Q 5-2 a. E = Efficiency score wi = Weight applied to i ’s input and output resources by the composite hospital
Dec 31, 2015
Q 5-2 a.
E = Efficiency score
wi = Weight applied to i ’s input and output
resources by the composite hospital
Q 5-2 a. cont’d
Min E
s.t. wa + wb + wc + wd + we + wf + wg = 1
55.31 wa + 37.64 wb + 32.91 wc + 33.53 wd + 32.48 we + 48.78 wf + 58.41 wg ≧ 33.53
49.52 wa + 55.63 wb + 25.77 wc + 41.99 wd + 55.3 we + 81.92 wf + 119.7 wg ≧ 41.99
291 wa + 156 wb + 141 wc + 160 wd + 157 we + 285 wf + 111 wg ≧ 160
47 wa + 3 wb + 26 wc + 21 wd + 82 we + 92 wf + 89 wg ≧ 21
250 E - 310 wa - 278.5 wb - 165.6 wc - 250 wd - 206.4 we - 384 wf - 530.1 wg ≧ 0
316 E - 134.6 wa - 114.3 wb - 131.3 wc - 316 wd - 151.2 we - 217 wf - 770.8 wg ≧ 0
94.4 E - 116 wa - 106.8 wb - 65.52 wc - 94.4 wd - 102.1 we - 153.7 wf - 215 wg ≧ 0
wa, wb, wc, wd, we, wf, wg ≧ 0
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*= wf*=wg* =0
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
Q 5-3 a.
E = Efficiency Score
wi = Weight applied to i ’s input and output
resources by the composite hospital
Q 5-3 a. cont’d
Min E
s.t. wa + wb + wc + wd + we + wf + wg = 1
55.31 wa + 37.64 wb + 32.91 wc + 33.53 wd + 32.48 we + 48.78 wf + 58.41 wg ≧ 32.48
49.52 wa + 55.63 wb + 25.77 wc + 41.99 wd + 55.3 we + 81.92 wf + 119.7 wg ≧ 55.3
291 wa + 156 wb + 141 wc + 160 wd + 157 we + 285 wf + 111 wg ≧ 157
47 wa + 3 wb + 26 wc + 21 wd + 82 we + 92 wf + 89 wg ≧ 82
260 E - 310 wa - 278.5 wb - 165.6 wc - 250 wd - 206.4 we - 384 wf - 530.1 wg ≧ 0
151.2 E - 134.6 wa - 114.3 wb - 131.3 wc - 316 wd - 151.2 we - 217 wf - 770.8 wg ≧ 0
102.1 E - 116 wa - 106.8 wb - 65.52 wc - 94.4 wd - 102.1 we - 153.7 wf - 215 wg ≧ 0
wa, wb, wc, wd, we, wf, wg ≧ 0
206.4 E
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
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.
Q 5-4 a.
Let, E = Efficiency Score
wb = Weight applied to Bardstown’s input and output
resources by the composite restaurant
wc = Weight applied to Clarksville’s input and output
resources by the composite restaurant
wj = Weight applied to Jeffersonville’s input and output
resources by the composite restaurant
wn = Wight applied to New Albany’s input and output
resources by the composite restaurant
ws = weight applied to St. Matthews’s input and output
resources by the composite restaurant
Q 5-4 a. cont’d
Min E
s.t. ws + wc + wj + wn + ws = 1
3800 ws + 4600 wc + 4400 wj + 6500 wn + 6000 ws ≧ 4600
25 ws + 32 wc + 35 wj + 30 wn + 28 ws ≧ 32
8 ws + 8.5 wc + 8 wj + 10 wn + 9 ws ≧ 8.5
110 E - 96 ws - 110 wc - 100 wj - 125 wn - 120 ws ≧ 0
22 E - 16 ws - 22 wc - 18 wj - 25 wn - 24 ws ≧ 0
1400 E - 850 ws - 1400 wc - 1200 wj - 1500 wn - 1600 ws ≧ 0
wb, wc, wj, wn, ws ≧ 0
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 0.009CONSTR 6 1.710 0.000CONSTR 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
Q 5-4 c.
E = 0.96 indicates the 96% level of
efficiency and 4% inefficiency.
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
Q 5-4 e.
wb = 0.175, wj = 0.575, and wn = 0.250.
So, from Bardstown, Jeffersonville, and
New Albany restaurants.
Game Theory
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.
Two key Assumptions:
(a) Both players are rational
(b) Both players choose their strategies solely
to increase their own welfare.
Payoff Table
StrategyPlayer 2
1 2 31 2 4 1 0 5 0 1 -1
Player 1123
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.
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.
StrategyPlayer 2
1 2 31 2 4 1 0 5 0 1 -1
Player 1123
For player 1,strategy 3 can be eliminated. ( 1 > 0, 2 > 1, 4 > -1)
1 2 31 2 4 1 0 5
12
For player 2,strategy 3 can be eliminated. ( 1 < 4, 1 < 5 )
1 21 2 1 0
12
For player 1,strategy 2 can be eliminated. ( 1 = 1, 2 < 0 )
1 21 2 1
For player 2,strategy 2 can be eliminated. ( 1 < 2 )
Consequently, both players should select their strategy 1.
A game that has a value of 0 is said to be
a fair game.
StrategyPlayer 2
1 2 3-3 -2 6 2 0 2 5 -2 -4
Player 1123
Minimum
Maximum: 5 0 6
-30-4
Minimax value Maximin value
Minimax criterion:
To minimize his maximum losses whenever resulting choice of strategy cannot be exploited by the opponent to then improve his position.
StrategyPlayer 2
1 2 3-3 -2 6 2 0 2 5 -2 -4
Player 1123
Minimum
Maximum: 5 0 6
-30-4
Saddle point
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.
StrategyPlayer 2
1 2 3 0 -2 2 5 4 -3 2 3 -4
Player 1123
Maximum: 5 4 2
-2-3-4
Minimum
There is no saddle point.
An unstable solution
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),
ix
jy
Expected payoff for player 1 =
m
i
n
jjiij yxp
1 1
,
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.
vvv
v
v
= maximin value
= minimax value
Solving by Linear Programming
Expected payoff for player 1 =
m
i
n
jjiij yxp
1 1
The strategy is optimal if),,,( 21 mxxx
vvyxpm
i
n
jjiij
1 1
,,,2,11
njforvxpm
iiij
),,,( 21 nyyy For each of the strategies where one and the rest equal 0. Substituting these values into the inequality yields
121 mxxx
mforixi ,,2,1,0
ixBecause the are probabilities,
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.
1mx 1mx
1mx
v
v
v
.m,,2,1ifor,0x
1xxx
0xxpxpxp
0xxpxpxp
0xxpxpxp .t.s
,xMaximize
i
m21
1mmmn2n21n1
1mm2m222112
1mm1m221111
1m
.n,,2,1jfor,0y
1yyy
0yypypyp
0yypypyp
0yypypyp .t.s
,yMinimize
j
n21
1nnmn22m11m
1nnn2222121
1nnn1212111
1n
1xx
0xx3x2
0xx4x2
0xx5.t.s
,xMaximize
21
321
321
32
3
Probability
Player 2
1 2 3 0 -2 2 5 4 -3 Player 1
12
1x
2x
PureStrategy
1y 2y 3yProbabilityExample
.0x,0x,0x 321
)11
2,
11
4,
11
7(),,( *
3*2
*1 xxx
1yyy
0yy3y4y5
0yy2y2.t.s
,yMinimize
321
4321
432
4
Probability
Player 2
1 2 3 0 -2 2 5 4 -3 Player 1
12
1x
2x
PureStrategy
1y 2y 3yProbability
.0y,0y,0y,0y 4321 )
11
2,
11
6,
11
5,0(
)y,y,y,y( *4
*3
*2
*1
The dual
Home Work
• Problem 5-13
• Problem 5-15
• Due Date: September 30
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.
1 2 3 41 5 0 3 1
Player 1 2 2 4 3 23 3 2 0 4
Player 2Strategy