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# Solution Manual For Quantitative Methods For Business 11th Edition By Anderson

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Solution Manual For Quantitative Methods For Business 11th Edition By Anderson ## Description

Chapter 12

Learning Objectives

1. Learn about applications of more advanced optimization models that are solved in practice.

1. Develop an appreciation for the diversity of problems that can be modeled as optimization programs.

1. Obtain practice and experience in formulating realistic optimization models.

1. Understand linear programming applications such as:

data envelopment analysis

revenue management

portfolio selection

Note to Instructor

The application problems of Chapter 12 are designed to give the student an understanding and appreciation of the broad range of problems that can be approached by optimization.   The interpretation of the solution to these problems will require the use of a software package such as Microsoft Excel‘s Solver or LINGO.

Solutions:

1. a.
 Min E s.t. wg + wu + wc + ws = 1 48.14wg + 34.62wu + 36.72wc + 33.16ws ³ 48.14 43.10wg + 27.11wu + 45.98wc + 56.46ws ³ 43.10 253wg + 148wu + 175wc + 160ws ³ 253 41wg + 27wu + 23wc + 84ws ³ 41 -285.2E + 285.2wg + 162.3wu + 275.7wc + 210.4ws £ 0 -123.80E + 1123.80wg + 128.70wu + 348.50wc + 154.10ws £ 0 -106.72E + 106.72wg + 64.21wu + 104.10wc + 104.04ws £ 0

wgwuwcws ³ 0

1. Since wg = 1.0, the solution does not indicate General Hospital is relatively inefficient.

1. The composite hospital is General Hospital. For any hospital that is not relatively inefficient, the composite hospital will be that hospital because the model is unable to find a weighted average of the other hospitals that is better.

1. a.
 Min E s.t. wa + wb + wc + wd + we + wf + wg = 1 55.31wa + 37.64wb + 32.91wc + 33.53wd + 32.48we + 48.78wf + 58.41wg ³ 33.53 49.52wa + 55.63wb + 25.77wc + 41.99wd + 55.30we + 81.92wf + 119.70wg ³ 41.99 281wa + 156wb + 141wc + 160wd + 157we + 285wf + 111wg ³ 160 47wa + 3wb + 26wc + 21wd + 82we + 92wf + 89wg ³ 21 -250E+310wa + 278.5wb + 165.6wc + 250wd + 206.4we + 384wf + 530.1wg £ 0 -316E+134.6wa + 114.3wb + 131.3wc + 316wd + 151.2we + 217wf + 770.8wg £ 0 -94.4E+116wa + 106.8wb + 65.52wc + 94.4wd + 102.1we + 153.7wf + 215wg £ 0

wawbwcwdwewfwg ³ 0

1. E   = 0.924

wa = 0.074

wc = 0.436

we = 0.489

All other weights are zero.

1. D is relatively inefficient

Composite requires 92.4 of D‘s resources.

1. 34.37 patient days (65 or older)

41.99 patient days (under 65)

1. Hospitals A, C, and E.

1. a. Make the following changes to the model in problem 27.

New Right-Hand Side Values for

Constraint 2                   32.48

Constraint 3                   55.30

Constraint 4                   157

Constraint 5                   82

New Coefficients for E in

Constraint 6                   -206.4

Constraint 7                   -151.2

Constraint 8                   -102.1

1. E = 1; we = 1; all other weights = 0

1. No; E = 1 indicates that all the resources used by Hospital E are required to produce the outputs of Hospital E.

1. Hospital E is the only hospital in the composite. If a hospital is not relatively inefficient, the hospital will make up the composite hospital with weight equal to 1.

1. a.
 Min E s.t. wb + wc + wj + wn + ws = 1 3800wb + 4600wc + 4400wj + 6500wn + 6000ws ³ 4600 25wb + 32wc + 35wj + 30wn + 28ws ³ 32 8wb + 8.5wc + 8wj + 10wn + 9ws ³ 8.5 – 110E + 96wb + 110wc + 100wj + 125wn + 120ws £ 0 –   22E + 16wb + 22wc + 18wj + 25wn + 24ws £ 0 -1400E + 850wb + 1400wc + 1200wj + 1500wn + 1600ws £ 0

wbwcwjwnws ³ 0

b.

OPTIMAL SOLUTION

Objective Function Value =             0.960

Variable             Value             Reduced Costs

————–     —————     ——————

E                      0.960                 0.000

WB                   0.175                 0.000

WC                   0.000                 0.040

WJ                   0.575                 0.000

WN                   0.250                 0.000

WS                   0.000                 0.085

1. Yes; E = 0.960 indicates a composite restaurant can produce Clarksville’s output with 96% of Clarksville’s available resources.

1. More Output (Constraint 2 Surplus) \$220 more profit per week.

Less Input

Hours of Operation 110E = 105.6 hours

FTE Staff 22E – 1.71 (Constraint 6 Slack) = 19.41

Supply Expense 1400E – 129.614 (Constraint 7 Slack) = \$1214.39

The composite restaurant uses 4.4 hours less operation time, 2.6 less employees and \$185.61 less supplies expense when compared to the Clarksville restaurant.

1. wb = 0.175, wj = 0.575, and wn = 0.250. Consider the Bardstown, Jeffersonville, and New Albany restaurants.

1. a. If the larger plane is based in Pittsburgh, the total revenue increases to \$107,849. If the larger plane is based in Newark, the total revenue increases to \$108,542. Thus, it would be better to locate the larger plane in Newark.

Note: The optimal solution to the original Leisure Air problem resulted in a total revenue of \$103,103. The difference between the total revenue for the original problem and the problem that has a larger plane based in Newark is \$108,542 – \$103,103 = \$5,439. In order to make the decision to change to a larger plane based in Newark, management must determine if the \$5,439 increase in revenue is sufficient to cover the cost associated with changing to the larger plane.

1. Using a larger plane based in Newark, the optimal allocations are:

PCQ = 33             PMQ = 23         POQ = 43

PCY = 16             PMY= 6                        POY = 11

NCQ = 26             NMQ = 56        NOQ = 39

NCY = 15             NMY = 7           NOY = 9

CMQ = 32           CMY = 8

COQ = 46                        COY = 10

The differences between the new allocations above and the allocations for the original Leisure Air problem involve the five ODIFs that are boldfaced in the solution shown above.

1. Using a larger plane based in Pittsburgh and a larger plane based in Newark, the optimal allocations are:

PCQ = 33             PMQ = 44         POQ = 45

PCY = 16             PMY= 6                        POY = 11

NCQ = 26             NMQ = 56        NOQ = 39

NCY = 15             NMY = 7           NOY = 9

CMQ = 37           CMY = 8

COQ = 44                        COY = 10

The differences between the new allocations above and the allocations for the original Leisure Air problem involve the four ODIFs that are boldfaced in the solution shown above. The total revenue associated with the new optimal solution is \$115,073, which is a difference of \$115,073 – \$103,103 = \$11,970.

1. In part (b), the ODIF that has the largest bid price is COY, with a bid price of \$443. The bid price tells us that if one more Y class seat were available from Charlotte to Myrtle Beach that revenue would increase by \$443. In other words, if all 10 seats allocated to this ODIF had been sold, accepting another reservation will provide additional revenue of \$443.
2. a. The calculation of the number of seats still available on each flight leg is shown below:

 ODIF ODIF   Code Original   Allocation Seats   Sold Seats   Available 1 PCQ 33 25 8 2 PMQ 44 44 0 3 POQ 22 18 4 4 PCY 16 12 4 5 PMY 6 5 1 6 POY 11 9 2 7 NCQ 26 20 6 8 NMQ 36 33 3 9 NOQ 39 37 2 10 NCY 15 11 4 11 NMY 7 5 2 12 NOY 9 8 1 13 CMQ 31 27 4 14 CMY 8 6 2 15 COQ 41 35 6 16 COY 10 7 3

Flight Leg 1: 8 + 0 + 4 + 4 + 1 + 2 = 19

Flight Leg 2: 6 + 3 + 2 + 4 + 2 + 1 = 18

Flight Leg 3: 0 + 1 + 3 + 2 + 4 + 2 = 12

Flight Leg 4: 4 + 2 + 2 + 1 + 6 + 3 = 18

Note: See the demand constraints for the ODIFs that make up each flight leg.

1. The calculation of the remaining demand for each ODIF is shown below:

 ODIF ODIF   Code Original   Allocation Seats   Sold Seats   Available 1 PCQ 33 25 8 2 PMQ 44 44 0 3 POQ 45 18 27 4 PCY 16 12 4 5 PMY 6 5 1 6 POY 11 9 2 7 NCQ 26 20 6 8 NMQ 56 33 23 9 NOQ 39 37 2 10 NCY 15 11 4 11 NMY 7 5 2 12 NOY 9 8 1 13 CMQ 64 27 37 14 CMY 8 6 2 15 COQ 46 35 11 16 COY 10 7 3

1. The LP model and solution are shown below:

MAX 178PCQ+268PMQ+228POQ+380PCY+456PMY+560POY+199NCQ+249NMQ+349NOQ+

385NCY+444NMY+580NOY+179CMQ+380CMY+224COQ+582COY

S.T.

1) 1PCQ+1PMQ+1POQ+1PCY+1PMY+1POY<19

2) 1NCQ+1NMQ+1NOQ+1NCY+1NMY+1NOY<18

3) 1PMQ+1PMY+1NMQ+1NMY+1CMQ+1CMY<12

4) 1POQ+1POY+1NOQ+1NOY+1COQ+1COY<18

5) 1PCQ<8

6) 1PMQ<1

7) 1POQ<27

8) 1PCY<4

9) 1PMY<1

10) 1POY<2

11) 1NCQ<6

12) 1NMQ<23

13) 1NOQ<2

14) 1NCY<4

15) 1NMY<2

16) 1NOY<1

17) 1CMQ<37

18) 1CMY<2

19) 1COQ<11

20) 1COY<3

OPTIMAL SOLUTION

Objective Function Value = 15730.000

Variable             Value             Reduced Costs

————–     —————     ——————

PCQ                     8.000                   0.000

PMQ                     1.000                   0.000

POQ                     3.000                   0.000

PCY                    4.000                   0.000

PMY                     1.000                   0.000

POY                     2.000                   0.000

NCQ                     6.000                   0.000

NMQ                     3.000                   0.000

NOQ                     2.000                   0.000

NCY                     4.000                   0.000

NMY                     2.000                   0.000

NOY                     1.000                   0.000

CMQ                     3.000                   0.000

CMY                     2.000                   0.000

COQ                     7.000                   0.000

COY                     3.000                   0.000

1. a. Let  CT = number of convention two-night rooms

CF = number of convention Friday only rooms

CS = number of convention Saturday only rooms

RT = number of regular two-night rooms

RF = number of regular Friday only rooms

RS = number of regular Saturday only room

b./c.   The formulation and output are shown below.

LINEAR PROGRAMMING PROBLEM

MAX 225CT+123CF+130CS+295RT+146RF+152RS

S.T.

1) 1CT<40

2) 1CF<20

3) 1CS<15

4) 1RT<20

5) 1RF<30

6) 1RS<25

7) 1CT+1CF>48

8) 1CT+1CS>48

9) 1CT+1CF+1RT+1RF<96

10) 1CT+1CS+1RT+1RS<96

OPTIMAL SOLUTION

Objective Function Value = 25314.000

Variable            Value             Reduced Costs

————–     —————     ——————

CT                   36.000                   0.000

CF                   12.000                   0.000

CS                    15.000                   0.000

RT                   20.000                   0.000

RF                   28.000                   0.000

RS                   25.000                   0.000

1. The shadow price for constraint 10 is 50 and shows an added profit of \$50 if this additional reservation is accepted.

1. To determine the percentage of the portfolio that will be invested in each of the mutual funds we use the following decision variables:

FS = proportion of portfolio invested in a foreign stock mutual fund

IB = proportion of portfolio invested in an intermediate-term bond fund

LG = proportion of portfolio invested in a large-cap growth fund

LV = proportion of portfolio invested in a large-cap value fund

SG = proportion of portfolio invested in a small-cap growth fund

SV = proportion of portfolio invested in a small-cap value fund

1. A portfolio model for investors willing to risk a return as low as 0% involves 6 variables and 6 constraints.

Max 12.03FS + 6.89IB + 20.52LG + 13.52LV + 21.27SG + 13.18SV

s.t.

 10.06FS + 17.64IB + 32.41LG + 32.36LV + 33.44SG + 24.56SV ≥ 0 13.12FS + 3.25IB + 18.71LG + 20.61LV + 19.40SG + 25.32SV ≥ 0 13.47FS + 7.51IB + 33.28LG + 12.93LV + 3.85SG – 6.70SV ≥ 0 45.42FS – 1.33IB + 41.46LG + 7.06LV + 58.68SG + 5.43SV ≥ 0 -21.93FS + 7.36IB – 23.26LG – 5.37LV – 9.02SG + 17.31SV ≥ 0 FS + IB + LG + LV + SG + SV = 1

FS, IB, LG, LV, SG, SV ≥ 0

1. The solution obtained is shown.

Objective Function Value = 18.499

Variable             Value             Reduced Costs

————–     —————     ——————

FS                     0.000                  13.207

IB                     0.000                    9.347

LG                     0.000                   5.125

LV                     0.000                   6.629

SG                     0.657                   0.000

SV                     0.343                   0.000

The recommended allocation is to invest 65.7% of the portfolio in a small-cap growth fund and 34.3% of the portfolio in a small-cap value fund. The expected return for this portfolio is 18.499%.

1. One constraint must be added to the model in part a. It is

FS ≥ .10

The solution is

Objective Function Value = 17.178

Variable             Value             Reduced Costs

————–     —————     ——————

FS                     0.100                   0.000

IB                   0.000                   9.347

LG                     0.000                   5.125

LV                     0.000                   6.629

SG                     0.508                   0.000

SV                     0.392                   0.000

The recommended allocation is to invest 10% of the portfolio in the foreign stock fund, 50.8% of the portfolio in the small-cap growth fund, and 39.2 percent of the portfolio in the small-cap value fund. The expected return for this portfolio is 17.178%. The expected return for this portfolio is 1.321% less than for the portfolio that does not require any allocation to the foreign stock fund.

1. To determine the percentage of the portfolio that will be invested in each of the mutual funds we use the following decision variables:

LS = proportion of portfolio invested in a large-cap stock mutual fund

MS = proportion of portfolio invested in a mid-cap stock fund

SS = proportion of portfolio invested in a small-cap growth fund

ES = proportion of portfolio invested in an energy sector fund

HS = proportion of portfolio invested in a health sector fund

TS = proportion of portfolio invested in a technology sector fund

RS = proportion of portfolio invested in a real estate sector fund

1. A portfolio model for investors willing to risk a return as low as 0% involves 7 variables and 6 constraints.

Max 9.68LS + 5.91MS + 15.20SS + 11.74ES + 7.34HS + 16.97TS + 15.44RS

s.t.

 35.3LS + 32.3MS + 20.8SS + 25.3ES + 49.1HS + 46.2TS + 20.5RS ≥ 2 20.0LS + 23.2MS + 22.5SS + 33.9ES + 5.5HS + 21.7TS + 44.0RS ≥ 2 28.3LS – 0.9MS + 6.0SS + 20.5ES + 29.7HS + 45.7TS – 21.1RS ≥ 2 10.4LS + 49.3MS + 33.3SS + 20.9ES + 77.7HS + 93.1TS + 2.6RS ≥ 2 -9.3LS – 22.8MS + 6.1SS – 2.5ES – 24.9HS – 20.1TS + 5.1RS ≥ 2 LS + MS + SS + ES + HS + TS + RS = 1

LS, MS, SS, ES, HS, TS, RS ≥ 0

1. The solution obtained is:

Objective Function Value = 15.539

Variable             Value             Reduced Costs

————–     —————     ——————

LS                     0.000                  6.567

MS                   0.000                 11.548

SS                     0.500                   0.000

ES                     0.000                   4.279

HS                     0.000                 10.095

TS                     0.143                   0.000

RS                     0.357                   0.000

The recommended allocation is to invest 50% of the portfolio in the small-cap stock fund, 14.3% of the portfolio in the technology sector fund, and 35.7% of the portfolio in the real estate sector fund. The expected portfolio return is 15.539%.

1. The portfolio model is modified by changing the right-hand side of the first 5 constraints from 2 to 0.

1. The solution obtained is:

Objective Function Value =         15.704

Variable             Value             Reduced Costs

————–     —————     ——————

LS                     0.000                 6.567

MS                  0.000                 11.548

SS                     0.255                 0.000

ES                     0.000                 4.279

HS                   0.000                 10.095

TS                     0.212                 0.000

RS                     0.533                 0.000

The recommended allocation is to invest 25.5% of the portfolio is the small-cap stock fund, 21.2% of the portfolio in the technology sector fund, and 53.3% of the portfolio in the real estate sector fund. The expected portfolio return is 15.704%. This is an increase of .165% over the portfolio that limits risk to a return of at least 2%. Most investors would conclude that the small increase in the portfolio return is not enough to justify the increased risk.

1. Using LINGO or Excel Solver, the optimal solution is X = 2, Y = -4, for an optimal solution value  of 0.

1. a. Using LINGO or Excel Solver, the optimal solution is X = 4.32, Y = 0.92, for an optimal solution value of 4.84

1. The shadow price on the constraint X + 4Y 8 is -0.88, so we expect the optimal objective function value to decrease by 0.88 if we increase the right-hand-side from 8 to 9.

1. If we resolve the problem with a new right-hand-side of 9 the new optimal objective function value is 4.0 so the actual decrease is only .84 rather than 0.88.

1. a. With \$1000 being spent on radio and \$1000 being spent on direct mail we can simply substitute those values into the sales function.

Sales of \$18,000 will be realized with this allocation of the media budget.

1. We simply add a budget constraint to the sales function that is to be maximized.

1. Using LINGO or Excel Solver, we find that the optimal solution is to invest \$2,500 in radio advertising and \$500 in direct mail advertising. The total sales generated will be \$37,000.

1. a. Here is the proper LINGO formulation. Note the extra use of parentheses needed because of the way LINGO uses the unary minus sign. For example, -X^2 is written as –(X^2)

MIN = 3*((1-X)^2)*@EXP(-(X^2) – (Y + 1)^2) -10*(X/5 – X^3 –

Y^5)*@EXP(-(X^2) – Y^2) – @EXP(-((X + 1)^2) – Y^2)/3;

@FREE(X);

@FREE(Y);

END

Minimizing this function without the global solver option turned on in LINGO gives X = 4.978 and Y = 1.402 for a value of 0.3088137E-08. This is a local minimum.

1. Go to LINGO>Options>Global Solver and place a check in the Global Solver box. When we solve this using LINGO with the Global Solver option turned on, the optimal solution (which is a global minimum) is X = 0.228 and Y = -1.626 for and objective function value of -6.551.

1. a. The optimization model is

1. Using LINGO or Excel Solver, the optimal solution to this is L = 750 and C = 750 for an optimal objective function value of 3750. If Excel Solver is used for this problem we recommend starting with an initial solution that has L > 0 and C > 0.

1. a. The optimization model is

1. Using LINGO or Excel Solver, the optimal solution to this problem is L = 2244.281 and C = 2618.328 for an optimal solution of \$374,046.9. If Excel Solver is used for this problem we recommend starting with an initial solution that has L > 0 and C > 0.

1. a. Let OT be the number of overtime hours scheduled. Then the optimization model is

1. Using LINGO or Excel Solver, the optimal solution is to schedule OT = 8.66667 overtime hours and produce= 3.66667 units of product 1 and = 3.00000 units of product 2 for a profit of 887.3333.

1. a. If X is the weekly production volume in thousand of units at the Dayton plant and Y is the weekly production volume in thousands of units at the Hamilton plant, then the optimization model is

1. Using LINGO or Excel Solver, the optimal solution is X = 4.75 and Y = 3.25 for an optimal objective value of 42.875.

MODEL:

TITLE   MARKOWITZ;

! MINIMIZE VARIANCE OF THE PORTFOLIO;

MIN = (1/5)*((R1 – RBAR)^2 + (R2 – RBAR)^2 + (R3 – RBAR)^2 + (R4 – RBAR)^2 + (R5 – RBAR)^2);

! SCENARIO 1 RETURN;

10.06*FS + 17.64*IB + 32.41*LG + 32.36*LV + 33.44*SG + 24.56*SV = R1;

! SCENARIO 2 RETURN;

13.12*FS + 3.25*IB + 18.71*LG + 20.61*LV + 19.40*SG + 25.32*SV = R2;

! SCENARIO 3 RETURN;

13.47*FS + 7.51*IB + 33.28*LG + 12.93*LV + 3.85*SG – 6.70*SV = R3;

! SCENARIO 4 RETURN;

45.42*FS – 1.33*IB + 41.46*LG + 7.06*LV + 58.68*SG + 5.43*SV = R4;

! SCENARIO 5 RETURN;

-21.93*FS + 7.36*IB – 23.26*LG – 5.37*LV – 9.02*SG + 17.31*SV = R5;

! MUST BE FULLY INVESTED IN THE MUTUAL FUNDS;

FS + IB + LG + LV + SG + SV = 1;

! DEFINE THE MEAN RETURN;

(1/5)*(R1 + R2 + R3+ R4 + R5) = RBAR;

! THE MEAN RETURN MUST BE AT LEAST 10 PERCENT;

RBAR > 10;

! SCENARIO RETURNS MAY BE NEGATIVE;

@FREE(R1);

@FREE(R2);

@FREE(R3);

@FREE(R4);

@FREE(R5);

END

Local optimal solution found.

Objective value:                             27.13615

Total solver iterations:                           17

Model Title: MARKOWITZ

Variable           Value       Reduced Cost

R1       18.95698           0.000000

RBAR       10.00000           0.000000

R2       11.51205           0.000000

R3       5.643902           0.000000

R4       9.728075           0.000000

R5       4.158993           0.000000

FS       0.1584074           0.000000

IB     0.5254795         -0.4148783E-07

LG       0.4206502E-01       0.000000

LV       0.000000           41.64136

SG       0.000000          15.60953

SV       0.2740480          -0.4062599E-07

1. Below is a screen capture of an Excel Spreadsheet Solver model for this problem.

In cell B1 is the smoothing parameter α. The forecasts in Column C are a function of the smoothing parameter, that is the forecast in time t + 1 is

where is the actual value of sales in period t and is the forecast of sales for period t.

In Cell E16 is the sum of squared errors. This is what we minimize by using solver. In the solver parameters dialog box, we use cell E16 as the target cell and cell B1 as the changing cell. After clicking Solve, the result is that the optimal value of α is 0.1743882 and the resulting sum of squared errors is 98.56.

This problem is also easily solved using LINGO. Here is a LINGO formulation. In the LINGO formulation, the Y variables correspond to observed sales. For example, in Week 1, Y1 = 17.

F1 = Y1;

F2 = F1 + ALPHA*( Y1 – F1);

F3 = F2 + ALPHA*( Y2 – F2);

F4 = F3 + ALPHA*( Y3 – F3);

F5 = F4 + ALPHA*( Y4 – F4);

F6 = F5 + ALPHA*( Y5 – F5);

F7 = F6 + ALPHA*( Y6 – F6);

F8 = F7 + ALPHA*( Y7 – F7);

F9 = F8 + ALPHA*( Y8 – F8);

F10 = F9 + ALPHA*( Y9 – F9);

F11 = F10 + ALPHA*( Y10 – F10);

F12 = F11 + ALPHA*( Y11 – F11);

MIN = (Y2 – F2)^2 + (Y3 – F3)^2 + (Y4 – F4)^2 + (Y5 – F5)^2 +

(Y6 – F6)^2 + (Y7 – F7)^2 + (Y8 – F8)^2 + (Y9 – F9)^2 +

(Y10 – F10)^2 + (Y11 – F11)^2 + (Y12 – F12)^2 ;

Y1 = 17;

Y2 = 21;

Y3 = 19;

Y4 = 23;

Y5 = 18;

Y6 = 16;

Y7 = 20;

Y8 = 18;

Y9 = 22;

Y10 = 20;

Y11 = 15;

Y12 = 22;

Solving this in LINGO also produces ALPHA = 0.1743882.

1. Here are the returns calculated from the Yahoo stock data.

 AAPL AMD ORCL AAPL AMD ORCL Date Adj. Close* Adj. Close* Adj. Close* Return Return Return 2-Jan-97 4.16 17.57 4.32 0.0962 -0.5537 -0.1074 2-Jan-98 4.58 10.1 3.88 0.8104 0.1272 0.8666 4-Jan-99 10.30 11.47 9.23 0.9236 0.4506 0.9956 3-Jan-00 25.94 18 24.98 -0.8753 0.3124 0.1533 2-Jan-01 10.81 24.6 29.12 0.1340 -0.4270 -0.5230 2-Jan-02 12.36 16.05 17.26 -0.5432 -1.1194 -0.3610 2-Jan-03 7.18 5.24 12.03 0.4517 1.0424 0.1416 2-Jan-04 11.28 14.86 13.86 1.2263 0.0613 -0.0065 3-Jan-05 38.35 15.8 13.77 0.6749 0.9729 -0.0912 3-Jan-06 75.51 41.8 12.57

Data Source: CSI

Web site: http://www.csidata.com

1. MODEL:

TITLE   MARKOWITZ;

! MINIMIZE VARIANCE OF THE PORTFOLIO;

MIN = (1/9)*((R1 – RBAR)^2 + (R2 – RBAR)^2 + (R3 – RBAR)^2 + (R4 – RBAR)^2 + (R5 – RBAR)^2   + (R6 – RBAR)^2 + (R7 – RBAR)^2 + (R8 – RBAR)^2 + (R9 – RBAR)^2);

! SCENARIO 1 RETURN;

0.0962*AAPL – 0.5537*AMD – 0.1074*ORCL = R1;

! SCENARIO 2 RETURN;

0.8104*AAPL + 0.1272*AMD + 0.8666*ORCL = R2;

! SCENARIO 3 RETURN;

0.9236*AAPL + 0.4506*AMD + 0.9956*ORCL = R3;

! SCENARIO 4 RETURN;

-0.8753*AAPL + 0.3124*AMD + 0.1533*ORCL = R4;

! SCENARIO 5 RETURN;

0.1340*AAPL – 0.4270*AMD – 0.5230*ORCL = R5;

! SCENARIO 6 RETURN;

-0.5432*AAPL – 1.1194*AMD – 0.3610*ORCL = R6;

!SCENARIO 7 RETURN;

0.4517*AAPL + 1.0424*AMD + 0.1416*ORCL = R7;

!SCENARIO 8 RETURN;

1.2263*AAPL + 0.0613*AMD – 0.0065*ORCL = R8;

!SCENARIO 9 RETURN;

0.6749*AAPL + 0.9729*AMD – 0.0912*ORCL = R9;

! MUST BE FULLY INVESTED IN THE MUTUAL FUNDS;

AAPL + AMD + ORCL = 1;

! DEFINE THE MEAN RETURN;

(1/9)*(R1 + R2 + R3+ R4 + R5 + R6 + R7 + R8 + R9) = RBAR;

! THE MEAN RETURN MUST BE AT LEAST 10 PERCENT;

RBAR > .12;

! SCENARIO RETURNS MAY BE NEGATIVE;

@FREE(R1);

@FREE(R2);

@FREE(R3);

@FREE(R4);

@FREE(R5);

@FREE(R6);

@FREE(R7);

@FREE(R8);

@FREE(R9);

END

Local optimal solution found.

Objective value:                0.1990478

Total solver iterations:              12

Model Title: MARKOWITZ

Variable           Value       Reduced Cost

R1     -0.1457056           0.000000

RBAR       0.1518649           0.000000

R2       0.7316081           0.000000

R3       0.8905417           0.000000

R4     -0.6823468E-02       0.000000

R5     -0.3873745           0.000000

R6     -0.5221017           0.000000

R7       0.3499810           0.000000

R8       0.2290317           0.000000

R9       0.2276271           0.000000

AAPL       0.1817734           0.000000

AMD       0.1687534           0.000000

ORCL       0.6494732           0.000000

1. MODEL:

TITLE   MATCHING S&P INFO TECH RETURNS;

! MINIMIZE SUM OF SQUARED DEVIATIONS FROM S&P INFO TECH RETURNS;

MIN = ((R1 – .2854)^2 + (R2 – .7814)^2 + (R3 – .7874)^2 + (R4 + .4090)^2 + (R5 + .2587)^2   + (R6 + .3741)^2 + (R7 – .4840)^2 + (R8 – .0256)^2 + (R9 – 0.0099)^2);

! SCENARIO 1 RETURN;

0.0962*AAPL – 0.5537*AMD – 0.1074*ORCL = R1;

! SCENARIO 2 RETURN;

0.8104*AAPL + 0.1272*AMD + 0.8666*ORCL = R2;

! SCENARIO 3 RETURN;

0.9236*AAPL + 0.4506*AMD + 0.9956*ORCL = R3;

! SCENARIO 4 RETURN;

-0.8753*AAPL + 0.3124*AMD + 0.1533*ORCL = R4;

! SCENARIO 5 RETURN;

0.1340*AAPL – 0.4270*AMD – 0.5230*ORCL = R5;

! SCENARIO 6 RETURN;

-0.5432*AAPL – 1.1194*AMD – 0.3610*ORCL = R6;

!SCENARIO 7 RETURN;

0.4517*AAPL + 1.0424*AMD + 0.1416*ORCL = R7;

!SCENARIO 8 RETURN;

1.2263*AAPL + 0.0613*AMD – 0.0065*ORCL = R8;

!SCENARIO 9 RETURN;

0.6749*AAPL + 0.9729*AMD – 0.0912*ORCL = R9;

! MUST BE FULLY INVESTED IN THE MUTUAL FUNDS;

AAPL + AMD + ORCL = 1;

! SCENARIO RETURNS MAY BE NEGATIVE;

@FREE(R1);

@FREE(R2);

@FREE(R3);

@FREE(R4);

@FREE(R5);

@FREE(R6);

@FREE(R7);

@FREE(R8);

@FREE(R9);

END

Local optimal solution found.

Objective value:                0.4120213

Total solver iterations:                8

Model Title: MATCHING S&P INFO TECH RETURNS

Variable           Value       Reduced Cost

R1     -0.5266475E-01       0.000000

R2       0.8458175           0.000000

R3       0.9716207           0.000000

R4     -0.1370104           0.000000

R5     -0.3362695           0.000000

R6     -0.4175977           0.000000

R7       0.2353628           0.000000

R8       0.3431437           0.000000

R9       0.1328016           0.000000

AAPL       0.2832558           0.000000

AMD       0.6577707E-02       0.000000

ORCL       0.7101665           0.000000

1. In order to measure the semi-variance it is necessary to measure only the downside (deviation below the mean). Do this by introducing two new variables for each scenario. For example, for scenario 1 define D1P as the deviation of return 1 above the mean, and D1N as the deviation of return 1 below the mean. That is

and D1P and D1N are required to be nonnegative. Then in the objective function, we minimize the average of the negative deviations squared, i.e. the semi-variance. The complete LINGO model is

TITLE MARKOWITZ WITH SEMIVARIANCE;

! MINIMIZE THE SEMI-VARIANCE

MIN = (1/5)*((D1N)^2 + (D2N)^2 + (D3N)^2 + (D4N)^2 + (D5N)^2);

! DEFINE THE EXPECTED RETURNS;

10.06*FS + 17.64*IB + 32.41*LG + 32.36*LV + 33.44*SG + 24.56*SV = R1;

13.12*FS + 3.25*IB + 18.71*LG + 20.61*LV + 19.40*SG + 25.32*SV = R2;

13.47*FS + 7.51*IB + 33.28*LG + 12.93*LV + 3.85*SG – 6.70*SV = R3;

45.42*FS – 1.33*IB + 41.46*LG + 7.06*LV + 58.68*SG + 5.43*SV = R4;

-21.93*FS + 7.36*IB – 23.26*LG – 5.37*LV – 9.02*SG + 17.31*SV = R5;

! INVESTMENT LEVELS SUM TO 1;

FS + IB + LG + LV + SG + SV = 1;

! DEFINE EXPECTED RETURN;

(1/5)*(R1 + R2 + R3+ R4 + R5) = RBAR;

! DEFINE THE MINIMUM ;

RBAR > RMIN;

RMIN = 10;

! MEASURE POSITIVE AND NEGATIVE DEVIATION FROM MEAN;

D1P – D1N = R1 – RBAR;

D2P – D2N = R2 – RBAR;

D3P – D3N = R3 – RBAR;

D4P – D4N = R4 – RBAR;

D5P – D5N = R5 – RBAR;

! MAKE THE RETURN VARIABLES UNRESTRICTED;

@FREE(R1);

@FREE(R2);

@FREE(R3);

@FREE(R4);

@FREE(R5);

Local optimal solution found.

Objective value:                             7.503540

Total solver iterations:                           18

Model Title: MARKOWITZ WITH SEMIVARIANCE

Variable           Value       Reduced Cost

D1N       0.000000           0.000000

D2N       0.8595142           0.000000

D3N       3.412762           0.000000

D4N       2.343876           0.000000

D5N       4.431505           0.000000

FS       0.000000           6.491646

IB       0.6908001           0.000000

LG       0.6408726E-01       0.000000

LV       0.000000           14.14185

SG       0.8613837E-01       0.000000

SV       0.1589743           0.000000

R1       21.04766           0.000000

R2       9.140486           0.000000

R3       6.587238           0.000000

R4       7.656124           0.000000

R5       5.568495           0.000000

RBAR       10.00000           0.000000

RMIN       10.00000           0.000000

D1P       11.04766           0.000000

D2P       0.000000           0.3438057

D3P       0.000000           1.365105

D4P       0.000000           0.9375505

D5P       0.000000           1.772602

The solution calls for investing 69.1% of the portfolio in the intermediate-term bond fund, 6.4% of the portfolio in the large-cap growth fund, 8.6% of the portfolio in the small-cap growth fund, and 15.9% of the portfolio in the small-cap value fund.

Excel Solver may have trouble with this problem, depending upon the starting solution that is used. A starting solution of each fund at .167 will produce the optimal value.

1. The objective is to minimize the Value at Risk at 1%. The selection of stocks that go into the portfolio determine the mean return of the portfolio and the standard deviation of portfolio returns. Let μ denote the mean return of the portfolio and σ the standard deviation. If returns are normally distributed this means that 99% of the returns will be above the value

Another way to state this is to say that there is a 99% probability that a portfolio return will exceed and a 1% probability that the portfolio return will be less than. The number is called the Value at Risk at 1%. The number 2.33 can be found in a Normal probability table, or by using LINGO and setting @PSN(-Z) = 0.01, or by using Excel Solver and setting NORMSDIST(-Z) = 0.01 and observing that Z = 2.33 (or more accurately, 2.32638013).

Although people commonly talk about minimizing the Value at Risk of a portfolio, this terminology is misleading. Minimizing the Value at Risk of a portfolio at 1 percent really means making the number as large as possible – in other words we want 99% of the returns to be as large as possible.

1. The LINGO model and solution is given below.

MAX = VaR;

10.06*FS + 17.64*IB + 32.41*LG + 32.36*LV + 33.44*SG + 24.56*SV = R1;

13.12*FS + 3.25*IB + 18.71*LG + 20.61*LV + 19.40*SG + 25.32*SV = R2;

13.47*FS + 7.51*IB + 33.28*LG + 12.93*LV + 3.85*SG – 6.70*SV = R3;

45.42*FS – 1.33*IB + 41.46*LG + 7.06*LV + 58.68*SG + 5.43*SV = R4;

-21.93*FS + 7.36*IB – 23.26*LG – 5.37*LV – 9.02*SG + 17.31*SV = R5;

FS + IB + LG + LV + SG + SV = 1;

(1/5)*(R1 + R2 + R3+ R4 + R5) = RBAR;

STD = ((1/5)*((R1 – RBAR)^2 + (R2 – RBAR)^2 + (R3 – RBAR)^2 + (R4 – RBAR)^2 + (R5 – RBAR)^2))^0.5;

MU = RBAR;

PROB = 0.01;

PROB = @PSN( -Z);

VaR = MU – Z*STD;

@FREE(R1);

@FREE(R2);

@FREE(R3);

@FREE(R4);

@FREE(R5);

@FREE(VaR);

END

The solution is

Local optimal solution found.

Objective value:                             2.118493

Total solver iterations:                           35

Model Title: MARKOWITZ

Variable           Value       Reduced Cost

VAR       -2.118493           0.000000

FS       0.1584074           0.000000

IB       0.5254795           0.000000

LG       0.4206502E-01       0.000000

LV       0.000000           9.298123

SG       0.000000           3.485460

SV       0.2740480           0.000000

R1       18.95698            0.000000

R2       11.51205           0.000000

R3       5.643902           0.000000

R4       9.728075           0.000000

R5        4.158993           0.000000

RBAR       10.00000           0.000000

RMIN       10.00000           0.000000

STD       5.209237           0.000000

MU       10.00000           0.000000

PROB       0.1000000E-01       0.000000

Z       2.326347           0.000000

1. No, minimizing risk is not the same thing as minimizing VaR. Minimizing σ is not the same thing as maximizing Value at Risk. If we maximize Value at Risk, the objective function is

and the objective has two variables, σ and μ.

1. If we fix mean return then it is a constant and

Finally observe that minimizing σ is the same as minimizing σ2 since the standard deviation is always nonnegative.

This problem may also be solved with Excel Solver.

1. Black Scholes Model

!PRICE OF CALL OPTION;

C = S*@PSN(Z) – X*@EXP(-R*T)*@PSN(Z – YSD *T^.5); !BLACK-SCHOLES OPTION FORMULA;

! THE PARAMETERS;

X = 60; !STRIKE OR EXERCISE PRICE;

S = 60.87; !CURRENT STOCK PRICE;

!YVAR = .3^2; !YEARLY VARIANCE;

!YSD = .3; !YEARLY STANDARD DEVIATION;

R = 0.0494; !YEARLY RISKFR-FREE RATE (THREE MONTH TBILL);

!TIME TO MATURITY IN YEARS ;

!WE ARE LOOKIN AT THE CLOSE ON AUGUST 25 UNTIL CLOSE ON SEPT 15;

T = 21/365;

! P&G WEEKLY VOLITILITY

! NORMALIZED STANDARD DEVIATION;

Z = (( R + YVAR/2)*T

+ @LOG( S/X))/( YSD * T^.5);

WVAR = 0.000479376;

YVAR = 52 * WVAR;

YSD = YVAR^.5;

@FREE(Z);

!BID – 1.35;

Variable           Value

C       1.524709

S       60.87000

Z       0.4741179

X       60.00000

R       0.4940000E-01

T       0.5753425E-01

YSD       0.1578846

YVAR       0.2492755E-01

WVAR      0.4793760E-03

This problem may also be solved with Excel Solver.

1. This is a nonlinear 0/1 integer programming problem. Let XIJ = 1 if tanker I is assigned loading dock J and 0 if not. First consider the constraints. Every tanker must be assigned to a loading dock. These constraints are as follows.

! EACH TANKER MUST BE ASSIGNED A DOCK;

X11 + X12 + X13 = 1;   !TANKER 1;

X21 + X22 + X23 = 1;   !TANKER 2;

X31 + X32 + X33 = 1;   !TANKER 3;

X11 + X21 + X31 = 1;   !DOCK 1;

X12 + X22 + X32 = 1;   !DOCK 2;

X13 + X23 + X33 = 1;   !DOCK 3;

The constraints that require each tanker to be assigned a loading dock, and each loading dock assigned a tanker form the constraint set for the assignment problem. The assignment problem was introduced in Chapter 6. However, unlike the assignment problem the objective function in this problem is nonlinear. Consider, for example, the result of assigning tanker 1 to dock 2 and tanker 3 to dock 1. The distance between loading docks 1 and 2 is 100 meters. Also, tanker 1 must transfer 80 tons of goods to tanker 3. This means that 80 tons must be moved 100 meters. To capture this in the objective function, we write 100*80*X12*X31. Since both X12 = 1 and X31 = 1 their product is 1 the product of distance and tonnage moved (100*80) is captured in the objective function. The complete objective function is

MIN = 100*60*X11*X22 + 150*60*X11*X23 + 100*80*X11*X32 + 150*80*X11*X33

+ 100*60*X12*X21 + 50*60*X12*X23 + 100*80*X12*X31 + 50*80*X12*X33

+ 150*60*X13*X21 + 50*60*X13*X22 + 150*80*X13*X31 + 50*80*X13*X32;

The solution to this model is

Global optimal solution found.

Objective value:                             10000.00

Extended solver steps:                               0

Total solver iterations:                           38

Variable           Value       Reduced Cost

X11       0.000000           0.000000

X22       0.000000           0.000000

X23       0.000000           0.000000

X32       0.000000          0.000000

X33       1.000000           0.000000

X12       1.000000           0.000000

X21       1.000000           0.000000

X31       0.000000           0.000000

X13       0.000000           0.000000

Thus tanker 1 should be assigned to dock 2, tanker 2 to dock 1 and tanker 3 to dock 3.

Depending on the starting point, Excel Solver will likely get stuck at a local optimum and not the find the optimal solution that LINGO finds.

1. The objective is to minimize total production cost. To minimize total product cost minimize the production cost at Aynor plus the production cost at Spartanburg. Minimize the production cost at the two plants subject to the constraint that total production of kitchen chairs is equal to 40. The model is:

The optimal solution to this model using either LINGO or Excel Solver is to produce 10 chairs at Aynor for a cost of \$1350 and 30 chairs at Spartanburg for a cost of \$3150. The total cost is \$4500.

28.

Part a.

Let        X = the x coordinate of the tool bin

Y = the y coordinate of the tool bin

Solution: X= 2.23, Y= 3.35

Part b.

Solution: X= 1.91, Y= 2.72

Part c.

Distance:

Demand-weighted Distance:

Using demand shifts the optimal location towards the paint cell (it has heavy demand).