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.

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