combined together to a reward, $r if served, and $0 else. We

assume customers arrive to the store with an exponential () process and are served

by the Soup Nazi with an exponential () process. We take that the Soup Nazi becomes

angry at customers with an inter-anger time following an exponential ()

distribution. We attempt to obtain analytical results throughout, but for comparison

purposes whenever we would provide some numerical result, we take the queueing input

parameters,1 1 1

, ,3 2 5

with all values in minutes-1. The motivation for taking

1The inspiration for the abstract comes from (Dixit, 2011)


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for the numerical computation is important and would be clear later. We take costs as1$0.1min , $8, $5d r c , i.e. waiting for 50minutes is monetarily equivalent to being

kicked out. With these as input we can numerically calculate the standard queue

output parameters.

Within FIFO we again make three distinctions:

2.1. Suffering for the SoupFirst we model the system as a simple M/M/1 queue with parameters mentioned above.

This is thus a birth death process- and the model dynamics are just like standard

queuing abandonment problems.

Figure 1 Transition rate diagram for ith state

From the balance equations we can find the steady state probabilities

0 0

1

( )...( ( 1) )( 1)

i i

i i

j

q q qi

j

(1.1)

We also define0

1

, 1

( ( 1) )

i

i i

j

j

and using the normalizing constraint 1iq we

get,

0 0

1, i

o i o i

j j

q q q

(1.2)

The queue is always stable.

We know that the Soup Nazi is an expert at soup making and customers are willing to

bear the costs associated with standing in line upto a certain amount of time with

the anticipation of being served. Next we assume that Newman and Kramer are

preparing a chart to help them decide on whether to join the queue or not. They would

need the probability of not being kicked out while at position i which we call as

Success Probability. Thus success occurs, for customer i, when one of (i-2) customers

ahead of him waiting in line is kicked out, or when the customer currently in service

finishes service -before this ith

customer is himself kicked out. Or, the kick-out time

for customer i must be larger than the minimum of the kick-out time for each of (i-2)

customers ahead of him in line and the service time for customer 1. We note that this

success probability is not the probability that he is served, but only the

probabilities of not being kicked out while at position i. It is also implicitly


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assumed that a customer can count the number already in line thus with this

probability known can decide whether to enter the queue or not. Let the sequence

1 2{ , ... }np p p denote this success probability. Further let iK be the kick-out time for the

ith

customer, andi

Tbe the waiting times at the ith

position alone. The latter definition

should be seen as associated with thei

p sequence as the amount of time a customer has

to wait at position i is the minimum of three quantities- i) service time of the

current customer, ii) the kick-out times of the (i-2) customers ahead of him, and iii) his

own kick-out time.

Thus:

1 1p (1.3)

since the first customer enters service directly.

In general we need the probability 1 2 3 1( min( , , ... ))i iP K S K K K .Since S1 and Ki are

exponentials and , the RHS is a min of (i-1) exponentials which is again anexponential with parameter ( 2)i . The final expression is thus obtained as:

( 2), 2

( 1)i

ip i

i

(1.4)

1 2 3min( , , ... ) ~ exp(( 1) )i iT S K K K i (1.5)

As a sanity check, we can easily verify the case when i.e. the kick out time is

instantaneous.

Since we have the{ }ip sequence, now we can also derive the probability that customer iis not kicked out until reaching the soup service. We define this as his Overall

Success probability. The Overall Success is the real probability of interest whereas

the success probability is only the perceived momentary success probability.

1

( )( 1)

i

i jP Overall Success pi

(1.6)

Similarly, we have a set of recursive equations for the expected waiting time for the

a customer, depending on the position he enters in line, iW . For that we would make use

of the identity,

[ ] [ | ]. [ | ].(1 )i i i i iE W E W not kicked out at i p E W kicked out at i p (1.7)

We note on the RHS of (1.7),the first term can be broken recursively as the time spent

in state i given that the customer is not kicked out and then the process repeating

itself but from state (i-1). The second term is the time spent at state i alone, if the

customer is kicked out. Now, we also have


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1 2 3 1 1 2 3 1[ ] [ | min( , , ... )]. [ | min( , , ... )].(1 )i i i i i i i i iE T E T K S K K K p E T K S K K K p (1.8)

From the memoryless property of exponentials, the conditional distribution of the

time of the first occurrence of a process(say kicked out or not kicked out) given that

this event happened before the first occurrence of the other process (say not kicked

out or kicked out) is the distribution of the first occurrence of the whole process.This gives us

1 2 3 1 1 2 3 1[ ] [ | min( , , ... )] [ | min( , , ... )]i i i i i i iE T E T K S K K K E T K S K K K (1.9)

Substituting (1.5)and (1.9)in (1.7) we get our required recursion as,

1[ ] ( [ ] [ ]) [ ](1 )i i i i i iE W E W E T p E T p (1.10)

which can be simplified again by substituting the expectation of the distribution in

(1.5) to,

1

1[ ] [ ]

( 1)i i iE W E W p

i

(1.11)

This equation has a simple intuitive feel too. The expected waiting time from the ith

stage is the time spent in that state plus the probability of moving forward to the

next stage times the expected waiting time from the (i-1)th

stage. This recursion can be

solved easily by setting [ ]i i iZ E W Y where1

( 1)iY

i

. Then,

1

1

1, 1

1

i iZ Z i

Z

and thus iZ i or

[ ]( 1)

i

iE W

i

(1.12)

Before we proceed, as a sanity check we verify Littles Law on the parameters above. It

is considerably easier to take parameter values and verify that, 10 [ ]i iiL q E W

and

1[ ]i iW q E W

are related as by Littles Law. The indices on L deserve attention as a

customer arriving at the i

th

position sees (i-1) people in the system.

2

We note a few things about the expression for E[Wi]. The expression, given by (1.12)can

be verified to be monotonically decreasing and convex when and monotonically

increasing and concave for .When the kick-out rate equals the service rate i.e.

2A significant amount of time was spent on this verification alone and the author cansafely say that all credit goes to Dr. Hasenbein for this.


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the waiting time is a constant for all entering positions, namely1

. In our

example we take the latter case, i.e. customer farther along the queue are expected to

wait longer in the system.

We present Kramer and Newmans chart in Figure 2, using the same parameter values asabove.

Figure 2 Success and Overall Success probabilities for different entering positions

The expected reward for the ith

customer can be defined by using the costs and

rewards defined above as:

[ ] (1 ) [ ]( )

( 1 )i i i i

E R rdi c

P c P dE Wi r

i

(1.13)

Note that the Pi in (1.13)is the overall success probability as defined in Expanding we

note that,

[ ] 0,

[ ] 0,

[ ] 0,

i

i

i

c rE R if i

d c

c rE R if i

d c

c rE R if id c

(1.14)

This is interesting because there exists a threshold for queue entry positions for

some rewards (and inter-anger time) beyond which the expected reward is always

positive (or negative). This is plotted inFigure 3. We also note from (1.15) that the first

and second derivatives of the expected reward are negative and positive respectively,

for i.e the expected reward is convex and decreasing.

at entry

overall

5 10 15 20Successprobability

0.2

0.4

0.6

0.8

1.0

ntry position


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'

2

''

3

( ) ( )[ ]

(( 1 ) )

2 (( ) ( ))[ ]

(( 1 ) )

i

i

d c rE R

i

c r dE R

i

(1.15)

Another interesting quantity is the expected waiting cost upto position j, whenentered at i. This is particularly of interest when a customer enters at i, and is

kicked out at j, j i , as in(2.2) to calculate his loss. It is plotted inFigure 4

( )( ),

(( 1 ) )([ ] ( [ ] [ ]

1 ))ij i j

d i jiE C d W E W j

jE

i

(2.1)

[ ] ( [( )( )

,(( 1 ) )( )

] [ ])1

ij i j

d i jE KO c d iE W E W c j

i j

(2.2)

Figure 3 Expected reward for different entering positions

5 10 15 20EntryPosition

4

2

2

4

6

8

ExpectedReward


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Figure 4 Expected waiting costs up to a certain position in queue, for different entering positions

2.2. Soup ModeWe observed from the episode that a large number of people gather outside the store

on the sidewalk waiting in line to be served soup, before they get inside the store

and into Soup Mode. Since the Soup Nazi cannot see these people he cannot be angry

at them either. Now we model this case with his power of kicking people out confinedto k-1 people alone (more precisely, k in shop implies k-1 in line).

This is again a birth death process as inFigure 5,

Figure 5 Transition rate diagram for ith stateAgain from the detailed balance equations (using the same variable symbols),

i =

15

i 15

i 3

2 4 6 8 10 12 14Exit position0.00

0.05

0.10

0.15

ExpectedCost upto


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0 0 ,

( )...( ( 1) )

( ) ,

i

i i

i k

i k

q q q i k i

q q i k

(2.3)

where

is defined as ( 1)k

. Rewriting,

( ) , 0ik i o k q q i

(2.4)

Using the normalizing we can solve (2.3)and (2.4)

1

0 0

0 0

( ) 1k

i

i kq q

(2.5)

This equation can be solved foro

q . The second half of the above equation represents a

geometric progression which is stable when the ratio is less than 1, and the first

half is a truncated expression of (1.2) and is always stable. Thus the stability

conditions lead to,

( 1)k (2.6)

(1 )k

(2.7)

We note that if we have 0 , as happens usually, then since at least 1 person

would be in the store, the queue is always stable. If then for stability there is a

threshold defining at least how many people can be allowed in the store.

We can numerically calculate the queue parameters, with the same input parameters.

However we are more interested in calculating the success probabilities. We note that

for the customers in the Soup Mode the success probabilities are the same as in the

previous section and for customers outside the shop success always happens. Thus,

( 2),

( 1)

1, 1

i

i

ip i k

i

p i k

(2.8)

The overall success probabilities are correspondingly defined as in(1.6) and are,

,( 1)

, 1( 1)

i

i

P i ki

P i kk

(2.9)


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Similarly we have the waiting time at the ith

position using the definition given

in(1.5),

~ exp(( 1) ),

~ exp(( 1) ),

i

i

T i i k

T k i k

(2.10)

This takes us back to the waiting time definition which we calculate as in(1.11). The

solution is the piecewise continuous function for [ ]i

E W ,

[ ] ,( 1)

[ ] ,( 1)

i

i

iE W i k

i

iE W i k

k

(2.11)

The associated rewards can also be calculated again using the same definition.

( ),( 1 )

(

[ ]

[ ])

,( 1 )

i

i

E R

E

di c i r i ki

di c k r kR i

k

(2.12)

Note that the expected reward is continuous at i=k due to the continuity in the Pi as

defined in(2.9). These are plotted inFigure 6.

Figure 6 Expected reward for different entry positions

2.3. I think you forgot my breadWe know that the Soup Nazi also had a cashier where customers were routed after

being served by him. This can be modeled in multiple ways as in (Hunt, 1965). However we

study this as a simple M/E2/1 queue. One change we make from the previous model

k=1

k=2

k=5

k=3

k=4

5 10 15 20EntryPosition

2

4

6

8

xpectedReward


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parameters is that now for both the Soup Nazi and the cashier the service rate are

exponentials with parameters 2, so that the overall service rate is Erlang with

parameter . We define the number of phases left in the system, m, when there are n

customers in the system as

2( 1) , {0,1}m n j j

(3.1)

Figure 7 Transition rate diagram for ith phase, i2

The transition rates are as follows (also represented inFigure 7:

12 : , 2

2

1: 2 , 12 : , 1

1: , 0

ii i i

i i ii i i

i i i

(3.2)

Note that an odd number of phases imply the customer in service is being served by

the cashier. The balance equations can be written as, with ri denoting the steady

state probability of i phases in the system,

1 2

1 1 2

1 2 2

2

2 2

(2 ) 2 , 2

o

i i i i

r r r

r r r

r r r r i

(3.3)

The solution to the above is messy but is computable in Mathematica. Using the

mapping process (3.1) we can get the steady state probabilities and the queue

parameters. So computing the phase parameters is sufficient. We can do the same

analysis for the rewards and waiting times as above, but now for each phase. Using

similar arguments as before, we have the time spent at the ith

phase alone, and the

success probability at the ith

phase as.


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1

[ ] , 31

22

iE T i

i

(3.4)

12 ( 1)

2

, 3122

i

i

p ii

(3.5)

Then using(1.10) we can model the expected reward at the ith

position.

3. Do I know you?When Jerry and his girlfriend Sheila are standing in line, with Sheila one position

ahead of Jerry, Sheila kisses Jerry but Jerry does not reciprocate. We take it that to

increase his overall success probability (and thus his reward) Jerry creates a

strategy to get Sheila kicked out and move to her spot3. This strategy could have

easily backfired as the Soup Nazi may have kicked both of them out, or instead onlyJerry out. Sheila, on the other hand, had no incentive in Jerry s strategy4. Thus in

general a customer who entered the queue at the ith

position and is currently at the

jth

position (with j i , and implicitly assuming that he is not kicked out between i and

j) may indulge in a game with the (j-1)th

customer in line. Of course, any customer has

an option to ignore anyone disturbing him/her, or refuse to play. We attempt to find

if both these types of groups can coexist or not and if so at what frequencies

(proportions). Before we proceed we provide some further definitions and some

simplifying characteristics

1. Player: A customer who takes part in this strategy and attempts to move aheadin line

a. Games between Players are very conspicuous and end only when one Playergets kicked out by the Soup Nazi (the one kicked out first is the one we

count).

2. Not Player: A customer who does not play and is contented to stay in line athis current position and wait for his turn

a. When a Not Player meets a Not Player, casual and discrete interactionfollows. None of them have an intention of throwing people out but their

talks may get noticed (although less conspicuously than the Player-Player)

The game ends when the Soup Nazi observes the one closer to him(the

opponent) and kicks him out or both maintain their stay.3. Assumptions on the game itself:

a. There is no cost to play the game itself

3As Elaine says later, So essentially you chose soup over a woman.4Although, this model may be grossly incorrect for most people.


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b. All Games are defined w.r.t. a player (without a capital P) at position j, whois one behind the opponent at (j-1). No games are initiated with a customer

behind our player.5

c. Every customer is either a Player or a Not Playerd. Games are instantaneous, so they do not interfere with the queueing input

output parameterse. A player can only be in one game at a timef. A player does not know beforehand who his opponent would beg. We still maintain that the expected kick out rate is lesser than the

expected service rate, . Later we would briefly discuss the implications

of relaxing this assumption

h. Probability of winning for a Player for a Player-Player game is zi. Probability of winning for a Player for a Player-Not Player game is y, y>zj. Probability of winning for a Not Player for a Not Player-Not Player game is

x

k. Probability of winning to a Not Player for a Not Player-Player game isundefined as he simply does not play.

4. We only model the Suffering for the Soup case, although the analysis is easilyextendable to other cases.

We define the corresponding payoff matrix (for a better understanding of these

definitions we refer the interested reader to (Prestwich) , -the web-page which was

immensely useful for this study- with Player denoting the jth

customer and Opponent

the (j-1)th

:

[ , ] [ , ']

[ ', ] [ ', ']

Player Opponent

Play NotPlayPlay E P P E P P

NotPlay E P P E P P

(3.6)

where P denotes Play and P denotes not Play. We attempt to model this as a hawk-dove

game (Alexander, 2009)where hawks behave aggressively and doves retreat immediately

if the opponent displays aggression. Thus hawk-hawk games end when one wins, hawk-

dove games win with the hawk, dove-hawk with the dove retreating and dove-dove with

sharing. We keep Players as the hawks and Not Players as the doves but make some

changes. For us, three events can happen at each position- move forward (Move),

maintain current position (Stay) or get kicked out (KO).

5. If a Player meets with a Player he risks being kicked out (KO) with themotivation of having a chance of moving forward (Move) quicker.

5There is a caveat here - if a customer is behind someone he is also ahead of someone.For large enough system size the assumption should be justified.


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6. If a Player meets with a Not Player then since the Not Player, realizing thatthe Player is cunning, would ignore him. Thus the reward is staying in this

case for the Player and the loss being a kick-out.

7. A Not Player being attracted to a Not Player would irresistibly andunknowingly start communicating with him, and if the Soup Nazi finds them he

would kick-out the opponent. Thus the reward is moving in this case for thePlayer and the loss being staying, although this may hurt our Not Player.

8. But our Not Player simply doesnt interact with a Player, then there is nospecial chance of being kicked out, other than the usual kick-out process and

he just stays there (Stay).

Thus we have the payoff matrix in(3.7), with the first entry denoting a win and the

second the loss.

/// /

Player Opponent

Play NotPlay

Play Stay KOMove KO

NotPlay Stay Stay Move Stay

(3.7)

Note that because of our assumption, if the customer is a Not Player then no matter

what the Opponent is, he stays. The expected payoff from a strategy (here Play and Not

play) is defined as:

[ ] ( ) ( ) ( )win loss

E Strategy p E Win p E Loss (3.8)

Here we should note that each of the four entries have different winning

expectations as well as probabilities. Specifically,

1

*

[ ] [ ]

[ ]

( ) ( )

(( 2 ) )(( 1 ) )

( )( )

(( 1 ) )(( 1 ) )[ ]

[ ] 0

j j

ij

c r d

j jE Move E R R

E KO E C c

E Sta

d i j

i j

y

(3.9)

where*ij

C is defined as the cost of exiting at j*if a customer entered at i. This is

equivalent to the insult cost plus the time spent waiting between i and j,

[ ] [ ]ij i jE C dE W W . (This is plotted inFigure 4using the parameter values we chose.) Note

again from(3.9) and also from Figure 8that for the Move reward is positive (thusmoving forward increases the expected reward). Also we observe from Figure 4that the

Kick-Out cost is greater as j is further away from i(thus if immediately kicked out

upon entry the wait is not so bad, as if kicked out at a position farther away).

Using (3.8)we have,


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1

1

[ , ] [ ] (1 ) [ ]

[ , '] (1 ) [ ]

[ ', ] 0

[ ', '] [ ]

j j ij

ij

j j

E P P zE R R z E C

E P P y E C

E P P

E P P xE R R

(3.10)

Figure 8 Expected Move reward for different initial positions

Further we let u be the frequency of people who Play and (1-u) that of those who do

not Play. Then the fitness of each strategy is defined as

( ) [ , ] (1 ) [ , ']

( ) [ ', ] (1 ) [ ', ']

W Play uE P P u E P P

W NotPlay uE P P u E P P

(3.11)

We can check if the Play and Not Play strategies are pure evolutionary stable

strategies (ESS) or not (Hamilton, 1967). For Not Play to be an ESS we should have,

( ) ( )W Play W NotPlay (3.12)

Expanding this using (3.10)and(3.11),

1 1( [ ] (1 ) [ ]) (1 )((1 ) [ ]) (1 )( [ ])j j ij ij j ju zE R R z E C u y E C u xE R R (3.13)

1 1( [ ] (1 ) [ ]) (1 )( [ ] (1 ) [ ])j j ij j j iju zE R R z E C u xE R R y E C (3.14)

Now we note that if 1[ ] (1 ) [ ] 0j j ijzE R R z E C then the above equation always holds and

Not Play is a pure ESS. This means that if the condition holds then regardless of itsproportion Not Play is immune to invasions by Players, and if Players arise in the

queue they would eventually be extinct. We can examine this condition further. For

Not Play to be a pure ESS,

1

[ ]

1 [ ]

ij

j j

E Cz

z E R R

(3.15)

5 10 15 20Movefrom

0.5

1.0

1.5

2.0

Move Reward


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Lets define the RHS of (3.15)as the adios-muchacho ratio, using the Soup-Nazis phrase

again (with the negative sign to make the whole term positive) at position j, ( , )r i j . For

brevity, we henceforth refer to it as simply adios ratio. We note that adios ratio

greater than unity implies the potential kick-out cost at position j exceeds the

potential move reward at j given an entry position i. Then (3.15)holds when

( , )

1 ( , )

r i jz

r i j

(3.16)

( , )1

zr i j

z

(3.17)

We can check when this equation holds true i.e. given an entry position i, at what

position j(if any) does Not Play become immune to Play(or Not Play becomes an ESS).

The mixed ESS case is also interesting i.e. 1[ ] (1 ) [ ] 0j j ijzE R R z E C , or,1

ij

zr

z

We

should note that the adios ratio is a function of the queue input parameters and

cost rates alone i.e. it does not depend on the game probabilities. Under our parameter

choices the ratio is always larger than 1 for any choice of i and j6. Then the above

condition yields that 0.5z which may not be very realistic as a Player-Player game

should have symmetric consequences.7

We do mention that if we take 0.5z (then normatively we should also take 0.5x to

have at least some sense of symmetry) then a solution to (3.14) with equality (equal

fitness of both Play and Not Play) may exist. For that, there must be a frequency, u,

such that

1

1

[ ] (1 ) [ ]

1 [ ] (1 ) [ ]

j j ij

j j ij

xE R R y E Cu

u zE R R z E C

(3.18)

which can be simplified using the adios ratio definition again.

(1 )

1 (1 )

ij

ij

x y ru

u z z r

(3.19)

For instance, if 0.5x z then some simple algebra leads to a conclusion that a

solution to u must exist. We can even specify bounds of u in this case satisfying the

condition that the adios ratio exceeds unity and is less than the odds of winning aPlayer-Player game. We have,

6The author does not expect this to be a general result. A simple excel goal seekexercise, while maintaining , r c shows that it is possible to construct many

example with the ratio less than 1.7It is possible though to extend to include cases where games are not symmetric.


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1 1

3 1.5

y yu

z y y

(3.20)

Note again that for 1ij

r (or rather1

ij

zr

z

) no solution exists to (3.19) since the RHS is

negative and the LHS is positive.

4. Pack it up. No more soup for you. Next!We conclude the following, for while remembering that our result maybe heavily

dependent on the queue parameters:

1. The Not-Play strategy is an ESS whenever the adios ratio exceeds the odds ofwinning in a Player-Player game.

2. A mixed ESS always exists when the adios ratio is less than the odds ofwinning the Player-Player game with player concentrations as defined above.

For the reader interested in the case of a brief discussion is provided in the

appendix.

References

Alexander, J. M. (2009). Evolutionary Game Theory. Retrieved from The Stanford Encyclopedia of Philosophy:

http://plato.stanford.edu/archives/fall2009/entries/game-evolutionary

Dixit, A. (2011). An Option Value Problem from Seinfeld. Economic Inquiry.

Hamilton, W. (1967). Extraordinary sex ratios. Science, 477-488.

Hunt, G. C. (1965). Sequential Arrays of Waiting Lines. Operations Research, 674-683.

Prestwich, K. N. (n.d.). Game Theory and EvolutionarilyStable Strategies . Retrieved from

http://college.holycross.edu/faculty/kprestwi/behavior/ESS/game_defs.html

APPENDIXAdios ratio Plots for entry positions i=2:15 versus exit positions with given

parameters. The x axis shows exiting positions and the y-axis shows the adios ratio.

Note that the ratio exceeds unity always and has a larger slope for exiting valuescloser to the entry position (visible more easily for larger i)


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,

,

,

,

2.2 2.4 2.6 2.8 3.0

1.4

1.6

1.8

2.0

2.2

2.0 2.5 3.0 3.5 4.0

1.5

2.0

2.5

3.0

3.5

2.5 3.0 3.5 4.0 4.5 5.0

2

3

4

5

3 4 5 6

2

3

4

5

6

7

3 4 5 6 7

2

4

6

8

3 4 5 6 7 8

2

4

6

8

10

12

3 5 6 7 8 9

2

4

6

8

10

12

14

4 6 8 10

5

10

15


18/20

,

,

Figure 9 Adios ratio for different entering positions

We can compare this with when as inFigure 10. We keep the same parameters except

interchange the , values (so that now 0.5, 0.2 ). The ratio does go below unity

and is more convex than the corresponding graphs inFigure 9.

4 6 8 10

5

10

15

20

4 6 8 10 12

5

10

15

20

25

4 6 8 10 12

5

10

15

20

25

30

4 6 8 10 12 14

5

10

15

20

25

30

4 6 8 10 12 14

10

20

30


19/20

,

,

,

,

2.0 2.2 2.4 2.6 2.8 3.0

1.0

1.5

2.0

2.5

3.0

2.5 3.0 3.5 4.0

2

4

6

8

2.5 3.0 3.5 4.0 4.5 5.0

2

4

6

8

0

2

4

3 4 5 6

5

10

15

20

3 4 5 6 7

5

10

15

20

25

30

3 4 5 6 7 8

10

20

30

40

3 5 6 7 8 9

10

0

0

0

0

0

4 6 8 10

20

40

60


20/20

,

,

Figure 10 Adios ratio for different entering positions

4 6 8 10

20

40

60

80

4 6 8 10 12

20

40

60

80

100

4 6 8 10 12

20

40

60

80

100

120

140

4 6 8 10 12 14

50

100

150

4 6 8 10 12 14

50

100

150

Documents

A Game-Theoretic Queueing Exercise From Seinfeld