solution to topic 2 lp

Q7.14

Solution:

Let X1 = number of air conditioner to be produced

X2 = number of large fans produced

Maximize profit Z = $25X1 + $15X2

Subject to

3 X1+2 X2≤ 240 (available wiring)

2 X1+ 1X2≤ 140 (available drilling)

X1, X2 ≥ 0 (non-negativity)

Graphic solution

3 X1+2 X2 = 240, X1 = 0, X2=120

X2 = 0, X1 = 80

2 X1+ X2 = 140, X1 = 0, X2= 140

X2 = 0, X1 = 70

Isoprofit line method

Z = 1800, X1 = 0, X2 = 120

X2 = 0, X1 = 72

Compare corner points

Point a: X1=0, X2=0 Z = $25*0 + $15*0 = $0

Point b: X1 = 0, X2=120 Z = $25*0+ $15*25 = $1800

Point c: X1 =40, X2 = 60 Z = $25*40 + $15*60 = $1900

Point d: X1=70, X2=0 Z=$25*70+$15*0 =1750

Optimal solution:

X1=40, X2=60

Z = $25×40 + $15×60 = $1900

Q7.17

Solution:

Let B = number of benches to be produced

T = number of picnic tables to be produced

Max profit Z = $9B + $20T

Subject to

4B + 6T ≤ 1200 (available labor hours)

10B + 35T ≤ 3500 (available stock)

B, T ≥ 0 (non-negativity)

Graphic solution

4B + 6T = 1200, B = 0, T = 200

T = 0, B = 300

10B + 35T = 3500, B = 0, T = 100

T = 0, B = 350

Isoprofit line method

Z = 1800, B = 0, T = 90

T = 0, B = 200

Compare corner points

Point a: B = 0, T = 100 Z = $9*0 + $20*100 = $2000

Point b: B = 262.5, T = 25 Z = $9*262.5 + $20*25 =$2862.5

Point c: B =3000, T = 0 Z = $9*300 + $20*0 = $2700

Optimal solution:

B = 262.5, T = 25

Z = $9×262.5 + $20×25 = $2862.5

Q8.2

Solution: (a)

Let L be dollars invested in Los Angeles municipal bonds

T be dollars invested in Thompson Electronics

U be dollars invested in United Aerospace

P be dollars invested in Palmer Drugs

H be dollars invested in Happy Days Nursing Homes

Maximize return Z = 0.053L + 0.068T + 0.049U + 0.084P + 0.118H

Subject to

1. Municipal bonds

L ≥ 0.2(L+T+U+P+H)

0.8L−0.2T−0.2U−0.2P−0.2H ≥ 0

2. Combination of electronics, aerospace, and drugs

T+U+P ≥ 0.4(L+T+U+P+H)

−0.4L+0.6T+0.6U+0.6P −0.4H ≥ 0

3. Nursing home as percent of bonds

H ≤ 0 .5L

H−0.5L 0

4. Fund to be invested

L+T+U+P+H ≤ 250,000

5. Non-negativity

L, T, U, P, H ≥ 0

(b)

Solved by Excel:

Heinlein and Krampf Brokerage

L T U P H

Los Angeles Thompson United Palmer Happy Days

$ invested $50,000.00 $0.00 $0.00 $175,000.00 $25,000.00

Return 5.30% 6.80% 4.90% 8.40% 11.80% $20,300.00 <- Objective

Constraints 　　　Total 1 1 1 1 1 250000.0 <= 250000

Nursing -0.5 0 0 0 1 0.0 <= 0

Electr, Aero, Drug -0.4 0.6 0.6 0.6 -0.4 75000.0 >= 0

Muni Bonds 0.8 -0.2 -0.2 -0.2 -0.2 0.0 >= 0

LHS Sign RHS

Total returns = $20,300

$50,000 invested in Los Angeles municipal bonds(L = 50,000)

No investment in Thompson Electronics and United Aerospace (T, U = 0)

$175,000 invested in Palmer Drugs (P =175,000)

$25,000 invested in Happy Days Nursing Homes (H = 25,000)

Q8.3

Solution:

(LP Formulation)

Let Xi be number of workers reporting for starts of work at period i

(for i = 1,2,3,4,5,6)

Minimize staff size Z = X1 +X2 +X3 +X4+X5 +X6

Subject to

1. Number of waiters and busboys required for each period

X6 + X1 ≥ 3 (3am – 7am)

X1 + X2 ≥ 12 (7am – 11am)

X2 + X3 ≥ 16 (11am – 3pm)

X3 + X4 ≥ 9 (3pm – 7pm)

X4 + X5 ≥ 11 (7pm – 11pm)

X5 + X6 ≥ 4 (11pm – 3am)

2. Non-negativity

Xi ≥ 0 (for i = 1,2,3,4,5,6)

Q8.4

7am

11am

3am11pm

7 pm

3pm

3

12

16

9

11

4

Solution:

(LP Formulation)

Let A be the number of pounds of oat product per horse each day

G be the number of pounds of enriched grain per horse each day

M be the number of pounds of mineral product per horse each day

Minimize cost Z = 0.09A +0.014G + 0.17M

Subject to

1. Min daily requirement on ingredients

2A + 3G + 1M ≥ 6 (ingredient A)

0.5A+ 1G + 0.5M ≥ 2 (ingredient B)

3A + 5G + 6M ≥ 9 (ingredient C)

1A + 1.5G + 2M ≥ 8 (ingredient D)

0.5A + 0.5G + 1.5M ≥ 5 (ingredient E)

2. Maximum feed per day

A + G + M 6

3. Non-negativity

A, G, M ≥ 0

Solved by computer:

Battery Park Stable

A G M

Oat Grain Mineral

Number of Pounds 1.333 0.000 3.333

Cost $0.09 $0.14 $0.17 0.6867 <- Objective

Constraints: 　　　Ingredient A 2 3 1 6.0 >= 6

Ingredient B 0.5 1 0.5 2.3 >= 2

Ingredient C 3 5 6 24.0 >= 9

Ingredient D 1 1.5 2 8.0 >= 8

Ingredient E 0.5 0.5 1.5 5.7 >= 5

Max feed 1 1 1 4.7 <= 6

LHS Sign RHS

Total Cost = $0.6867

The optimal daily mix consists of 1.3333 pounds of oat product, no enriched grain and

3.3333 pounds of mineral product per house each day.

Q8.14

Solution:

(LP Formulation)

(a)

Let Xij be the acres of crop i planted on parcel j

(for i = W: wheat, A: alfalfa, B: barley)

(for j =1 : SE, 2: N, 3: NW, 4: W, 5: SW)

Maximize profit Z= $2(50) (XW1 + XW2 + XW3 + XW4 + XW5)

+ $40(1.5) (XA1 + XA2 + XA3 + XA4 + XA5)

+ $50(2.2) (XB1 + XB2 + XB3 + XB4 + XB5)

Subject to:

1. Irrigation limits (in term of acre-feet):

1.6XW1 + 2.9XA1 + 3.5XB1 3,200 (acre-feet in SE)

1.6XW2 + 2.9XA2 + 3.5XB2 3,400 (acre-feet in N)

1.6XW3 + 2.9XA3 + 3.5XB3 800 (acre-feet in NW)

1.6XW4 + 2.9XA4 + 3.5XB4 500 (acre-feet in W)

1.6XW5 + 2.9XA5 + 3.5XB5 600 (acre-feet in SW)

1.6(XW1 + XW2 + XW3 + XW4 + XW5)

+ 2.9(XA1 + XA2 + XA3 + XA4 + XA5)

+3.5(XB1 + XB2 + XB3 + XB4 + XB5) 7,400 (Total water acre-feet)

2. Sales limited (in term of acres):

XW1 + XW2 + XW3 + XW4 + XW5 2,200 (110,000 bushels/ 50 bushels per acre)

XA1 + XA2 + XA3 + XA4 + XA5 1,200 (1,800 tons/1.5 tons per acre)

XB1 + XB2 + XB3 + XB4 + XB5 1,000 (2,200 tons/2.2 tons per acre)

3. Area availability (in term of acres):

XW1 + XA1 + XB1 2,000 (acre in SE)

XW2 + XA2 + XB2 2,300 (acre in N)

XW3 + XA3 + XB3 600 (acre in NW)

XW4 + XA4 + XB4 1,100 (acre in W)

XW5 + XA5 + XB5 500 (acre in SW)

4. Non-negativity

Xij ≥ 0 for i = W, A, B and j = 1, 2, 3, 4, 5

Profit from Wheat

Profit from Alfalfa

Profit from Barley

Margaret Black’s Farmland

X11 X12 X13 X14 X15 X21 X22 X23 X24 X25 X31 X32 X33 X34 X35

　Wheat

SE

Wheat

N

Wheat

NW

Wheat

W

Wheat

SW

Alfalfa

SE

Alfalfa

N

Alfalfa

NW

Alfalfa

W

Alfalfa

SW

Barley

SE

Barley

N

Barley

NW

Barley

W

Barley

SW

Number of acres 1762.5 437.5 0.0 0.0 0.0 131.0 0.0 0.0 0.0 0.0 0.0 771.4 228.6 0.0 0.0

Profit $100 $100 $100 $100 $100 $60 $60 $60 $60 $60 $110 $110 $110 $110 $110 $337,862.07 <-- Objective

Constraints: 　　　SE acreage 1 1 1 1893.5 <= 2000

N acreage 1 1 1 1208.9 <= 2300

NW acreage 1 1 1 228.6 <= 600

W acreage 1 1 1 0.0 <= 1100

SW acreage 1 1 1 0.0 <= 500

Wheat acreage 1 1 1 1 1 2200.0 <= 2200

Alfalfa acreage 1 1 1 1 1 131.0 <= 1200

Barley acreage 1 1 1 1 1 1000.0 <= 1000

SE water 1.6 2.9 3.5 3200.0 <= 3200

N water 1.6 2.9 3.5 3400.0 <= 3400

NW water 1.6 2.9 3.5 800.0 <= 800

W water 1.6 2.9 3.5 0.0 <= 500

SW water 1.6 2.9 3.5 0.0 <= 600

Total water 1.6 1.6 1.6 1.6 1.6 2.9 2.9 2.9 2.9 2.9 3.5 3.5 3.5 3.5 3.5 7400.0 <= 7400

LHS Sign RHS

(b)

Total Profit: $337,862.07

Plant 1,762.5 acres of wheat in SE parcel (XW1 = 1,762.5)

Plant 437.5 acres of wheat in N parcel (XW2 = 437.5)

Plant 131 acres of alfalfa in SE parcel (XA1 = 131)

Plant 771.4 acres of barley in N parcel (XB2 = 771.4)

Plant 228.6 acres of barley in NW parcel (XB3 = 228.6)

Multiple optimal solution exist

(c)

Yes, need only 500 more water-feet.

Q8.12

Solution:

(a)

Let I be the number of units of internal modems produced per week

E be the number of units of external modems produced per week

C be the number of units of circuit boards produced per week

F be the number of units of floppy disk drives modems produced per week

H be he number of units of hard drives produced per week

M be the number of units of memory boards produced per week

Maximize profit = Revenue – material cost – test cost

Max Z = 200I + 120E + 180C + 130F + 430H + 260M

− 35I – 25E – 40C – 45F – 170H – 60M

− (60

15) (7I + 3E + 12C + 6F + 18H + 17M)

− (60

12) (2I + 5E + 3C + 2F + 15H + 17M)

− (60

18) (5I + 1E + 3C + 2F + 9H + 2M)

( i.e. 161.35I + 92.95E + 135.5C + 82.5F +249.8H + 191.75M )

Subject to

Revenue

Material cost

Test cost(in terms of minutes)

1. Time limit (in terms of minutes)

7I + 3E + 12C + 6F + 18H + 17M 7,200 (= 120*60mins) (Test device 1)



2. Non-negativity

I, E, C, F, H, M ≥ 0

(b)

Quitmeyer Electronics

I E C F H M

　Internal

modems

External

modems

Circuit

boards

Floppy

drives

Hard

drives

Memory

boards

Solution value 496.55 1241.38 0.00 0.00 0.00 0.00

Selling price per unit $200 $120 $180 $130 $430 $260 $248,275.86 <-- Revenue

Material cost per unit $35 $25 $40 $45 $170 $60 $48,413.79 <-- Matl Cost

Labor cost per unit $3.65 $2.05 $4.50 $2.50 $10.20 $8.25 $4,357.24

<-- Labor

Cost

Profit $161.35 $92.95 $135.50 $82.50 $249.80 $191.75 $195,504.83 <-- Objective

Constraints 　　　 Cost

Test device 1 7 3 12 6 18 17 7200.00 <= 7200 $15

Tets device 2 2 5 3 2 15 17 7200.00 <= 7200 $12

Test device 3 5 1 3 2 9 2 3724.14 <= 6000 $18

LHS Sign RHS

Total Profit = $195,504.83

Produce 496.55 internal modems (I = 496.55)

Produce 1,241.38 external modems (E = 1,241.38)

Do not produce any circuit boards, floppy drives, hard drives and memory boards (C,F,H,M=0)

(c)

Microsoft Excel 10.0 Answer Report

Worksheet: [P3-10.xls]P3-10

Target Cell (Max)

Cell Name Original Value Final Value

$H$9 Profit $0.00 $195,504.83

Adjustable Cells


$B$5 Solution value Internal modems 0.00 496.55

$C$5 Solution value External modems 0.00 1241.38

$D$5 Solution value Circuit boards 0.00 0.00

$E$5 Solution value Floppy drives 0.00 0.00

$F$5 Solution value Hard drives 0.00 0.00

$G$5 Solution value Memory boards 0.00 0.00

Constraints

Cell Name Cell Value Formula Status Slack

$H$11 Test device 1 7200.00 $H$11<=$J$11 Binding 0

$H$12 Tets device 2 7200.00 $H$12<=$J$12 Binding 0

$H$13 Test device 3 3724.14 $H$13<=$J$13 Not Binding 2275.862069

Microsoft Excel 10.0 Sensitivity Report


Report Created: 29/1/2004 11:31:03

Adjustable Cells

　　 Final Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease

$B$5 Solution value Internal modems 496.55 0.00 161.35 55.53333333 69.40833333

$C$5 Solution value External modems 1241.38 0.00 92.95 310.425 23.8

$D$5 Solution value Circuit boards 0.00 -138.64 135.5 138.637931 1E+30

$E$5 Solution value Floppy drives 0.00 -57.44 82.5 57.44137931 1E+30

$F$5 Solution value Hard drives 0.00 -221.73 249.8 221.7275862 1E+30

$G$5 Solution value Memory boards 0.00 -269.86 191.75 269.8586207 1E+30

Constraints

　　 Final Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease

$H$11 Test device 1 7200.00 21.41 7200 2869.565217 2880

$H$12 Tets device 2 7200.00 5.74 7200 4800 5142.857143

$H$13 Test device 3 3724.14 0.00 6000 1E+30 2275.862069

The value of an additional minute in test device 1 is $ 21.41, test device 2 is 5.74 and

test device 3 is 0. Therefore, Quitmeyer Electronics should no more than 2869.56 minutes

to test device 1, 4800 minutes to test device 2 only.

Q4.13

Solutions:

LP model

Let T be number of 1-minute TV spots selected each week.

N be number of full-page daily newspaper ads selected each week.

P be number of 30-second prime-time radio spots selected each week.

A be number of 1-minute afternoon radio spots selected each week.

Maximize audience coverage Z = 5000 T + 8500 N + 2400 P +2800 A

subject to:

T 12 (maximum TV spots/week)

N 5 (maximum newspaper ads/week)

P 24 (maximum 30-second radio

spots/week)

A 20 (maximum 1-minute radio spots/week)

800T + 925N + 290P + 380A 8000 (budget)

P + A 5 (min radio spots contracted)

290P + 380A 1800 (max dollars spent on radio)

T, N, P, A 0 (non-negativity)

Use the Sensitivity Report for this LP model to answer the following questions. Each

question is independent of the others.

Win Big Gambling Club

T N P A

TV spotsNewspape

r

Part-time

radio

Afternoon

radio Number of Units 1.97 5.00 6.21 0.00 Audience 5000 8500 2400 2800 67240.3<-objectiveConstraints Max TV 1 1.97 <= 12Max Newspaper 1 5.00 <= 5Max Prime-Time Radio 1 6.21 <= 25Max Afternoon Radio 1 0.00 <= 20Budget $800.00 $925.00 $290.00 $380.00 8000.00 <= $8,000Max Radio $ $290.00 $380.00 1800.00 <= $1,800Min Radio Spots 1 1 6.21 >= 5 LHS Sign RHS

Adjustable Cells

Final Reduced Objective Allowable Allowable


$B$5 Number of Units TV spots 1.97 0.00 5000.00 1620.69 5000.00

$C$5 Number of Units Newspaper ads 5.00 0.00 8500.00 1E+30 2718.75

$D$5 Number of Units Prime-time radio spots 6.21 0.00 2400.00 1E+30 263.16

$E$5 Number of Units Afternoon radio spots 0.00 -344.83 2800.00 344.83 1E+30

Constraints

Final Shadow Constraint Allowable Allowable


$F$1

4 Min Radio Spots 6.21 0.00 5.00 1.21 1E+30

$F$8 Max TV 1.97 0.00 12.00 1E+30 10.03

$F$9 Max Newspaper 5.00 2718.75 5.00 1.70 5.00

$F$1

0 Max Prime-Time Radio 6.21 0.00 25.00 1E+30 18.79

$F$11 Max Afternoon Radio 0.00 0.00 20.00 1E+30 20.00

$F$1

2 Budget 8,000.00 6.25 8000.00 8025.00 1575.00

$F$1

3 Max Radio $ 1,800.00 2.03 1800.00 1575.00 350.00


Worksheet: [LP_2.xls]media

Adjustable Cells

　　 Final Reduced Objective Allowable Allowable


$B$5 Number of Units TV spots 1.97 0.00 5000 1620.689655 5000

$C$5 Number of Units Newspaper ads 5.00 0.00 8500 1E+30 2718.75

$D$5 Number of Units Prime-time radio spots 6.21 0.00 2400 1E+30 263.1578947

$E$5 Number of Units Afternoon radio spots 0.00 -344.83 2800 344.8275862 1E+30

Constraints

　　 Final Shadow Constraint Allowable Allowable


$F$14 Min Radio Spots 6.21 0.00 5 1.206896552 1E+30

$F$8 Max TV 1.97 0.00 12 1E+30 10.03125

$F$9 Max Newspaper 5.00 2718.75 5 1.702702703 5

$F$10 Max Prime-Time Radio 6.21 0.00 25 1E+30 18.79310345

$F$11 Max Afternoon Radio 0.00 0.00 20 1E+30 20

$F$12 Budget $8,000.00 $6.25 8000 8025 1575

$F$13 Max Radio $ $1,800.00 $2.03 1800 1575 350

Microsoft Excel 10.0 Answer ReportWorksheet: [LP_2.xls]media

Target Cell (Max)Cell Name Original Value Final Value

$F$6 Audience 0.00 67240.30

Adjustable CellsCell Name Original Value Final Value

$B$5 Number of Units TV spots 0.00 1.97$C$5 Number of Units Newspaper ads 0.00 5.00$D$5 Number of Units Prime-time radio spots 0.00 6.21$E$5 Number of Units Afternoon radio spots 0.00 0.00

ConstraintsCell Name Cell Value Formula Status Slack

$F$14 Min Radio Spots 6.21 $F$14>=$H$14 Not Binding 1.21$F$8 Max TV 1.97 $F$8<=$H$8 Not Binding 10.03125$F$9 Max Newspaper 5.00 $F$9<=$H$9 Binding 0$F$10 Max Prime-Time Radio 6.21 $F$10<=$H$10 Not Binding 18.79310345$F$11 Max Afternoon Radio 0.00 $F$11<=$H$11 Not Binding 20$F$12 Budget $8,000.00 $F$12<=$H$12 Binding 0$F$13 Max Radio $ $1,800.00 $F$13<=$H$13 Binding 0

(a)

Spending $200 more on radio advertising is within the allowable increase (1575). The

shadow price would remain unchanged. The audience coverage would increase the

by(=2.03*200) to 67646.3

(b)

No. Since we are already placing 6.21 radio spots, this contractual agreement is not a

binding constraint.

(c)

The optimal solution will not change since 3100 audience reached per ad (increase of

300 of the objective function coefficient value) is within the allowable increase (344.83)

The audience reached for each afternoon radio spot would have to increase to at least

3144.83(=2800+344.83) in order for these spot to become attractive.

(refer to reduce cost definition)

(d)

Currently 5000 audience reached per TV spot. (the objective function

coefficient=5000).

Optimal number of TV spots would remain unchanged if the number of audience

reach per TV spot is between 0 and 6620.69. Obviously, if the objective function

coefficient is 0, it is not worthwhile to use any TV spots.

Q4.16

Solutions

Battery Park Stable

A G M

Oat Grain Mineral

Number of Pounds 1.333 0.000 3.333

Cost $0.09 $0.14 $0.17 $0.69 <- Objective

Constraints:

Ingredient A 2 3 1 6.0 >= 6

Ingredient B 0.5 1 0.5 2.3 >= 2

Ingredient C 3 5 6 24.0 >= 9

Ingredient D 1 1.5 2 8.0 >= 8

Ingredient E 0.5 0.5 1.5 5.7 >= 5

Max feed 1 1 1 4.7 <= 6

LHS Sign RHS

Microsoft Excel 9.0 Answer Report


Target Cell (Min)


$E$6 Cost $0.00 $0.69

Adjustable Cells


$B$5 Number of Pounds Oat 0.000 1.333

$C$5 Number of Pounds Grain 0.000 0.000

$D$5 Number of Pounds Mineral 0.000 3.333

Constraints

Cell Name Cell Value Formula Status Slack

$E$13 Max feed 4.67$E$13<=$G$13 Not Binding 1.333

$E$8 Ingredient A 6.00$E$8>=$G$8 Binding 0.000

$E$9 Ingredient B 2.33$E$9>=$G$9 Not Binding 0.333

$E$10 Ingredient C 24.00$E$10>=$G$10 Not Binding 15.000

$E$11 Ingredient D 8.00$E$11>=$G$11 Binding 0.000

$E$12 Ingredient E 5.67$E$12>=$G$12 Not Binding 0.667



Adjustable Cells

Final Reduced Objective Allowable Allowable


$B$5 Number of Pounds Oat 1.33 0.000 0.09 0.003 0.005

$C$5 Number of Pounds Grain 0.00 0.005 0.14 1E+30 0.005

$D$5 Number of Pounds Mineral 3.33 0.000 0.17 0.010 0.125

Constraints

Final Shadow Constraint Allowable Allowable


$E$13 Max feed 4.67 0.000 6.00 1E+30 1.333

$E$8 Ingredient A 6.00 0.003 6.00 4.000 2.000

$E$9 Ingredient B 2.33 0.000 2.00 0.333 1E+30

$E$10 Ingredient C 24.00 0.000 9.00 15.000 1E+30

$E$11 Ingredient D 8.00 0.083 8.00 4.000 0.800

$E$12 Ingredient E 5.67 0.000 5.00 0.667 1E+30

(a)

The decrease of $0.01 per pound is outside the allowable decrease (0.005). Therefore

the optimal solution will change. We can resolve the LP model to determine the new cost.

(b)The constraints prescribing the minimum daily requirement for ingredient A and

ingredient D are binding (i.e 2A +3G +1M ≥ 6 & 1A +0.5G +2M ≥ 8 )

For each additional unit of ingredient A required in the mix, the cost will increase by

$0.0033.

For each additional unit of ingredient D required in the mix, the cost will increase by

$0.0833.

(c)

A 20% decrease in the cost of mineral implies the cost will decrease by $0.034

(=0.17*0.2), which is less the allowable decrease (0.125). The optimal solutions remain

unchanged. The revised cost will be decrease by 0.1133 (=0.034*3.3333) to $0.5734

(=0.6867-0.1133)

(d)

The price of oats can fluctuate between $0.085 and $0.0933 per pound for the current

solution to remain optimal.

Q4.18

Solutions:

Quitmeyer Electronics

I E C F H M

Internal

modems

External

modems

Circuit

boards

Floppy

drives

Hard

drives

Memory

boards

Solution value 496.55 1241.38 0.00 0.00 0.00 0.00

Selling price per unit $200 $120 $180 $130 $430 $260 $248,275.86 <-- Revenue

Material cost per unit $35 $25 $40 $45 $170 $60 $48,413.79 <-- Matl Cost

Labor cost per unit $3.65 $2.05 $4.50 $2.50 $10.20 $8.25 $4,357.24 <-- Labor Cost

Profit $161.35 $92.95 $135.50 $82.50 $249.80 $191.75 $195,504.83 <-- Objective

Constraints Cost

Test device 1 0.117 0.050 0.200 0.100 0.300 0.283 120.00 <= 120 $15

Tets device 2 0.033 0.083 0.050 0.033 0.250 0.283 120.00 <= 120 $12

Test device 3 0.083 0.017 0.050 0.033 0.150 0.033 62.07 <= 100 $18

LHS Sign RHS

Adjustable Cells

FinalReduce

d Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$5 Solution value Internal modems 496.55 0.00 161.35 55.53333333 69.40833333$C$5 Solution value External modems 1241.38 0.00 92.95 310.425 23.8$D$5 Solution value Circuit boards 0.00 -138.64 135.5 138.637931 1E+30$E$5 Solution value Floppy drives 0.00 -57.44 82.5 57.44137931 1E+30$F$5 Solution value Hard drives 0.00 -221.73 249.8 221.7275862 1E+30

$G$5 Solution value Memory boards 0.00 -269.86 191.75 269.8586207 1E+30

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$H$1

1 Test device 1 120.00 1284.52 120 47.82608696 48$H$1

2 Test device 2 120.00 344.69 120 80 85.71428571$H$1

3 Test device 3 62.07 0.00 100 1E+30 37.93103448

(a)Four products (Circuit boards, Floppy drives, Hard drives, Memory boards)

The reduced costs indicate the minimum amount by which the profit contributions of

these products must increase before they would be included in the production mix.

For example, the profit contribution of a circuit board should increase by at least

$138.64 (from the current $ 135.5 to at least $274.14) before it becomes a viable product.

(b)The current production plan required only 3724.14 minutes of the available 6000

minutes on test device 3. Also, taking 2100 minutes (=35 hours*60) on test device 3

(decrease of 2100) is within the allowable decrease (2275.86). The optimal solution

would not be affected.

(c)Since the additional of 1200 minutes (= 20 hours*60) on test device 1 is within the

allowable increase (2275.86), the shadow price remains unchanged. (i.e. for each

additional minutes of time on test device 1 increase, the profit would increase by $21.41)

Refer to Q8.12, the time on test device 1 cost only $0.25 per minutes (i.e. $15 per

hour). The 20 additional hours at a cost of $25 per hour will therefore increase the profit

by $25692 (= $21.41*60*20) less than the cost of $500(= *60*20). The new profit

will be $220,696.83 ($195,504.83 + $21.41 – $500).

Q7.14

The Electrocomp Corporation manufactures two electrical products: air conditioners and

large fans. The assembly process for each is similar in that both require a certain amount

of wiring and drilling. Each air conditioner takes 3 hours of wiring and 2 hours of

drilling. Each fan must go through 2 hours of wiring and 1 hour of drilling. During the

next production period, 240 hours of wiring time are available and up to 140 hours of

drilling time may be used. Each air conditioner sold yields a profit of $25. Each fan

assembled may be sold for a $15 profit. Formulate and solve this LP production mix

situation to find the best combination of air conditioners and fans that yields the highest

profit. Use the corner point graphical approach.

Q7.17

The Outdoor Furniture Corporation manufactures two products, benches and picnic

tables, for use in yards and parks. The firm has two main resources: its carpenters (labor

force) and a supply of redwood for use in the furniture. During the next production cycle,

1,200 hours of labor are available under a union agreement. The firm also has a stock of

3,500 feet of good-quality redwood. Each bench that Outdoor Furniture produces

requires 4 labor hours and 10 feet of redwood; each picnic table takes 6 labor hours and

35 feet of redwood. Completed benches will yield a profit of $9 each, and tables will

result in a profit of $20 each. How many benches and tables should Outdoor Furniture

produce to obtain the largest possible profit? Use the graphical LP approach.

Q8.2

(Investment decision problem) The Heinlein and Krampf Brokerage firm has just been

instructed by one of its clients to invest $250,000 for her money obtained recently

through the sale of land holdings in Ohio. The client has a good deal of trust in the

investment house, but she also has her own ideas about the distribution of the funds being

invested. In particular, she requests that the firm selects whatever stocks and bonds they

believe are well rated, but within the following guidelines:

1. Municipal bonds should constitute at least 20% of the investment.

2. At least 40% of the funds should be placed in a combination of electronics

firms, aerospace firms, and drug manufacturers.

3. No more than 50% of the amount invested in municipal bonds should be

placed in a high-risk, high-yield nursing home stock.

Subject to these restraints, the client’s goal is to maximize projected return on

investments. The analysts at Heinlein and Krampf, aware of these guidelines, prepare a

list of high-quality stocks and bonds and their corresponding rates of return.

Investment Projected Rate of Return (%)

Los Angeles municipal bonds 5.3

Thompson Electronics, Inc. 6.8

United Aerospace Corp. 4.9

Palmer Drugs 8.4

Happy Days Nursing Homes 11.8

(a) Formulate this portfolio selection problem using LP.

(b) Solve this problem.

Q8.3

(Restaurant work scheduling problem) The famous Y. S. Chang Restaurant is open 24

hours a day. Waiters and busboys report for duty at 3am, 7am, 11am, 3pm, 7pm, or

11pm, and each works an 8-hour shift. The following table shows the minimum number

of workers needed during the six periods into which the day is divided. Chang’s

scheduling problem is to determine how many waiters and busboys should report for

work at the start of each time period to minimize the total staff required for one day’s

operation. (Hint: Let Xi equal the number of waiters and busboys beginning work in time

period i, where i = 1, 2, 3, 4, 5, 6)

Period Time Number of Waiters & Busboys Required

1 3am – 7am 3

2 7am – 11am 12

3 11am – 3pm 16

4 3pm – 7pm 9

5 7pm – 11pm 11

6 11pm – 3am 4

Time period 3am-7am 7am-11am 11am-3pm 3pm-7pm 7pm-11pm 11pm-3am

X1

X2

X3

X4

X5

X6

Min

requirement

3 12 16 9 11 4

Q8.4

(Animal feed mix problem) The Battery Park Stable feeds and houses the horses used to

pull tourist-filled carriages through the streets of Charleston’s historic waterfront area.

The stable owner, an ex-racehorse trainer, recognizes the need to set a nutritional diet for

the horses in his care. At the same time, he would like to keep the overall daily cost of

feed to a minimum.

The feed mixes available for the horses’ diet are an oat product, a highly enriched

grain, and a mineral product. Each of these mixes contains a certain amount of five

ingredients needed daily to keep the average horse healthy. The table shows these

minimum requirements, units of each ingredient per pound of feed mix, and costs for the

three mixes.

In addition, the stable owner is aware that an overfed horse is a sluggish worker.

Consequently, he determines that 6 pounds of feed per day is the most that any horse

needs to function properly. Formulate this problem and solve for the optimal daily mix of

the three feeds.

Diet Requirement

(Ingredients)

Feed Mix

Oat Product

(Units / lb)

Enriched Grain

(Units / lb)

Mineral Product

(Units / lb)

Minimum Daily

Requirement (Units)

A 2 3 1 6

B 0.5 1 0.5 2

C 3 5 6 9

D 1 1.5 2 8

E 0.5 0.5 1.5 5

Cost / lb $0.09 $0.14 $0.17

Q8.14

(Agricultural Production Planning) Margaret Black’s family owns five parcels of

farmland broken into a southeast sector, north sector, northwest sector, west sector, and

southwest sector. Margaret is involved primarily in growing wheat, alfalfa, and barley

crops and is currently preparing her production plan for next year. The Pennsylvania

Water Authority has just announced its yearly water allotment, with the Black farm

receiving 7,400 acre-feet. Each parcel can only tolerate a certain amount of irrigation per

growing season, as specified in the following table:

Parcel Area (Acres) Water Irrigation Limit (Acre-Feet)

Southeast 2,000 3,200

North 2,300 3,400

Northwest 600 800

West 1,100 500

Southwest 500 600

Each of Black’s crops needs a minimum amount of water per acre, and there is a

projected limit on sales of each crop. Crop data follow:

Crop Maximum Sales Water Needed per Acre (Acre-Feet)

Wheat 110,000 bushels 1.6

Alfalfa 1,800 tons 2.9

Barley 2,200 tons 3.5

Margaret’s best estimate is that she can sell wheat at a net profit of $2 per bushel,

alfalfa at $40 per ton, and barley at $50 per ton. One acre of land yields an average of 1.5

tons of alfalfa and 2.2 tons of barley. The wheat yield is approximately 50 bushels per

acre.

(a) Formulate Margaret’s production plan.

(b) What should the crop plan be, and what profit will it yield?

(c) The Water Authority informs Margaret that for a special fee of $6,000 this year,

her farm will qualify for an additional allotment of 600 acre-feet of water. How

should she respond?

Q8.12

(High tech production problem) Quitmeyer Electronics Incorporated manufactures the

following six micro-computer peripheral devices: internal modems, external modems,

graphics circuit boards, floppy disk drives, hard disk drives, and memory expansion

boards. Each of these technical products requires time, in minutes, on three types of

electronic testing equipment, as shown in the table.

Internal Modem

External Modem

Circuit Board

Floppy Drives

Hard Drives

Memory Boards

Test device 1

7 3 12 6 18 17

Test device 2

2 5 3 2 15 17

Test device 3

5 1 3 2 9 2

The first two test devices are available 120 hours per week. The third (device 3)

requires more preventive maintenance and may be used only 100 hours each week. The

market for all six computer components is vast, and Quitmeyer Electronics believes that

it can sell as many units of each product as it can manufacture. The table that follows

summarizes the revenues and material costs for each product:

Device Revenue per Unit Sold ($) Material Cost per Unit ($)Internal Modem 200 35External Modem 120 25

Graphics Circuit Board 180 40Floppy Disk Drives 130 45Hard Disk Drives 430 170

Memory Expansion Boards 260 60In addition, variable labor costs are $15 per hour for test device 1, $12 per hour for

test device 2, and $18 per hour for test device 3. Quitmeyer Electronics wants to

maximize its profits.

(a) Formulate this problem as an LP model.

(b) Solve the problem by computer. What is the best product mix?

(c) What is the value of an additional minute of time per week on test device 1? Test

device 2? Test device 3? Should Quitmeyer Electronics add more test device time?

If so, on which equipment?

*4.13

(Media Selection) The Win Big Gambling Club promotes gambling junkets from a large

Midwestern city to casinos in the Bahamas. The club has budgeted up to $8,000 per

week for local advertising. The money is to be allocated among four promotional media:

TV spots, newspaper ads, and two types of radio advertisements. Win Big’s goal is to

reach the largest possible high-potential audience through the various media. The

following table presents the number of potential gamblers reached by making use of an

advertisement in each of the four media. It also provides the cost per advertisement

placed and the maximum number of ads that can be purchased per week.

MediumAudience Reached

per Ad

Cost per

Ad ($)

Maximum Ads

per Week

TV spot (1 minute) 5,000 800 12

Daily newspaper (full-page ad) 8,500 925 5

Radio spot (30 seconds, prime time) 2,400 290 25

Radio spot (1 minute, afternoon) 2,800 380 20

Win Big’s contractual arrangements require that at least five radio spots be placed

each week. To ensure a broad-scoped promotional campaign, management also insists

that no more than $1,800 be spent on radio advertising every week.

(a) What would be the impact if management approves spending $200 more on radio

advertising each week?

(b) Would it help Win Big if it can get out of the contractual agreement to place at

least five radio spots each week?

(c) The radio station manager agrees to run the afternoon radio spots during some of

their more popular programs. He thinks this will increase the audience reached per ad to

3,100. Will this change the optimal solution? Why or why not?

(d) There is some uncertainty in the audience reached per TV spot. For what range of

values for this OFC will the current solution remain optimal?

*4.16

Consider Battery Park Stables’ animal feed problem first presented in Q8.4. Use Solver

to create the Answer and Sensitivity Reports for this LP problem. Now answer the

following questions using these reports. Each question is independent of the others.

(a) If the price of grain decreases by $0.01 per pound, will the optimal solution

change?

(b) Which constraints are binding? Interpret the shadow price for the binding

constraints.

(c) What would happen to the total cost if the price of mineral decreased by 20% from

its current value?

(d) For what price range of oats is the current solution optimal?

*4.18

Consider Quitmeyer Electronics’ product mix problem first presented in Q8.12. Use

Solver to create the Answer and Sensitivity Reports for this LP problem. Now answer the

following questions using these reports. Each question is independent of the others.

(a) Interpret the reduced costs for the products that are not currently included in the

optimal production plan.

(b) Another part of the corporation wants to take 35 hours of time on test device 3.

How does this affect Quitmeyer’s optimal solution?

(c) Quitmeyer has the opportunity to obtain 20 additional hours on test device 1 at a

cost of $25 per hour. Is this deal worthwhile?

solution to topic 2 lp

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