Data Envelopment Analysis for Hospital Efficiency Evaluation

WESTERN MICHIGAN UNIVERSITY CCE-6100-100 - Civil Systems Analysis HW.5 Done by: Aws Tarawneh

1. Data envelopment analysis can measure the relative efficiency of a group of hospitals. The following data from a particular study involving seven teaching hospitals include three input measures and four output measures: a. Formulate a linear programming model so that data envelopment analysis can be used to evaluate the performance of hospital D. E= efficiency of hospital D HA= hospital A HB= hospital B HC= hospital C HD= hospital D HE= hospital E HF= hospital F HG= hospital G Min E s.t. HA+HB+HC+HD+HE+HF+HG=1 310HA+278.5HB+165.6HC+250HD+206.4HE+384HF+530HG-250E<=0 134HA+114.3HB+131.3HC+316HD+151.2HE+217HF+770.8HG-316E<=0 116HA+106.8HB+65.52HC+94.4HD+102.1HE+153.7HF+215HG-94.4E<=0 55.31HA+37.64HB+32.91HC+33.53HD+32.48HE+48.78HF+58.41HG>=33.53 49.52HA+55.63HB+25.77HC+41.99HD+55.3HE+81.92HF+119.7HG>=41.99 291HA+156HB+141HC+160HD+157HE+285HF+111HG>=160 47HA+3HB+26HC+21HD+82HE+92HF+89HG>=21 b. Solve the model. Global optimal solution found. Objective value: 0.9235235 Infeasibilities: 0.000000 Total solver iterations: 6 Model Class: LP Total variables: 8 Nonlinear variables: 0 Integer variables: 0 Total constraints: 9 Nonlinear constraints: 0 Total nonzeros: 60 Nonlinear nonzeros: 0 Variable Value Reduced Cost

E 0.9235235 0.000000 HA 0.7446598E-01 0.000000 HB 0.000000 0.4737552E-01 HC 0.4361526 0.000000 HD 0.000000 0.7647654E-01 HE 0.4893814 0.000000 HF 0.000000 0.1252398E-01 HG 0.000000 0.4825028 Row Slack or Surplus Dual Price 1 0.9235235 -1.000000 2 0.000000 -0.1492373 3 34.82291 0.000000 4 150.5937 0.000000 5 0.000000 0.1059322E-01 6 0.8376040 0.000000 7 0.000000 -0.1224079E-01 8 0.000000 -0.1626847E-02 9 33.96914 0.000000 E=0. 9235235 HA=0.0744 HC=0.4361 HE=0.4893 c. Is hospital D relatively inefficient? What is the interpretation of the value of the objective function? Hospital F is relatively inefficient Composite requires 92.4 of D’s resources d. How many patient-days of each type are produced by the composite hospital? 33.53 patient days (65 or older) 41.99 patient days (under 65) e. Which hospitals would you recommend hospital D consider emulating to improve the efficiency of its operation? Hospitals A, C and E.

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2. With the data in Problem 1 a. Formulate a linear programming model that can be used to perform data envelopment analysis for hospital E. Min E s.t. HA+HB+HC+HD+HE+HF+HG=1 310HA+278HB+165HC+250HD+206.4HE+384HF+530HG-206.4E<=0 134HA+114.3HB+131.3HC+316HD+151.2HE+217HF+770.8HG-151.2E<=0 116HA+106.8HB+65.52HC+94.4HD+102.1HE+153.7HF+215HG-102.1E<=0 55.31HA+37.64HB+32.91HC+33.53HD+32.48HE+48.78HF+58.41HG>=32.48 49.52HA+55.63HB+25.77HC+41.99HD+55.3HE+81.92HF+119.7HG>=55.3 291HA+156HB+141HC+160HD+157HE+285HF+111HG>=157 47HA+3HB+26HC+21HD+82HE+92HF+89HG>=82 b. Solve the model. Global optimal solution found. Objective value: 1.000000 Infeasibilities: 0.000000 Total solver iterations: 1 Model Class: LP Total variables: 8 Nonlinear variables: 0 Integer variables: 0 Total constraints: 9 Nonlinear constraints: 0 Total nonzeros: 60 Nonlinear nonzeros: 0 Variable Value Reduced Cost E 1.000000 0.000000 HA 0.000000 0.9287673 HB 0.000000 1.310314 HC 0.000000 0.4823454 HD 0.000000 0.9551427 HE 1.000000 0.000000 HF 0.000000 0.7385139 HG 0.000000 1.482464 Row Slack or Surplus Dual Price 1 1.000000 -1.000000 2 0.000000 0.000000 3 0.000000 0.4844961E-02 4 0.000000 0.000000 5 0.000000 0.000000 6 0.000000 0.000000 7 0.000000 0.000000 8 0.000000 0.000000 9 0.000000 -0.1219512E-01

E=1 HE=1 c. Is hospital E relatively inefficient? What is the interpretation of the value of the objective function? No, E=1 indicates that hospital E is efficient d. Which hospitals are involved in making up the composite hospital? Can you make a general statement about which hospitals will make up the composite unit associated with a unit that is not inefficient? Hospital E is the only hospital in the composite because the efficiency is 1 3. Hanson Inn is a 96-room hotel a. Define the decision variables and state the objective function. CT = number of convention two night rooms CF= number of convention Friday night only rooms CS= number of convention Saturday night only rooms RT = number of regular two-night rooms RF= number of regular Friday night only rooms RS= number of regular Saturday night only rooms b. Formulate a linear programming model for this revenue management application. Max 225CT+123CF+130CS+295RT+146RF+152RS s.t. 1CT<40 1CF<20 1CS<15 1RT<20 1RF<30 1RS<25 1CT+1CF>48 1CT+1CS>48 1CT+1CF+1RT+RF<96 1CT+1CS+1RT+1RS<96

c. What are the optimal allocation and the anticipated total revenue? Global optimal solution found. Objective value: 25314.00 Infeasibilities: 0.000000 Total solver iterations: 3 Model Class: LP Total variables: 6 Nonlinear variables: 0 Integer variables: 0 Total constraints: 11 Nonlinear constraints: 0 Total nonzeros: 24 Nonlinear nonzeros: 0 Variable Value Reduced Cost CT 36.00000 0.000000 CF 12.00000 0.000000 CS 15.00000 0.000000 RT 20.00000 0.000000 RF 28.00000 0.000000 RS 25.00000 0.000000 Row Slack or Surplus Dual Price 1 25314.00 1.000000 2 4.000000 0.000000 3 8.000000 0.000000 4 0.000000 28.00000 5 0.000000 47.00000 6 2.000000 0.000000 7 0.000000 50.00000 8 0.000000 -23.00000 9 3.000000 0.000000 10 0.000000 146.0000 11 0.000000 102.0000 Maximum revenue =$25,314.00 CT = 36 CF= 12 CS= 15 RT = 20 RF= 28 RS= 25 d. Suppose that one week before the convention the number of regular customers/ Saturday night only rooms that were made available sell out. If another nonconvention customer calls and requests a Saturday night only room, what is the value of accepting this additional reservation? The dual price for regular Saturday night only room is 50 which mean accepting this reservation will increase the maximum revenue $50

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4. Two opposing armies, Red and Blue, must each decide whether to attack or defend. These decisions are made without knowledge of the opposing army’s decision. The payoff table, in terms of value of property gained or lost for the Red Army, appears below. Any gains for the Red Army are losses for the Blue Army. min maximin 30 30 0 a. What is the optimal mixed strategy for the Red Army? For red army p= probability of attack (1-p) = probability of defense 30p+40(1-p)=50p+0(1-p) 40-10p=50p p=40/60=2/3 probability to attack =2/3 probability to defense =1/3 b. What is the optimal mixed strategy for the Blue Army? For blue army p= probability of attack (1-p) = probability of defense 30p+50(1-p)=40p+0(1-p) 50-20p=40p p=50/60=5/6 probability to attack =5/6 probability to defense =1/6 40 50 Minimax 40

5. Two television stations compete with each other for viewing audience. Local programming options for the 5:00 p.m. weekday time slot include a sitcom rerun, an early news program, or a home improvement show. Each station has the same programming options and must make its preseason program selection before knowing what the other television station will do. The viewing audience gains in thousands of viewers for Station A are shown in the payoff table. Determine the optimal strategy for each station. What is the value of the game? For station A Max GAINA s.t. 10PA1+8PA2+4PA3- GAINA>=0 -5PA1+7PA2+8PA3- GAINA>=0 3PA1+6PA2+7PA3- GAINA>=0 PA1+PA2+PA3=1 Global optimal solution found. Objective value: 6.400000 Infeasibilities: 0.000000 Total solver iterations: 4 Model Class: LP Total variables: 4 Nonlinear variables: 0 Integer variables: 0 Total constraints: 5 Nonlinear constraints: 0 Total nonzeros: 16 Nonlinear nonzeros: 0 Variable Value Reduced Cost GAINA 6.400000 0.000000 PA1 0.000000 2.000000 PA2 0.6000000 0.000000 PA3 0.4000000 0.000000 Row Slack or Surplus Dual Price 1 6.400000 1.000000 2 0.000000 -0.2000000 3 1.000000 0.000000 4 0.000000 -0.8000000 5 0.000000 6.400000 The optimal strategy is for Station A to implement strategy a 2 with probability 0.6 and strategy a 3 with probability 0.4.

For station B from the dual value the optimal for Station B to implement strategy b 1 with probability 0.2 and strategy b 3 with probability 0.8. The value of the game is 6.4 which means the expected increase for station A is 6400 viewers. 6. In a gambling game, Player A and Player B both have a $1 and a $5 bill. Each player selects one of the bills without the other player knowing the bill selected. Simultaneously they both reveal the bills selected. If the bills do not match, Player A wins Player B’s bill. If the bills match, Player B wins Player A’s bill. a. Develop the game theory table for this game. The values should be expressed as the gains (or losses) for Player A. B1 B2 A1 -1 5 A2 1 -5 b. Is there a pure strategy? Why or why not? B1 B2 min Maximin A1 -1 5 -1 -1 A2 1 -5 -5 max 1 5 Minimax 1 Since the maximum of the minimums and the minimum of the maximum is not the same; there is no pure strategy and should use a mixed strategy to obtain the optimal results c. Determine the optimal strategies and the value of this game. Does the game favor one player over the other? Player A -1p+1(1-p)=5p-5(1-p) -2p+1=10p-5 6=12p P=0.5 Player B -1p+5(1-p)=1p-5(1-p) -6p+5=6p-5 10=12p p=0.8333

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EV (Payoff) = 0.5(-1) + 0.5(1) = 0The value of the game is 0 so the game is fair. d. Suppose Player B decides to deviate from the optimal strategy and begins playing each bill 50% of the time. What should Player A do to improve Player A’s winnings? Comment on why it is important to follow an optimal game theory strategy. EV (A1) = 0.5(-1) + 0.5(5) = 2 EV (A2) = 0.5(1) + 0.5(-5) = -2 Player A should choose only the $1 to maximize winnings since the EV for such is greater. 7. The offensive coordinator for the Chicago Bears professional football team is preparing a game plan for the upcoming game against the Green Bay Packers. A review of game tapes from previous Bears–Packers games provides data on the yardage gained for run plays and pass plays. Data show that when the Bears run against the Packers’ run defense, the Bears gain an average of 2 yards. However, when the Bears run against the Packers’ pass defense, the Bears gain an average of 6 yards. A similar analysis of pass plays reveals that if the Bears pass against the Packers’ run defense, the Bears gain an average of 11 yards. However, if the Bears pass against the Packers’ pass defense, the Bears average a loss of 1 yard. This loss, or negative gain of –1, includes the lost yardage due to quarterback sacks and interceptions. Develop a payoff table that shows the Bears’ average yardage gain for each combination of the Bears’ offensive strategy to run or pass and the Packers’ strategy of using a run defense or a pass defense. What is the optimal strategy for the Chicago Bears during the upcoming game against the Green Bay Packers? What is the expected value of this strategy? Green Bay Packers Pass Run Chicago Bears Pass Run -1 11 6 2 PB1= probability of strategy Pass for Bears PB2= probability of strategy Run for Bears max GAINBEARS s.t. -1PB1+6PB2- GAINBEARS>=0 11PB1+2PB2- GAINBEARS>=0 PB1+PB2=1

Global optimal solution found. Objective value: 4.250000 Infeasibilities: 0.000000 Total solver iterations: 2 Model Class: LP Total variables: 3 Nonlinear variables: 0 Integer variables: 0 Total constraints: 4 Nonlinear constraints: 0 Total nonzeros: 9 Nonlinear nonzeros: 0 Variable Value Reduced Cost GAINBEARS 4.250000 0.000000 PB1 0.2500000 0.000000 PB2 0.7500000 0.000000 Row Slack or Surplus Dual Price 1 4.250000 1.000000 2 0.000000 -0.5625000 3 0.000000 -0.4375000 4 0.000000 4.250000 The optimal mixed strategy is for the Bears to run with a 0.75 probability and pass with a 0.25 probability. Using the absolute value of the dual prices, the optimal mixed strategy for the Packers is to use a run defense with a 0.437 probability and a pass defense with a 0.563 probability. The expected value of the game shows that with the mixed strategy solution, the Bears average 4.25 yards per play.

HW5 CCE 6100

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