Section 14.2
Solution to Problem 10(b):
From Problem 9, probability of pass = (a-b)/(a+d-b-c). We want to
show that an increase in c decreases the probability that the
offense will pass. The partial derivative of (probability of pass)
with respect to c = (a-b)(a+d-b-c)-2<0, so improving your pass
performance against run defense causes you to pass less. This is
because improving pass performance against run defense causes
defense to choose pass defense more often, forcing offense to pass
less!
Section 14.3
Solution to Problem 2:
Row 3 is dominated by row 1 or row 2 and column 4 is dominated by
column 1. So we may deal with the following reward matrix:
----------4
5 1
___________
2
1 6
----------The row and column player's LP's are as follows:
Row LP
Column LP
max v
min w
st v4x1 + 2x2
st w4y1 + 5y2 + y3
v5x1 + x2
w2y1 + y2 + 6y3
vx1 + 6x2
y1 + y2 + y3 = 1
x1 + x2 = 1
y1, y2, y30
x1 , x20
Substituting x2 = 1 - x1 and y3 = 1 - y1 - y2 we obtain
Row LP
Column LP
max v
min w
st (1) v2 + 2x1
st (4) w1 + 3y1 + 4y2
(2) v1 + 4x1
(5) w6 - 4y1 - 5y2
(3) v6 - 5x1
All variables 0
All Variables 0
Assume that (1) and (2) hold with no slack. This yields x1 =
x2 = 1/2, v = 3, which satisfies (3) with slack. If this is to be
optimal, complementary slackness implies that y3 = 0(or y1 + y2 =
1) and both (4) and (5) hold with no slack. This yields w = 3, y1
= 2, y2 = -1, which is infeasible. Now assume that (1) and (3)
hold with no slack. This yields v = 22/7, x1 = 4/7, x2 = 3/7. This
satisfies (2) with slack. If this solution is optimal, then
complementary slackness implies that y2 = 0 and (4) and (5) hold
with no slack. This yields w = 22/7, y1 = 5/7, y2 = 0, y3 = 2/7.
This solution is feasible for the column player so we have found
the value of the game and optimal strategies for each player:
Value to row player = 22/7
Row's optimal strategy: x1 = 4/7, x2 = 3/7, x3 = 0
Column's optimal strategy: y1 = 5/7, y2 = 0, y3 = 2/7, y4 = 0
Section 14.3
Solution to Problem 4:
Let strategy (i, j) mean a player sends i regiments to city 1 and
j regiments to city 2. The appropriate reward matrix is as
follows:
Custard
Peabody
(3, 0)
(0, 3)
(2, 1) (1,2)
----------------------------------(4, 0)
4
0
2
1
----------------------------------(0, 4)
0
4
1
2
----------------------------------(3, 1)
1
-1
3
0
----------------------------------(1, 3)
-1
1
0
3
----------------------------------(2, 2)
-2
-2
2
2
----------------------------------Let xi = probability that Custard chooses strategy specified by
row i and yj = probability that Peabody chooses strategy specified
by column j. Since there is no difference between the two cities
we conjecture that x1 = x2, x3 = x4, y1 = y2, and y3
= y4 will hold for each player's optimal strategies. We therefore
define
x1 = x2 = a, x3 = x4 = b x5 = c , y1 = y2 = d, y3 = y4 = e
Then 2a + 2b + c = 1 and 2d +2e = 1. The row and column problems
are as follows:
Row player’s LP:
max v
st v4a -2(1 - 2a - 2b) = 8a + 4b - 2
v4a -2 (1 - 2a - 2b) = 8a + 4b - 2
v3a + 3b + 2(1 - 2a - 2b) = -a - b + 2
v3a + 3b + 2(1 - 2a - 2b) = -a - b + 2
a, b0
Column player’s LP:
min w
st (1) w4d + 3(1/2 - d) = 3/2 + d
(2) w4d + 3(1/2 - d) = 3/2 + d
(3) w3(1/2 - d) = 3/2 - 3d
(4) w3(1/2 - d) = 3/2 - 3d
(5) w-4d + 4(1/2 - d) = 2 - 8d
d0
If we try (1) and (5) with no slack we obtain d = 1/18, e =
4/9, w = 14/9. This satisfies (3) and (4) with slack and (2) with
no slack. For this solution to be optimal for column player, b = 0
must hold and all row player constraints must hold with equality.
This yields a = 4/9, b = 0, c = 1/9, v = 14/9. This is a feasible
strategy for the row player. Thus by the dual theorem we have
found optimal strategies. Peabody should play mixed strategy
(1/18, 1/18, 4/9, 4/9) while Custard should play (4/9, 4/9, 0, 0,
1/9). Value to Custard = 14/9. Observe that the stronger player
(Custard) gambles more often in that he usually sends all his
forces to one city while the weaker player (Peabody) usually
splits his forces between the two cities.