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Transcript of 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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12
13
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)
1
2
3
4
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7
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9
10
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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)
1
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11
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
x2 = level of activity 2
x3 = level of activity 3
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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2
3
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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 $2,904.76
(per week) <=
Sales Potential 20
Data cells: B2:D2, B5:D7, G5:G7, and D12
Changing cells: B10:D10
Target cell: G10
Output cells: E5:E7
3
4
5
6
7
E
Hours
Used
=SUMPRODUCT(B5:D5,$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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2
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5
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7
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9
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 Usage Resource Resource
per Unit of Activity Used Available
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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2
3
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9
10
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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
x2 = level of activity 2
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
Activity 1 Activity 2 Activity 3 Activity 4
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
Benefit Contribution per
Unit of Each Activity
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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2
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5
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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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5
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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 $7,650,000
Data cells: B4:P8, B10:P10, and S4:S8
Changing cells: B13:P13
Target cell: S13
Output cells: Q4:Q8
1
2
3
4
5
6
7
8
Q
Total
Leased
(sq. f t.)
=SUMPRODUCT(B4:P4,$B$13:$P$13)
=SUMPRODUCT(B5:P5,$B$13:$P$13)
=SUMPRODUCT(B6:P6,$B$13:$P$13)
=SUMPRODUCT(B7:P7,$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
1
2
3
4
5
6
7
8
9
10
11
A B C D E F G H
Activity 1 Activity 2 Activity 3 Activity 4
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
Benefit Contribution per
Unit of Each Activity
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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3
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5
6
7
8
9
10
11
12
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)
1
2
3
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5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
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
xM2S1 = number of units shipped from Mine 2 to Storage 1
xM2S2 = number of units shipped from Mine 2 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)
1
2
3
4
5
6
7
8
9
10
11
12
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
20x2L + 15x2M + 12x2S ≤ 12,000 square feet
20x3L + 15x3M + 12x3S ≤ 5,000 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
Resource Usage Resource Resource
per Unit of Activity Used Available
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
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
3-20
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
Shipments1 2 3 4 Total Shipped Supply
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
Shipments1 2 3 4 Total Shipped Supply
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
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 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
Unit Cost Distribution Center1 2 3
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