Q 4 – 1 a. Let T = number of TV advertisements R = number of radio advertisements N = number of newspaper advertisements Max 100,000T + 18,000R 40,000N s.t. 2,000T 300R 600N ≦ 18,200 Budget T 10 Max TV R 20 Max Radio N Max News -0.5T 0.5R - 0.5N Max 50% Radio 0.9T 0.1R 0.1N ≧ Min 10% TV T, R, N, ≧ 0

Q 4 – 1 a. cont’d Optimal Solution: T = 4, R = 14, N = 10 Allocation: TV 2,000(4) = $8000 Radio 300(14) = $4,200 News 600(10) = $6,000 Objective Function Value (Expected number of audience): 100,000(4) + 18,000(14) + 40,000(10) =1,052,000

Q 4 – 1 b. Computer Results OPTIMAL SOLUTION Objective Function Value = 1052000 Variable Value Reduced Costs T 4.000 0.000 R 14.000 N 10.000 Constraint Slack/Surplus Dual Prices 1 51.304 2 6.000 3 4 11826.087 5 5217.391 6 1.200

Q 4 – 1 b. cont’d RIGHT HAND SIDE RANGE Constraint Lower Limit Current Value Upper Limit 1 14750.000 18200.000 31999.996 2 4.000 10.000 No Upper Limit 3 14.000 20.000 4 0.000 12.339 5 -8.050 2.936 6 No Lower Limit 1.200 The dual price for the budget constraint is 51.30. Thus, a $100 increase in budget should provide an increase in audience coverage of approximately 5,130. The RHS range for the budget constraint will show this interpretation is correct.

Q 4 – 10 a. Let S = the proportion of funds invested in stocks B = the proportion of funds invested in bonds M = the proportion of funds invested in mutual funds C = the proportion of funds invested in cash Max 0.1S + 0.03B 0.04M 0.01C s.t. 1S 1B 1M 1C = 1 0.8S 0.2B 0.3M ≦ 0.4 0.75 - ≧ 0.1 0.3 S, B, M, C ≧ 0

Q 4 – 10 a. cont’d OPTIMAL SOLUTION Objective Function Value = 0.054 Variable Value Reduced Costs S 0.409 0.000 B 0.145 M C 0.300 Constraint Slack/Surplus Dual Prices 1 0.005 2 0.118 3 0.341 4 -0.001 5 0.200 6

Q 4 – 10 a. cont’d OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit S 0.090 0.100 No Upper Limit B 0.028 0.030 0.036 M No Lower Limit 0.040 0.042 C 0.005 0.010 RIGHT HAND SIDE RANGES Constraint 1 0.800 1.000 1.900 2 0.175 0.400 0.560 3 0.409 0.750 4 -0.267 0.000 0.320 5 0.300 6 0.500

Q 4 – 10 a. cont’d From computer results, the optimal allocation among the four investment alternatives is Stocks 40.0% Bonds 14.5% Mutual Funds 14.5% Cash 30.0% The annual return associated with the optimal portfolio is 5.4% The total risk = 0.409(0.8) + 0.145(0.2) + 0.145(0.3) + 0.300(0.0) = 0.4

Q 4 – 10 b. Changing the RHS value for constraint 2 to 0.18 and resolving using computer, we obtain the following optimal solution: Stocks 0.0% Bonds 36.0% Mutual Funds 36.0% Cash 28.0% The annual return associated with the optimal portfolio is 2.52% The total risk = 0.0(0.8) + 0.36(0.2) + 0.36(0.3) + 0.28(0.0) = 0.18

Q 4 – 10 c. Changing the RHS value for constraint 2 to 0.7 and resolving using computer, we obtain the following optimal solution: Stocks 75.0% Bonds 0.0% Mutual Funds 15.0% Cash 10.0% The annual return associated with the optimal portfolio is 8.2% The total risk = 0.75(0.8) + 0.0(0.2) + 0.15(0.3) + 0.10(0.0) = 0.65

Q 4 – 10 d. Note that a maximum risk of 0.7 was specified for this aggressive investor, but that the risk index for the portfolio is only 0.67. Thus, this investor is willing to take more risk than the solution shown above provides. There are only two ways the investor can become even more aggressive: increase the proportion invested in stocks to more than 75% or reduce the cash requirement of at least 10% so that additional cash could be put into stocks. For the data given here, the investor should ask the investment advisor to relax either or both of these constraints.

Q 4 – 10 e. Defining the decision variables as proportions means the investment advisor can use the linear programming model for any investor, regardless of the amount of the investment. All the investor advisor needs to do is to establish the maximum total risk for the investor and resolve the problem using the new value for maximum total risk.

A – 1 (a) & (b)

A – 1 (c)

A – 2 (a) Let S = Tablespoons of Strawberry C = Tablespoons of Cream V = Tablespoons of Vitamin A = Tablespoons of Artificial sweetener T = Tablespoons of Thickening agent

A – 2 (a) Min 10S + 8C 25V 15A 6T s.t. 50S 100C 120A 80T ≧ 380 ≦ 420 - 24A 14T 20S 50V 2T 50 1S 2A 3S 1V 25T = 15 All variables ≧0

A – 2 (b) ≤ Max 380u1 - 420u2 + 0u3 50u4 0u5 15u6 s.t. 50u1 50u2 9u3 10 100u1 100u2 55u3 8u6 8 1u6 25 120u1 120u2 24u3 2u5 2u6 15 80u1 80u2 14u3 2u4 25u6 6 u1 ~ u5 ≧ 0, u6 : URS

A – 2 (c) Since u1*, u5*, u6* > 0, the 1st, 5th, and 6th constraints are binding.

Data Envelopment Analysis The Langley County School District is trying to determine the relative efficiency of its three high schools. In particular, it wants to evaluate Roosevelt High. The district is evaluating performances on SAT scores, the number of seniors finishing high school, and the number of students who enter college as a function of the number of teachers teaching senior classes, the prorated budget for senior instruction, and the number of students in the senior class.

Roosevelt Lincoln Washington Input Roosevelt Lincoln Washington Senior Faculty 37 25 23 Budget ($100,000's) 6.4 5.0 4.7 Senior Enrollments 850 700 600

Roosevelt Lincoln Washington Output Roosevelt Lincoln Washington Average SAT Score 800 830 900 High School Graduates 450 500 400 College Admissions 140 250 370

Data Envelopment Analysis Decision Variables E = Fraction of Roosevelt's input resources required by the composite high school w1 = Weight applied to Roosevelt's input/output resources by the composite high school w2 = Weight applied to Lincoln’s input/output resources by the composite high school w3 = Weight applied to Washington's input/output resources by the composite high school

Data Envelopment Analysis Objective Function Minimize the fraction of Roosevelt High School's input resources required by the composite high school: MIN E

Data Envelopment Analysis Constraints Sum of the Weights is 1: (1) w1 + w2 + w3 = 1 Output Constraints: Since w1 = 1 is possible, each output of the composite school must be at least as great as that of Roosevelt: (2) 800w1 + 830w2 + 900w3 > 800 (SAT Scores) (3) 450w1 + 500w2 + 400w3 > 450 (Graduates) (4) 140w1 + 250w2 + 370w3 > 140 (College Admissions)

Input Constraints: The input resources available to the composite school is a fractional multiple, E, of the resources available to Roosevelt. Since the composite high school cannot use more input than that available to it, the input constraints are: (5) 37w1 + 25w2 + 23w3 < 37E (Faculty) (6) 6.4w1 + 5.0w2 + 4.7w3 < 6.4E (Budget) (7) 850w1 + 700w2 + 600w3 < 850E (Seniors) Non-negativity of variables: E, w1, w2, w3 > 0

Data Envelopment Analysis OBJECTIVE FUNCTION VALUE = 0.765 VARIABLE VALUE E 0.765 W1 0.000 W2 0.500 W3 0.500

Conclusion The output shows that the composite school is made up of equal weights of Lincoln and Washington. Roosevelt is 76.5% efficient compared to this composite school when measured by college admissions (because of the 0 slack on this constraint (#4)). It is less than 76.5% efficient when using measures of SAT scores and high school graduates (there is positive slack in constraints 2 and 3.)

Data Envelopment Analysis (1) Relative Comparison (2) Multiple Inputs and Outputs (3) Efficiency Measurement (0%-100%) (4) Avoid the Specification Error between Inputs and Outputs (5) Production/Cost Analysis

Case : 1 input – 1 output Table 1.1 : 1 input – 1 output Case

Efficiency Frontier E G Output F C A H B D Employees Figure 1.1:Comparison of efficiencies in 1 input–1 output case

Efficiency Frontier E G Output F C A Regression Line H B D Employees Figure 1.2 : Regression Line and Efficiency Frontier

Table 1.2 : Efficiency 1 = C > G > A> B > E > D = F > H = 0.4

Efficiency Frontier Output C D2 D1 D Employee Figure 1.3 : Improvement of Company D

Case : 2 inputs – 1 output Table 1.3 : 2 inputs – 1 output Case

Production Possibility Set G C F Offices/Sales A I E D B Efficiency Frontier H Employees/Sales Figure 1.4 : 2 inputs – 1 output Case

C A Offices/Sales A1 A2 B Employees/Sales Figure 1.5 : Improvement of Company A

Case : 1 input – 2 outputs Table 1.4 : 1 input – 2 outputs Case

A1 B C A Efficiency Frontier D F Sales/Office Production Possibility Set E1 G E Customers/Office Figure 1.6 : 1 input – 2 outputs Case

Case : Multiple inputs – Multiple outputs Table 1.5 : Example of Multiple inputs–Multiple outputs Case

Ratio model

R : A Reference Set

Primal Problem

Dual Problem

Slack

Reference Set:

Example Problem Table 1.6 : 2 inputs – 1 output Case

Primal Problem

Dual Problem

Efficiency Frontier E D A A1 F C Figure 1.7 : Efficiency of DMU A

Original Ratio Model

Model Under Variable RTS

Dual Problem

Efficiency Frontier of Ratio model (B) d c Output Efficiency Frontier of VRTS model b (C) a (A) Input Figure 2.1 : Efficiency Frontier and Production Possibility Set

Data Envelopment Analysis MIN E s.t. Weighted outputs > Unit k’s output (for each measured output) Weighted inputs < E [Unit k’s input] (for each measured input) Sum of weights = 1 E, weights > 0

Tim Dekker MEMGT New Mexico Institute of Mining and Technology Final Project: International Competitiveness in the Semiconductor Industry: An Application of DEA Tim Dekker MEMGT New Mexico Institute of Mining and Technology

DEA Efficiency Results From Data Rev 2

Weights used in the Analysis (2003)

Final Ranking for Firms in 2004

Home Work Problem 5-2 Problem 5-3 Problem 5-4 Due Day: Sep 23

“Complementary slackness Conditions” are obtained from (4) ( c - y*A ) x* = 0 y*( b - Ax* ) = 0 xj* > 0 y*aj = cj , y*aj > cj xj* = 0 yi* > 0 aix* = bi , ai x* < bi yi* = 0 (5) (6)

Fundamental Insight Z RHS Z 1 Row0 Row1~N

18-1 BV x1 x2 x3 s1 s2 s3 RHS 2 1 220 4 -2 7 80 -1 3 30 20 Max 5x1+6x2+4x3+0s1+0s2+0s3 CB = [0, 4, 5]

18-1 a. Replace “5” by “c1”. Since x1 is basic, then cBB-1A2 – c2 ≥ 0 becomes

18-1 a. cont. cBB-1As1 – cs1 ≥ 0 becomes

18-1 a. cont. cBB-1As2 – cs2 ≥ 0 becomes

18-1 b. Replace “6” by “c2”. Since x2 is non-basic, then cBB-1A2 – c2 ≥ 0 becomes

18-1 c. Replace “4” by “cs1”. Since s1 is non-basic, then cBB-1As1 – cs1 ≥ 0 becomes

18-3 a, b, c BV x1 x2 x3 s1 s2 s3 RHS 2 1 220 4 -2 7 80 -1 3 30 20 The dual prices for the first, second, third constraints = 1 = 2 = 0

18-3 d Let , then x1 x2 x3 s1 s2 s3 RHS 2 1 220 4 -2 7 80 -1 3 30 20 BV x1 x2 x3 s1 s2 s3 RHS 2 1 220 4 -2 7 80 -1 3 30 20

18-3 d cont.

18-3 e

The optimal simplex tableau BV x1 x2 x3 s1 s2 s3 RHS 2.5 7.5 15 75 1 4 0.25 -0.25 0.5 0.75 -0.75 -0.5 1.5 The optimal solution: x1 = 4, x2 = 0.5 The value of the objective function: 75

The optimal simplex tableau 18-15 c The optimal simplex tableau BV x1 x2 x3 s1 s2 s3 RHS 2.5 7.5 15 75 1 4 0.25 -0.25 0.5 0.75 -0.75 -0.5 1.5 If a dual variable is positive, then its corresponding constraint is binding. So, the first and second constrains binding (due to CSC).

18-15 d The redundant (non-binding) is Constraint 3. x1+x2+2x3+s3=6 is replaced by x1=4, x2=0.5, x3=0 as follows: 4+0.5+2(0)+s3=6. Hence, s3=1.5.

The optimal simplex tableau BV x1 x2 x3 s1 s2 s3 RHS 2.5 7.5 15 75 1 4 0.25 -0.25 0.5 0.75 -0.75 -0.5 1.5 Dual prices: = 7.5 = 15 = 0

18-15 e Increasing the RHS of constraint 2 would have the greatest positive effect on the objective function.

18-15 f Replace “15” by “c1”. Since x1 is basic, then cBB-1A3 – c3 ≥ 0 becomes

18-15 f. cont. cBB-1As1 – cs1 ≥ 0 becomes

18-15 f. cont. cBB-1As2 – cs2 ≥ 0 becomes

18-15 f cont. Replace “30” by “c2”. Since x2 is basic, then cBB-1A3 – c3 ≥ 0 becomes

18-15 f cont. cBB-1As1 – cs1 ≥ 0 becomes

18-15 f cont. cBB-1As2 – cs2 ≥ 0 becomes

18-15 f cont. Replace “20” by “c3”. Since x2 is non-basic, then cBB-1A3 – c3 ≥ 0 becomes

18-15 f cont. The optimal solution will not change as long as the objective function coefficients stay in these intervals.

18-15 g

18-15 g For b1 4 + ∆b1(1) ≥ → ∆b1 -4 0.5 ∆b1(-0.25) ≤ 2 1.5 ∆b1(-0.75) → ∆b1 -4 0.5 ∆b1(-0.25) ≤ 2 1.5 ∆b1(-0.75) Therefore -4 ≤ ∆b1 ≤ 2 Range: (4-4 ≤ b1 ≤ 4+2) So, 0 ≤ b1 ≤ 6

18-15 g

18-15 g cont. For b2 4 + ∆b2(0) ≥ → no restriction 0.5 ∆b2(0.5) ∆b2 -1 → no restriction 0.5 ∆b2(0.5) ∆b2 -1 1.5 ∆b2(-0.5) ≤ 3 Therefore -1 ≤ ∆b2 ≤ 3 Range: (3-1≤ b2 ≤ 3+3) So, 2 ≤ b2 ≤ 6

18-15 g

18-15 g cont. For b3 4 + ∆b3(0) ≥ → no restriction 0.5 1.5 ∆b3(1) ∆b3 → no restriction 0.5 1.5 ∆b3(1) ∆b3 -1.5 Therefore -1.5 ≤ ∆b3 Range: (6-1.5≤ b3) So, 4.5 ≤ b3

18-17 a Min 550y1 + 700y2 200y3 s.t. 1.5y1 4y2 2y3 ≥ 4 2y1 1y2 3y3 6 4y1 2y2 1y3 3 3y1 23 1 y1, y2 , y3 ≥ 0

The optimal simplex tableau BV y1 y2 y3 s1 s2 s3 s4 RHS 425 25 125 525 1 -0.4 0.2 1.8 -3.25 -0.65 -0.05 0.05 0.5 -0.5 3.5 0.1 -0.3 0.3 Optimal solution: y1 = 0.3, y2 = 0, y3 = 1.8 x1 = 0, x2 = 25, x3 = 125, x4 = 0.

18-17 c Dual variables y1 = 0.3, y2 = 0, y3 = 1.8 indicate the increase level of the objective function by one unit increase in each RHS. If we increase Machine A by one hour, the profit will be increased by 0.3 ($/hour). Similarly, a profit increase of Machine C will be 1.8 ($/hour).