1. Homework Solutions
MGMT E-5070
Game Theory
Computer-Based
Manual

2. Players ‘X’ and ‘Y’ Problem 1
REQUIREMENT:
1. Determine the strategies for players ‘X’ and ‘Y’
given the following game.
2. What is the value of the game?

3. Players ‘X’ and ‘Y’ 4 6 6 10 Player ‘X’ plays strategy X2 all the time
Player ‘X’ plays strategy X2 all the time
Player ‘Y’ plays strategy Y1 all the time
The value of the game is “6”
Each time the game is played,
‘X’ wins “6” and ‘Y’ loses “6”
saddle
point

4. Players ‘A’ and ‘B’ Problem 2
REQUIREMENT:
1. Determine the strategies for players ‘A’ and ‘B’ given
the following game.
2. What is the value of the game?

5. Players ‘A’ and ‘B’ Strategies B1 B2 A1 19 20 A2 5 - 4 19 - 4 19 20
Player ‘A’ plays strategy ‘A1’ all the time
Player ‘B’ plays strategy ‘B1’ all the time
The value of the game is “19”
Each time the game is played,
‘A’ wins “19” and ‘B’ loses “19”
saddle
point

6. Players ‘X’ and ‘Y’ Problem 3
REQUIREMENT:
1. Determine the strategies for players ‘X’ and ‘Y’ given
the following game.
2. What is the value of the game?

7. Players ‘X’ and ‘Y’
Strategies Y1 Y2 X1 86 42 X2 36
106

8. Players ‘X’ and ‘Y’ Player ‘X’ strategy
86Q + 36(1-Q) = 42Q (1-Q)
Q = 35/57 or 61.4% ; (1-Q) = 22/57 or 38.6%
Player ‘X’ plays strategy ‘X1’ 61.4% of the time
Player ‘X’ plays strategy ‘X2’ 38.6% of the time

9. Players ‘X’ and ‘Y’ Player ‘Y’ strategy
86P + 42(1-P) = 36P (1-P)
P = 32/57 or 56.14% ; (1-P) = 25/57 or 43.86%
Player ‘Y’ plays strategy ‘Y1’ 56.14% of the time
Player ‘Y’ plays strategy ‘Y2’ 43.86% of the time

10. Players ‘X’ and ‘Y’ .5614 .4386 .6141 .3860 The value of the game
.6141
.3860
The value of the game
86 ( ) + 36 ( ) = 66.70
86 ( ) + 42 ( ) = 66.70

11. Players ‘X’ and ‘Y’ Problem 4 REQUIREMENT:
1. Solve the following game:

12. Players ‘X’ and ‘Y’ This game can be reduced to a 2 x 2 game.
Player ‘X’ would never play strategy ‘X1’ or ‘X4’,
since player ‘X’ stands to lose in every case under
those two strategies.

13. Players ‘X’ and ‘Y’ This game can be reduced to a 2 x 2 game.
Player ‘X’ would never play strategy ‘X1’ or ‘X4’,
since player ‘X’ stands to lose in every case under
those two strategies.

14. Players ‘X’ and ‘Y’ Player ‘X’ strategy 12Q + 4 (1-Q) = 8Q + 12 (1-Q)
Q = 2/3 or 66.67% ; (1-Q) = 1/3 or 33.33%
Player ‘X’ plays strategy ‘X1’ 66.67% of the time
Player ‘X’ plays strategy ‘X2’ 33.33% of the time

15. Players ‘X’ and ‘Y’ Player ‘Y’ strategy 12P + 8 (1-P) = 4P + 12 (1-P)
P = 1/3 or 33.33% ; (1-P) = 2/3 or 66.67%
Player ‘Y’ plays strategy ‘Y1’ 33.33% of the time
Player ‘Y’ plays strategy ‘Y2’ 66.67% of the time

16. Players ‘X’ and ‘Y’ .3333 .6667 .6667 .3333 The value of the game
.6667
.3333
The value of the game
12 ( ) + 4 ( ) = 9.333
12 ( ) + 8 ( ) = 9.333

17. Shoe Town & Fancy Foot Problem 5
Shoe Town and Fancy Foot are both vying for more share of
the market. If Shoe Town does no advertising, it will not lose
any share of the market if Fancy Foot does nothing.
It will lose 2% of the market if Fancy Foot invests $10,000.00
in advertising, and it will lose 5% of the market if Fancy Foot
invests $20, in advertising.
On the other hand, if Shoe Town invests $15, in adver-
tising, it will gain 3% of the market if Fancy Foot does nothing.
It will gain 1% of the market if Fancy Foot invests $10,000.00
in advertising. It will lose 1% if Fancy Foot invests $20,000.00
in advertising.

19. Shoe Town
Fancy Foot
FANCY FOOT
( Y )
SHOE TOWN
( X )
THE PAYOFF TABLE

20. To Solve Via
“ GAME THEORY”
Module

23. THE DATA INPUT TABLE
IS EXACTLY THE SAME
AS THE
PAYOFF TABLE

27. THE SADDLE POINT
( pure strategy )
Shoe Town will always
play strategy X2
Fancy Foot will always
play strategy Y3
Expected Game Value
( -1 )

30. To Solve Via
“Linear Programming”
Module

33. Linear Equalities Suitable for Inclusion in the Simplex Matrix :
The Data Input Table
Shows The
Objective Function
and the
2 Constraints
Maximize 1Y1 + 1Y2 + 1Y3
subject to: 0Y1 - 2Y2 - 5Y3 <= 1
3Y1 + 1Y2 - 1Y3 <= 1
non-negativity
constraint: Y1, Y2, Y3 => 0
Linear Equalities Suitable
for Inclusion in the Simplex Matrix :
0Y1 - 2Y2 - 5Y3 + 1X1 + 0X2 = 1
3 Y1 + 1Y2 - 1Y3 + 0X1 + 1X2 = 1

34. The Solution Is “UNBOUNDED”
THIS MEANS THAT THIS IS A PURE STRATEGY
GAME
THE PROGRAM CANNOT TELL US WHAT PERCENTAGE OF TIME
EACH OF THE STRATEGIES SHOULD BE PLAYED

59. Homework Solutions
MGMT E-5070
Game Theory
Computer-Based
Manual