A Game-theoretic Queueing Exercise from Seinfeld ...

A Game-theoretic Queueing Exercise from Seinfeld Bismark Singh

Abstract “No abstract for you. Next1!” 1.

Introduction

In 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 soup’s 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 the following rewards and costs associated with the queue from the customer’s perspective:  Cost to purchase the soup, if served  Happiness reward at 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 (λ) process and are served by the Soup Nazi with an exponential (µ) process. We take that the Soup Nazi becomes angry at customers with an inter-anger time following an exponential (ϴ) distribution. We attempt to obtain analytical results throughout, but for comparison purposes whenever we would provide some numerical result, we take the queueing input 1

1

1

3

2

5

parameters,   ,   ,  

1

-1

with all values in minutes . The motivation for taking   

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

for the numerical computation is important and would be clear later. We take costs as d  $0.1min 1 , r  $8, c  $5 , i.e. waiting for 50minutes is monetarily equivalent to being

kicked output Within 2.1.

out. With these as input we can numerically calculate the standard queue parameters. FIFO we again make three distinctions: Suffering for the Soup

First 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 qi 





i

 (    )...(   (i  1) )

q0 

i

i

   ( j  1)

q0

(1.1)

j 1

We also define i 

i i

 (  ( j  1) )

, 0  1 and using the normalizing constraint  qi  1 we

j 1

get, qo 

1

, qi 



 0

j

i

 qo i





(1.2)

j

0

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 th finishes service -before this i 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

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 { p1 , p2 ... pn } denote this success probability. Further let K i be the kick-out time for the th

i

th

customer, and Ti be the waiting times at the i

position alone. The latter definition

should be seen as associated with the pi 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:

p1  1

(1.3)

since the first customer enters service directly. In

general

we

need

the

probability P( Ki  min(S1 , K2 , K3 ...Ki 1 )) .Since

S1

and

Ki are

exponentials µ and ϴ, the RHS is a min of (i-1) exponentials which is again an exponential with parameter (i  2)   . The final expression is thus obtained as: pi 

(i  2)   ,i  2 (i  1)  

(1.4)

Ti  min(S1 , K2 , K3 ...Ki ) ~ exp((i  1)   )

(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 { pi } sequence, now we can also derive the probability that customer i is 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. i

P(Overall Success)i   p j  1

 (i  1)  

(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, Wi . For that we would make use of the identity,

E[Wi ]  E[Wi | not kicked out at i]. pi  E[Wi | kicked out at i].(1  pi )

(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

E[Ti ]  E[Ti | Ki  min(S1 , K2 , K3 ...Ki 1 )]. pi  E[Ti | Ki  min(S1 , K2 , K3 ...Ki 1 )].(1  pi )

(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 E[Ti ]  E[Ti | Ki  min(S1 , K2 , K3 ...Ki 1 )]  E[Ti | Ki  min(S1 , K2 , K3 ...Ki 1 )]

(1.9)

Substituting (1.5)and (1.9)in (1.7) we get our required recursion as, E[Wi ]  ( E[Wi 1 ]  E[Ti ]) pi  E[Ti ](1  pi )

(1.10)

which can be simplified again by substituting the expectation of the distribution in (1.5) to, E[Wi ]  E[Wi 1 ] pi 

1   (i  1)

(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 Zi  E[Wi ]Yi where Yi 

1 . Then,   (i  1)

Zi  Zi 1  1, i  1 Z1  1

and thus Zi  i or E[Wi ] 

i   (i  1)

(1.12)

Before we proceed, as a sanity check we verify Little’s Law on the parameters above. It is considerably easier to take parameter values and verify that, L   i 0 qi 1 E[Wi ] and 

W  1 qi E[Wi ] are related as by Little’s Law. The indices on L deserve attention as a 

customer arriving at the ith 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.

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

   the waiting time is a constant for all entering positions, namely

1



. 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 Newman’s chart in Figure 2, using the same parameter values as above.

Entry position 1.0 at entry 0.8 aa aaa aaa aaa aaa

0.6 0.4

overall

0.2

5

10

15

20

Successprobability

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: E[ Ri ]  rPi  c(1  Pi )  dE[Wi ] 

di  c(  i )  r  (1  i)  

(1.13)

Note that the Pi in (1.13)is the overall success probability as defined in Expanding we note that, c   r d  c c   r E[ Ri ]  0, if i  d  c c   r E[ Ri ]  0, if i  d  c E[ Ri ]  0, if i 

(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 in Figure 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.

d (   )  (c  r ) ((1  i)   ) 2 2 ((c  r )  d (   )) E[ R ''i ]  ((1  i)   )3 E[ R 'i ] 

(1.15)

Another interesting quantity is the expected waiting cost upto position j, when entered 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 in Figure 4 E[Cij ]  d ( E[Wi ]  E[W j ]) 

d (i  j )(    ) ,i  j ((1  i)   )(1  j )  

E[ KOij ]  c  d ( E[Wi ]  E[W j ])  c 

d (i  j )(    ) ,i  j ((1  i)   )(1  j )  

ExpectedReward 8 6 4 2 5

10

15

20

2 4 Figure 3 Expected reward for different entering positions

EntryPosition

(2.1)

(2.2)

ExpectedCost upto

0.15

0.10

0.05 i= 15

i

3 i

0.00

2

4

6

8

10

12

15 14

Exit position

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

2.2. “Soup Mode” We 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 confined to k-1 people alone (more precisely, k in shop implies k-1 in line). This is again a birth death process as in Figure 5,

Figure 5 Transition rate diagram for ith state

Again from the detailed balance equations (using the same variable symbols),

i q0  q0 i , i  k  (    )...(  (i  1) )  qi  qk ( )i  k , i  k  qi 

(2.3)

where α is defined as     (k  1) . Rewriting,

 

qk i  qo k ( )i , i  0

(2.4)

Using the normalizing we can solve (2.3)and (2.4) k 1



0

0

 

q0  i  q0 k  ( )i  1

(2.5)

This equation can be solved for qo . 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,

      (k  1) k  (1 

 ) 

(2.6) (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, (i  2)   , i  k (i  1)   pi  1, i  k  1 pi 

(2.8)

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

 (i  1)  

, i  k

 , i  k  1 (k  1)  

(2.9)

Similarly we have the waiting time at the i in(1.5),

th

position using the definition given

Ti ~ exp((i  1)   ), i  k

(2.10)

Ti ~ exp((k  1)   ), i  k

This takes us back to the waiting time definition which we calculate as in(1.11). The solution is the piecewise continuous function for E[Wi ] , i ,i  k   (i  1) i E[Wi ]  ,i  k   (k  1) E[Wi ] 

(2.11)

The associated rewards can also be calculated again using the same definition. di  c(  i )  r  ,i  k (1  i )   di  c(  k )  r  E[ Ri ]  ,i  k (1  k )   E[ Ri ] 

(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 in Figure 6. ExpectedReward 8

6 k=1 4 k=2 2 k=3 k=4 5

10

15

20

EntryPosition

k=5

Figure 6 Expected reward for different entry positions

2.3. “I think you forgot my bread” We 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

parameters is that now for both the Soup Nazi and the cashier the service rate are exponentials with parameters 2µ, so that the overall service rate is Erlang with parameter µ. We define the number of phases left in the system, m, when there are n customers in the system as m  2(n  1)  j, j  {0,1}

(3.1)

Figure 7 Transition rate diagram for ith phase, i≥2

The transition rates are as follows (also represented in Figure 7: i 1 ,i  2 2 i  i  1: 2 , i  1 i  i  2 : , i  1 i i2:

(3.2)

i  i  1:  , i  0

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 r i denoting the steady state probability of i phases in the system, ro   r1 2  r2 r1 2  r1  r2 2 ri (2     )  ri 1 2  ri  2  ri  2  , i  2

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 th similar arguments as before, we have the time spent at the i phase alone, and the th success probability at the i phase as.

(3.3)

E[Ti ] 

pi 

1 ,i  3 i 1 2   2

i 1  1) 2 ,i  3 i 1 2   2

2  (

Then using(1.10) we can model the expected reward at the i 3.

(3.4)

(3.5)

th

position.

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 only Jerry 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.

2.

3.

3 4

Player: A customer who takes part in this strategy and attempts to move ahead in line a. Games between Players are very conspicuous and end only when one Player gets kicked out by the Soup Nazi (the one kicked out first is the one we count). Not Player: A customer who does not play and is contented to stay in line at his current position and wait for his turn a. When a Not Player meets a Not Player, casual and discrete interaction follows. 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. Assumptions on the game itself: a. There is no cost to play the game itself

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

b.

All Games are defined w.r.t. a player (without a capital ‘P’) at position j, who is 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 Player d. Games are instantaneous, so they do not interfere with the queueing input output parameters e. A player can only be in one game at a time f. A player does not know beforehand who his opponent would be g. We still maintain that the expected kick –out rate is lesser than the expected service rate,    . Later we would briefly discuss the implications

4.

of relaxing this assumption h. Probability of winning for a Player for a Player-Player game is z i. Probability of winning for a Player for a Player-Not Player game is y, y>z j. 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 is undefined as he simply does not play. We only model the Suffering for the Soup case, although the analysis is easily extendable 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 th the (j-1) : Player Play NotPlay

Opponent Play

NotPlay

E[ P, P] E[ P ', P]

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, hawkdove 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 the motivation of having a chance of moving forward (Move) quicker.

There 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. 5

6.

7.

8.

If a Player meets with a Not Player then since the Not Player, realizing that the Player is cunning, would ignore him. Thus the reward is staying in this case for the Player and the loss being a kick-out. A Not Player being attracted to a Not Player would irresistibly and unknowingly 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 the Player and the loss being staying, although this may hurt our Not Player. But our Not Player simply doesn’t interact with a Player, then there is no ‘special’ 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 Move / KO Stay / 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:

E[Strategy]  pwin E(Win)  ( ploss ) E( Loss)

(3.8)

Here we should note that each of the four entries have different winning expectations as well as probabilities. Specifically, (c  r )  d (   ) ((2  j )   )((1  j )   ) d (    )(i  j ) E[ KO]  E[Cij * ]  c  ((1  i )   )((1  j )   ) E[ Stay ]  0 E[ Move]  E[ R j 1  R j ] 

(3.9)

*

where Cij* 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, E[Cij ]  dE[Wi  W j ] . (This is plotted in Figure 4 using the parameter values we chose.) Note

again from(3.9) and also from Figure 8that for    the Move reward is positive (thus moving forward increases the expected reward). Also we observe from Figure 4 that 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,

E[ P, P]  zE[ R j 1  R j ]  (1  z ) E[Cij ] E[ P, P ']  (1  y ) E[Cij ]

(3.10)

E[ P ', P]  0 E[ P ', P ']  xE[ R j 1  R j ] Move Reward 2.0

1.5

1.0

0.5

5

10

15

20

Movefrom

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 W ( Play)  uE[ P, P]  (1  u ) E[ P, P '] W ( NotPlay)  uE[ P ', P]  (1  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), u( zE[ R j 1  R j ]  (1  z) E[Cij ])  (1  u)((1  y) E[Cij ])  (1  u)( xE[ R j 1  R j ])

(3.13)

 u( zE[ R j 1  R j ]  (1  z) E[Cij ])  (1  u)( xE[ R j 1  R j ]  (1  y) E[Cij ])

(3.14)

Now we note that if zE[ R j 1  R j ]  (1  z) E[Cij ]  0 then the above equation always holds and Not Play is a pure ESS. This means that if the condition holds then regardless of its proportion 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,  E[Cij ] z  1  z E[ R j 1  R j ]

(3.15)

Let’s define the RHS of (3.15)as the adios-muchacho ratio, using the Soup-Nazi’s 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 z

r (i, j ) 1  r (i, j )

 r (i, j ) 

z 1 z

(3.16)

(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). z  rij We 1 z 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 6 choices the ratio is always larger than 1 for any choice of i and j . Then the above condition yields that z  0.5 which may not be very realistic as a Player-Player game should have symmetric consequences.7

The mixed ESS case is also interesting i.e. zE[ R j 1  R j ]  (1  z) E[Cij ]  0 , or,

We do mention that if we take z  0.5 (then normatively we should also take x  0.5 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 xE[ R j 1  R j ]  (1  y ) E[Cij ] u  1  u zE[ R j 1  R j ]  (1  z ) E[Cij ]

(3.18)

which can be simplified using the adios ratio definition again. x  (1  y )rij u  1  u z  (1  z )rij

(3.19)

For instance, if x  z  0.5 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 a Player-Player game. We have,

The author does not expect this to be a general result. A simple excel goal seek exercise, while maintaining    , r  c shows that it is possible to construct many example with the ratio less than 1. 7 It is possible though to extend to include cases where games are not symmetric. 6

1 y 1 y u 3z  y 1.5  y

Note again that for rij  1 (or rather

(3.20)

z  rij ) no solution exists to (3.19) since the RHS is 1 z

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. 2.

The Not-Play strategy is an ESS whenever the adios ratio exceeds the odds of winning in a Player-Player game. A mixed ESS always exists when the adios ratio is less than the odds of winning 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

APPENDIX Adios 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 values closer to the entry position (visible more easily for larger i)

3.5 2.2

3.0 2.0

2.5 1.8

2.0 1.6

1.5

1.4 2.2

2.4

2.6

2.8

2.0

3.0

2.5

3.0

3.5

4.0

,

3

4

5

6

,

8

,

10

,

7 5

6 4

5

4 3

3 2

2 2.5

3.0

3.5

4.0

4.5

5.0

12

8

10

8 6

6 4

4

2

2 3

4

5

6

3

7

4

5

6

7

14 15

12 10

10

8 6

5

4 2

3

4

5

6

7

8

9

4

6

8

25 20

20 15

15 10

10

5

5

4

6

8

4

10

6

8

10

12

,

14

,

30

30

25

25

20

20

15 15

10 10

5

5

4

6

8

10

12

4

6

8

10

12

30

20

10

4

6

8

10

12

14

Figure 9 Adios ratio for different entering positions

We can compare this with when    as in Figure 10. We keep the same parameters except interchange the µ,ϴ values (so that now   0.5,   0.2 ). The ratio does go below unity and is “more” convex than the corresponding graphs in Figure 9.

8 3.0

6 2.5

2.0

4

1.5

2 1.0

2.0

2.2

2.4

2.6

2.8

3.0

2.5

3.0

3.5

4.0

,

3

4

5

6

,

8

,

10

,

14

20 12 10

15

8

10

6 4

5 2

2.5

3.0

3.5

4.0

4.5

5.0

30

40

25 30 20 15

20

10 10 5

3

4

5

6

7

3

4

5

6

7

60

50 60

40 40

30

20 20

10

3

4

5

6

7

8

9

4

6

8

100

80

80

60 60

40 40

20

20

4

6

8

4

10

6

8

10

12

,

14

,

140 150

120 100

100

80 60

50

40 20

4

6

8

10

12

4

6

8

10

150

100

50

4

6

8

10

12

14

Figure 10 Adios ratio for different entering positions

12