Q 5-2 a.

1
Q 5-2 a.
E Efficiency score wi Weight applied to i
s input and output resources by the composite
hospital
2
Q 5-2 a. contd
3
Q 5-2 b.
OPTIMAL SOLUTION - SUMMARY REPORT (Nonzero
Variables) Variable Value
Cost VAR 1(E) 0.9235
1.0000 VAR 2(wa) 0.0745
0.0000 VAR 4(wc) 0.4362
0.0000 VAR 6(we) 0.4894
0.0000 Slack Variables CONSTR 2
0.8376 0.0000 CONSTR 5
33.9691 0.0000 CONSTR 6
34.5612 0.0000 CONSTR 7
150.549 0.0000 Objective Function Value
0.923523
E 0.924, wa 0.075, wc 0.436, we
0.489 wb wd wfwg 0
4
Q 5-2 c. d. e
(c) Hospital D is relatively inefficient
(7.6). Efficiency score (Objective) is
92.4 . (d) Patient-days (65 or older)
55.31(0.0745) 32.91(0.4362)
32.48(0.4894) 34.37
Patient-days (under 65) 49.52(0.0745)
25.77(0.4362)
55.3(0.4894) 41.99 (e) Hospitals A, C,
and E
5
Q 5-3 a.
E Efficiency Score wi Weight applied to i s
input and output resources by the composite
hospital
6
Q 5-3 a. contd
206.4 E
7
Q 5-3 b.
OPTIMAL SOLUTION - SUMMARY REPORT (Nonzero
Variables) Variable Value
Cost VAR 1(E) 1.0000
1.0000 VAR 6(we) 1.0000
0.0000 Slack Variables CONSTR 6
53.6000 0.0000 Objective Function Value
1
E 1, we 1 All the other weights 0
8
Q 5-3 c. d.
(c) Hospital E is efficient. (d)
Hospital E is the only hospital in the composite.
If a hospital is relatively efficient, the
hospital will make up the composite hospital with
weight equal to 1.
9
Q 5-4 a.
Let, E Efficiency Score wb Weight
applied to Bardstowns input and output resources
by the composite restaurant wc Weight
applied to Clarksvilles input and output
resources by the composite restaurant wj
Weight applied to Jeffersonvilles input and
output resources by the composite restaurant
wn Wight applied to New Albanys input and
output resources by the composite restaurant
ws weight applied to St. Matthewss input
and output resources by the composite restaurant
10
Q 5-4 a. contd
11
Q 5-4 b.
OPTIMAL SOLUTION - SUMMARY REPORT (Nonzero
Variables) Variable Value
Cost VAR 1(E) 0.960
1.0000 VAR 2(wb) 0.175
0.0000 VAR 4(wj) 0.575
0.0000 VAR 5(wn) 0.250
0.0000 Constraint Slack/Surplus Dual Price
CONSTR 1 0.000 0.200
CONSTR 2 220.000 0.000
CONSTR 3 0.000 -0.004 CONSTR
4 0.000 -0.123 CONATR 5
0.000 0.009 CONSTR 6 1.710
0.000 CONSTR 7 129.614 0.000 Objective
Function Value 0.960
E 0.960, wb 0.175, wj 0.575, wn
0.250, wc, ws 0
12
Q 5-4 c.
E 0.96 indicates the 96 level of
efficiency and 4 inefficiency.
13
Q 5-4 d.
More Output (220 more profit per week) is
needed. Less Input Hours of Operation
110(0.96) 105.6 FTE Staff 22(0.96) 1.71
19.41 Supply Expense 1400(0.96) 129.614
1214.39
14
Q 5-4 e.
wb 0.175, wj 0.575, and wn 0.250. So, from
Bardstown, Jeffersonville, and New Albany
restaurants.
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Game Theory
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Game theory is a mathematical theory that deals
with the general features of competitive
situations. The final outcome depends primarily
upon the combination of strategies selected by
the adversaries.
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Two key Assumptions (a) Both players are
rational (b) Both players choose their strategies
solely to increase their own welfare.
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Payoff Table
Player 2
Strategy
1 2 3
1 2 4 1 0 5 0 1 -1
1 2 3
Player 1
Each entry in the payoff table for player 1
represents the utility to player 1 (or the
negative utility to player 2) of the outcome
resulting from the corresponding strategies used
by the two players.
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A strategy is dominated by a second strategy if
the second strategy is always at least as good
regardless of what the opponent does. A
dominated strategy can be eliminated immediately
from further consideration.
Player 2
Strategy
1 2 3
1 2 4 1 0 5 0 1 -1
1 2 3
Player 1
For player 1, strategy 3 can be eliminated. ( 1
gt 0, 2 gt 1, 4 gt -1)
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1 2 3
1 2 4 1 0 5
1 2
For player 2, strategy 3 can be eliminated. ( 1
lt 4, 1 lt 5 )
1 2
1 2 1 0
1 2
For player 1, strategy 2 can be eliminated. ( 1
1, 2 lt 0 )
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1 2
1 2
1
For player 2, strategy 2 can be eliminated. ( 1
lt 2 )
Consequently, both players should select their
strategy 1. A game that has a value of 0 is said
to be a fair game.
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Minimax criterion To minimize his maximum losses
whenever resulting choice of strategy cannot be
exploited by the opponent to then improve his
position.
Player 2
Strategy
Minimum
1 2 3
-3 0 -4
-3 -2 6 2 0 2 5 -2 -4
1 2 3
Player 1
5 0 6
Maximum
Minimax value
Maximin value
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The value of the game is 0, so this is fair
game Saddle Point A Saddle point is an entry
that is both the maximin and minimax.
Player 2
Strategy
Minimum
1 2 3
-3 0 -4
-3 -2 6 2 0 2 5 -2 -4
1 2 3
Player 1
5 0 6
Maximum
Saddle point
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There is no saddle point. An unstable
solution
Player 2
Strategy
Minimum
1 2 3
-2 -3 -4
0 -2 2 5 4 -3 2 3 -4
1 2 3
Player 1
5 4 2
Maximum
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Mixed Strategies
probability that player 1 will use strategy i
( i 1,2,,m), probability that player 2 will
use strategy j ( j 1,2,,n),
Expected payoff for player 1
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Minimax theorem If mixed strategies are allowed,
the pair of mixed strategies that is optimal
according to the minimax criterion provides a
stable solution with (the
value of the game), so that neither player can do
better by unilaterally changing her or his
strategy.
maximin value minimax value
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Solving by Linear Programming
Expected payoff for player 1
The strategy is optimal if
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For each of the strategies
where one and the rest equal 0. Substituting
these values into the inequality yields
Because the are probabilities,
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The two remaining difficulties are (1) is
unknown (2) the linear programming problem has
no objective function. Replacing the
unknown constant by the variable and
then maximizing , so that
automatically will equal at the optimal
solution for the LP problem.
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Player 2
Example
Probability
Pure Strategy
Probability
1 2 3
0 -2 2 5 4 -3
1 2
Player 1
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Player 2
The dual
Probability
Pure Strategy
Probability
1 2 3
0 -2 2 5 4 -3
1 2
Player 1
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Home Work

• Problem 5-13
• Problem 5-15
• Due Date September 30

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Question 1 (Optional Not Home Work) Consider the
game having the following payoff table.
(a) Formulate the problem of finding optimal
mixed strategies according to the minimax
criterion as a linear programming problem. (b)
Use the simplex method to find these optimal
mixed strategies.