Solution Qa Case

103

TEACHING SUGGESTIONS

Teaching Suggestion 8.1: Importance of Formulating Large LP Problems.Since computers are used to solve virtually all business LP prob-lems, the most important thing a student can do is to get experi-ence in formulating a wide variety of problems. This chapter pro-vides such a variety.

Teaching Suggestion 8.2: Note on Production Scheduling Problems.The Greenberg Motor example in this chapter is the largest prob-lem in the book in terms of constraints, so it provides a good prac-tice environment. An interesting feature to point out is that LPconstraints 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 canclearly be solved more quickly using QM for Windows’ assign-ment program than by the LP program. Students should be askedwhy anyone would choose to use the LP approach. There are twoanswers: (1) many commercial LP programs do not contain as-signment algorithms (which are more popular in academic soft-ware such as QM for Windows); and (2) the LP program can pro-vide more sensitivity analysis and economic interpretation than isavailable in the assignment module. The assignment problem istreated in Chapter 10.

Teaching Suggestion 8.4: Labor Planning Problem—ArlingtonBank.This example is a good practice tool and lead-in for the ChaseManhattan Bank case at the end of the chapter. Without this exam-ple, 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 thischapter. First, both the diet and fuel blending problems presentedhere are tiny compared to huge real-world blending problems. Butthey do provide some sense of the issues to be faced.

Second, diet problems that are missing the constraints thatforce variety into the diet can be terribly embarrassing. It has beensaid that a hospital in New Orleans ended up with an LP solutionto feed each patient only castor oil for dinner because analysts ne-glected to add constraints forcing a well-rounded diet.

ALTERNATIVE EXAMPLES

Alternative Example 8.1: Natural Furniture Company manu-factures three outdoor products, chairs, benches, and tables. Eachproduct must pass through the following departments before it is

shipped: sawing, sanding, assembly, and painting. The time re-quirements (in hours) are summarized in the tables below.

The production time available in each department each weekand the minimum weekly production requirement to fulfill con-tracts are as follows:

The production manager has the responsibility of specifying pro-duction 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 producesthree grades, A, B, and C, which cost the firm $40, $50, and $60per kilogram, respectively. The products require the labor and ma-terials per batch that are shown on the following page.

8C H A P T E R

Linear Programming Modeling Applications: WithComputer Analyses in Excel and QM for Windows

MinimumCapacity Production

Department (In Hours) Product Level

Sawing 450 Chairs 100Sanding 400 Benches 50Assembly 625 Tables 50Painting 550

UnitProduct Sawing Sanding Assembly Painting Profit

Chairs 1.5 1.0 2.0 1.5 $15Benches 1.5 1.5 2.0 2.0 $10Tables 2.0 2.0 2.5 2.0 $20

Hours Required

M08_REND6289_10_IM_C08.QXD 5/7/08 2:26 PM 03REVISED

104 CHAPTER 8 LINEAR PROGRAMMING MODEL ING APPL ICAT IONS

What mix of products would yield minimum cost?Objective function

Minimize cost � 40A � 50B � 60C

Constraints

Labor: 4A � 4B � 5C � 80Raw material #1 200A � 300B � 300C � 6,000Raw material #2 600A � 400B � 500C � 5,000

SOLUTIONS TO PROBLEMS

8-1. Since the decision centers about the production of the twodifferent cabinet models, we let

X1 � number of French Provincial cabinets producedeach day

X2 � number of Danish Modern cabinets produced eachday

Objective: maximize revenue � $28X1 � $25X2

subject to

3X1 � 2X2 � 360 hours (carpentry department)

11–2 X1 � 1X2 � 200 hours (painting department)3–4 X1 � 3–

4 X2 � 125 hours (finishing department)

X1 � 60 units (contract requirement)

X2 � 60 units (contract requirement)

X1, X2 � 0

Problem 8-1 solved by computer:

Produce 60 French Provincial cabinets (X1) per dayProduce 90 Danish Modern cabinets (X2) per dayRevenue � $3,930

8-2. Let X1 � dollars invested in Los Angeles municipal bonds

X2 � dollars invested in Thompson Electronics

X3 � dollars invested in United Aerospace

X4 � dollars invested in Palmer Drugs

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)

X1 � .2 (X1 � X2 � X3 � X4 � X5) (bonds)

or

� .8X1 � .2X2 � .2X3 � .2X4 � .2X5 � 0

X2 � X3 � X4 � .4 (X1 � X2 � X3 � X4 � X5) (combination of electronics, aerospace, and drugs)

or

�0.4X1 � 0.6X2 � 0.6X3 � 0.6X4 � 0.4X5 � 0(X5 � 0.5X1) rewritten as

�0.5X1 � X5 � 0 (nursing home as percent of bonds)

X1, X2, X3, X4, X5 � 0


$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)

$25,000 invested in Happy Days (X5)

This produces an annual return on investment of $20,300.

8-3. Minimize staff size � X1 � X2 � X3 � X4 � X5 � X6

where

Xi � number of workers reporting for start of work at period i(with i � 1, 2, 3, 4, 5, or 6)

X1 � X2 � 12

X2 � X3 � 16

X3 � X4 � 9

X4 � X5 � 11

X5 � X6 � 4

X1 � X6 � 3

All variables � 0The computer solution is to hire 30 workers:

16 begin at 7 A.M.

9 begin at 3 P.M.

2 begin at 7 P.M.

3 begin at 11 P.M.

An alternative optimum is

3 begin at 3 A.M.

9 begin at 7 A.M.

7 begin at 11 A.M.

2 begin at 3 P.M.

9 begin at 7 P.M.

0 begin at 11 P.M.

8-4. Let X1 � number of pounds of oat product per horse eachday

X2 � number of pounds of enriched grain per horseeach day

X3 � number of pounds of mineral product per horseeach day

Minimize cost � 0.09X1 � 0.14X2 � 0.17X3

subject to

2X1 � 3X2 � 1X3 � 6 (ingredient A)1–2X1 � 1X2 � 1–

2X3 � 2 (ingredient B)

3X1 � 5X2 � 6X3 � 9 (ingredient C)

1X1 � 11–2X2 � 2X3 � 8 (ingredient D)

1–2X1 � 1–

2X2 � 11–2X3 � 5 (ingredient E)

X1 � X2 � X3 � 6 (maximum feed/day)

All variables � 0

Solution: X1 � 11–3

X2 � 0

X3 � 31–3

cost � 0.687

Grade Grade Grade AvailableA B C Resources

Labor hours 4 4 5 80 hrRaw material #1 200 300 300 6,000 kgRaw material #2 600 400 500 5,000 kg

M08_REND6289_10_IM_C08.QXD 5/7/08 2:26 PM Page 104REVISED

CHAPTER 8 LINEAR PROGRAMMING MODEL ING APPL ICAT IONS 105

8-5. Let Xij � 1 if pitcher i is scheduled to go against opponent j,0 otherwise

where i � 1, 2, 3, 4 stands for Jones, Baker, Parker, andWilson, respectively, and

j � 1, 2, 3, 4 stands for Des Moines, Davenport,Omaha, and Peoria, respectively.

Objective: maximize sum of ratings �

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

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)

X13 � X23 � X33 � X43 � 1 (Omaha)

X14 � X24 � X34 � X44 � 1 (Peoria)

Solution: X12 � 1, X23 � 1, X34 � 1, X41 � 1, Total P � 2.9

8-6. Let

T � number of TV ads

R � number of radio ads

B � number of billboard ads

N � number of newspaper ads

Maximize total audience � 30,000T � 22,000R � 24,000B �8,000N

Subject to

800T � 400R � 500B � 100N � 15,000

� 10

R 10

10

� 10

� � R � 6

500B � 100N 800T

�, R, , � � 0

Solution: T � 6.875; R � 10; B � 9; N � 10; Audience reached� 722,250.If integer solutions are necessary, integer programming (see Chapter 11) could be used.

8-7. Let: X1 � number of newspaper ads placed

X2 � number of TV spots purchased

Minimize cost � $925X1 � $2,000X2

subject to 0.04X1 � 0.05X2 � 0.40 (city exposure)

0.03X1 � 0.03X2 � 0.60 (exposure innorthwest suburbs)

X1, X2 � 0

Note that the problem is not limited to unduplicated exposure(e.g., one person seeing the Sunday newspaper three weeks in arow counts for three exposures).


Buy 20 Sunday newspaper ads (X1)

Buy 0 TV ads (X2)

This has a cost of $18,500. Perhaps the paint store should considera blend of TV and newspaper, not just the latter.

8-8. Let Xij � number of new leases in month i for j-months, i � 1, . . . , 6; j � 3, 4, 5

Minimize cost � 1260X13 � 1260X23 � 1260X33 � 1260X43

� 840X53 � 420X63 � 1600X14 � 1600X24

� 1600X34 � 1200X44 � 800X54 � 400X64

� 1850X15 � 1850X25 � 1480X35 � 1110X45

� 740X55 � 370X65

subject to: X13 � X14 � X15 � 420 � 390

X13 � X14 � X15 � X23 � X24

� X25 � 400 � 270

X13 � X14 � X15 � X23 � X24 � X25 � X33

� X34 � X35 � 430 � 130

X14 � X15 � X23 � X24 � X25 � X33 � X34

� X35 � X43 � X44 � X45 � 460

X15 � X24 � X25 � X33 � X34 � X35 � X43

� X44 � X45 � X53 � X54 � X55 � 470

X25 � X34 � X35 � X43 � X44 � X45 � X53

� X54 � X55 � X63 � X64 � X65 � 440

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)

All variables � 0

Solving this on the computer results in the following solution:

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.

As a result of this, there are 440 cars remaining at the end of August.

8-9. The linear program has the same constraints as in problem8-8. The objective function changes and is now:

Minimize cost � 1260(X13 � X23 � X33 � X43 � X53 � X63)

� 1600(X14 � X24 � X34 � X44 � X54 � X64)

� 1850(X15 � X25 � X35 � X45 � X55

� X65)

Solving this on the computer results in the following solution:

X15 � 30 5-month leases in March

X25 � 100 5-month leases in April





X43 � 160 3-month leases in June

X53 � 10 3-month leases in July


Total cost � $752,950.

8-10. Let Xij � number of students bused from sector i to school j

Objective: minimize total travel miles �

5XAB � 8XAC � 6XAE

� 0XBB � 4XBC � 12XBE

� 4XCB � 0XCC � 7XCE

� 7XDB � 2XDC � 5XDE

� 12XEB � 7XEC � 0XEE

subject to

XAB � XAC � XAE � 700 (number of students in sector A)

XBB � XBC � XBE � 500 (number of students in sector B)

XCB � XCC � XCE � 100 (number of students in sector C)

XDB � XDC � XDE � 800 (number of students in sector D)

XEB � XEC � XEE � 400 (number students in sector E)

XAB � XBB � XCB � XDB � XEB � 900 (school B capacity)

XAC � XBC � XCC � XDC � XEC � 900 (school C capacity)

XAE � XBE � XCE � XDE � XEE � 900 (school E capacity)

All variables � 0

Solution: XAB � 400

XAE � 300

XBB � 500

XCC � 100

XDC � 800

XEE � 400

Distance � 5,400 “student miles”

8-11. Maximize number of rolls of Supertrex sold �20X1 � 6.8X2 � 12X3 � 65,000X4

where X1 � dollars spent on advertising

X2 � dollars spent on store displays

X3 � dollars in inventory

X4 � percent markup

subject to

X1 � X2 � X3 � $17,000 (budgeted)

X1 � $3,000 (advertising constraint)

X2 � 0.05X3 (or X2 � 0.05X3 � 0)(ratio of displays to inventory)

(markup ranges)

X1, X2, X3, X4 � 0Problem 8-11 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.

X

X4

4

0 20

0 45

�

.

.

⎫⎬⎪

⎭⎪

This solution implies that no on-hand inventory or displaysare 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 needed.

8-12. 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

16X1 � 81X2 � 74X3 � 83X4 � 7X5

� 14X6 � 8X7 � 26

22X1 � 28X5 � 19X6 � 63X7 � 50

All Xi � 0


The meal plan for the evening is

No milk (X1 � 0)

0.499 pound of ground meat (X2)

0.173 pound of chicken (X3)

No fish (X4 � 0)

No beans (X5 � 0)

0.105 pound of spinach (X6)

0.762 pound of white potatoes (X7)

Each meal has a cost of $1.75.

The meal is fairly well-balanced (two meats, a green veg-etable, and a potato). The weight of each item is realistic. Thisproblem is very sensitive to changing food prices.

Sensitivity analysis when prices change:

Milk increases 10 cents/lb: no change in price or dietMilk decreases 10 cents/lb: no change in price or dietMilk decreases 30 cents/lb (to 30 cents): potatoes drop out and

milk enters, price � $1.42/mealGround meat increases from $2.35 to $2.75: price � $1.93 and

spinach leaves the optimal solutionGround meat increases to $5.25/lb: price � $2.07 and meat

leaves; milk, chicken, and potatoes in solutionFish decreases from $2.25 to $2.00/lb: no changeChicken 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 lbmilk � 1.891 lbpotatoes � 0.133 lb

If chicken and meat are omitted;

fish � 0.679 lbspinach � 0.0988 lbmilk � 2.188 lb



8-13. a. Let X1 � no. of units of internal modems produced perweek

X2 � no. of units of external modems producedper week

X3 � no. of units of circuit boards produced perweek

X4 � no. of units of floppy disk drives producedper week

X5 � no. of units of hard drives produced perweek

X6 � no. of units of memory boards produced perweek

Objective function analysis: First find the time used on each testdevice:

hours on test device 1



Thus, the objective function is

maximize profit � revenue � material cost � test cost

� 200X1 � 120X2 � 180X3 � 130X4 � 430X5

� 260X6

� 35X1 � 25X2 � 40X3 � 45X4 � 170X5 � 60X6

This can be rewritten as

maximize profit � $161.35X1 � 92.95X2 � 135.50X3

� 82.50X4 � 249.80X5 � 191.75X6

subject to

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

5 1 3 2 9 2

601001 2 3 4 5 6X X X X X X� � � � �

hours

2 5 3 2 15 17

601201 2 3 4 5 6X X X X X X� � � � �

hours

7 3 12 6 18 17

601201 2 3 4 5 6X X X X X X� � � � �

hours

��

185 1 3 2 9 2

601 2 3 4 5 6X X X X X X

��

122 5 3 2 15 17

601 2 3 4 5 6X X X X X X

��

157 3 12 6 18 17

601 2 3 4 5 6X X X X X X

�� 5 1 3 2 9 2

601 2 3 4 5 6X X X X X X

�� 2 5 3 2 15 17

601 2 3 4 5 6X X X X X X

=7 3 12 6 18 17

601 2 3 4 5 6X X X X X X� � � � �

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.

8-14. a. Let Xi � no. of trained technicians available at start ofmonth 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

subject to

130X1 � 90Y1 � 40,000 (Aug. need, hours)

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)

X4 � X3 � Y3 � 0.05X3 (staff on Nov. 1)

X5 � X4 � Y4 � 0.05X4 (staff on Dec. 1)

All Xi, Yi � 0

b. The computer-generated results are:

Total salaries paid over the five-month period � $3,627,279.

8-15. a. Let Xij � acres of crop i planted on parcel j

where i � 1 for wheat, 2 for alfalfa, 3 for barley

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

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)

1 6 2 9 3 5 7 4001 21

5

1

5

31

5

. . . ,, ,X X Xj jjj

jj

� � �� ∑∑ ∑

TrainedTechnicians Trainees

Month Available Beginning

Aug. 350 13.7 (actually 14)Sept. 346.2 0Oct. 328.8 72.2 (actually 72)Nov. 384.6 0Dec. 365.4 0



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

Objective function:

maximize profit

b. The solution is to plant

X12 � 1,250 acres of wheat in N parcel

X13 � 500 acres of wheat in NW parcel

X14 � 3121–2 acres of wheat in W parcel

X15 � 1371–2 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.

8-16. Amalgamated’s blending problem will have eight variablesand 11 constraints. The eight variables correspond to the eight materi-als available (three alloys, two irons, three carbides) that can be se-lected for the blend. Six of the constraints deal with maximum andminimum quality limits, one deals with the 2,000 pound total weightrestriction, 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 poundsof 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 tomanganese 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)

� � ($5$40(1.5 tons) X ,j2 00 2 2 31

5

1

5

)( . ) ,tons X jjj ��∑∑

� $ ( )2 50 bushels X1,j1

5

j �∑

Availability by weight:

� X2 � 300

X6 � 50

X7 � 200

� X8 � 100

One-ton weight:

� X1 � X2 � X3 � X4 � X5 � X6 � X7 � X8 � 2,000

The solution is infeasible.

8-17. This problem refers to Problem 8-16’s infeasibility. Someinvestigative work is needed to track down the issues. From a finalsimplex tableau, we find that constraints 5 and 11 still have artifi-cial variables in the final solution. The two issues are:

1. Requiring at least 5.05% carbon is not possible.2. Producing 1 ton from the materials is not possible.

If constraints 5 and 11 are relaxed (or removed), one solutionis X2 � $83.6 (alloy 2), X6 � 50 lb (carbide 1), X7 � $83.6 (car-bide 2), and X8 � 100 lb (carbide 3). Cost � $34.91.

Each student may take a different approach and other recom-mendations may result.

8-18. 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


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, whichyields 61 medical beds and 29 surgical beds. (Note that an alterna-tive approach would be to formulate with X1, X2 as number of beds.)

See the printout on the next page for the solution and sensi-tivity analysis.

8-19. This problem, suggested by Professor C. Vertullo, is anexcellent exercise in report writing. Here is a chance for studentsto present management science results in a management format.Basically, the following issues need to be addressed in any report:

(a) As seen in Problem 8-18, there should be 61 medicaland 29 surgical beds, yielding $9,551,659 per year.



Printout for Problems 8-18 and 8-19

Sensitivity Analysis Printout for Problems 8-18 and 8-19



(b) Referring to the QM for Windows printout, there areno empty beds.(c) There are 876 lab tests of unused capacity.(d) The x-ray is used to its maximum and has a $65.45shadow price.(e) The operating room still has 695 operations available.

8-20. 8-20. Let

Si � 1 if Smith is assigned to Job i for i � 1, 2, 3, 4

� 0 otherwise

Ji � 1 if Jones is assigned to Job i for i � 1, 2, 3, 4

� 0 otherwise

Di � 1 if Davis is assigned to Job i for i � 1, 2, 3, 4

� 0 otherwise

Ni � 1 if Nguyen is assigned to Job i for i � 1, 2, 3, 4

� 0 otherwise

Minimize days � 4S1 � 10S2 � 8S3 � 9S4 � 5J1 � 14J2

� 8J3 � 10J4 � 4D1 � 13D2 � 9D3 � 12D4 � 5N1 �11N2 � 7N3 � 11N4

Subject to

S1 � S2 � S3 � S4 � 1

J1 � J2 � J3 � J4 � 1

D1 � D2 � D3 � D4 � 1

N1 � N2 � N3 � N4 � 1

S1 � J1 � D1 � N1 � 1

S2 � J2 � D2 � N2 � 1

S3 � J3 � D3 � N3 � 1

S4 � J4 � D4 � N4 � 1

All variables 0, 1There are multiple optimal solutions. All of these require a total of31 days. One solution is to assign Smith to Job 2, Jones to Job 4,Davis to Job 1, and Nguyen to Job 3.

8-21. a. 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

X2 � tons shipped to Builder’s Home from plant 2

Y1 � tons shipped to Homeowners’ Headquarters fromplant 1

Y2 � tons shipped to Homeowners’ Headquarters fromplant 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 mustequal units shipped out of plant 1

A2 � B2 � X2 � Y2 � Z2 units shipped into plant 2 mustequal units shipped out of plant 2

All variables � 0

b. Solving this on the computer, we find the following solution:

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 � 200 tons shipped to Builder’s Home from plant 1

Y1 � 240 tons shipped to Homeowners’ Headquartersfrom plant 1

Z1 � 60 tons shipped to Hardware City from plant 1

Z2 � 270 tons shipped to Hardware City from plant 2


Minimum total cost � $19,160

8-22. a. The formulation is the same as the formulation in prob-lem 8-21 except for a change in the objective function. We add theprocessing cost in the objective function, and the new objectivefunction is:

Minimize cost � 28A1 � 30A2 � 25B1 � 28B2 � 13X1

� 19X2 � 17Y1 � 22Y2 � 20Z1 � 21Z2

All the constraints are the same as in the previous problem.

b. The solution is the same as problem 8-21 except thevalue of the objective function is $34,300.

8-23. 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 � 200XG1 � XG2 � XG3 � XG4 � XG5 � XG6 � 225XS1 � XS2 � XS3 � XS4 � XS5 � XS6 � 275

XA1 � XG1 � XS1 � 80XA2 � XG2 � XS2 � 120XA3 � XG3 � XS3 � 150XA4 � XG4 � XS4 � 210XA5 � XG5 � XS5 � 60XA6 � XG6 � XS6 � 80

All variables � 0


A � FA � 36 maximum amount of fuel board whenleaving Atlanta

L � FL � 15 minimum amount of fuel board whenleaving Los Angeles

L � FL � 23 maximum amount of fuel board whenleaving Los Angeles

H � FH � 9 minimum amount of fuel board whenleaving Houston

H � FH � 17 maximum amount of fuel board whenleaving Houston

N � FN � 11 minimum amount of fuel board whenleaving New Orleans

N � FN � 20 maximum amount of fuel board whenleaving New Orleans

FL � A � FA � (12 � 0.05(A � FA � 24))

This says that the fuel on board when the plane lands in Los Angeleswill equal the amount on board at take-off minus the fuel consumedon that flight. The fuel consumed is 12 (thousand gallons) plus 5% ofthe excess above 24 (thousand gallons). This simplifies to:

0.95A � 0.95 FA � FL � 10.8Similarly,

FH � L � FL � (7 � 0.05(L � FL � 15)) becomes 0.95L � 0.95FL � FH � 6.25

FN � H � FH � (3 � 0.05(H � FH � 9)) becomes 0.95H � 0.95FH � FN � 2.55

FA � N � FN � (5 � 0.05(N � FN � 11)) becomes 0.95N � 0.95FN � FA � 4.45

All variables � 0

b. The optimal solution is

A � 18 (1,000 gallons of fuel to purchase in Atlanta)

FA � 6 (1,000 gallons of fuel remaining when plane lands inAtlanta)

L � 3 (1,000 gallons of fuel to purchase in Los Angeles)

FL � 12 (1,000 gallons of fuel remaining when plane landsin Los Angeles)

H � 1 (1,000 gallons of fuel to purchase in Houston)

FH � 8 (1,000 gallons of fuel remaining when plane lands inHouston)

N � 5 (1,000 gallons of fuel to purchase in New Orleans)

FN � 6 (1,000 gallons of fuel remaining when plane lands inNew Orleans)

Total cost � 112.45 ( 1,000)

SOLUTIONS TO INTERNET HOMEWORK PROBLEMS

8-26. To formulate this problem, we first add an activity G torepresent the end of the project:

Objective � minimize XG

subject to: XA � 2

XB � 3

XC � 1

XD � XA � 4

XF � XB � 1


8-24. Let

Xi � proportion of investment invested in stock i for i � 1, 2, . . . , 5

Minimize beta � 1.2X1 � 0.85X2 � 0.55X3 � 1.40X4

� 1.25X5

subject to

X1 � X2 � X3 � X4 � X5 � 1 total of the proportionsmust add to 1

0.11X1 � 0.09X2 � 0.065X3 � 0.15X4 � 0.13X5 � 0.11return should be at least 11%

X1 � 0.35 no more than 35% in any single stock

X2 � 0.35

X3 � 0.35

X4 � 0.35

X5 � 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

Return � 0.11(0) � 0.09(0.10625) � 0.065(0.35)� 0.15(0.35) � 0.13(0.19375) � 0.11

8-25. Let

A � 1,000 gallons of fuel to purchase in Atlanta

L � 1,000 gallons of fuel to purchase in Los Angeles

H � 1,000 gallons of fuel to purchase in Houston

N � 1,000 gallons of fuel to purchase in New Orleans

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

FN � fuel remaining when plane lands in New Orleans

Minimize cost � 4.15A � 4.25L � 4.10H � 4.18N

subject to

A � FA � 24 minimum amount of fuel board whenleaving Atlanta

Source Destination Number of(Station) (Wing) Trays

5A 5 605A 6 805A 3 603G 1 803G 3 903G 4 551S 4 1551S 2 120

Optimal cost � 4,825 minutes. Multiple optimalsolutions exist.

Solution:



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

Z � 11

8-27. 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

1X1 �4X3 � 52 oz (bourbon limit)

1X1 �1X2 � 38 oz (brandy limit)

1X1 �22–3X4 � 64 oz (vodka limit)

1X2 �11–3X4 � 24 oz (dry vermouth limit)

1X1 �2X2 � 36 oz (sweet vermouth limit)

All variables � 0

Because a Chauncey (X1) is 1–4 sweet vermouth, it requires 1 oz of

that resource (each drink totals 4 oz).


Mix 25.99 (or 26) Chaunceys (X1)

Mix 5.00 (or 5) Sweet Italians (X2)

Mix 6.50 (or 61–2) bourbon on the rocks (X3)

Mix 14.25 (or 141–4) Russian martinis (X4)

This is a total of 51.75 drinks (in five iterations).

8-28. Minimize 6X11 � 8X12 � 10X13 � 7X21 � 11X22 � 11X23

� 4X31 � 5X32 � 12X33

subject to

X11 � X12 � X13 � 150

X21 � X22 � X23 � 175

X31 � X32 � X33 � 275

X11 � X21 � X31 � 200

X12 � X22 � X32 � 100

X13 � X23 � X33 � 300

All variables � 0

The solution is:

X11 � 25, X13 � 125, X23 � 175, X31 � 175, X32 � 100

Cost � $4,525.

8-29. 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 doesJob 1, and Nguyen does Job 3. The total time is 31 days.

8-30. Let Xi � number of BR54 produced in month i, for i � 1,2, 3.

Yi � number of BR49 produced in month i, for i � 1, 2, 3.

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

IX3 � 100 ending inventory of BR54

IY3 � 150 ending inventory of BR49

X1 � Y1 � 1,100 maximum production level in August

X2 � Y2 � 1,100 maximum production level in September

X3 � Y3 � 1,100 maximum production level in October

X1 � IX0 � 320 � IX1 BR54 requirements for August

X2 � IX1 � 740 � IX2 BR54 requirements for September

X3 � IX2 � 500 � IX3 BR54 requirements for October

Y1 � IY0 � 450 � IY1 BR49 requirements for August

Y2 � IY1 � 420 � IY2 BR49 requirements for September

Y3 � IY2 � 480 � IY3 BR49 requirements for OctoberAll variables � 0

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. Thetotal cost � $267,028.50.



SOLUTION TO RED BRAND CANNERS CASE

1. The main issue in this case is how to allocate 3 million poundsof tomatoes. The overall objective is to maximize total sales lessvariable costs. These costs include production and selling ex-penses. Twenty percent of the crop was grade A and the rest wasgrade B. In setting up the constraints, the amount of grade A toma-toes cannot exceed 20% of 3 million pounds. Thus not more than600,000 pounds of grade A tomatoes can be used. Similarly, notmore than 2,400,000 pounds of grade B tomatoes can be used.Furthermore, the demand for 50,000 cases of tomato juice and80,000 cases of tomato paste should be met. The demand forwhole tomatoes is not a constraint in this problem. Finally, mini-mum quality requirements should be met. This includes an aver-age of 8 points per pound for whole tomatoes and 6 points perpound for tomato juice. There is no constraint for tomato paste.

Another issue is whether or not to buy 80,000 additionalpounds of grade A tomatoes. This would increase the amount ofavailable grade A tomatoes from 600,000 pounds to 680,000pounds. To answer this question, a new formulation can be madeusing the new 680,000-pound constraint and a price of 8.5 centsper pound for the 80,000 additional pounds of grade A tomatoes inthe objective function. A faster way to resolve this issue is to usepostoptimality analysis, or shadow prices. Using this approach,you compare the value of the 80,000 additional tomatoes with thecost, which is 8.5 cents per pound.

2. 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

subject to

1X1 � 1X2 � 14,400,000

1X3 � 1X4 � 1,000,000

1X5 � 1X6 � 2,000,000

1X1 � 1X3 � 1X5 � 600,000

1X2 � 1X4 � 1X6 � 2,400,000

1X1 � 3X2 � 0

3X3 � 1X4 � 0

All variables � 0

The first constraint refers to the 14 million pounds of wholetomatoes—800,000 cases at 18 pounds per case—that constitutesmaximum demand. Similarly, the maximum demand for tomatojuice 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 25pounds per case or 2 million pounds, and these are constraints 2and 3. Constraints 4 and 5 reflect the availability of grade A andgrade B tomatoes, respectively, and the last two constraints are thequality constraints. The requirements that canned tomatoes mustaverage at least 8 points means that at least three-fourths of thetomatoes must be grade A:

X1 � 0.75(X1 � X2) �� X1 � 3X2 � 0

Similarly, the requirements that tomato juice must average at least6 points means that at least one-fourth of the tomato juice must begrade A, and that is the last constraint.

The coefficients in the objective function are the unit profits. Acase of whole tomatoes (grade A and grade B) sells for $4. The vari-able cost (less the tomatoes) is $2.52. Since the tomatoes are alreadyon hand (and no salvage appears to be possible), they represent asunk cost and are not part of the decision process. Since there are 18pounds per case, the unit profit is (4.00 � 2.52)/18 � 0.0822. Simi-lar analyses hold for the other terms in the objective function.

The solution of the linear programming problem is

X1 � 525,000 X2 � 175,000X3 � 75,000 X4 � 225,000X5 � 0 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 poundof grade A tomatoes costing 8.5 cents will increase profits by0.093 � 0.0850 � 0.0053. A sensitivity analysis indicates that upto an additional 600,000 pounds of grade A tomatoes could bepurchased without affecting the solution basis.

SOLUTION TO CHASE MANHATTAN BANK CASE

This very advanced and challenging scheduling problem can besolved most expeditiously using linear programming, preferablyinteger programming. Let F denote the number of full-time em-ployees. Some number, F1, of them will work 1 hour of overtimebetween 5 P.M. and 6 P.M. each day and some number, F2, of thefull-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 thenumber of part-time employees who begin their workday at hour j,j � 1, 2, . . . , 7, with P1 being the number of workers beginning at9 A.M., P2 at 10 A.M., . . . , P7 at 3 P.M. Note that because part-timeemployees must work a minimum of 4 hours, none can start after3 P.M. since the entire operation ends at 7 P.M. Similarly, somenumber 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:

F � P1 � 14

F � P1 � P2 � 25

At 11 A.M. half of the full-time employees go to lunch; the remain-ing 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 � P4F � P1 � P2 � P5 � P6 � Q4 � Q5 � 60

F � P1 � P2 � P3 � P4 � P5F � P1 � P6 � P7 � Q4 � Q5 � Q6 � 51

F � P1 � P2 � P3 � P4 � P5 � P6F � P1 � P7 � Q4 � Q5 � Q6 � Q7 � 29

M08_REND6289_10_IM_C08.QXD 5/7/08 2:26 PM 3REVISED


For the two overtime hours:

F1 � P1 � P2 � P3 � P4 � P5 � P6F1 � P1 � P2 � P7 � Q4 � Q5 � Q6 � Q7 � Q8 � 14

F2 � P1 � P2 � P3 � P4 � P5 � P6 � P7F1 � P1 � P2 � Q4 � Q5 � Q6 � Q7 � Q8 � Q9 � 9

If the left-hand sides of these 10 constraints are added, one findsthat 7F hours of full-time labor are used in straight time (although 8Fare paid for), F1 � F2 full-time labor hours are used and paid for atovertime 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 orless, an average of 1 hour or less per day per employee. Thus thenumber of overtime hours worked per day cannot exceed the num-ber 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 thefifth hour, those leaving must be drawn from the P1 � Q4 remain-ing 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

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 workmore 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 con-straints. If integer programming software is not available, the linearprogramming problem can be solved and the solution rounded,making certain that none of the constraints have been violated. Notethat the integer programming solution might also need to be ad-justed—if F is an odd integer, 0.5F will not be an integer and the re-quirement that “half” of the full-time employees go to lunch at 11A.M. and the other half at noon will have to be altered by assigningthe extra employee to the appropriate hour.

1. The least-cost solution requires 29 full-time employees, 9 ofwhom work two hours of overtime per day. In actuality, 18 of thefull-time employees would work overtime on two different daysand 9 would work overtime on one day. Fourteen of the full-timeworkers would take lunch at 11 A.M. and the other 15 would take itat noon. Eleven part-timers would begin at 11 A.M., with 9 of themleaving at 3 P.M. and the other 2 at 4 P.M. Fifteen part-time em-ployees would work from noon until 4 P.M., and 5 would workfrom 2 P.M. until 6 P.M. The resulting cost of 232 hours of straighttime, 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 befound that result in this same cost.

2. The same staffing would be used every day. In fact, onewould expect different patterns to present themselves on differentdays; for example, Fridays are usually much busier bank days thanthe others. In addition, the person-hours required for each hour ofthe day are assumed to be deterministic. In a real situation, widefluctuations will be experienced in a stochastic manner.

The optimal solution results in a considerable amount of idletime, partly caused by the restriction that employees can start atthe beginning of an hour and leave at the end. Eliminating this re-striction might yield better results at the risk of increasing theproblem size.


Solution Qa Case

Documents

Transcript of Solution Qa Case