84 CHAPTER 10. NETWORK PROBLEMS
Minimum-cost ( s, t ) -flow problem
Primal Dual
min
∑
(i,j)∈E
cijxij
∑
j∈V (i,j)∈E
xij −
∑
j∈V (j,i)∈E
xji =
f i=s
−f i=t 0 else
for alli∈V
0 ≤ xij ≤ uij for all(i,j)∈E
maxf ys−f yt−
∑
(i,j)∈E
uijvij
yi − yj − vij ≤ cij for all(i,j)∈E vij ≥ 0 for all(i,j)∈E yiunrestricted for alli∈V Obstruction (to feasibility): setS⊆Vwiths∈Sandt∈Ssuch that f >
∑
(i,j)∈E(S,S)
uij
Transshipment problem
Primal Dual
min
∑
(i,j)∈E
cijxij
∑
j∈V (i,j)∈E
xij −
∑
j∈V (j,i)∈E
xji = bi
|{z}
net supply
for alli∈V
xij ≥ 0 for all(i,j)∈E
max
∑
i∈V
biyi
yi − yj ≤ cij for all(i,j)∈E yiunrestricted for alli∈V
Obstruction (to feasibility): setS⊆Vsuch that
∑
i∈S
bi>0andE(S,S) =∅
Minimum-cost network flow problem
Primal Dual
min
∑
(i,j)∈E
cijxij
∑
j∈V (i,j)∈E
xij −
∑
j∈V (j,i)∈E
xji = bi for alli∈V
0 ≤ xij ≤ uij for all(i,j)∈E
max
∑
i∈V
biyi−
∑
(i,j)∈E
uijvij
yi − yj − vij ≤ cij for all(i,j)∈E vij ≥ 0 for all(i,j)∈E yiunrestricted for alli∈V Obstruction (to feasibility): setS⊆Vsuch that
∑
i∈S
bi >
∑
(i,j)∈E(S,S)
uij
11
Game Theory
During the 8PM to 9PM time slot, two TV networks compete for an audience of 100 million viewers. The networks announce their schedule ahead of time and do not know of each other’s decision until the show time. Based on that a certain number of people will tune to Network 1 while the rest will watch Network 2. The market research revealed the following expected number of viewers of Network 1.
Network 2
Network 1 Western Soap Opera Comedy
Western 35 15 60
Soap Opera 45 58 50
Comedy 38 14 70
For instance, if Network 1 shows a Western while Network 2 shows a Comedy, then
60million viewers will watch Network 1, and 100−60=40million watch Network 2.
Question: What strategy should the two networks use to maximize their viewership?
Terminology:
• Network 1 is a row player.
• Network 2 is a column player.
• The above matrix is called a payoff matrix.
• This is a constant-sum game (the outcome for both players always sums up to a constant 100 million).
How to solve this game? Let us look at the structure of outcomes. For instance, if Network 1 chooses to show Western, then it can get as many as 60 million viewers (if Network 2 chooses to show a Comedy) but also as little as 15 million (if Network 2 shows a Soap Opera). Thus this choice cannot guarantee more than 15 milion viewers for Network 1 (in the worst case). If the network instead chooses to show a Comedy, the situation is even worse, since then we can guarantee only 14 million viewers (the minimum in the 3rd row). The best therefore for Network 1 is to choose to show a Soap Opera in which case 45 million or more viewers will tune to Network 1 regardless of what Network 2 does. Note that in this strategy Network 1 (being the row player) simply calculates the row minimum of each row and then chooes the row with largest row minimum.
By the same argument, Network 2 (being the column player) can maximize its vieweship by calculating each column maximum and choosing column with the smallest column maximum.
It is easy to see that the two outcomes will satisfy the following inequality all rowsmax (row minimum) ≤ min
all columns (column maximum)
In this example, Network 1 chooses Soap Opera and Network 2 chooses Western whereby 45 million viewers will watch Network 1 and 55 million will watch Network 2. Note that this choice is simultaneously best for both Network
85
86 CHAPTER 11. GAME THEORY 1 and Network 2 (we have equalitymax(row minimum) = min(col maximum)). This is called a saddle point and the common value of both sides of the equation is called the value of the game.
• An equilibrium point of the game: choice of strategies for both players such that neither player can improve their outcome by changing his strategy.
• A saddle point of a game is an example of an equilibrium.
11.1 Pure and Mixed strategies
Note the in the above example each player’s strategy was deterministic; they each examined possible outcomes and made a specific single choice to follow. This is called a pure strategy.
Unfortunately, there are games with no saddle points and following a pure strategy may not always give the players the best outcome. (Consider for instance Poker – see next section.) We need to generalize our definition of strategy.
Instead of choosing one fixed move, the player considers all moves and chooses randomly according to a distribution.
The outcome of the game will depend on the chosen distribution.
• a mixed (randomized) strategy(x1,x2, . . . ,xn)is a probability distribution overnpossible moves of the player (i.e. satisfiesx1+x2+. . .+xn=1)
• optimal strategy is a strategy maximizing the expected gain for the player
Note that since we are now using chance we can sometimes gain more, sometimes less; thus we will settle for maxi- mizing the average outcome (if, say, we play the game repeatedly for many rounds).
Betting Game
Player 1 draws a card from a card deck (and hides it from the other player). Then he decides to either pass in which case he discards the card and pays$1to Player 2, or he will bet in which case it is 2nd player’s turn. Player 2 can either fold in which case she pays$1to Player 1, or she will call and the card is revealed.
If the revealed card is high (10, Jack, Queen, King, Ace), then Player 2 pays$2to Player 1. Otherwise, the card is low (2 through 9) and Player 1 pays$2to Player 2.
Let us analyze the possible strategies for Player 1. Given the card, the card can be either high or low. Based on that the player can either pass or bet. So there are 4 possible strategies: pass on both high and low (PP), pass on high and bet on low (PB), bet on high and pass on low (BP), and bet on both high and low (BB). Player 2 can either Call or Fold.
The possible expected outcomes of the game are then as follows.
Player 2
Player 1 Call Fold Row
Minimum
PP −1 −1 −1
PB −1138 133 −1138
BP 132 −133 −133
BB −136 1 −136
Column Maximum
132 1
Thus, for instance, suppose that Player 1 plays strategy BP (bet on high, pass on low) and Player 2 calls. Then either the card is high, Player 1 bets and wins$2, or the card is low and Player 1 folds and loses$1. The chance of getting a high card is5/13and thus getting a low card has probability of8/13. So on average Player 1 gains
5
13$2+ 8
13(−$1) = 2
13 ≈$0.15
On the other hand, if Player 1 playsBPbut Player 2 folds, then the expected gain of Player 1 is only−133 ≈ −$0.23. So clearly, the intuitive strategy BP (bet on high, pass on low) is not necessarily best against a worst-case opponent.
Note that this is a zero-sum game since either Player 1 pays Player 2 or vice-versa (the sum of the players’ gains is zero). Looking at the table above, we see that largest row minimum is−133 while smalest column maximum is 132. So this game does not have a saddle point (unlike the first game we discussed).
11.1. PURE AND MIXED STRATEGIES 87 Notice that some strategies are better then others in all possible cases. For instance,
playing BP instead of PP always gives better outcome for Player 1. We say that the strategy PP is dominated by another strategy (in this case BP). Clearly, if a strategy is dominated, the player can always do better by playing another strategy. Thus we can safely remove any dominated strategy from consideration without changing the problem (its optimal solution). In our case, this eliminates the strategyPPas well as the strategyPBwhich is dominated by the strategyBB. This leaves us with only two strategies and the pay-off matrix shown on the right.
Player 2 Player 1 Call Fold
BP 132 −133
BB −136 1
Now we are ready to determine the best mixed strategy for Player 1. The player chooses to playBPwith probabilityx1or playsBBwith probabilityx2, wherex1+x2=1. We denote this mixed strategy as(x1,x2). An expected outcome of this strategy is132x1−136x2if Player 2 calls and is−133x1+x2if Player 2 folds. So the worst-case outcome is simply the minimum of the two
(xmin1,x2){132x1−136x2,−133x1+x2} Sincex1+x2=1, we can simplify this to
outcome=min
x1 {138x1−136, 1− 1613x1}
1@ 16
A B C1
8
AB C1@ 6
A B
1
0 1
C1
eC D eEF eG
g H IJKL
PMHNe O 1
PMHNe OQL KMGR
PMHNe OQEHMMR
minimum
S
We can plot the possible outcomes (based on the choices ofx1) as shown on the right. From this we determine that the best mixed strategy for Player 1 is the pointEcorresponding to strategy(x1,x2)wherex1=19/24andx2=5/24.
This guarantees the player expected gain of1/39≈$0.025.
Answer: The best strategy for Player 1 is to randomly choose between betting on high, passing on low with probability 7916%or always bet with probability2056%. This gives him expected gain of¢2.5.
Similarly, we can determine the best mixed strategy for Player 2. Let(y1,y2)be the probabilities of Player 2 calling (y1) or folding (y2) wherey1+y2 = 1. If Player 1 playsBP, then the expected outcome (loss) for Player 2 is 132y1−133y2. If Player 1 playBB, then the outcome is−136y1+y2. Thus the worst-case loss for Player 2 is the maximum of the two.
(ymax1,y2){132y1−133y2,−136y1+y2} Using the fact thaty1+y2=1, this simplifies to
maxy1 {135y1−133, 1−1913y1}
We can plot the possible outcomes as shown on the right. The best strategy for Player 2 is the pointFcorresponding to strategy(y1,y2)wherey1= 23 andy2=
13with expected loss of 391 ≈$0.025. Thus the best strategy for Player 2 is to call with6623%probability and to fold with3313%probability.
1T 1U VW
X1
Y
VW X1T W
VW
1
0 1
X1
Z [\Z] ^Z _
l `aa`d
flh XZj 2
flh XZjV\lh Xakk
flh XZj 1
\lh Xakf
maximum
m
Linear programing
We can now see what are the best strategies for the two players. Player 1 tries to choose probabilitiesx1,x2so as to maximize themin{132x1−136x2,−133x1+x2}where probabilitiesx1,x2sum up to 1. Similarly, Player 2 chooses his probabilitiesy1,y2so as to minimizemax{132y1−133y2,−136y1+y2}wherey1+y2= 1. Using the tricks we learnt, we see that both these problem can be transformed into linear programs as follows.
88 CHAPTER 11. GAME THEORY maxz
subject to 132x1 − 136x2 ≥ z
−133x1 + x2 ≥ z x1 + x2 = 1 x1,x2 ≥ 0
minw
subject to 132y1 − 133y2 ≤ w
−136y1 + y2 ≤ w y1 + y2 = 1 y1,y2 ≥ 0 which we can rewrite as follows:
max z
subject to −132x1 + 136x2 + z ≤ 0
133x1 − x2 + z ≤ 0
x1 + x2 = 1
x1,x2 ≥ 0 zunrestricted
min w
subject to −132y1 + 133y2 + w ≥ 0
136y1 − y2 + w ≥ 0
y1 + y2 = 1
y1,y2 ≥ 0 wunrestricted
Notice that the two programs are duals of each other. This is not a coincidence. In fact, it is always the case in zero-sum (constant-sum) games. This tells us that the optimal solutions to the two programs have the same value by Strong Duality, which defines the value of the game. Moreover, the optimal strategies for the two players satisfy Complementary Slackness (verify this for the solutions we found above). In other words, the solutions to the two problems form an equilibrium point (neither play can do better by changing his/hers strategy). In the literature, this is often called Nash equilibrium.