Chap003

3-1

CHAPTER 3

LINEAR PROGRAMMING: FORMULATION AND APPLICATIONS

Review Questions

3.1-1 Determine which levels should be chosen of different advertising media to obtain the most

effective advertising mix for the new cereal.

3.1-2 The expected number of exposures.

3.1-3 TV commercials are not being used and that is the primary method of reaching young

children.

3.1-4 They need to check the assumption that fractional solutions are allowed and the assumption

of proportionality.

3.2-1 Each functional constraint in the linear programming model is a resource constraint.

3.2-2 Amount of resource used ≤ Amount of resource available.

3.2-3 1) The amount available of each limited resource.

2) The amount of each resource needed by each activity. Specifically, for each

combination of resource and activity, the amount of resource used per unit of activity

must be estimated.

3) The contribution per unit of each activity to the overall measure of performance.

3.2-4 The four activities in the examples are determining the most profitable mix of production

rates for two new products, purchase quantities for airplanes, capital budgeting, and

choosing the mix of advertising media.

3.2-5 The resources in the examples are available production capacities of different plants,

investment capital available, a maximum on small airplane purchases, cumulative

investment capital available by certain times, financial allocations for advertising and for

planning purposes, and TV commercial spots available for purchase.

3.3-1 For resource-allocation problems, limits are set on the use of various resources, and then

the objective is to make the most effective use of these given resources. For cost-benefit-

tradeoff problems, management takes a more aggressive stance, prescribing what benefits

must be achieved by the activities under consideration, and then the objective is to achieve

all these benefits with minimum cost.

3.3-2 The identifying feature for a cost-benefit-tradeoff problem is that each functional constraint

is a benefit constraint.

3.3-3 Level achieved ≥ Minimum acceptable level.

3.3-4 1) The minimum acceptable level for each benefit (a managerial policy decision).

3-2

2) For each benefit, the contribution of each activity to that benefit (per unit of the

activity).

3) The cost per unit of each activity.

3.3-5 The activities for the two examples are choosing the mix of advertising media and

personnel scheduling.

3.3-6 The benefits for the two examples are increased market share and minimizing total

personnel costs while meeting service requirements.

3.4-1 Mixed problems may contain all three types of functional constraints: resource constraints,

benefit constraints, and fixed-requirement constraints.

3.4-2 Two new goals need to be incorporated into the model. The first is that the advertising

should be seen by at least 5 million young children. The second is that the advertising

should be seen by at least 5 million parents of young children.

3.4-3 Two benefit constraints and a fixed-requirement constraint are included in the new linear

programming model.

3.4-4 Management decided to adopt the new plan because it does a much better job of meeting all

of management’s goals for the campaign.

3.5-1 Transportation problems deal with transporting goods through a distribution network at

minimum cost.

3.5-2 An identifying feature for a transportation problem is that each functional constraint is a

fixed-requirement constraint.

3.5-3 In contrast to the ≤ form for resource constraints and the ≥ form for benefit constraints,

fixed-requirement constraints have an = form.

3.5-4 Factory 1 must ship 12 lathes, Factory 2 must ship 15 lathes, Customer 1 must receive 10

lathes, Customer 2 must receive 8 lathes, and Customer 3 must receive 9 lathes.

3.6-1 Assignment problems involve making assignments.

3.6-2 Pure assignment problems have all fixed-requirement constraints.

3.6-3 The changing cells for a pure asignment problem give a value of 1 when the corresponding

assignment is made, and a value of 0 otherwise.

3.7-1 A linear programming model must accurately reflect the managerial view of the problem.

3.7-2 Large linear programming models generally are formulated by management science teams.

3.7-3 The line of communication between the management science team and the manager is vital.

3.7-4 Model validation is a testing process used on an initial version of a model to identify the

errors and omissions that inevitably occur when constructing large models.

3.7-5 The process of model enrichment involves beginning with a relatively simple version of the

model and then using the experience gained with this model to evolve toward more

elaborate models that more nearly reflect the complexity of the real problem.

3-3

3.7-6 What-if analysis is an important part of a linear programming study because an optimal

solution can only be solved for with respect to one specific version of the model at a time.

Management may have “what-if” questions about how the solution will change given

changes in the model formulation.

Problems

3.1 a)

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A B C D E F G H

TV Spots Magazine Ads Radio Ads SS Ads

Exposures per Ad 1300 600 900 500

(thousands)

Budget Budget

Cost per Ad ($thousands) Spent Available

Ad Budget 300 150 200 100 4000 <= 4000

Planning Budget 90 30 50 40 1000 <= 1000

Total Exposures

TV Spots Magazine Ads Radio Ads SS Ads (thousands)

Number of Ads 0 10 10 5 17,500

<= <=

Max TV Spots 5 10 Max Radio Spots

Data cells: B2:E2, B6:E7, H6:H7, B13, and D13

Changing cells: B11:E11

Target cell: H11

4

5

6

7

F

Budget

Spent

=SUMPRODUCT(B6:E6,$B$11:$E$11)

=SUMPRODUCT(B7:E7,$B$11:$E$11)

9

10

11

H

Total Exposures

(thousands)

=SUMPRODUCT(B2:E2,B11:E11)

b) This is a linear programming model because the decisions are represented by changing

cells that can have any value that satisfy the constraints. Each constraint has an output

cell on the left, a mathematical sign in the middle, and a data cell on the right. The

overall level of performance is represented by the target cell and the objective is to

maximize that cell. Also, the Excel equation for each output cell is expressed as a

SUMPRODUCT function where each term in the sum is the product of a data cell and a

changing cell.

c) Let T = number of commercials on TV

M = number of advertisements in magazines

R = number of commercials on radio

S = number of advertisements in Sunday supplements.

Maximize Exposures (thousands) = 140T + 60M + 90R + 50S

subject to 300T + 150M + 200R + 100S ≤ 4,000 ($thousands)

90T + 30M + 50R + 40S ≤ 1,000 ($thousands)

T ≤ 5 spots

R ≤ 10 spots

and T ≥ 0, M ≥ 0, R ≥ 0, S ≥ 0.

3-4

3.2 a & c)

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A B C D E F

Activity 1 Activity 2

Contribution per unit $20 $30

Resource Resource

Used Available

Resource 1 2 1 10 <= 10

Resource 2 3 3 20 <= 20

Resource 3 2 4 20 <= 20

Activity 1 Activity 2 Total Contribution

Level of Activity 3.333 3.333 $166.67

Resource Usage

per Unit of Activity

b)

(x1, x2) Feasible? Total Contribution

(2,2) Yes $100

(3,3) Yes $150

(2,4) Yes $160 Best

(4,2) Yes $140

(3,4) No

(4,3) No

d) Let x1 = level of activity 1

x2 = level of activity 2

Maximize Contribution = $20x1 + $30x2

subject to 2x1 + x2 ≤ 10

3x1 + 3x2 ≤ 20

2x1 + 4x2 ≤ 20

and x1 ≥ 0, x2 ≥ 0.

e) Optimal Solution: (x1, x2) = (3.333, 3.333) and Total Contribution = $166.67.

3-5

3.3 a)

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A B C D E F G

Activity 1 Activity 2 Activity 3

Contribution per unit $50 $40 $70

Resource Usage Resource Resource

per Unit of Activity Used Available

Resource A 30 20 0 500 <= 500

Resource B 0 10 40 600 <= 600

Resource C 20 20 30 783.333 <= 1,000

Activity 1 Activity 2 Activity 3 Total Contribution

Level of Activity 16.667 0 15 $1,883.33

b) Let x1 = level of activity 1



Maximize Contribution = $50x1 + $40x2 + $70x3

subject to 30x1 + 20x2 ≤ 500

10x2 + 40x3 ≤ 600

20x1 + 20x2 + 30x3 ≤ 1,000

and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0.

3.4 a & c)

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A B C D E F G H

Activity 1 Activity 2 Activity 3 Activity 4

Contribution per unit $11 $9 $8 $9

Resource Resource

Used Available

Resource P 3 5 -2 4 400 <= 400

Resource Q 4 -1 3 2 300 <= 300

Resource R 6 3 2 -1 400 <= 400

Resource S -2 2 5 3 300 <= 300

Activity 1 Activity 2 Activity 3 Activity 4 Total Contribution

Level of Activity 39.421 41.953 37.071 36.528 $1,436.53

Resource Usage

per Unit of Activity

b) Below are five possible guesses (many answers are possible).

(x1, x2, x3, x4) Feasible? P

(30,30,30,30) Yes $1110

(40,40,40,40) No

(35,39,30,40) Yes $1336

(35,39,34,40) Yes $1368

(37,39,35,40) Yes $1398 Best

3.5 a) The activities are the production rates of products 1, 2, and 3. The limited resources are

hours available per week on the milling machine, lathe, and grinder.

b) The decisions to be made are how many of each product should be produced per week.

The constraints on these decisions are the number of hours available per week on the

milling machine, lathe, and grinder as well as the sales potential of product 3. The

overall measure of performance is total profit, which is to be maximized.

3-6

c) milling machine: 9(# units of 1) + 3(# units of 2) + 5(# units of 3) ≤ 500

lathe: 5(# units of 1) + 4(# units of 2) ≤ 350

grinder: 3(# units of 1) + 2(# units of 3) ≤ 150

sales: (# units of 3) ≤ 20

Nonnegativity: (# units of 1) ≥ 0, (# units of 2) ≥ 0, (# units of 3) ≥ 0

Profit = $50(# units of 1) + $20(# units of 2) + $25(# units of 3)

d)

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A B C D E F G

Product 1 Product 2 Product 3

Unit Profit $50 $20 $25

Hours Hours

Machine Hours Used per Unit of Product Used Available

Mil ling machine 9 3 5 500 <= 500

Lathe 5 4 0 350 <= 350

Grinder 3 0 2 118.571429 <= 150

Product 1 Product 2 Product 3 Total Profit

Production Rate 26.190 54.762 20 $2,904.76

(per week) <=

Sales Potential 20

Data cells: B2:D2, B5:D7, G5:G7, and D12

Changing cells: B10:D10

Target cell: G10

Output cells: E5:E7

3

4

5

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7

E

Hours

Used

=SUMPRODUCT(B5:D5,$B$10:$D$10)

=SUMPRODUCT(B6:D6,$B$10:$D$10)

=SUMPRODUCT(B7:D7,$B$10:$D$10) 9

10

G

Total Profit

=SUMPRODUCT(B2:D2,B10:D10)

e) Let x1 = units of product 1 produced per week

x2 = units of product 2 produced per week

x3 = units of product 3 produced per week

Maximize Profit = $50x1 + $20x2 + $25x3

subject to 9x1 3x2 + 5x3 ≤ 500 hours

5x1 + 4x2 ≤ 350 hours

3x1 + 2x3 ≤ 150 hours

x3 ≤ 20

and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0.

3.6 a) The activities are the production quantities of parts A, B, and C. The limited resources

are the hours available on machine 1 and machine 2.

3-7

b & d)

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A B C D E F G

Part A Part B Part C

Unit Profit $50 $40 $30

Hours Hours

Processing Time (hours per unit) Used Available

Machine 1 0.02 0.03 0.05 40 <= 40

Machine 2 0.05 0.02 0.04 40 <= 40

Part A Part B Part C Total Profit

Production 363.636 1090.909 0 $61,818.18

c) Below are three possible guesses (many answers are possible).

(x1, x2, x3) Feasible? P

(500,500,300) No

(350,1000,0) Yes $57,500

(400,1000,0) Yes $60,000 Best

e) Let A = number of part A produced

B = number of part B produced

C = number of part C produced

Maximize Profit = $50A + $40B + $30C

subject to 0.02A + 0.03B + 0.05C ≤ 40 hours

0.05A + 0.02B + 0.04C ≤ 40 hours

and A ≥ 0, B ≥ 0, C ≥ 0.

3.7

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A B C D E F G

Activity 1 Activity 2 Activity 3

Unit Profit 20 40 30



Resource 1 3 5 4 400 <= 400

Resource 2 1 1 1 100 <= 100

Resource 3 1 3 2 200 <= 200

Activity 1 Activity 2 Activity 3 Total Profit

Level of Activity 50 50 0 3,000

3-8

3.8 a & c)

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A B C D E F

Activity 1 Activity 2

Unit Cost $60 $50

Minimum

Level Acceptable

Achieved Level

Benefit 1 5 3 60 >= 60

Benefit 2 2 2 31 >= 30

Benefit 3 7 9 126 >= 126

Activity 1 Activity 2 Total Cost

Level of Activity 6.75 8.75 $842.50

Benefit Contribution per

Unit of Each Activity

b)

(x1, x2) Feasible? C

(7,7) No

(7,8) No

(8,7) No

(8,8) Yes $880 Best

(8,9) Yes $930

(9,8) Yes $940

d) Let x1 = level of activity 1


Minimize Cost = $60x1 + $50x2

subject to 5x1 + 3x2 ≥ 60

2x1 + 2x2 ≥ 30

7x1 + 9x2 ≥ 126

and x1 ≥ 0, x2 ≥ 0.

e) Optimal Solution: (x1, x2) = (6.75, 8.75) and Total Cost = $842.50.

3-9

3.9 a & c)

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A B C D E F G H


Unit Cost $400 $600 $500 $300

Minimum

Level Acceptable

Achieved Level

Benefit P 2 -1 4 3 80 >= 80

Benefit Q 1 4 -1 2 60 >= 60

Benefit R 3 5 4 -1 110 >= 110

Activity 1 Activity 2 Activity 3 Activity 4 Total Cost

Level of Activity 32.5 3.75 0 6.25 $17,125



b) Below are five possible guesses (many answers are possible).

(x1, x2, x3, x4) Feasible? C

(32,4,0,6) No

(33,4,0,6) Yes $17,400 Best

(33,5,0,6) No

(33,4,1,6) Yes $17,900

(33,4,1,7) Yes $18,200

3.10 a & d)

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A B C D E F G

Corn Tankage Alfalfa

Unit Cost $0.84 $0.72 $0.60

(per kg) Minimum

Level Daily

Nutritional Contents (per kg) Achieved Requirement

Carbohydrates 90 20 40 200 >= 200

Protein 30 80 60 180 >= 180

Vitamins 10 20 60 157.142857 >= 150

Corn Tankage Alfalfa Total Cost

Diet (kg) 1.143 0 2.429 $2.42

b) (x1, x2, x3) = (1,2,2) is a feasible solution with a daily cost of $3.48. This diet will

provide 210 kg of carbohydrates, 310 kg of protein, and 170 kg of vitamins daily.

c) Answers will vary.

e) Let C = kg of corn to feed each pig

T = kg of tankage to feed each pig

A = kg of alfalfa to feed each pig

Minimize Cost = $0.84C + $0.72T + $0.60A

subject to 90C + 20T + 40A ≥ 200

30C + 80T + 40A ≥ 180

10C + 20T + 60A ≥ 150

and C ≥ 0, T ≥ 0, A ≥ 0.

3-10

3.12 a) The activities are leasing space in each month for a number of months. The benefit is

meeting the space requirements for each month.

b) The decisions to be made are how much space to lease and for how many months. The

constraints on these decisions are the minimum required space. The overall measure of

performance is cost which is to be minimized.

c) Month 1: (M1 1mo lease) + (M1 2mo lease) + (M1 3mo lease) + (M1 4mo lease) + (M1

5 mo lease) ≥ 30,000 square feet.

Month 2: (M1 2mo lease) + M1 3 mo lease) + (M1 4 mo lease) + (M1 5mo lease) +

(M2 1 mo lease) + (M2 2 mo lease) + (M2 3 mo lease) + (M2 4 mo lease) ≥ 20,000

square feet.

Month 3: (M1 3mo lease) + (M1 4mo lease) + (M1 5mo lease) + (M2 2mo lease) + (M2

3mo lease) + (M2 4mo lease) + (M3 1mo lease) + (M3 2mo lease) + (M3 3mo lease) ≥

40,000 square feet.

Month 4: (M1 4mo lease) + (M1 5mo lease) + (M2 3mo lease) + (M2 4mo lease) + (M3

2 mo lease) + (M3 3mo lease) + (M4 1mo lease) + (M4 2mo lease) ≥ 10,000 square

feet.

Month 5: (M1 5mo lease) + (M2 4mo lease) + (M3 3mo lease) + (M4 2 mo lease) +

(M5 1mo lease) ≥ 50,000 square feet.

Nonnegativity: (M1 1mo lease) ≥ 0, (M1 2mo lease) ≥ 0, (M1 3 mo lease) ≥ 0, (M1 4

mo lease) ≥ 0, (M1 5mo lease) ≥ 0, (M2 1mo lease) ≥ 0, (M2 2mo lease) ≥ 0, (M2 3 mo

lease) ≥ 0, (M2 4mo lease) ≥ 0, (M3 1mo lease) ≥ 0, (M3 2mo lease) ≥ 0, (M3 3mo

lease) ≥ 0, (M4 1mo lease) ≥ 0, (M4 2mo lease) ≥ 0, (M5 1mo lease) ≥ 0.

Cost = ($650)[(M1 1mo lease) + (M2 1mo lease) + (M3 1mo lease) + (M4 1mo lease) +

(M5 1mo lease)] + ($1,000)[(M1 2mo lease) + (M2 2mo lease) + (M3 2mo lease) +

(M4 2mo lease)] + ($1,350)[(M1 3mo lease) + (M2 3mo lease) + (M3 3mo lease)] +

($1,600)[(M1 4mo lease) + (M2 4mo lease)] + ($1,900)[M1 5mo lease]

3-11

d)

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A B C D E F G H I J K L M N O P Q R S

Month Covered by Lease? Total Space

Month of Lease: 1 1 1 1 1 2 2 2 2 3 3 3 4 4 5 Leased Required

Length of Lease: 1 2 3 4 5 1 2 3 4 1 2 3 1 2 1 (sq. ft.) (sq. ft.)

Month 1 1 1 1 1 1 30,000 >= 30,000

Month 2 1 1 1 1 1 1 1 1 30,000 >= 20,000

Month 3 1 1 1 1 1 1 1 1 1 40,000 >= 40,000

Month 4 1 1 1 1 1 1 1 1 30,000 >= 10,000

Month 5 1 1 1 1 1 50,000 >= 50,000

Cost of Lease $65 $100 $135 $160 $190 $65 $100 $135 $160 $65 $100 $135 $65 $100 $65

(per sq. ft.)

Total Cost

Lease (sq. ft.) 0 0 0 0 30,000 0 0 0 0 10,000 0 0 0 0 20,000 $7,650,000

Data cells: B4:P8, B10:P10, and S4:S8

Changing cells: B13:P13

Target cell: S13

Output cells: Q4:Q8

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5

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Q

Total

Leased

(sq. f t.)

=SUMPRODUCT(B4:P4,$B$13:$P$13)




=SUMPRODUCT(B8:P8,$B$13:$P$13) 12

13

S

Total Cost

=SUMPRODUCT(B10:P10,B13:P13)

e) Let xij = square feet of space leased in month i for a period of j months.

for i = 1, ... , 5 and j = 1, ... , 6-i.

Minimize C = $650(x11 + x21 + x31 + x41 + x51) + $1,000(x12 + x22 + x32 + x42)

+$1,350(x13 + x23 + x33) + $1,600(x14 + x24) + $1,900x15

subject to x11 + x12 + x13 + x14 + x15 ≥ 30,000 square feet

x12 + x13 + x14 + x15 + x21 + x22 + x23 + x24 ≥ 20,000 square feet

x13 + x14 + x15 + x22 + x23 + x24 + x31 + x32 + x33 ≥ 40,000 sq. feet

x14 + x15 + x23 + x24 + x32 + x33 + x41 + x42 ≥ 10,000 square feet

x15 + x24 + x33 + x42 + x51 ≥ 50,000 square feet

and xij ≥ 0, for i = 1, ... , 5 and j = 1 , ... , 6-i.

3.13

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A B C D E F G H


Unit Cost 2 1 -1 3

Minimum

Level Acceptable

Achieved Level

Benefit 1 3 2 -2 5 80 >= 80

Benefit 2 1 -1 0 1 10 >= 10

Benefit 3 1 1 -1 2 32.857 >= 30

Activity 1 Activity 2 Activity 3 Activity 4 Total Cost

Level of Activity 0 4.286 0 14.286 47.14



3-12

3.15 a) This is a distribution-network problem because it deals with the distribution of goods

through a distribution network at minimum cost.

b)

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A B C D E F G

Shipping Cost Customer 1 Customer 2 Customer 3

Factory 1 $600 $800 $700

Factory 2 $400 $900 $600

Total

Shipped

Units Shipped Customer 1 Customer 2 Customer 3 Out Output

Factory 1 0 200 200 400 = 400

Factory 2 300 0 200 500 = 500

Total To Customer 300 200 400

= = = Total Cost

Order Size 300 200 400 $540,000

c) Let xij = number of units to ship from Factory i to Customer j (i = 1,2; j = 1, 2, 3)

Minimize Cost = $600x11 + $800x12 + $700x13 + $400x21 + $900x22 + $600x23

subject to x11 + x12 + x13 = 400

x21 + x22 + x23 = 500

x11 + x21 = 300

x12 + x22 = 200

x13 + x23 = 400

and x11 ≥ 0, x12 ≥ 0, x13 ≥ 0, x21 ≥ 0, x22 ≥ 0, x23 ≥ 0.

3.16 a) Requirement 1: The total amount shipped from Mine 1 must be 40 tons.

Requirement 2: The total amount shipped from Mine 2 must be 60 tons.

Requirement 3: The total amount shipped to the Plant must be 100 tons.

Requirement 4: For Storage 1, the amount shipped out = the amount in.

Requirement 5: For Storage 2, the amount shipped out = the amount in.

3-13

b)

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A B C D E F

Shipping

Cost S1 S2

M1 $2,000 $1,700

M2 $1,600 $1,100

P $400 $800

Capacity S1 S2

M1 30 30

M2 50 50

P 70 70

Units Total Shipped

Shipped S1 S2 Out of M1,M2 Output

M1 30 10 40 = 40

M2 10 50 60 = 60

Total Into S1,S2 40 60

= = Total Shipped

Total Out of S1,S2 40 60 Into P Needed

P 40 60 100 = 100

Total Cost

Units Shipped ² Capacity $212,000

c) Let xM1S1 = number of units shipped from Mine 1 to Storage 1

xM1S2 = number of units shipped from Mine 1 to Storage 2



xS1P = number of units shipped from Storage 1 to the Plant

xS2P = number of units shipped from Storage 2 to the Plant

Minimize Cost = $2,000xM1S1 + $1,700xM1S2 + $1,600xM2S1 + $1,100xM2S2

+$400xS1P + $800xS2P

subject to xM1S1 + xM1S2 = 40

xM2S1 + xM2S2 = 60

xM1S1 + xM2S1 = xS1P

xM1S2 + xM2S2 = xS2P

xS1P + xS2P = 100

xM1S1 ≤ 30, xM1S2 ≤ 30, xM2S1 ≤ 50, xM2S2 ≤ 50, xS1P ≤ 70, xS2P ≤ 70

and xM1S1 ≥ 0, xM1S2 ≥ 0, xM2S1 ≥ 0, xM2S2 ≥ 0, xS1P ≥ 0, xS2P ≥ 0.

3.17 a) A1 + B1 + R1 = $60,000

A2 + B2 + C2 + R2 = R1

A3 + B3 + R3 = R2 + 1.40A1

A4 + R4 = R3 + 1.40A2 + 1.70B1

A5 + D5 + R5 = R4 + 1.40A3 + 1.70B2

3-14

b) Let At = amount invested in investment A at the beginning of year t.

Bt = amount invested in investment B at the beginning of year t.

Ct = amount invested in investment C at the beginning of year t.

Dt = amount invested in investment D at the beginning of year t.

Rt = amount not invested at the beginninf of year t.

Maximize Return = 1.40A4 + 1.70B3 + 1.90C2 + 1.30D5 + R5

subject to A1 + B1 + R1 = $60,000

A2 + B2 + C2 – R1 + R2 = 0

–1.40A1 + A3 + B3 – R2 + R3 = 0

–1.40A2 + A4 – 1.70B1 – R3 + R4 = 0

–1.40A3 – 1.70B2 + D5 – R4 + R5 = 0

and At ≥ 0, Bt ≥ 0, Ct ≥ 0, Dt ≥ 0, Rt ≥ 0.

c)

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A B C D E F G H I J K L M N O P Q R

Investment A A A A B B B C D R R R R R Total Available

Year 1 2 3 4 1 2 3 2 5 1 2 3 4 5 Invested to Invest

Year 1 1 1 1 $60,000 = $60,000

Year 2 1 1 1 -1 1 $0 = $0

Year 3 -1.4 1 1 -1 1 $0 = $0

Year 4 -1.4 1 -1.7 -1 1 $0 = $0

Year 5 -1.4 -1.7 1 -1 1 $0 = $0

Total Return

Return in Year 6 1.4 1.7 1.9 1.3 1 $152,880

Dollars Invested $60,000 $0 $84,000 $0 $0 $0 $0 $0 $117,600 $0 $0 $0 $0 $0

3.18 a) Let xi = percentage of alloy i in the new alloy (i = 1, 2, 3, 4, 5).

(60%)x1 + (25%)x2 + (45%)x3 + (20%)x4 + (50%)x5 = 40%

(10%)x1 + (15%)x2 + (45%)x3 + (50%)x4 + (40%)x5 = 35%

(30%)x1 + (60%)x2 + (10%)x3 + (30%)x4 + (10%)x5 = 25%

x1 + x2 + x3 + x4+ x5 = 100%

b)

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A B C D E F G H I

Alloy 1 Alloy 2 Alloy 3 Alloy 4 Alloy 5

Cost ($/lb.) $22 $20 $25 $24 $27

New Alloy Desired

Alloy Composition Composition Composition

Tin 60% 25% 45% 20% 50% 40% = 40%

Zinc 10% 15% 45% 50% 40% 35% = 35%

Lead 30% 60% 10% 30% 10% 25% = 25%

Alloy 1 Alloy 2 Alloy 3 Alloy 4 Alloy 5 Total Blend

New Alloy Blend 4.3% 28.3% 67.4% 0.0% 0.0% 100% = 100%

Total Cost

$23.46

3-15

c) Let xi = percentage of alloy i in the new alloy (i = 1, 2, 3, 4, 5).

Minimize Cost = $22x1 + $20x2 + $25x3 + $24x4 + $27x5

subject to (60%)x1 + (25%)x2 + (45%)x3 + (20%)x4 + (50%)x5 = 40%

(10%)x1 + (15%)x2 + (45%)x3 + (50%)x4 + (40%)x5 = 35%

(30%)x1+ (60%)x2 + (10%)x3 + (30%)x4 + (10%)x5 = 25%

and x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0, x5 ≥ 0.

3.19 a)

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A B C D E F G H I J K

Large Medium Small

Unit Profit $420 $360 $300

Space Required 20 15 12

(sq.ft. per unit)

Total Space Space

Production Large Medium Small Produced Capacity Required Available

Plant 1 516.67 177.78 0 694.4 <= 750 13,000 <= 13,000

Plant 2 0 666.67 166.67 833.3 <= 900 12,000 <= 12,000

Plant 3 0 0 416.67 416.7 <= 450 5,000 <= 5,000

Total Produced 516.67 844.44 583.33

<= <= <= Total Profit

Sales Forecast 900 1200 750 $696,000

Percentage of Plant 1 Capacity 93% = 93% Percentage of Plant 2 Capacity

Percentage of Plant 1 Capacity 93% = 93% Percentage of Plant 3 Capacity

b) Let xij = number of units produced at plant i of product j (i = 1, 2, 3; j = L, M, S).

Maximize Profit = $420(x1L + x2L + x3L) + $360(x1M + x2M + x3M) + $300(x1S + x2S + x3S)

subject to x1L + x1M + x1S ≤ 750

x2L + x2M + x2S ≤ 900

x3L + x3M + x3S ≤ 450

20x1L + 15x1M + 12x1S ≤ 13,000 square feet



x1L + x2L + x3L ≤ 900

x1M + x2M + x3M ≤ 1,200

x1S + x2S + x3S ≤ 750

(x1L + x1M + x1S) / 750 = (x2L + x2M + x2S) / 900

(x1L + x1M + x1S) / 750 = (x3L + x3M + x3S) / 450

and x1L ≥ 0, x1M ≥ 0, x1S ≥ 0, x2L ≥ 0, x2M ≥ 0, x2S ≥ 0, x3L ≥ 0, x3M ≥ 0, x3S ≥ 0.

3-16

3.21 a)

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A B C D E F G

Men's Women's Children's

Gross Profit $8 $10 $6

(per glove)

Full Time Part Time

Labor Cost (per hour) $13 $10

Hours worked per week 40 20

Labor Cost (per week) $520 $200



Material (sq. ft.) 2 1.5 1 5000 <= 5000

Labor (minutes) 30 45 40 75000 <= 75,000

Men's Women's Children's

Production (per week) 2500 0 0

Gross Profit $20,000

Full Time Part Time Labor Cost $15,500

Employees 25 12.5 Net Profit $4,500

>=

Minimum Full Time 20

2

Full Time 25 >= 25 Times Part-Time

b) Let M =number of men’s gloves to produce per week,

W = number of women’s gloves to produce per week,

C = number of children’s gloves to produce per week,

F = number of full-time workers to employ,

PT = number of part-time workers to employ.

Maximize Profit = $8M + $10W + $6C – $13(40)F – $10(20)PT

subject to 2M + 1.5W + C ≤ 5,000 square feet

30M + 45W + 40C ≤ 40(60)F + 20(60)PT hours

F ≥ 20

F ≥ 2PT

and M ≥ 0, W ≥ 0, C ≥ 0, F ≥ 0, PT ≥ 0.

3.22

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A B C D E F G H I J

Hours Available

Wage Rate Monday Tuesday Wednesday Thursday Friday

K.C. $10.00 6 0 6 0 6

D.H. $10.10 0 6 0 6 0

H.B. $9.90 4 8 4 0 4

S.C. $9.80 5 5 5 0 5

K.S. $10.80 3 0 3 8 0

N.K. $11.30 0 0 0 6 2

Hours

Hours Worked Monday Tuesday Wednesday Thursday Friday Worked Output

K.C. 4 0 2 0 3 9 >= 8

D.H. 0 2 0 6 0 8 >= 8

H.B. 4 7 4 0 4 19 >= 8

S.C. 5 5 5 0 5 20 >= 8

K.S. 1 0 3 3 0 7 >= 7

N.K. 0 0 0 5 2 7 >= 7

Hours Worked 14 14 14 14 14

= = = = = Total Cost

Hours Needed 14 14 14 14 14 $710

Hours Worked <= Hours Available

3-17

3.23 a) Resource Constraints:

Calories must be no more than 420.

No more than 20% of total calories from fat.

Benefit Constraints:

Calories must be at least 380

There must be at least 50 mg of vitamin content.

There must be at least 2 times as much strawberry flavoring as sweetener.

Fixed-Requirement Constraints:

There must be 15 mg of thickeners.

b)

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A B C D E F G H I J K

Strawberry Cream Vitamin Sweetener Thickener

Unit Cost $0.10 $0.08 $0.25 $0.15 $0.06

(per tbsp)

Level

Nutritional Contents (per tbsp) Achieved Minimum Maximum

Total Calories 50 100 0 120 80 380 >= 380 <= 420

Vitamin Content (mg) 20 0 50 0 2 64.167 >= 50

Thickeners (mg) 3 8 1 2 25 15 = 15

Calories from Fat 1 75 0 0 30 23.521 <= 76

20%

Strawberry Cream Vitamin Sweetener Thickener Total Cost of Total Calories

Contents (tbsp) 3.208 0.271 0 1.604 0 $0.58

>=

3.208 2 times Sweetener

c) Let S = Tablespoons of strawberry flavoring,

CR = Tablespoons of cream,

V = Tablespoons of vitamin supplement,

A = Tablespoons of artificial sweetener,

T = Tablespoons of thickening agent,

Minimize C = $0.10S + $0.08CR + $0.25V + $0.15A + $0.06T

subject to 50S + 100CR + 120A + 80T ≥ 380 calories

50S + 100CR + 120A + 80T ≤ 420 calories

S + 75CR + 30T ≤ 0.2(50S + 100C + 120A + 80T)

20S + 50V + 2T ≥ 50 mg Vitamins

S ≥ 2A

3S + 8CR + V + 2A + 25T = 15 mg Thickeners

and S ≥ 0, CR ≥ 0, V ≥ 0, A ≥ 0, T ≥ 0.

3-18

3.24 a) Resource Constraints:

Calories must be no more than 600.

No more than 30% of total calories from fat.

Benefit Constraints:

Calories must be at least 400

There must be at least 60 mg of vitamin C.

There must be at least 12 g of protein.

There must be at least 2 times as much peanut butter as jelly.

There must be at least 1 cup of liquid

Fixed-Requirement Constraints:

There must be 2 slices of bread.

b)

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A B C D E F G H I J K L

Peanut Strawberry Graham

Bread Butter Jelly Cracker Milk Juice

(sl ice) (tbsp.) (tbsp.) (tbsp.) (cup) (cup)

Unit Cost $0.05 $0.04 $0.07 $0.08 $0.15 $0.35

Level

Nutritional Contents Achieved Minimum Maximum

Total Calories 70 100 50 60 150 100 400 >= 400 <= 600

Vitamin C (mg) 0 0 3 0 2 120 60 >= 60

Protein (g) 3 4 0 1 8 1 13.949 >= 12

Calories from Fat 10 75 0 20 70 0 120 <= 120

30%

Peanut Strawberry Graham of Total Calories

Bread Butter Jelly Cracker Milk Juice

(sl ice) (tbsp.) (tbsp.) (tbsp.) (cup) (cup) Total Cost

Contents (tbsp) 2 0.575 0.287 1.039 0.516 0.484 $0.47

=

2

Peanut Butter 0.575 >= 0.575 2 Times Strawberry Jelly

Total Liquid 1 >= 1

c) Let B = slices of bread,

P= Tablespoons of peanut butter,

S = Tablespoons of strawberry jelly,

G = graham crackers,

M = cups of milk,

J = cups of juice.

Minimize C = $0.05B + $0.04P + $0.07S + $0.08G + $0.15M + $0.35J

subject to 70B + 100P + 50S + 60G + 150M + 100J ≥ 400 calories

70B + 100P + 50S + 60G + 150M + 100J ≤ 600 calories

10B + 75P + 20G + 70M

≤ 0.3(70B + 100P + 50S + 60G + 150M + 100J)

3S + 2M + 120J ≥ 60mg Vitamin C

3B + 4P + G + 8M + J ≥ 12mg Protein

B = 2 slices

P ≥ 2S

M + J ≥ 1 cup

and B ≥ 0, P ≥ 0, S ≥ 0, G ≥ 0, M ≥ 0, J ≥ 0.

3-19

6.3 a)

Unit Cost ($)

Destination (Retail Outlet)

1 2 3 4 Supply

1 500 600 400 200 10

Source 2 200 900 100 300 20

(Plant) 3 300 400 200 100 20

4 200 100 300 200 10

Demand 20 10 10 20

b)

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A B C D E F G H I

Unit Cost1 2 3 4

1 $500 $600 $400 $200

Plant 2 $200 $900 $100 $3003 $300 $400 $200 $1004 $200 $100 $300 $200

Shipments1 2 3 4 Total Shipped Supply

1 0 0 0 10 10 = 10Plant 2 20 0 0 0 20 = 20

3 0 0 10 10 20 = 204 0 10 0 0 10 = 10

Total Received 20 10 10 20

= = = = Total Cost

Demand 20 10 10 20 $10,000

Retail Outlet

Retail Outlet

3.25

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A B C D E F G H I

Unit Cost1 2 3 4

1 $500 $600 $400 $200

Plant 2 $200 $900 $100 $3003 $300 $400 $200 $1004 $200 $100 $300 $200


1 0 0 0 10 10 = 10Plant 2 20 0 0 0 20 = 20

3 0 0 10 10 20 = 204 0 10 0 0 10 = 10

Total Received 20 10 10 20

= = = = Total Cost

Demand 20 10 10 20 $10,000

Retail Outlet

Retail Outlet

3-20

3.26

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A B C D E F G H I

Distance (miles)1 2 3 4

1 800 1,300 400 700Plant 2 1,100 1,400 600 1,000

3 600 1,200 800 900

Fixed Cost $100Cost per Mile $0.50

Unit Cost1 2 3 4

1 $500 $750 $300 $450Plant 2 $650 $800 $400 $600

3 $400 $700 $500 $550


1 0 0 2 10 12 = 12Plant 2 0 9 8 0 17 = 17

3 10 1 0 0 11 = 11Total Received 10 10 10 10

= = = = Total Cost

Demand 10 10 10 10 $20,200

Distribution Center

Distribution Center

Distribution Center

3.27

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A B C D E F G H I

Unit Profit1 2 3 4

1 $800 $700 $500 $200

Plant 2 $500 $200 $100 $3003 $600 $400 $300 $500


1 0 60 0 0 60 = 60Plant 2 40 0 0 40 80 = 80

3 0 0 20 20 40 = 40Total Received 40 60 20 60

= = >= Total Cost

Commitment 40 60 20 $90,000

Customer

Customer

3-21

3.28

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A B C D E F G H

Unit Cost Distribution Center1 2 3

Plant A $800 $700 $400B $600 $800 $500

Shipments Distribution Center1 2 3 Total Shipped Supply

Plant A 0 20 20 40 <= 50B 20 0 0 20 <= 50

Total Received 20 20 20

= = = Total Cost

Demand 20 20 20 $34,000

3.29

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A B C D E F G H


Plant A $800 $700 $400B $600 $800 $500

Shipments Distribution Center1 2 3 Total Shipped Supply

Plant A 0 10 30 40 <= 50B 20 0 0 20 <= 50

10 10 10<= <= <= Total

Total Received 20 10 30 60 = 60<= <= <=

Demand 30 30 30 Total Cost

$31,000

3-22

3.30

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A B C D E F G H

Unit Cost Job1 2 3

A $5 $7 $4Person B $3 $6 $5

C $2 $3 $4

Assignments Job Total1 2 3 Assignments Supply

A 0 0 1 1 = 1Person B 1 0 0 1 = 1

C 0 1 0 1 = 1Total Assigned 1 1 1

= = = Total Cost

Demand 1 1 1 $10

3.31 a) This problem fits as an assignment problem with ships as assignees and ports as

assignments.

b)

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A B C D E F G H I

Unit Cost1 2 3 4

1 $500 $400 $600 $700

Ship 2 $600 $600 $700 $5003 $700 $500 $700 $6004 $500 $400 $600 $600

Assignments Total1 2 3 4 Assignments Supply

1 0 1 0 0 1 = 1Ship 2 0 0 0 1 1 = 1

3 0 0 1 0 1 = 14 1 0 0 0 1 = 1

Total Assigned 1 1 1 1

= = = = Total Cost

Demand 1 1 1 1 $2,100

Port

Port

3-23

3.32

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A B C D E F G H


Plant A $800 $700 $400B $600 $800 $500

Demand 10 20 30

Cost of Assignment Distribution Center1 2 3

Plant A $8,000 $14,000 $12,000B $6,000 $16,000 $15,000

Shipments Distribution Center1 2 3 Total Assignments Supply

Plant A 0 1 1 2 <= 2B 1 0 0 1 <= 2

Total Assigned 1 1 1

= = = Total Cost

Demand 1 1 1 $32,000

3.33 a) Let T = the number of tow bars to produce

S = the number of stabilizer bars to produce

Maximize Profit = $130T + $150S

subject to 3.2T + 2.4S ≤ 16 hours

2T + 3S ≤ 15 hours

and T ≥ 0, S ≥ 0

T, S are integers.

b)

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A B C D E F

Tow Bars Stabil izer BarsUnit Profit $130 $150

Hours HoursUsed Available

Machine 1 3.2 2.4 12 <= 16Machine 2 2 3 15 <= 15

Tow Bars Stabil izer Bars Total Profit

Units Produced 0 5 $750

Hours Used Per Unit Produced

3-24

3.34 a)

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A B C D E F

Model A Model B(high speed) (lower speed)

Unit Cost $6,000 $4,000

Total CapacityCapacity Needed

Capacity 20,000 10,000 80,000 >= 75,000

Model A Model B

(high speed) (lower speed) Total Total Cost

Purchase 2 4 6 $28,000

>= >=1 Min Needed 6

Copies per Day

b) Let A = the number of Model A (high-speed) copiers to buy

B = the number of Model B (lower-speed) copiers to buy

Minimize Cost = $6,000A + $4,000B

subject to A + B ≥ 6 copiers

A ≥ 1 copier

20,000A + 10,000B ≥ 75,000 copies/day

and A ≥ 0, B ≥ 0

A, B are integers.

3.35 a)

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A B C D E F G

Long-Range Medium-Range Short-RangeJets Jets Jets

Annual Profit ($mil lion) 4.2 3 2.3Resource Resource

Resource Used Per Unit Produced Used AvailableBudget 67 50 35 1498 <= 1500

Maintenance Capacity 1.667 1.333 1 39.333 <= 40Pilot Crews 1 1 1 30 <= 30

Long-Range Medium-Range Short-Range Total Annual

Jets Jets Jets Profit ($mil lion)

Purchase 14 0 16 95.6

b) Let L = the number of long-range jets to purchase

M = the number of medium-range jets to purchase

S = the number of short-range jets to purchase

Maximize Annual Profit ($millions) = 4.2L + 3M + 2.3S

subject to 67L + 50M + 35S ≤ 1,500 ($million)

(5/3)L + (4/3)M + S ≤ 40 (maintenance capacity)

L + M + S ≤ 30 (pilot crews)

and L ≥ 0, M ≥ 0, S ≥ 0

Documents

Chap003