Hillier and Lieberman Problem 14.42 Page 746

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Hillier and Lieberman Problem 14.4-2 Page 746
Consider the game having the following pay-off
(to A) table
Player B Strategy 1
2 1 3
-2
Player A
2 -1 2
2
Use the graphical procedure to determine the
value of the game and the optimal strategy for
each player according to the minimax criterion.
When B plays strategy As Expected pay-off
1 3 x1-1(1-x1)
-1 4x1
2 -2 x12(1-x1)
2 4x1
3
B2
(3/8, 1/2)
x1
1
0
B1
4
Thus A should play his strategies 1, 2 with
probabilities 3/8, 5/8. And the value of the game
1/2. Now if Bs optimal strategies are y1 and
(1-y1), then 3 y1 - 2(1- y1 ) 1/2 Or 5 y1
5/2 , i.e., y1 1/2 y2 Thus B should
play his strategies 1, 2 with probabilities 1/2,
1/2.
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Hillier and Lieberman Problem 14.4-3(a) Page 746
Consider the game having the following pay-off
(to A) table
Player B Strategy 1 2
3 1 4 3
1
Player A
2 0 1 2
6
Use the graphical procedure to determine the
value of the game and the optimal strategy for
each player according to the minimax criterion
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Hillier and Lieberman Problem 14.4-3(b) Page 746
Consider the game having the following pay-off
(to A) table
Player B
Strategy 1 2 3
1 1 -1 3

2 0
4 1
Player A
3 3 -2
5
4 -3
6 -2
8
Use the graphical procedure to determine the
value of the game and the optimal strategy for
each player according to the minimax criterion.
It is clear that the strategy 3 of player B is
dominated by the strategy 1 of player B, in the
sense that Bs pay-off to A is less whatever
strategy A plays. (Note 1 lt 3, 0 lt 1, 3 lt 5, -3 lt
-2). Thus B should never play the strategy 3 and
the pay-off matrix becomes
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Player B
Strategy 1 2
1 1 -1
2 0 4
Player A
3 3 -2
4 -3
6

Thus we first solve Bs problem graphically.
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Let B play the strategies 1, 2 with probabilities
y1 and (1-y1). Thus
When A plays strategy Bs expected pay-off
(to A)
1 y1
(1-y1) -1 2y1
2
4(1-y1) 4 - 4y1
3 3y1
2(1-y1) -2 5y1
4 -3y1 6
(1-y1) 6 - 9y1
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A4
A2
(2/3, 4/3)
y1
1
A1
A3
Thus B should play strategy 1 with prob 2/3 and
strategy 2 with prob 1/3 and the value of the
game 4/3.
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Now we find As prob distribution. From the graph
we see that A should never play the strategy 1.
Thus x1 0. If x2, x3, x4 are the optimal
prob with which he plays the strategies 2, 3, 4,
then we find x2(4 4y1) x3(-25y1)
x4(6-9y1) ? 4/3 for all y1 0 ? y1 ? 1. This
becomes an equality when y1 2/3. Hence
(4/3)x2 (4/3) x3 (0)x4 4/3 Hence x2
x3 1 which implies x4 0
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Now 3 x3 4/3 or x3 4/9 and so x2 5/9
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Hillier and Lieberman problem 14.5-3 Page 747
Consider the game having the following pay-off
matrix. Solve it by LPP method
Player B
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We now add 3 to each entry so that we get the new
pay-off matrix with all entries non-negative.
Player B
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Putting u1 x1/?, u2 x2/?, u3 x3/? , As
problem is Minimize z u1 u2 u3 Subject to
7u1 2u2 5u3 ? 1 5u1 3u2
6u3 ? 1 6u2
u3 ? 1
u1 , u2 , u3 ? 0
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Putting t1 y1/?, t2 y2/?, t3 y3/? , Bs
problem is Maximize w t1 t2 t3 Subject to
7t1 5t2 ? 1 2t1
3t2 6t3 ? 1 5t1 6t2
t3 ? 1
t1 , t2 , t3 ? 0
Solving we get t1 8/91, t2 1/13, t3 9/91
t1 t2 t3 1/? 24/91.
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Thus y1 t1 ? ?
y2 t2 ? ?
y3 t3 ? ?
And the value of the game ? - 3
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Also from the optimal tableau of Bs LPP, we
read out optimal values of u1, u2, u3 as
u1
u2
u3
Hence x1 u1 ? ?
x2 u2 ? ?
x3 u3 ? ?
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Problem 1 Set 14.4C Page 541
On a picnic outing, 2 two-person teams are
playing hide-and-seek. There are four hiding
locations (A, B, C, and D), and the two members
of the hiding team can hide separately in any two
of the four locations. The other team will then
have the chance to search any two locations. The
searching team gets a bonus point if they find
both members of the hiding team.If they miss
both, they lose a point. Otherwise, the game is a
draw.
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• Set up the problem as a two-person zero-sum
  game.
• Determine the optimal strategy and the value of
  the game.

Thus each team has 6 strategies, namely,
AB, AC, AD, BC, BD, CD The pay-off matrix (to the
searching team) is given below
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Hiding team
Searching Team
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Since max (Row minima) -1 and min (Col
maxima) 1,
the game has no saddle point. We go in for mixed
strategies. Adding 1 to each entry, the new
pay-off matrix is
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The hiding teams problem is Maximize w t1
t2 t3 t4 t5 t6 Subject to 2t1 t2 t3
t4 t5 ? 1
t1 2t2 t3 t4 t6 ? 1
t1 t2 2t3 t5
t6 ? 1
t1 t2 2t4 t5 t6 ? 1
t1 t3 t4 2t5 t6 ? 1
t2 t3 t4 t5 2t6 ? 1
t1, t2, t3, t4, t5, t6 ? 0
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Solving, using TORA, we get t1 0.5, t2 0.0, t3
0.0, t4 0.0, t5 0.0, t6 0.5
The optimal tableau also gives u1 0.0, u2 0.0,
u3 0.5, u4 0.5, u5 0.0, u6 0.0 Also ? tj
?ui 1. Hence x1 0.0, x2 0.0, x3 0.5, x4
0.5, x5 0.0, x6 0.0 y1 0.5, y2 0.0, y3 0.0,
y4 0.0, y5 0.0, y6 0.5 And the value of the
game is 1-1 0.