5-11(a) Approximate cash receipts = Yield per acre  1,000  Price per bushel

ACT

EventCornTomatoesBeetsAsparagusCauliflower

Dry$20,000$20,000$22,500$30,000$20,000

Moderate 35,000 40,000 30,000 25,000 20,000

Damp 40,000 20,000 45,000 20,000 20,000

(b)Inadmissible acts: cauliflower (dominated by every other act). The last column in the payoff table may be eliminated.

CornTomatoesBeetsAsparagus

Column minimum$20,000$20,000$22,500$20,000

The maximin payoff act is to plant beets.

Payoff  Probability

EventProbabilityCornTomatoesBeetsAsparagus

Dry .3$ 6,000$ 6,000$ 6,750$ 9,000

Moderate .5 17,500 20,000 15,000 12,500

Damp .2 8,000 4,000 9,000 4,000

Expected payoff:$31,500$30,000$30,750$25,500

The act having the maximum expected payoff, $31,500, is to choose corn.

5-22See the following decision tree. The optimal course of action is to first test market. Then if the

results are favorable, market; but if they are unfavorable, abandon.


6-1(a)$1,000(1/200) + $0(199/200) = $5

(b)Since her insurance payment provides a payoff of $10, Shirley maximizes expected payoff

($5.00) by not buying the insurance.

(c)Her certainty equivalent for the stolen car risk is $10. Since her net worth would change by

$9 by buying the insurance and since $9 > $10, Shirley would buy.

6-2(a)(b)(c)(d)

(1) Expected payoff$35,000$19,300$2,500$ 0

(2) Risk premium 4,000 1,000 200 20,000

(3) Certainty equivalent 31,000 18,300 2,30020,000

6-5(a) Certainty equivalent = $10

(b) Risk premium = $5  ($10) = $5

Exercises for Chp. 7

7-6(a) See the following.


(b)

 = 0.20MSE = 954.4516

PeriodActualForecast Error

tYear Yt FtYt Ft

11990205.00

21991206.00205.00 1.00

31992223.00205.2017.80

41993231.00208.7622.24

51994234.00213.2120.79

61995241.00217.3723.63

71996267.00222.0944.91

81997268.00231.0736.93

91998277.00238.4638.54

101999290.00246.1743.83

112000254.93

(c)  = 0.30  = 0.20MSE = 297.1306

PeriodActualTrendSlopeForecast Error

tYear Yt Tt bt FtYt Ft

11990205.00

21991206.00205.001.00

31992223.00211.102.02206.0011.01

41993231.00218.483.09213.1217.88

51994234.00225.303.84221.5812.42

61995241.00232.704.55229.1411.86

71996267.00246.176.33237.2529.75

81997268.00257.167.26252.5115.49

91998277.00268.198.02264.4212.58

101999290.00280.358.85276.2113.79

112000289.20

(d)The twoparameter exponential smoothing is better because (1) it reflects trend and (2) it has the smaller MSE.

7-7 (a) See the following figure.


(b)  = 0.30 = 0.20 = 0.40MSE = 14.48276

PeriodActualTrendSlope SeasonalForecast Error

tQuarter Yt Tt bt St Ft Yt Ft

11995W 3.90

2S 6.10 3.902.201.56

3S 4.30 5.562.090.77

4F10.80 8.602.281.26

51996W 7.80 9.952.100.78

6S10.6010.471.781.3418.858.25

7S 6.9011.251.580.71 9.472.57

8F13.5012.211.461.2016.122.62

91997W12.9014.501.620.8310.70 2.20

10S15.2014.681.331.2221.666.46

11S10.3015.571.240.6911.361.06

12F18.7016.461.171.1720.111.41

131998W13.9017.391.130.8214.560.66

14S14.4016.500.721.0822.598.19

15S10.2016.490.580.6611.891.69

16F17.3016.370.441.1320.012.71

171999W13.5016.740.420.8113.710.21

18S18.2017.060.401.0818.550.35

19S14.2018.660.640.7011.55 2.65

20F20.7019.030.591.1121.741.04

(c) 2000W 15.93

S 21.73

S 14.58

F 23.75

7-11(a) See the following figure.


(b)

X Y XY X2

14 68 952 196

23105 2,415 529

9 40 360 81

17 79 1,343 289

10 81 810 100

22 95 2,090 484

5 31 155 25

12 72 864 144

6 45 270 36

16 93 1,488 256

13470910,7472,140

=X=Y=XY= X2

a = 70.9  3.619(13.4) = 22.405

c) extra question. Consider the excel report given below.

SUMMARY OUTPUT
Regression Statistics
Multiple R / 0,895347
R Square / 0,801646
Adjusted R Square / 0,776852
Standard Error / 11,81164
Observations / 10
ANOVA
df / SS / MS / F / Significance F
Regression / 1 / 4510,781 / 4510,781 / 32,3319 / 0,000462
Residual / 8 / 1116,119 / 139,5149
Total / 9 / 5626,9
Coefficients / Standard Error / t Stat / P-value / Lower 95% / Upper 95% / Lower 95,0% / Upper 95,0%
Intercept / 22,40476 / 9,31077 / 2,406328 / 0,042752 / 0,934089 / 43,87544 / 0,934089 / 43,87544
X Variable 1 / 3,619048 / 0,636471 / 5,686115 / 0,000462 / 2,151343 / 5,086753 / 2,151343 / 5,086753

-What can you say about the performance of regression?

Answer: r-square is 0.80, close to 1, showing that there is a significant linear relation between X and Y. This regression seems to be appropriate.

-Is the trend significant?

In the ANOVA table F-significance is 0.000462, almost 0. Since it is less than 5%, at a 95% confidence level, we are quite sure that there is a significant trend.

7-17(a)

(1)(2)(3) (4)(5) (6)

Sum ofPercentage

12Month 12Month Moving of Moving

MonthCashTotal Totals AverageAverage

1995J 2.7

F 5.4

M 9.3

A 2.4

M 6.1

J 7.3 81.9

J 6.5 82.1 164.06.8 95.6

A 9.7 83.1 165.26.9140.6

S13.4 83.9 167.07.0191.4

O10.6 85.6 169.57.1149.3

N 5.1 87.3 172.97.2 70.8

D 3.4 87.4 174.77.3 46.6

1996J 2.9 86.4 173.87.2 40.3

F 6.4 86.3 172.77.2 88.9

M10.1 86.4 172.77.2140.3

A 4.1 86.5 172.97.2 56.9

M 7.8 86.2 1.72.77.2108.3

J 7.4 85.6 171.87.2102.8

J 5.5 86.2 171.87.2 76.4

A 9.6 87.1 173.37.2133.3

S13.5 88.3 175.47.3184.9

O10.7 88.0 176.37.3146.6

N 4.8 88.3 176.37.3 65.8

D 2.8 87.8 176.17.3 38.4

1997J 3.5 88.8 176.67.4 47.3

F 7.3 88.1 176.97.4 98.6

M11.3 88.9 177.07.4152.7

A 3.8 89.7 178.67.4 51.4

M 8.1 91.4 181.17.5108.0

J 6.9 92.5 183.97.7 89.6

J 6.5 91.5 184.07.7 84.4

A 8.9 92.3 183.87.7115.6

S14.3 88.9 181.27.6188.2

O11.5 90.3 179.27.5153.3

N 6.5 91.4 181.77.6 85.5

D 3.9 92.6 184.07.7 50.6

1998J 2.5 93.7 186.37.8 32.1

F 8.1 94.1 187.87.8103.8

M 7.9 95.6 189.77.9100.0

A 5.2 96.7 192.38.0 65.0

M 9.2 97.4 194.18.1113.6

J 8.1 97.6 195.08.1100.0

J 7.6 99.3 196.98.2 92.7

A 9.3100.8 200.18.3112.0

S15.8105.3 206.18.6183.7

O12.6106.3 211.68.8143.2

N 7.2105.8 212.18.8 81.8

D 4.1106.0 211.88.8 46.6

(1)(2)(3) (4)(5) (6)

Sum ofPercentage

12Month 12Month Moving of Moving

MonthCashTotal Totals AverageAverage

1999J 4.2105.0 211.08.8 47.7

F 9.6105.5 210.58.8109.1

M12.4106.0 211.58.8140.9

A 6.2106.8 212.88.9 69.7

M 8.7106.5 213.38.9 97.8

J 8.3108.5 215.09.0 92.2

J 6.6

A 9.8

S16.3

O13.4

N 6.9

D 6.1

(b)

Percentages of Moving Averages

Seasonal Cash

19951996199719981999Median IndexRequired

J 40.3 47.3 32.1 47.7 43.80 43.0$ 430,000

F 88.9 98.6103.8109.1101.20 99.4 994,000

M140.3152.7100.0140.9140.60138.21,382,000

A 56.9 51.4 65.0 69.7 60.95 59.9 599,000

M108.3108.0113.6 97.8108.15106.31,063,000

J102.8 89.6100.0 92.2 96.10 94.4 944,000

J 95.6 76.4 84.4 92.7 88.55 87.0 870,000

A140.6133.3115.6112.0124.45122.31,223,000

S191.4184.9188.2183.7186.55183.31,833,000

O149.3146.6153.3143.2147.95145.41,454,000

N 70.8 65.8 85.5 81.8 76.30 75.0 750,000

D 46.6 38.4 50.6 46.6 46.60 45.8 458,000

1,221.20 1,200.0

Seasonal index = (1 ,200/1,221.20)  Median

92 Letting XA = number of model A's to be made

XB = number of model B's to be made

XC = number of model C's to be made

XD = number of model D's to be made

Maximize P = 2.00XA + 1.50XB + 2.00XC + 2.50XD

Subject to 5XA + 6XB + 6XC + 7XD 100(plastic)

10XA + 12XB + 15XC + 15XD 500(beads)

4XA + 5XB + 5XC + 0XD 600(nylon)

0XA + 2XB + 3XC + 4XD 200(teflon)

.5XA + .4XB + .5XC + .8XD 300(labor)

XA XBXC XD0 (mixture)

XB 50(quantity B)

XD 20 (quantity D)

where XA, XB, XC, XD 0

96(a) LettingXA = dollar investment in bond A

XB = dollar investment in bond B

XC = dollar investment in bond C

XD = dollar investment in bond D

XE = dollar investment in bond E

Maximize P = .08XA + .09XB + .09XC + .l0XD + .09XE

(b)(1) XA + XB + XC + XD + XE = 100,000

(2) XB + XD + XE 50,000

(3)XA + XC + XE 30,000

(4)XA + XE 25,000

(5) XA + XB  XC + XD 0 or XB + XD XA + XC

(6) .06XA + .0225XB + .0225XC + .025XD .0675XE 0

or .08XA + .09XE ¼(.08XA + .09XB + .09XC + .10XD + .09XE)

97Letting XH = pounds of hog bellies

XP = pounds of pork

XT = pounds of tripe

XC = pounds of chicken

XB = pounds of beef

Minimize C = .30XH + .20XT + .70XB + .60XP + .45XC

Subject to XH + XT .10(restriction)

XC .25(chicken)

XB .30 (beef)

3XH + 5XT + 4XB + 3XP + 3XC 3(protein)

5XH + 3XT + 2XB + 4XP + 3XC 4(fat)

6XH + 4XT + 5XB + 9XP + 4XC 8(water)

XH + XT + XB + XP + XC= 1 (total weight)

where all Xs 0

915LettingXij = fraction of time that type i persons are assigned to job j (if fractional assignment is not possible, all Xij must be {0,1} binary variables)

i = C, T, or S

j = F, B, or R

MinimizeC = 20XCF + 25XCB + 35XCR + 25XTF + 20XTB + 30XTR + 30XSF + 25XSB + 25XSR

Subject to

(clerk assmnt.) XCF + XCB + XCR= 1

(typist assmnt.) XTF + XTB + XTR= 1

(steno. assmnt.) XSF + XSB + XSR= 1

(fil. assmnt.) XCF + XTF + XSF = 1

(book. assmnt.) XCB + XTB + XSB = 1

(rep. assmnt.) XCR + XTR + XSR = 1

where all Xs 0

918 Letting Xij = quantity shipped from plant i to warehouse j

i = A or B

j = C, D, or E

Minimize C = 11XAC + 12XAD + 13XAE + 13XBC + 12XBD + 13XBE

Subject to XAC + XAD + XAE = 1,000(A capacity)

XBC + XBD + XBE= 500(B capacity)

XAC + XBC= 500(C demand)

XAD + XBD= 500(D demand)

XAE + XBE= 500(E demand)

where all Xs 0

10-17

Adjustable Cells
Final / Reduced / Objective / Allowable / Allowable
Cell / Name / Value / Cost / Coefficient / Increase / Decrease
$C$9 / Xt / 9.142857143 / 0 / 20 / 30 / 5
$D$9 / Xc / 2.857142857 / 0 / 15 / 5 / 9
$E$9 / Xb / 0 / -5 / 15 / 5 / 1E+30
Constraints
Final / Shadow / Constraint / Allowable / Allowable
Cell / Name / Value / Price / R.H. Side / Increase / Decrease
$I$5 / wood / 100 / 0.714285714 / 100 / 20 / 64
$I$6 / labor / 60 / 2.571428571 / 60 / 106.6666667 / 10

(a)$ 5/7 = $ .714 for wood

$18/7 = $2.571 for labor

(b)You may ignore this question.

(c) The value of the resources used in making one desk is

30(5/7) + 10(18/7) = 330/7 = $47.14

Since this is smaller than the unit profit of $50, desks should be made.

10-36(a) The problem is formulated in the text. The solution is:

X1 = 300X6 = 0

X2 = 300X7 = 500

X3 = 300X8 = 300

X4 = X9 = 200

X5 = 0X10 = 0

P = 4,810

(b)stated in the output report (1) (800, 1,066.67)(2) (8, 100, )(3) (1,900, )(4) (887, )

(c)stated in the output report (1) (1.19, 2.01)(2) (2.49, 3.16)(3) (.94, 1.76)(4) (, 3.65)

stated in the output report (5) (, 3.45)(6) (, 3.29)(7) 2.40, 2.91)(8) (2.60, 3.16)

(9) (2.70, 3.71)(10) (, 3.29)

(d)UM = 3.85UGA = .183

UST = 0UP1 = .452

UMO = 0UP2 = .295

UC = .002UOL = .405

USA = 0

U1 = 0U6 = .44

U2 = 0U7 = 0

U3 = 0U8 = 0THIS BLUE PART IS NOT COVERED IN THE LECTURE, JUST SKIP!

U4 = 2.15U9 = 0

U5 = 1.40U10 = 1.34

(e) Um=3.85 is greatest so, the greatest increase in total profit would occur if the available gallons of milk could be increased.

10-31Primal

Letting Xi = number of ads to be placed in publication i

Maximize P =

1.5X1 + 1.4X2 + .7X3 + .8X4 + 1.1X5 + 1.5X6 + .3X7 + .9X8 + 2.1X9 + 1.9X10 + 1.4X11 + 1.6X12 + 2.0X13 + 1.6X14 + .9X15

Subject to

(budget) (UB)2.0X1 + 2.5X2 + 1.5X3 + 1.0X4 + 2.0X5 + 2.0X6 + .8X7 + 1.2X8 + 3.0X9 + 1.8X10 + 1.5X11 + 1.5X12 + 3.0X13 + 2.0X14 + .8X15 100,000

(age mix) (UM1) 1X1 + 1X2 1X3 1X4 + 1X5 + 0X6 1X7 + 0X8  1X9 1X10 + 1X11 + 1X12 + 1X13 + 0X14  1X15 0

(gen. mix) (UM2).5X1 .5X2 .5X3 .5X4 .5X5 + 1X6 .5X7 + 1X8  .5X9 .5X10 .5X11  .5X12 .5X13 + 1X14  .5X15 0

(teen lim.) (UT)2.0X1 + 2.5X2 + 0X3 + 0X4 + 2.0X5 + 0X6 + 0X7 + 0X8 + 0X9 + 0X10 + 1.5X11+ 1.5X12 + 3.0X13 + 0X14 + 0X15 40,000

(pub. 1) (UM1) 1X1 10

(pub. 2) (UM2) 1X2 2

(pub. 3) (UM3) 1X3 10

(pub. 4) (UM4) 1X4 3

(pub. 5) (UM5) 1X5 2

(pub. 6) (UM6) 1X6 5

(pub. 7) (UM7) 1X7 4

(pub. 8) (UM8) 1X8 6

(pub. 9) (UM9) 1X9 7

(pub. 10) (UM10) 1X10 5

(pub. 11) (UM11) 1X11 5

(pub. 12) (UM12) 1X12 4

(pub. 13) (UM13) 1X13 2

(pub. 14) (UM14)1X14 2

(pub. 15) (UM15) 1X15 3

where all X’s 0

Dual

LettingUB= marginal value per dollar of budget

UM1= marginal value for teenyouth mix

UM2= marginal value for general mix

UT= marginal value per dollar of teen limit

Ui = marginal value per ad limit for publication i

(a) dual formulation was not covered in the lecture, just skip.

(b)as stated in the solution

UB = 0U1 = 1.5U5 = 1.1U9 = 2.1U13 = 2

UM1= 0U2 = 1.4U6 = 1.5U10 = 1.9U14 =1.6

UM2 = 0U3 = .7U7 = .3U11 = 1.4U15 = .9

UT = 0U4 = .8U8 = .9U12 = 1.6

C=91

(c)$0

(d)$0

11-9(a) 5 ⅓ regulars, 6 ⅔ deluxes, P = 153 ⅓

(b) 0 regulars, 10 deluxes, R = 600

11-10Let XR, XD denote the number of regular and deluxe models, respectively. The subscripts R, B, P, and M relate to the (1) revenue, (2) budget, (3) profit, and (4) mixture goals.

(a)Minimize C = 0XR + 0XD + 0YR+ + 1YR + 1YB+ + 0YB + 0YP+ + 1YP + 0YM+ + 2YM

Subject to:

(labor): 5XR + 8XD 80

(frames): 1XR + 1XD 12

GI (revenue): 30XR + 60XD  (YR+ YR) = 500

G2 (budget):20XR + 45XD  (YB+ YB) = 400

G3 (profit):10XR + 15XD (YP+ YP) = 140

G4 (mixture: 1XR 1XD  (YM+ YM) = 0

where all variables 0

(b) when XR = 4 and XD = 7,

YR+ YR = 30(4) + 60(7)  500 = 40, YR+ = 40 and YR = 0

YB+ YB = 20(4) + 45(7)  400 =  5, YB+ = 0 and YB = 5

YP+ YP = 10(4) + 15(7)  140 = + 5, YP+ = 5 and YP = 0

YM+ YM =1(4)  1(7)  0 = 3, YM+ = 0 and YM = 3

C = 0(4) + 0(7) + 0(40) + 1(0) + 1(0) + 0(5) + 0(5) + 1(0) + 0(0) + 2(3) = 6