McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., 2003 4.1 Table of Contents Chapter 4 (Linear...

© The McGraw-Hill Companies, Inc., 20034.1McGraw-Hill/Irwin

Table of ContentsChapter 4 (Linear Programming: Formulation and Applications)

Super Grain Corp. Advertising-Mix Problem (Section 4.1)

4.2–4.5Resource Allocation Problems & Think-Big Capital Budgeting (Section 4.2)

4.6–4.10Cost-Benefit-Trade-Off Problems & Union Airways (Section 4.3)

4.11–4.15Distribution-Network Problems & Big M Co. (Section 4.4)

4.16–4.20Continuing the Super Grain Corp. Case Study (Section 4.5)

4.21–4.24Mixed Formulations & Save-It Solid Waste Reclamation (Section 4.6)

4.25–4.30


Super Grain Corp. Advertising-Mix Problem

• Goal: Design the promotional campaign for Crunchy Start.

• The three most effective advertising media for this product are– Television commercials on Saturday morning programs for children.

– Advertisements in food and family-oriented magazines.

– Advertisements in Sunday supplements of major newspapers.

• The limited resources in the problem are– Advertising budget ($4 million).

– Planning budget ($1 million).

– TV commercial spots available (5).

• The objective will be measured in terms of the expected number of exposures.

Question: At what level should they advertise Crunchy Start in each of the three media?


Cost and Exposure Data

Costs

Cost CategoryEach

TV CommercialEach

Magazine AdEach

Sunday Ad

Ad Budget $300,000 $150,000 $100,000

Planning budget 90,000 30,000 40,000

Expected number of exposures

1,300,000 600,000 500,000


Spreadsheet Formulation

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B C D E F G HTV Spots Magazine Ads SS Ads

Exposures per Ad 1,300 600 500(thousands)

Budget BudgetCost per Ad ($thousands) Spent Available

Ad Budget 300 150 100 4,000 <= 4,000Planning Budget 90 30 40 1,000 <= 1,000

Total ExposuresTV Spots Magazine Ads SS Ads (thousands)

Number of Ads 0 20 10 17,000<=

Max TV Spots 5


Algebraic Formulation

Let TV = Number of commercials for separate spots on televisionM = Number of advertisements in magazines.SS = Number of advertisements in Sunday supplements.

Maximize Exposure = 1,300TV + 600M + 500SSsubject to

Ad Spending: 300TV + 150M + 100SS ≤ 4,000 ($thousand)Planning Cost: 90TV + 30M + 30SS ≤ 1,000 ($thousand)Number of TV Spots: TV ≤ 5

andTV ≥ 0, M ≥ 0, SS ≥ 0.


Think-Big Capital Budgeting Problem

• Think-Big Development Co. is a major investor in commercial real-estate development projects.

• They are considering three large construction projects– Construct a high-rise office building.

– Construct a hotel.

– Construct a shopping center.

• Each project requires each partner to make four investments: a down payment now, and additional capital after one, two, and three years.

Question: At what fraction should Think-Big invest in each of the three projects?


Financial Data for the Projects

Investment Capital Requirements

Year Office Building Hotel Shopping Center

0 $40 million $80 million $90 million

1 60 million 80 million 50 million



Net present value $45 million $70 million $50 million



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B C D E F G HOffice Shopping

Building Hotel CenterNet Present Value 45 70 50

($millions) Cumulative CumulativeCapital Capital

Cumulative Capital Required ($millions) Spent AvailableNow 40 80 90 25 <= 25

End of Year 1 100 160 140 44.757 <= 45End of Year 2 190 240 160 60.583 <= 65End of Year 3 200 310 220 80 <= 80

Office Shopping Total NPVBuilding Hotel Center ($millions)

Participation Share 0.00% 16.50% 13.11% 18.11



Let OB = Participation share in the office building,H = Participation share in the hotel,SC = Participation share in the shopping center.

Maximize NPV = 45OB + 70H + 50SCsubject to

Total invested now: 40OB + 80H + 90SC ≤ 25 ($million)Total invested within 1 year: 100OB + 160H + 140SC ≤ 45 ($million)Total invested within 2 years: 190OB + 240H + 160SC ≤ 65 ($million)Total invested within 3 years: 200OB + 310H + 220SC ≤ 80 ($million)

andOB ≥ 0, H ≥ 0, SC ≥ 0.


Summary of Formulation Procedure for Resource-Allocation Problems

1. Identify the activities for the problem at hand.

2. Identify an appropriate overall measure of performance (commonly profit).

3. For each activity, estimate the contribution per unit of the activity to the overall measure of performance.

4. Identify the resources that must be allocated.

5. For each resource, identify the amount available and then the amount used per unit of each activity.

6. Enter the data in steps 3 and 5 into data cells.

7. Designate changing cells for displaying the decisions.

8. In the row for each resource, use SUMPRODUCT to calculate the total amount used. Enter ≤ and the amount available in two adjacent cells.

9. Designate a target cell. Use SUMPRODUCT to calculate this measure of performance.


Union Airways Personnel Scheduling

• Union Airways is adding more flights to and from its hub airport and so needs to hire additional customer service agents.

• The five authorized eight-hour shifts are– Shift 1: 6:00 AM to 2:00 PM

– Shift 2: 8:00 AM to 4:00 PM

– Shift 3: Noon to 8:00 PM

– Shift 4: 4:00 PM to midnight

– Shift 5: 10:00 PM to 6:00 AM

Question: How many agents should be assigned to each shift?


Schedule Data

Time Periods Covered by Shift

Time Period 1 2 3 4 5

MinimumNumber of

Agents Needed

6 AM to 8 AM √ 48

8 AM to 10 AM √ √ 79

10 AM to noon √ √ 65

Noon to 2 PM √ √ √ 87

2 PM to 4 PM √ √ 64

4 PM to 6 PM √ √ 73

6 PM to 8 PM √ √ 82

8 PM to 10 PM √ 43

10 PM to midnight √ √ 52

Midnight to 6 AM √ 15

Daily cost per agent $170 $160 $175 $180 $195



3456789101112131415161718192021

B C D E F G H I J6am-2pm 8am-4pm Noon-8pm 4pm-midnight 10pm-6am

Shift Shift Shift Shift ShiftCost per Shift $170 $160 $175 $180 $195

Total MinimumTime Period Shift Works Time Period? (1=yes, 0=no) Working Needed

6am-8am 1 0 0 0 0 48 >= 488am-10am 1 1 0 0 0 79 >= 79

10am- 12pm 1 1 0 0 0 79 >= 6512pm-2pm 1 1 1 0 0 118 >= 872pm-4pm 0 1 1 0 0 70 >= 644pm-6pm 0 0 1 1 0 82 >= 736pm-8pm 0 0 1 1 0 82 >= 82

8pm-10pm 0 0 0 1 0 43 >= 4310pm-12am 0 0 0 1 1 58 >= 52

12am-6am 0 0 0 0 1 15 >= 15

6am-2pm 8am-4pm Noon-8pm 4pm-midnight 10pm-6amShift Shift Shift Shift Shift Total Cost

Number Working 48 31 39 43 15 $30,610



Let Si = Number working shift i (for i = 1 to 5),

Minimize Cost = $170S1 + $160S2 + $175S3 + $180S4 + $195S5

subject toTotal agents 6AM–8AM: S1 ≥ 48Total agents 8AM–10AM: S1 + S2 ≥ 79Total agents 10AM–12PM: S1 + S2 ≥ 65Total agents 12PM–2PM: S1 + S2 + S3 ≥ 87Total agents 2PM–4PM: S2 + S3 ≥ 64Total agents 4PM–6PM: S3 + S4 ≥ 73Total agents 6PM–8PM: S3 + S4 ≥ 82Total agents 8PM–10PM: S4 ≥ 43Total agents 10PM–12AM: S4 + S5 ≥ 52Total agents 12AM–6AM: S5 ≥ 15

andSi ≥ 0 (for i = 1 to 5)


Summary of Formulation Procedure forCost-Benefit-Tradeoff Problems

1. Identify the activities for the problem at hand.

2. Identify an appropriate overall measure of performance (commonly cost).

3. For each activity, estimate the contribution per unit of the activity to the overall measure of performance.

4. Identify the benefits that must be achieved.

5. For each benefit, identify the minimum acceptable level and then the contribution of each activity to that benefit.

6. Enter the data in steps 3 and 5 into data cells.

7. Designate changing cells for displaying the decisions.

8. In the row for each benefit, use SUMPRODUCT to calculate the level achieved. Enter ≤ and the minimum acceptable level in two adjacent cells.

9. Designate a target cell. Use SUMPRODUCT to calculate this measure of performance.


The Big M Distribution-Network Problem

• The Big M Company produces a variety of heavy duty machinery at two factories. One of its products is a large turret lathe.

• Orders have been received from three customers for the turret lathe.

Question: How many lathes should be shipped from each factory to each customer?


Some Data

Shipping Cost for Each Lathe

To Customer 1 Customer 2 Customer 3

From Output

Factory 1 $700 $900 $800 12 lathes

Factory 2 800 900 700 15 lathes

Order Size 10 lathes 8 lathes 9 lathes


The Distribution Network

F1

C2

C3

C1

F2

12 latheproduced

15 lathesproduced

10 lathesneeded

8 lathesneeded

9 lathesneeded

$700/lathe

$900/lathe

$800/lathe

$800/lathe $900/lathe

$700/lathe



3456789101112131415

B C D E F G HShipping Cost

(per Lathe) Customer 1 Customer 2 Customer 3Factory 1 $700 $900 $800Factory 2 $800 $900 $700

TotalShipped

Units Shipped Customer 1 Customer 2 Customer 3 Out OutputFactory 1 10 2 0 12 = 12Factory 2 0 6 9 15 = 15

Total To Customer 10 8 9= = = Total Cost

Order Size 10 8 9 $20,500



Let Sij = Number of lathes to ship from i to j (i = F1, F2; j = C1, C2, C3).

Minimize Cost = $700SF1-C1 + $900SF1-C2 + $800SF1-C3 + $800SF2-C1 + $900SF2-C2 + $700SF2-C3

subject toFactory 1: SF1-C1 + SF1-C2 + SF1-C3 = 12Factory 2: SF2-C1 + SF2-C2 + SF2-C3 = 15Customer 1: SF1-C1 + SF2-C1 = 10Customer 2: SF1-C2 + SF2-C2 = 8Customer 3: SF1-C3 + SF2-C3 = 9

andSij ≥ 0 (i = F1, F2; j = C1, C2, C3).


Continuing the Super Grain Case Study

• David and Claire conclude that the spreadsheet model needs to be expanded to incorporate some additional considerations.

• In particular, they feel that two audiences should be targeted — young children and parents of young children.

• Two new goals– The advertising should be seen by at least five million young children.

– The advertising should be seen by at least five million parents of young children.

• Furthermore, exactly $1,490,000 should be allocated for cents-off coupons.


Benefit and Fixed-Requirement Data

Number Reached in Target Category (millions)

EachTV Commercial

EachMagazine Ad

EachSunday Ad

MinimumAcceptable

Level

Young children 1.2 0.1 0 5

Parents of young children 0.5 0.2 0.2 5

Contribution Toward Required Amount

EachTV Commercial

EachMagazine Ad

EachSunday Ad

RequiredAmount

Coupon redemption 0 $40,000 $120,000 $1,490,000



3456789101112131415161718192021

B C D E F G HTV Spots Magazine Ads SS Ads

Exposures per Ad 1,300 600 500(thousands)

Cost per Ad ($thousands) Budget Spent Budget AvailableAd Budget 300 150 100 3,775 <= 4,000

Planning Budget 90 30 40 1,000 <= 1,000

Number Reached per Ad (millions) Total Reached Minimum AcceptableYoung Children 1.2 0.1 0 5 >= 5

Parents of Young Children 0.5 0.2 0.2 5.85 >= 5

TV Spots Magazine Ads SS Ads Total Redeemed Required AmountCoupon Redemption per Ad 0 40 120 1,490 = 1,490

($thousands)Total Exposures

TV Spots Magazine Ads SS Ads (thousands)Number of Ads 3 14 7.75 16,175

<=Maximum TV Spots 5



Let TV = Number of commercials for separate spots on televisionM = Number of advertisements in magazines.SS = Number of advertisements in Sunday supplements.

Maximize Exposure = 1,300TV + 600M + 500SSsubject to

Ad Spending: 300TV + 150M + 100SS ≤ 4,000 ($thousand)Planning Cost: 90TV + 30M + 30SS ≤ 1,000 ($thousand)Number of TV Spots: TV ≤ 5

Young children: 1.2TV + 0.1M ≥ 5 (millions)Parents: 0.5TV + 0.2M + 0.2SS ≥ 5 (millions)

Coupons: 40M + 120SS = 1,490 ($thousand)

andTV ≥ 0, M ≥ 0, SS ≥ 0.


Types of Functional Constraints

Type Form* Typical Interpretation Main Usage

Resource constraint LHS ≤ RHSFor some resource, Amount used ≤ Amount available

Resource-allocation problems and mixed problems

Benefit constraint LHS ≥ RHSFor some benefit, Level achieved ≥ Minimum Acceptable

Cost-benefit-trade-off problems and mixed problems

Fixed-requirement constraint

LHS = RHSFor some quantity, Amount provided = Required amount

Distribution-network problems and mixed problems

* LHS = Left-hand side (a SUMPRODUCT function). RHS = Right-hand side (a constant).


Save-It Company Waste Reclamation

• The Save-It Company operates a reclamation center that collects four types of solid waste materials and then treats them so that they can be amalgamated into a salable product.

• Three different grades of product can be made: A, B, and C (depending on the mix of materials used).

Question: What quantity of each of the three grades of product should be produced from what quantity of each of the four materials?


Product Data for the Save-It Company

Grade SpecificationAmalgamation Cost per Pound

Selling Price per Pound

A

Material 1: Not more than 30% of totalMaterial 2: Not less than 40% of totalMaterial 3: Not more than 50% of totalMaterial 4: Exactly 20% of total

$3.00 $8.50

BMaterial 1: Not more than 50% of totalMaterial 2: Not less than 10% of the totalMaterial 4: Exactly 10% of the total

2.50 7.00

C Material 1: Not more than 70% of the total 2.00 5.50


Material Data for the Save-It Company

MaterialPounds/Week

AvailableTreatment Cost

per Pound Additional Restrictions

1 3,000 $3.00 1. For each material, at least half of the pounds/week available should be collected and treated.

2. $30,000 per week should be used to treat these materials.

2 2,000 6.00

3 4,000 4.00

4 1,000 5.00



345678910111213141516171819202122232425262728

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

Unit Amalg. Cost $3.00 $2.50 $2.00 Total Treatment Cost $30,000Unit Selling Price $8.50 $7.00 $5.50 =

Unit Profit $5.50 $4.50 $3.50 Treatment Funds Available $30,000

Material Allocation Unit Total(pounds of material used for each product grade) Treament Minimum Material AmountGrade A Grade B Grade C Cost to Treat Treated Available

Material 1 412.3 2,587.7 0 $3 1,500 <= 3,000 <= 3,000Material 2 859.6 517.5 0 $6 1,000 <= 1,377 <= 2,000Material 3 447.4 1,552.6 0 $4 2,000 <= 2,000 <= 4,000Material 4 429.8 517.5 0 $5 500 <= 947 <= 1,000

Total Products 2,149.1 5,175.4 0Mixture

Mixture Specifications PercentsGrade A, Material 1 412.3 <= 644.7 30% of Grade A

Total Profit $35,110 Grade A, Material 2 859.6 >= 859.6 40% of Grade AGrade A, Material 3 447.4 <= 1,074.6 50% of Grade AGrade A, Material 4 429.8 = 429.8 20% of Grade A

Grade B, Material 1 2,587.7 <= 2,587.7 50% of Grade BGrade B, Material 2 517.5 >= 517.5 10% of Grade BGrade B, Material 4 517.5 = 517.5 10% of Grade B

Grade C, Material 1 0.0 <= 0.0 70% of Grade C



Let xij = Pounds of material j allocated to product i per week (i = A, B, C; j = 1, 2, 3, 4).Maximize Profit = 5.5(xA1 + xA2 + xA3 + xA4) + 4.5(xB1 + xB2 + xB3 + xB4) + 3.5(xC1 + xC2 + xC3 + xC4)subject to Mixture Specifications: xA1 ≤ 0.3 (xA1 + xA2 + xA3 + xA4)

xA2 ≥ 0.4 (xA1 + xA2 + xA3 + xA4) xA3 ≤ 0.5 (xA1 + xA2 + xA3 + xA4) xA4 = 0.2 (xA1 + xA2 + xA3 + xA4) xB1 ≤ 0.5 (xB1 + xB2 + xB3 + xB4)xB2 ≥ 0.1 (xB1 + xB2 + xB3 + xB4)xB4 = 0.1 (xB1 + xB2 + xB3 + xB4)xC1 ≤ 0.7 (xC1 + xC2 + xC3 + xC4)

Availability of Materials: xA1 + xB1 + xC1 ≤ 3,000xA2 + xB2 + xC2 ≤ 2,000xA3 + xB3 + xC3 ≤ 4,000xA4 + xB4 + xC4 ≤ 1,000

Restrictions on amount treated: xA1 + xB1 + xC1 ≥ 1,500xA2 + xB2 + xC2 ≥ 1,000xA3 + xB3 + xC3 ≥ 2,000xA4 + xB4 + xC4 ≥ 500

Restriction on treatment cost: 3(xA1 + xB1 + xC1) + 6(xA2 + xB2 + xC2)+ 4(xA3 + xB3 + xC3) + 5(xA4 + xB4 + xC4) = 30,000

and xij ≥ 0 (i = A, B, C; j = 1, 2, 3, 4).


Formulating an LP Spreadsheet Model

• Enter all of the data into the spreadsheet. Color code (blue).

• What decisions need to be made? Set aside a cell in the spreadsheet for each decision variable (changing cell). Color code (yellow with border).

• Write an equation for the objective in a cell. Color code (orange with heavy border).

• Put all three components (LHS, ≤/=/≥, RHS) of each constraint into three cells on the spreadsheet.

• Some Examples:– Production Planning

– Diet / Blending

– Workforce Scheduling

– Transportation / Distribution

– Assignment


George Dantzig’s Diet

• Stigler (1945) “The Cost of Subsistence”– heuristic solution. Cost = $39.93.

• Dantzig invents the simplex method (1947)– Stigler’s problem “solved” in 120 man days. Cost = $39.69.

• Dantzig goes on a diet (early 1950’s), applies diet model:– ≤ 1,500 calories

– objective: maximize (weight minus water content)

– 500 food types

• Initial solutions had problems– 500 gallons of vinegar

– 200 bouillon cubes

For more details, see July-Aug 1990 Interfaces article “The Diet Problem”, available for download at www.mhhe.com/hillier2e/articles


Least-Cost Menu Planning Models in Food Systems Management

• Used in many institutions with feeding programs: hospitals, nursing homes, schools, prisons, etc.

• Menu planning often extends to a sequence of meals or a cycle.

• Variety important (separation constraints).

• Preference ratings (related to service frequency).

• Side constraints (color, categories, etc.)

• Generally models have reduced cost about 10%, met nutritional requirements better, and increased customer satisfaction compared to traditional methods.

• USDA uses these models to plan food stamp allotment.

For more details, see Sept-Oct 1992 Interfaces article “The Evolution of the Diet Model in Managing Food Systems”, available for download at www.mhhe.com/hillier2e/articles

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