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8
C H A P T E R
Linear Programming Modeling Applications: With
Computer Analyses in Excel and QM for Windows
TEACHING SUGGESTIONS
Teaching Suggestion 8.1: Importance of Formulating Large
LP Problems.
Since computers are used to solve virtually all business LP problems, the most important thing a student can do is to get experience in formulating a wide variety of problems. This chapter provides such a variety.
Teaching Suggestion 8.2: Note on Production
Scheduling Problems.
The Greenberg Motor example in this chapter is the largest problem in the book in terms of constraints, so it provides a good practice environment. An interesting feature to point out is that LP
constraints are capable of tying one production period to the next.
Teaching Suggestion 8.3: Solving Assignment Problems by LP.
The example of the law firm of Ivan and Ivan in this chapter can
clearly be solved more quickly using QM for Windows’ assignment program than by the LP program. Students should be asked
why anyone would choose to use the LP approach. There are two
answers: (1) many commercial LP programs do not contain assignment algorithms (which are more popular in academic software such as QM for Windows); and (2) the LP program can provide more sensitivity analysis and economic interpretation than is
available in the assignment module. The assignment problem is
treated in Chapter 10.
Teaching Suggestion 8.4: Labor Planning Problem—Arlington
Bank.
This example is a good practice tool and lead-in for the Chase
Manhattan Bank case at the end of the chapter. Without this example, the case would probably overpower most students.
Teaching Suggestion 8.5: Ingredient Blending Applications.
Three points can be made about the two blending examples in this
chapter. First, both the diet and fuel blending problems presented
here are tiny compared to huge real-world blending problems. But
they do provide some sense of the issues to be faced.
Second, diet problems that are missing the constraints that
force variety into the diet can be terribly embarrassing. It has been
said that a hospital in New Orleans ended up with an LP solution
to feed each patient only castor oil for dinner because analysts neglected to add constraints forcing a well-rounded diet.
ALTERNATIVE EXAMPLES
Alternative Example 8.1: Natural Furniture Company manufactures three outdoor products, chairs, benches, and tables. Each
product must pass through the following departments before it is
shipped: sawing, sanding, assembly, and painting. The time requirements (in hours) are summarized in the tables below.
The production time available in each department each week
and the minimum weekly production requirement to fulfill contracts are as follows:
Capacity
(In Hours)
Department
Sawing
Sanding
Assembly
Painting
450
400
625
550
Product
Sawing
Chairs
Benches
Tables
1.5
1.5
2.0
Product
Minimum
Production
Level
Chairs
Benches
Tables
100
50
50
Hours Required
Sanding Assembly
1.0
1.5
2.0
2.0
2.0
2.5
Painting
Unit
Profit
1.5
2.0
2.0
$15
$10
$20
The production manager has the responsibility of specifying production levels for each product for the coming week. Let
X1 Number of chairs produced
X2 Number of benches produced
X3 Number of tables produced
The objective function is
Maximize profit 15X1 10X2 20X3
Constraints
1.5X1 1.5X2 2.0X3 450 hours of sawing available
1.0X1 1.5X2 2.0X3 400 hours of sanding available
2.0X1 2.0X2 2.5X3 625 hours of assembly available
1.5X1 2.0X2 2.0X3 550 hours of painting available
X1 2.0X2 2.0X3 100 chairs
X2 2.0X3 50 benches
X3 50 tables
X1, X2, X3 0
Alternative Example 8.2: A phosphate manufacturer produces
three grades, A, B, and C, which cost the firm $40, $50, and $60
per kilogram, respectively. The products require the labor and materials per batch that are shown on the following page.
109
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Labor hours
Raw material #1
Raw material #2
LINEAR PROGRAMMING MODELING APPLICATIONS
Grade
A
Grade
B
Grade
C
Available
Resources
4
200
600
4
300
400
5
300
500
80 hr
6,000 kg
5,000 kg
Problem 8-2 solved by computer:
$50,000 invested in Los Angeles municipal bonds (X1)
$0 invested in Thompson Electronics (X2)
$0 invested in United Aerospace (X3)
$175,000 invested in Palmer Drugs (X4)
What mix of products would yield minimum cost?
Objective function
Minimize cost 40A 50B 60C
Constraints
$25,000 invested in Happy Days (X5)
This produces an annual return on investment of $20,300.
Minimize staff size X1 X2 X3 X4 X5 X6
8-3.
where
4A 4B 5C 80
200A 300B 300C 6,000
600A 400B 500C 5,000
Labor:
Raw material #1
Raw material #2
Xi number of workers reporting for start of work at period i
(with i 1, 2, 3, 4, 5, or 6)
X1 X2 12
SOLUTIONS TO PROBLEMS
X2 X3 16
8-1. Since the decision centers about the production of the two
different cabinet models, we let
X4 X5 11
X3 X4 9
X5 X6 4
X1 number of French Provincial cabinets produced
each day
X2 number of Danish Modern cabinets produced each
day
X1 X6 3
All variables 0
The computer solution is to hire 30 workers:
Objective: maximize revenue $28X1 $25X2
16 begin at 7 A.M.
subject to
9 begin at 3 P.M.
3X1 2X2 360 hours
(carpentry department)
112X1 1X2 200 hours
(painting department)
X1 X2 125 hours
(finishing department)
3
4
X1
3
4
2 begin at 7 P.M.
3 begin at 11 P.M.
An alternative optimum is
60 units
(contract requirement)
3 begin at 3 A.M.
X2 60 units
(contract requirement)
9 begin at 7 A.M.
X1, X2 0
7 begin at 11 A.M.
Problem 8-1 solved by computer:
2 begin at 3 P.M.
Produce 60 French Provincial cabinets (X1) per day
Produce 90 Danish Modern cabinets (X2) per day
Profit $3,930 (in four iterations)
9 begin at 7 P.M.
8-2.
Let X1 dollars invested in Los Angeles municipal bonds
X2 dollars invested in Thompson Electronics
0 begin at 11 P.M.
8-4.
Let X1 number of pounds of oat product per horse each
day
X2 number of pounds of enriched grain per horse
each day
X3 dollars invested in United Aerospace
X4 dollars invested in Palmer Drugs
X3 number of pounds of mineral product per horse
each day
X5 dollars invested in Happy Days Nursing Homes
Maximize return 0.053X1 0.068X2 0.049X3 0.084X4 0.118X5
subject to X1 X2 X3 X4 X5 $250,000 (funds to be
invested)
X1 .2 (X1 X2 X3 X4 X5) (municipal
bonds)
or
.8X1 .2X2 .2X3 .2X4 .2X5 0
X2 X3 X4 .4 (X1 X2 X3 X4 X5) (combination
of electronics, aerospace, and drugs)
(X5 0.5X1) rewritten as
0.5X1 X5 0 (nursing home as percent of bonds)
X1, X2, X3, X4, X5 0
Minimize cost 0.09X1 0.14X2 0.17X3
subject to
2X1 3X2 1X3 6 (ingredient A)
1
2
X1 1X2 12X3 2 (ingredient B)
3X1 5X2 6X3 9 (ingredient C)
1X1 112 X2 2X3 8 (ingredient D)
1
2
X1 21 X2 112 X3 5 (ingredient E)
X1 X2 X3 6 (maximum feed/day)
All variables 0
Solution: X1 113
X2 0
X3 313
cost 0.687
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8-5.
LINEAR PROGRAMMING MODELING APPLICATIONS
Let Xij 1 if pitcher i is scheduled to go against opponent j,
0 otherwise
X14 X15 X23 X24 X25 X33 X34
X35 X43 X44 X45 460
i 1, 2, 3, 4 stands for Jones, Baker, Parker, and
Wilson, respectively, and
X15 X24 X25 X33 X34 X35 X43
X44 X45 X53 X54 X55 470
j 1, 2, 3, 4 stands for Des Moines, Davenport,
Omaha, and Peoria, respectively.
X25 X34 X35 X43 X44 X45 X53
X54 X55 X63 X64 X65 440
where
X15 X25 X35 X45 X55 X65
0.50(X13 X14 X15 X23 X24 X25
X33 X34 X35 X43 X44 X45
X53 X54 X55 X63 X64 X65)
Objective: maximize overall probability of winning sum of
probability of winning each game 0.6X11 0.8X12 0.5X13 0.4X14
0.7X21 0.4X22 0.8X23 0.3X24
0.9X31 0.8X32 0.7X33 0.8X34
0.5X41 0.3X42 0.4X43 0.2X44
All variables 0
Solving this on the computer results in the following solution:
subject to
X11 X12 X13 X14 1 (“Dead-Arm” Jones)
X21 X22 X23 X24 1 (“Spitball” Baker)
X31 X32 X33 X34 1 (“Ace” Parker)
X41 X42 X43 X44 1 (“Gutter” Wilson)
X11 X21 X31 X41 1 (Des Moines)
X12 X22 X32 X42 1 (Davenport)
X15 30
5-month leases in March
X25 100
5-month leases in April
X35 170
5-month leases in May
X45 160
5-month leases in June
X55 10
5-month leases in July
All other variables equal 0.
Total cost $677,100.
X13 X23 X33 X43 1 (Omaha)
As a result of this, there are 440 cars remaining at the end of
August.
X14 X24 X34 X44 1 (Peoria)
8-8.
Solution: X12 1, X23 1, X34 1, X41 1, Total
P 2.9
8-6.
Let: X1 number of newspaper ads placed
The linear program has the same constraints as in problem
8-7. The objective function changes and is now:
Minimize cost Minimize cost $925X1 $2,000X2
subject to
0.04X1 0.05X2 0.40 (city exposure)
0.03X1 0.03X2 0.60 (exposure in
northwest suburbs)
1260(X13 X23 X33 X43 X53 X63)
1600(X14 X24 X34 X44 X54
X64)
X2 number of TV spots purchased
1850(X15 X25 X35 X45 X55
X65)
Solving this on the computer results in the following solution:
X1, X2 0
X15 30
5-month leases in March
X25 100
5-month leases in April
Note that the problem is not limited to unduplicated exposure
(e.g., one person seeing the Sunday newspaper three weeks in a
row counts for three exposures).
X34 65
4-month leases in May
X35 105
5-month leases in May
X43 160
3-month leases in June
Problem 8-6 solved by computer:
X53 10
3-month leases in July
Buy 20 Sunday newspaper ads (X1)
Buy 0 TV ads (X2)
This has a cost of $18,500. Perhaps the paint store should consider
a blend of TV and newspaper, not just the latter.
8-7.
111
Let Xij number of new leases in month i for j-months,
i 1, . . . , 6; j 3, 4, 5
Minimize cost subject to:
All other variables equal 0.
Total cost $752,950.
8-9.
Let Xij number of students bused from sector i to school j
Objective: minimize total travel miles 5XAB 8XAC 6XAE
0XBB 4XBC 12XBE
1260X13 1260X23 1260X33 1260X43
840X53 420X63 1600X14 1600X24
1600X34 1200X44 800X54 400X64
1850X15 1850X25 1480X35 1110X45
740X55 370X65
X13 X14 X15 420 390
4XCB 0XCC 7XCE
7XDB 2XDC 5XDE
12XEB 7XEC 0XEE
subject to
XAB XAC XAE 700 (number students in sector A)
X13 X14 X15 X23 X24
X25 400 270
XBB XBC XBE 500 (number students in sector B)
X13 X14 X15 X23 X24 X25 X33
X34 X35 430 130
XDB XDC XDE 800 (number students in sector D)
XCB XCC XCE 100 (number students in sector C)
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XEB XEC XEE 400 (number students in sector E)
16X1 81X2 74X3 83X4 7X5
XAB XBB XCB XDB XEB 900 (school B capacity)
14X6 8X7 26
XAC XBC XCC XDC XEC 900 (school C capacity)
XAE XBE XCE XDE XEE 900 (school E capacity)
All variables 0
22X1 28X5 19X6 63X7
Problem 8-11 solved by computer:
Solution: XAB 400
The meal plan for the evening is
XAE 300
No milk (X1 0)
XBB 500
0.499 pound of ground meat (X2)
XCC 100
0.173 pound of chicken (X3)
XDC 800
No fish (X4 0)
XEE 400
No beans (X5 0)
Distance 5,400 “student miles”
8-10.
Maximize number of rolls of Supertrex sold 20X1 6.8X2 12X3 65,000X4
X1 dollars spent on advertising
where
X2 dollars spent on store displays
X3 dollars in inventory
subject to
X1 X2 X3 $17,000 (budgeted)
$3,000 (advertising constraint)
X2
0.105 pound of spinach (X6)
0.762 pound of white potatoes (X7)
Each meal has a cost of $1.75 (in six iterations).
The meal is fairly well-balanced (two meats, a green vegetable, and a potato). The weight of each item is realistic. This
problem is very sensitive to changing food prices.
Sensitivity analysis when prices change:
X4 percent markup
X1
50
All Xi 0
0.05X3 (or X2 0.05X3 0)
(ratio of displays to inventory)
X4 0.20 
 (markup ranges)
X4 0.45 
X1, X2, X3, X4 0
Problem 8-10 solved by computer:
Spend $17,000 on advertising (X1).
Spend nothing on in-store displays or on-hand inventory
(X2 and X3).
Take a 20% markup.
The store will sell 327,000 rolls of Supertrex (in six iterations).
This solution implies that no on-hand inventory or displays
are needed to sell the product, probably due to an oversight on
Mr. Kruger’s part. Perhaps a constraint indicating that X3 $3,000 of inventory should be held might be added.
8-11. Minimize total cost $0.60X1 2.35X2 1.15X3 2.25X4 0.58X5 1.17X6 0.33X7
subject to
295X1 1,216X2 394X3 358X4 128X5
118X6 279X7 1,500
295X1 1,216X2 394X3 358X4 128X5
118X6 279X7 900
.2X1 121.2X2 .4.3X3 3.2X4 3.2X5
14.1X6 2.2X7 4
16X1 1,296X2 .4.9X3 0.5X4 0.8X5
1.4X6 0.5X7 50
Milk increases 10 cents/lb: no change in price or diet
Milk decreases 10 cents/lb: no change in price or diet
Milk decreases 30 cents/lb (to 30 cents): potatoes drop out and
milk enters, price $1.42/meal
Ground meat increases from $2.35 to $2.75: price $1.93 and
spinach leaves the optimal solution
Ground meat increases to $5.25/lb: price $2.07 and meat
leaves; milk, chicken, and potatoes in solution
Fish decreases from $2.25 to $2.00/lb: no change
Chicken increases to $3.00/lb: price $1.91 and meat, fish,
spinach, and potatoes in solution
If meat and fish are omitted from the problem, the solution is
chicken 0.774 lb
milk
1.891 lb
potatoes 0.133 lb
If chicken and meat are omitted;
fish
0.679 lb
spinach 0.0988 lb
milk
2.188 lb
8-12.
a. Let X1 no. of units of internal modems produced per
week
X2 no. of units of external modems produced
per week
X3 no. of units of circuit boards produced per
week
X4 no. of units of floppy disk drives produced
per week
X5 no. of units of hard drives produced per
week
X6 no. of units of memory boards produced per
week
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Objective function analysis: First find the time used on each test
device:
subject to
130X1 90Y1 40,000 (Aug. need, hours)
hours on test device 1
7 X 3 X2 12 X3 6 X4 18 X5 17 X6
= 1
60
hours on test device 2
2 X 5 X2 3 X3 2 X4 15 X5 17 X6
1
60
hours on test device 3
130X2 90Y2 45,000 (Sept. need)
130X3 90Y3 35,000 (Oct. need)
130X4 90Y4 50,000 (Nov. need)
130X5 90Y5 45,000 (Dec. need)
X1 350 (starting staff on Aug. 1)
X2 X1 Y1 0.05X1 (staff on Sept. 1)
X3 X2 Y2 0.05X2 (staff on Oct. 1)
5 X1 1X2 3 X3 2 X4 9 X5 2 X6
60
Thus, the objective function is
maximize profit revenue material cost test cost
X4 X3 Y3 0.05X3 (staff on Nov. 1)
X5 X4 Y4 0.05X4 (staff on Dec. 1)
All Xi, Yi 0
200X1 120X2 180X3 130X4 430X5
b.
The computer-generated results are:
260X6
35X1 25X2 40X3 45X4 170X5 60X6
7 X 3 X2 12 X3 6 X4 18 X5 17 X6
15 1
60
2 X1 5 X2 3 X3 2 X4 15 X5 17 X6
12
60
5 X1 1X2 3 X3 2 X4 9 X5 2 X6
18
60
This can be rewritten as
maximize profit $161.35X1 92.95X2 135.50X3
82.50X4 249.80X5 191.75X6
subject to
7 X1 3 X2 12 X3 6 X4 18 X5 17 X6
120 hours
60
2 X1 5 X2 3 X3 2 X4 15 X5 17 X6
120 hours
60
5 X1 1X2 3 X3 2 X4 9 X5 2 X6
100 hours
60
All variables 0
b.
The solution is
X1 496.55 internal modems
X2 1,241.38 external modems
X3 through X6 0
profit $195,504.80
c. The shadow prices, as explained in Chapters 7 and 9,
for additional time on the three test devices are $21.41,
$5.75, and $0, respectively, per minute.
8-13.
a.
Let Xi no. of trained technicians available at start of
month i
Yi no. of trainees beginning in month i
Minimize total salaries paid $2,000X1
2,000X2 2,000X3 2,000X4 2,000X5
900Y1 900Y2 900Y3 900Y4 900Y5
113
LINEAR PROGRAMMING MODELING APPLICATIONS
Month
Trained
Technicians
Available
Trainees
Beginning
350
346.2
328.8
384.6
365.4
13.7 (actually 14)
0
72.2 (actually 72)
0
0
Aug.
Sept.
Oct.
Nov.
Dec.
Total salaries paid over the five-month period $3,627,279.
8-14.
a.
Let Xij acres of crop i planted on parcel j
i 1 for wheat, 2 for alfalfa, 3 for barley
where
j 1 to 5 for SE, N, NW, W, and SW parcels
Irrigation limits:
1.6X11 2.9X21 3.5X31 3,200 acre-feet in SE
1.6X12 2.9X22 3.5X32 3,400 acre-feet in N
1.6X13 2.9X23 3.5X33 800 acre-feet in NW
1.6X14 2.9X24 3.5X34 500 acre-feet in W
1.6X15 2.9X25 3.5X35 600 acre-feet in SW
5
∑
5
1.6 X1 j j 1
∑
j 1
5
2.9 X2, j ∑ 3.5X
3, j 7, 400
j 1
water acre-feet total
Sales limits:
X11 X12 X13 X14 X15 2,200 wheat in acres
( 110,000 bushels)
X21 X22 X23 X24 X25 1,200 alfalfa in acres
( 1,800 tons)
X31 X32 X33 X34 X35 1,000 barley in acres
( 2,200 tons)
Acreage availability:
X11 X21 X31 2,000 acres in SE parcel
X12 X22 X32 2,300 acres in N parcel
X13 X23 X33 600 acres in NW parcel
X14 X24 X34 1,100 acres in W parcel
X15 X25 X35 500 acres in SW parcel
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Objective function:
The solution is infeasible.
5
maximize profit 8-16. This problem refers to Problem 8-15’s infeasibility. Some
investigative work is needed to track down the issues. From a final
simplex tableau, we find that constraints 5 and 11 still have artificial variables in the final solution. The two issues are:
∑ $2(50 bushels) X
1, j
j 1
5
∑
5
$40(1.5 tons) X2, j j 1
∑ ($50)(2.2 tons) X
3, j
j 1
b. The solution is to plant
X12 1,250 acres of wheat in N parcel
X13 500 acres of wheat in NW parcel
X14 31212 acres of wheat in W parcel
X15 13712 acres of wheat in SW parcel
X25 131 acres of alfalfa in SW parcel
X31 600 acres of barley in SE parcel
X32 400 acres of barley in N parcel
Profit will be $337,862.10. Multiple optimal solutions exist.
c. Yes, need only 500 more water-feet.
8-15. Amalgamated’s blending problem will have eight variables and 11 constraints. The eight variables correspond to the
eight materials available (three alloys, two irons, three carbides)
that can be selected for the blend. Six of the constraints deal with
maximum and minimum quality limits, one deals with the 2,000
pound total weight restriction, and four deal with the weight availability limits for alloy 2 (300 lb), carbide 1 (50 lb), carbide 2 (200
lb), and carbide 3 (100 lb).
Let X1 through X8 represent pounds of alloy 1 through pounds
of carbide 3 to be used in the blend.
Minimize cost 0.12X1 0.13X2 0.15X3 0.09X4
0.07X5 0.10X6 0.12X7 0.09X8
subject to
manganese quality:
0.70X1 0.55X2 0.12X3 0.01X4 0.05X5 42
(2.1% of 2,000)
0.70X1 0.55X2 0.12X3 0.01X4 0.05X5 46
(2.3% of 2,000)
silicon quality:
0.15X1 0.30X2 0.26X3 0.10X4 0.025X5
0.24X6 0.25X7 0.23X8 86 (4.3% of 2,000)
0.15X1 0.30X2 0.26X3 0.10X4 0.025X5
0.24X6 0.25X7 0.23X8 92 (4.6% of 2,000)
carbon quality:
0.03X1 0.01X2 0.03X4 0.18X6 0.20X7
0.25X8 101 (5.05% of 2,000)
0.03X1 0.01X2 0.03X4 0.18X6 0.20X7
0.25X8 107 (5.35% of 2,000)
Availability by weight:
X2 300
X6 50
X7 200
X8 100
One-ton weight:
X1 X2 X3 X4 X5 X6 X7 X8 2,000
1.
2.
Requiring at least 5.05% carbon is not possible.
Producing 1 ton from the materials is not possible.
If constraints 5 and 11 are relaxed (or removed), one solution
is X2 84 lb (alloy 2), X6 50 lb (carbide 1), X7 104 lb (carbide 2), and X8 100 lb (carbide 3). Cost $37.31.
Each student may take a different approach and other recommendations may result.
8-17.
X1 number of medical patients
X2 number of surgical patients
Maximize revenue $2,280X1 $1,515X2
subject to
8X1 2.5X2 32,850 (patient-days available
365 days 90 new beds)
3.1X1 2.6X2 15,000 (lab tests)
1X1 2.2X2 7,000 (x-rays)
X2 2,800 (operations/surgeries)
X1, X2 0
Problem 8-17 solved by computer:
X1 2,791 medical patients
X2 2,105 surgical patients
revenue $9,551,659 per year
To convert X1 and X2 to number of medical versus surgical beds,
find the total number of hospital days for each type of patient:
medical (2,791 patients)(8 days/patient)
22,328 days
surgical (2,105 patients)(5 days/patient)
10,525 days
total 32,853 days
This represents 68% medical days and 32% surgical days, which
yields 61 medical beds and 29 surgical beds. (Note that an alternative approach would be to formulate with X1, X2 as number of
beds.)
See the printout on the next page for the solution and sensitivity analysis.
8-18. This problem, suggested by Professor C. Vertullo, is an
excellent exercise in report writing. Here is a chance for students
to present management science results in a management format.
Basically, the following issues need to be addressed in any report:
(a) As seen in Problem 8-17, there should be 61 medical
and 29 surgical beds, yielding $9,551,659 per year.
(b) Referring to the QM for Windows printout, there are
no empty beds.
(c) There are 876 lab tests of unused capacity.
(d) The x-ray is used to its maximum and has a $65.45
shadow price.
(e) The operating room still has 695 operations available.
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Printout for Problems 8-17 and 8-18
Sensitivity Analysis Printout for Problems 8-17 and 8-18
LINEAR PROGRAMMING MODELING APPLICATIONS
115
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8-19.
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a.
LINEAR PROGRAMMING MODELING APPLICATIONS
Let
A1 tons of ore from mine A to plant 1
A2 tons of ore from mine A to plant 2
B1 tons of ore from mine B to plant 1
B2 tons of ore from mine B to plant 2
X1 tons shipped to Builder’s Home from plant 1
b. The solution is the same as problem 8-19 except the
value of the objective function is $34,300.
8-21. Minimize time 12XA1 11XA2 8XA3 9XA4 6XA5 6XA6 6XG1 12XG2 7XG3 7XG4 5XG5 8XG6 8XS1 9XS2 6XS3 6XS4 7XS5 9XS6
subject to
XA1 XA2 XA3 XA4 XA5 XA6 200
XG1 XG2 XG3 XG4 XG5 XG6 225
XS1 XS2 XS3 XS4 XS5 XS6 275
XA1 XG1 XS1
80
XA2 XG2 XS2
120
XA3 XG3 XS3
150
XA4 XG4 XS4
210
XA5 XG5 XS5
60
XA6 XG6 XS6
80
All variables 0
X2 tons shipped to Builder’s Home from plant 2
Y1 tons shipped to Homeowners’ Headquarters from
plant 1
Y2 tons shipped to Homeowners’ Headquarters from
plant 2
Z1 tons shipped to Hardware City from plant 1
Z2 tons shipped to Hardware City from plant 2
Minimize cost 6A1 8A2 7B1 10B2 13X1 19X2
17Y1 22Y2 20Z1 21Z2
subject to
A1 A2 320
supply at A
B1 B2 450
supply at B
A1 B1 500
capacity at plant 1
A2 B2 500
capacity at plant 2
X1 X2 200 demand at Builder’s Home
Y1 Y2 240
demand at Homeowners’ Headquarters
Z1 Z2 330
demand at Hardware City
A1 B1 X1 Y1 Z1 units shipped into plant 1 must
equal units shipped out of
plant 1
A2 B2 X2 Y2 Z2 units shipped into plant 2 must
equal units shipped out of
plant 2
Solution:
Source
(Station)
Destination
(Wing)
Number of
Trays
5A
5A
5A
3G
3G
3G
1S
1S
5
6
3
1
3
4
4
2
60
80
60
80
90
55
155
120
Optimal cost 4,825 minutes. Multiple optimal
solutions exist.
8-22.
Let
Xi proportion of investment invested in stock i for i 1,
2, . . . , 5
All variables 0
Minimize beta 1.2X1 0.85X2 0.55X3 1.40X4
1.25X5
b. Solving this on the computer, we find the following
solution:
subject to
A1 50
tons of ore from mine A to plant 1
A2 270
tons of ore from mine A to plant 2
B1 450
tons of ore from mine B to plant 1
X1 X2 X3 X4 X5 1
0.11X1 0.09X2 0.065X3 0.15X4 0.13X5 0.11
return should be at least 11%
X1 200
tons shipped to Builder’s Home from plant 1
X1 0.35
Y1 240
tons shipped to Homeowners’ Headquarters
from plant 1
X2 0.35
Z1 60
tons shipped to Hardware City from plant 1
Z2 270
X4 0.35
tons shipped to Hardware City from plant 2
X5 0.35
All other variables equal 0.
Minimum total cost $19,160
8-20. a. The formulation is the same as the formulation in problem 8-19 except for a change in the objective function. We add the
processing cost in the objective function, and the new objective
function is:
Minimize cost 28A1 30A2 25B1 28B2 13X1
19X2 17Y1 22Y2 20Z1 21Z2
All the constraints are the same as in the previous problem.
total of the proportions
must add to 1
no more than 35% in any single stock
X3 0.35
Xi 0 for i 1, 2, . . . , 5
b. Solving this on the computer, we have
X1 0
X2 0.10625
X3 0.35
X4 0.35
X5 0.19375
Minimum beta 1.015
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Return 0.11(0) 0.09(0.10625) 0.065(0.35)
0.15(0.35) 0.13(0.19375) 0.11
8-23.
Let
A 1,000 gallons of fuel to purchase in Atlanta
L 1,000 gallons of fuel to purchase in Los Angeles
N 0 (1,000 gallons of fuel to purchase in New Orleans)
FN 11 (1,000 gallons of fuel remaining when plane lands
in New Orleans)
Total cost $31.2211 ( 1,000)
H 1,000 gallons of fuel to purchase in Houston
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS
N 1,000 gallons of fuel to purchase in New Orleans
8-24. To formulate this problem, we first add an activity G to
represent the end of the project:
FA fuel remaining when plane lands in Atlanta
FL fuel remaining when plane lands in Los Angeles
FH fuel remaining when plane lands in Houston
Objective minimize XG
subject to:
XA 2
XB 3
FN fuel remaining when plane lands in New Orleans
XC 1
Minimize cost 1.15A 1.25L 1.10H 1.18N
XD XA 4
subject to
XF XB 1
A FA 24
minimum amount of fuel board when
leaving Atlanta
A FA 36
maximum amount of fuel board when
leaving Atlanta
L FL 15
minimum amount of fuel board when
leaving Los Angeles
L FL 23
maximum amount of fuel board when
leaving Los Angeles
H FH 9
minimum amount of fuel board when
leaving Houston
H FH 17
maximum amount of fuel board when
leaving Houston
N FN 11
minimum amount of fuel board when
leaving New Orleans
N FN 20
XE XC 5
XE XD 5
XG XE 0
XG XF 0
All variables 0
Solution with QM for Windows:
XA 2
XB 10
XC 6
XD 6
XE 11
XF 11
XG 11
maximum amount of fuel board when
leaving New Orleans
FL A FA (12 0.05(A FA 24))
This says that the fuel on board when the plane lands in Los Angeles
will equal the amount on board at take-off minus the fuel consumed
on that light. The fuel consumed is 12 (thousand gallons) plus 5% of
the excess above 24 (thousand gallons). This simplifies to:
0.95A 0.95 FA FL 10.8
Similarly,
Z 11
8-25.
Let X1 number of Chaunceys mixed
X2 number of Sweet Italians mixed
X3 number of bourbon on the rocks mixed
X4 number of Russian martinis mixed
Maximize total drinks X1 X2 X3 X4
subject to
FH L FL (7 0.05(L FL 15)) becomes 0.95L
0.95FL FH 6.25
1X1 4X3 52 oz (bourbon limit)
FN H FH (3 0.05(H FH 9)) becomes 0.95H
0.95FH FN 2.55
1X1 223X4 64 oz (vodka limit)
FL N FN (5 0.05(N FN 11)) becomes 0.95N
0.95FN FL 4.45
All variables 0
b.
117
The optimal solution is
A 21.1579 (1,000 gallons of fuel to purchase in Atlanta)
FA 6 (1,000 gallons of fuel remaining when plane lands in
Atlanta)
L 0 (1,000 gallons of fuel to purchase in Los Angeles)
FL 15 (1,000 gallons of fuel remaining when plane lands
in Los Angeles)
H 6.2632 (1,000 gallons of fuel to purchase in Houston)
FH 8 (1,000 gallons of fuel remaining when plane lands in
Houston)
1X1 1X2 38 oz (brandy limit)
1X2 113X4 24 oz (dry vermouth limit)
1X1 2X2 36 oz (sweet vermouth limit)
All variables 0
Because a Chauncey (X1) is 14 sweet vermouth, it requires 1 oz of
that resource (each drink totals 4 oz).
Problem 8-25 solved by computer:
Mix 25.99 (or 26) Chaunceys (X1)
Mix 5.00 (or 5) Sweet Italians (X2)
Mix 6.50 (or 612) bourbon on the rocks (X3)
Mix 14.25 (or 1414) Russian martinis (X4)
This is a total of 51.75 drinks (in five iterations).
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8-26. Minimize 6X11 8X12 10X13 7X21 11X22 11X23
4X31 5X32 12X33
subject to
X11 X12 X13 150
X21 X22 X23 175
X31 X32 X33 275
All variables 0
8-27. Let Si 1 if Smith is assigned to Job i, 0 otherwise, for i
1, 2, 3, 4
Ji 1 if Jones is assigned to Job i, 0 otherwise,
for i 1, 2, 3, 4
Di 1 if Davis is assigned to Job i, 0 otherwise,
for i 1, 2, 3, 4
Ni 1 if Nguyen is assigned to Job i, 0 otherwise,
for i 1, 2, 3, 4
Minimize days 4S1 5J1 4D1 5N1 10S2 14J2 13D2
11N2 8S3 8J3 9D3 7N3 9S4 10J4 12D4 11N4
subject to
S1 J1 D1 N1 1
S2 J2 D2 N2 1
S3 J3 D3 N3 1
S4 J4 D4 N4 1
S1 S2 S3 S4 1
J1 J2 J3 J4 1
D1 D2 D3 D4 1
N1 N2 N3 N4 1
All variables 0
Solving this with QM for Windows, we have S2 1, J4 1, D1 1, and N3 1. So, Smith does Job 2, Jones does Job 4, Davis does
Job 1, and Nguyen does Job 3. The total time is 31 days.
8-28.
2, 3.
Let Xi number of BR54 produced in month i, for i 1,
IXi number of BR54 units in inventory at end of month i,
for i 0, 1, 2, 3.
IYi number of BR49 units in inventory at end of month i,
for i 0, 1, 2, 3.
Minimize cost 80(X1 X2 X3) 95(Y1 Y2 Y3)
0.8(IX1 IX2 IX3) 0.95(IY1 IY2 IY3)
Subject to:
IX0 50
initial inventory of BR54
IY0 50
initial inventory of BR49
maximum production level in August
maximum production level in September
X3 Y3 1,100
maximum production level in October
Y1 IY0 450 IY1
BR49 requirements for August
Y2 IY1 420 IY2
BR49 requirements for September
Y3 IY2 480 IY3
All variables 0
BR49 requirements for October
A computer solution to this results in IX0 50, IX1 190, IX2 130, IX3 100, IY0 50, IY3 150, X1 460, X2 680, X3 470, Y1 400, Y2 420, Y3 630. All other variables 0. The
total cost $267,028.50.
SOLUTION TO RED BRAND CANNERS CASE
1. The main issue in this case is how to allocate 3 million pounds
of tomatoes. The overall objective is to maximize total sales less
variable costs. These costs include production and selling expenses. Twenty percent of the crop was grade A and the rest was
grade B. In setting up the constraints, the amount of grade A tomatoes cannot exceed 20% of 3 million pounds. Thus not more than
600,000 pounds of grade A tomatoes can be used. Similarly, not
more than 2,400,000 pounds of grade B tomatoes can be used.
Furthermore, the demand for 50,000 cases of tomato juice and
80,000 cases of tomato paste should be met. The demand for
whole tomatoes is not a constraint in this problem. Finally, minimum quality requirements should be met. This includes an average of 8 points per pound for whole tomatoes and 6 points per
pound for tomato juice. There is no constraint for tomato paste.
Another issue is whether or not to buy 80,000 additional
pounds of grade A tomatoes. This would increase the amount of
available grade A tomatoes from 600,000 pounds to 680,000
pounds. To answer this question, a new formulation can be made
using the new 680,000-pound constraint and a price of 8.5 cents
per pound for the 80,000 additional pounds of grade A tomatoes in
the objective function. A faster way to resolve this issue is to use
postoptimality analysis, or shadow prices. Using this approach,
you compare the value of the 80,000 additional tomatoes with the
cost, which is 8.5 cents per pound.
2.
Yi number of BR49 produced in month i, for i 1, 2, 3.
X1 Y1 1,100
X2 Y2 1,100
X3 IX2 500 IX3 BR54 requirements for October
X13 X23 X33 300
Cost $4,525.
ending inventory of BR49
X2 IX1 740 IX2 BR54 requirements for September
X12 X22 X32 100
X11 25, X13 125, X23 175, X31 175, X32 100
ending inventory of BR54
IY3 150
X1 IX0 320 IX1 BR54 requirements for August
X11 X21 X31 200
The solution is:
IX3 100
The problem can be formulated using LP as follows:
X1 pounds of whole A tomatoes
X2 pounds of whole B tomatoes
X3 pounds of juice A tomatoes
X4 pounds of juice B tomatoes
X5 pounds of paste A tomatoes
X6 pounds of paste B tomatoes
Maximize: 0.0822X1 0.0822X2 0.066X3 0.066X4 0.074X5 0.074X6
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subject to
1X1 1X2
14,400,000
1X3 1X4
1,000,000
1X5 1X6 2,000,000
1X1 1X3 1X2 1X5
1X4 1X1 3X2
600,000
1X6 2,400,000
0
3X3 1X4
0
All variables 0
The first constraint refers to the 14 million pounds of whole
tomatoes—800,000 cases at 18 pounds per case—that constitutes
maximum demand. Similarly, the maximum demand for tomato
juice is 50,000 cases at 20 pounds per case or 1 million pounds,
and the maximum demand for tomato paste is 80,000 cases at 25
pounds per case or 2 million pounds, and these are constraints 2
and 3. Constraints 4 and 5 reflect the availability of grade A and
grade B tomatoes, respectively, and the last two constraints are the
quality constraints. The requirements that canned tomatoes must
average at least 8 points means that at least three-fourths of the
tomatoes must be grade A:
X1 0.75(X1 X2) X1 3X2 0
Similarly, the requirements that tomato juice must average at least
6 points means that at least one-fourth of the tomato juice must be
grade A, and that is the last constraint.
The coefficients in the objective function are the unit profits.
A case of whole tomatoes (grade A and grade B) sells for $4. The
variable cost (less the tomatoes) is $2.52. Since the tomatoes are
already on hand (and no salvage appears to be possible), they represent a sunk cost and are not part of the decision process. Since
there are 18 pounds per case, the unit profit is (4.00 2.52)/18 0.0822. Similar analyses hold for the other terms in the objective
function.
The solution of the linear programming problem is
X1 525,000
X3 75,000
X5 0
X2 175,000
X4 225,000
X6 2,000,000
The maximum profit is $225,340.
All of the grade A tomatoes are used. The shadow price for
the slack variable in constraint 4 is 0.0903. Each additional pound
of grade A tomatoes costing 8.5 cents will increase profits by
0.093 0.0850 0.0053. A sensitivity analysis indicates that up
to an additional 600,000 pounds of grade A tomatoes could be
purchased without affecting the solution basis.
SOLUTION TO CHASE MANHATTAN BANK CASE
This very advanced and challenging scheduling problem can be
solved most expeditiously using linear programming, preferably
integer programming. Let F denote the number of full-time employees. Some number, F1, of them will work 1 hour of overtime
between 5 P.M. and 6 P.M. each day and some number, F2, of the
full-time employees will work overtime between 6 P.M. and 7 P.M.
There will be seven sets of part-time employees; Pj will be the
number of part-time employees who begin their workday at hour j,
j 1, 2, . . . , 7, with P1 being the number of workers beginning at
9 A.M., P2 at 10 A.M., . . . , P7 at 3 P.M. Note that because part-time
119
employees must work a minimum of 4 hours, none can start after 3
since the entire operation ends at 7 P.M. Similarly, some number of part-time employees, Qj, leave at the end of hour j, j 4, 5,
. . . , 9.
The workforce requirements for the first two hours, 9 A.M.
and 10 A.M., are:
P.M.
F P1
14
F P1 P2 25
At 11 A.M. half of the full-time employees go to lunch; the remaining half go at noon. For those hours:
0.5F P1 P2 P3
26
0.5F P1 P2 P3 P4 38
Starting at 1 P.M., some of the part-time employees begin to leave.
For the remainder of the straight-time day:
F P1 P2 P3 P4 P5 Q4 55
F P1 P2 P3 P4
F P1 P2 P5 P6 Q4 Q5 60
F P1 P2 P3 P4 P5
F P1 P6 P7 Q4 Q5 Q6 51
F P1 P2 P3 P4 P5 P6
F P1 P7 Q4 Q5 Q6 Q7 29
For the two overtime hours:
F1 P1 P2 P3 P4 P5 P6
F1 P1 P2 P7 Q4 Q5 Q6 Q7 Q8 14
F2 P1 P2 P3 P4 P5 P6 P7
F1 P1 P2 Q4 Q5 Q6 Q7 Q8 Q9 9
If the left-hand sides of these 10 constraints are added, one
finds that 7F hours of full-time labor are used in straight time (although 8F are paid for), F1 F2 full-time labor hours are used
and paid for at overtime rates, and the total number of part-time
hours is
10P1 9P2 8P3 7P4 6P5 5P6 4P7 6Q4
5Q5 4Q6 3Q7 2Q8 Q9 128.4
which is 40% of the day’s total requirement of 321 person-hours.
This also leads to the objective function. The total daily labor
cost which must be minimized is
Z 8(10.11)F 8.08(F1 F2) 7.82(10P1 9P2 8P3
7P4 6P5 5P6 4P7 6Q4 5Q5 4Q6
3Q7 2Q8 Q9)
Total overtime for a full-time employee is restricted to 5 hours or
less, an average of 1 hour or less per day per employee. Thus the
number of overtime hours worked per day cannot exceed the number of full-time employees:
F1 F2 F
Since part-time employees must work at least 4 hours per day,
Q4 P1
for those leaving at the end of the fourth hour. At the end of the
fifth hour, those leaving must be drawn from the P1 Q4 remaining plus the P2 that arrived at the start of the second hour:
Q5 P1 P2 Q4
Similarly, for the remainder of the day,
Q6 P1 P2 P3 Q4 Q5
Q7 P1 P2 P3 P4 Q4 Q5 Q6
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Q8 P1 P2 P3 P4 P5 Q4 Q5 Q6 Q7
Q9 P1 P2 P3 P4 P5 P6 Q4 Q5 Q6
Q7 Q8
To ensure that all part-timers who began at 9 A.M. do not work
more than 7 hours:
Q4 Q5 Q6 Q7 P1
Similarly,
Q4 Q5 Q6 Q7 Q8 P1 P2
Q4 Q5 Q Q7 Q8 Q9 P1 P2 P3
Finally, to ensure that all part-time employees leave at some time:
P1 P2 P3 P4 P5 P6 P7
Q4 Q5 Q6 Q7 Q8 Q9
The resulting problem has 16 integer variables and 22 constraints. If integer programming software is not available, the linear
programming problem can be solved and the solution rounded,
making certain that none of the constraints have been violated. Note
that the integer programming solution might also need to be adjusted—if F is an odd integer, 0.5F will not be an integer and the requirement that “half” of the full-time employees go to lunch at 11
A.M. and the other half at noon will have to be altered by assigning
the extra employee to the appropriate hour.
1. The least-cost solution requires 29 full-time employees, 9 of
whom work two hours of overtime per day. In actuality, 18 of the
full-time employees would work overtime on two different days
and 9 would work overtime on one day. Fourteen of the full-time
workers would take lunch at 11 A.M. and the other 15 would take it
at noon. Eleven part-timers would begin at 11 A.M., with 9 of them
leaving at 3 P.M. and the other 2 at 4 P.M. Fifteen part-time employees would work from noon until 4 P.M., and 5 would work
from 2 P.M. until 6 P.M. The resulting cost of 232 hours of straight
time, 18 hours of overtime, and 126 hours of part-time work is
$3,476.28 per day.
This solution is not unique—other work assignments can be
found that result in this same cost.
2. The same staffing would be used every day. In fact, one
would expect different patterns to present themselves on different
days; for example, Fridays are usually much busier bank days than
the others. In addition, the person-hours required for each hour of
the day are assumed to be deterministic. In a real situation, wide
fluctuations will be experienced in a stochastic manner.
The optimal solution results in a considerable amount of idle
time, partly caused by the restriction that employees can start at
the beginning of an hour and leave at the end. Eliminating this restriction might yield better results at the risk of increasing the
problem size.