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Page 1: 1 Chapter 3 Applications of Linear and Integer Programming Models - 1.

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Chapter 3

Applications of Linear and Applications of Linear and Integer ProgrammingInteger Programming

Models - 1Models - 1

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3.1 The Evolution of Linear Programming Models in Business and Government• Many examples are presented that demonstrate

the successful application of linear and integer programming.

• Our goals are to:– Examine potential application areas– Develop good modeling skills– Illustrate the use of spreadsheets to generate results– Interpret properly and analyze spreadsheet results

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Examples of Linear Programming Models in Business and Government

• An optimal portfolio• Optimal scheduling of personnel• An optimal blend of raw crude oils• A minimized cost diet• An operation and shipping pattern• The optimal production levels

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• The optimal assignment of fleets to flights• How to best expand a communication network• An efficient air-pollution control system• An agricultural resource allocation plan• The set of public projects to select

Examples of Linear Programming Models in Business and Government

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• Three important factors may affect the successful process of building good models:– Familiarity– Simplicity– Clarity

3.2 Building Good Linear and Integer Programming Models

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• Many times the use of summation variables (representing the sum of all or of part of the decision variables) along with the summation constraints associated with them, may simplify models’ formulation.

• See the following example.

Summation Variables / Constraints

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Summation Variables / Constraints

• Example– Three television models are to be produced.– Each model uses 2, 3, and 4 pounds of plastic

respectively.– 7000 pounds of plastic are available.– No model should exceed 40% of the total quantity

produced.– The profit per set is $23, $34, and $45 respectively.– Find the production plan that maximizes the profit.

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• Solution

Max 23X1 + 34X2 + 45X3

S.T. 2X1 + 3X2 + 4X3

Without summation variables

X1 .4(X1 + X2 + X3)X2 .4(X1 + X2 + X3)

X3 .4(X1 + X2 + X3)

X1, X2, X3

With summation variablesX1 + X2 + X3 = X4

X1 .4X4

X2 .4X4

X3 .4X4

X1, X2, X3, X4

Summation Variables / Constraints

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Summation variables/constraints – TV production spreadsheet

=SUM(B2:D2) Total production

Decision Variables

Percentage

ConstraintsPlastic

Constraint

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– Bring the expression to the form:(Expression) [Relation] (Constant)

A checklist for building linear models

A + 2BA + 2B 2A + B +10

- A + B 10

– Formulate a relationship / function in words

before formulating it in mathematical terms.

(Expression) [Has some relation to] (Another expression or constant)

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A checklist for building linear models– Keep the units on both sides of the expressions

consistent

– Use summation variables when appropriate

– Indicate which variables are

Non-negative or Free

Integers

Binary

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• The modeling of real problems is illustrated in this section.

• Examples include:– Production– Purchasing– Finance– Cash flow accounting

3.4 Applications of Linear Programming Models

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• The modeling of real problems is illustrated in this section.

• Examples include:– Production– Purchasing– Finance– Cash flow accounting

• Emphasis is given to:– Various application area, – Model development,– Spreadsheet design,– Analysis and

interpretation of the output.

3.4 Applications of Linear Programming Models

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• These models can assist managers in making decision regarding the efficient utilization of scarce resource.

• Applications include:– Determining production levels– Scheduling shifts– Using overtime– The cost effectiveness of adding resources

3.4.1 Production Scheduling Models

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Galaxy Industries Expansion Plan

• Galaxy Industries is planning to increase its production and include two new products

• Data– Up to 3000 pounds of plastic will be available.

– Regular time available will be 40 hours.

– Overtime available will be 32 hours .

– One hour of overtime costs $180 more than one hour of regular time.

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• Data - continued– Two new products will be introduced:

• Big Squirts• Soakers

– Marketing requirements:• Space Rays should account for exactly 50% of total

production.• No other product should account for more than 40% of total

production.• Total production should increase to at least 1000 dozen per

week.

Galaxy Industries Expansion Plan

The old products are:• Space rays• Zappers

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• Data - Continued Plastic ProductionProduct Profit (lbs) Time (min)

Space Rays $16 2 3Zappers $15 1 4Big Squirts $20 3 5Soakers $22 4 6

Plastic ProductionProduct Profit (lbs) Time (min)

Space Rays $16 2 3Zappers $15 1 4Big Squirts $20 3 5Soakers $22 4 6

• Management wants to maximize the Net Weekly Profit. • A weekly production schedule must be determined.

Galaxy Industries Expansion Plan

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• Decision Variables.

X1 = number of dozen Space Rays, to be produced weekly

X2 = number of dozen Zapper, to be produced weekly

X3 = number of dozen Big Squirts, to be produced weekly

X4 = number of dozen Soakers, to be produced weekly

X5 = number of hours of overtime to be scheduled weekly

Galaxy Industries Expansion Plan – Solution

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• Objective FunctionThe total net weekly profit from the sale of products,

less the extra cost of overtime, to be maximized.

– One hour of overtime costs $180 more than one hour of regular time.

Maximize 16X1 +15X2 +20X3+22X4 - 180X5

Galaxy Industries Expansion Plan – Solution

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• Constraints

200X2 :200produced] [Zappers

32X5 :32used] hours [Overtime

240060X5- 6X4+5X3+4X2+3x1

or 60X5,+24006X4+5X3+4X2+3X1

:available] minutes overtime) +meregular ti ( of[Number

minutes] production of[Number

30004X4+3X3+X2+2X

:3000used plastic ofAmount [

1

Galaxy Industries Expansion Plan – Solution

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Introduce the summation Variable X6, that helps in setting up the production mix constraints

X6 = total weekly production (in dozens ),X6 = X1+X2+X3+X4, or X1+X2+X3+X4 -X6 =0

Galaxy Industries Expansion Plan – Solution

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221000X61000least at is production Total

0.4X6X4production total of 40%Soakers of productionWeekly

0.4X6X3production total of 40%Squirts Big of productionWeekly

0.4X6X2production total of 40%Zappers of productionWeekly

0.5X6X1production total of 50%Rays Space of productionWeekly

Galaxy Industries Expansion Plan

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The Complete Mathematical Model

Max 16X1 + 15X2 + 20X3 + 22X4 – 180X5

S.T. 2X1 + 1X2 + 3X3 + 4X4 3X1+ 4X2 + 5X3 + 6X4 – 60X5 2400

X5 321X2 200

X1 + X2 + X3 + X4 - - X6 = 0X1 -.5X6 = 0

X2 -.4X6 = 0X3 -.4X6 = 0

X4 -.4X6 = 0X6 1000

Xj are non-negative

Galaxy Industries Expansion Plan

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=SUM(B4:E4)

PercentageConstraints

SUMPRODUCT($B$4:$F$4,B6,F6) Drag to G7:G10

Galaxy Industries Expansion Plan

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Microsoft Excel Sensitivity ReportWorksheet: [Galaxy Expansion.xls]Galaxy Expansion Input

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$4 Space Rays 565 0 16 4 12$C$4 Zappers 200 0 15 0.5 1E+30$D$4 Big Squirts 365 0 20 1E+30 0.571428571$E$4 Soakers 0 -2.5 22 2.5 1E+30$F$4 O/T Hours 32 0 -180 1E+30 90

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$C$4 Zappers 200 0 0 1E+30 252$D$4 Big Squirts 365 0 0 1E+30 87$E$4 Soakers 0 0 0 1E+30 452$B$4 Space Rays 565 5 0 486.6666667 91.57894737$G$7 Platistic Total 2425 0 3000 1E+30 575$G$8 Prod. Min. Total 2400 4.5 2400 920 520$G$9 Overtime Hours Total 32 90 32 15.33333333 8.666666667$G$10 Contract Total 200 -0.5 200 280 89.23076923$G$4 Total Production 1130 0 1000 130 1E+30

Microsoft Excel Sensitivity ReportWorksheet: [Galaxy Expansion.xls]Galaxy Expansion Input

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$4 Space Rays 565 0 16 4 12$C$4 Zappers 200 0 15 0.5 1E+30$D$4 Big Squirts 365 0 20 1E+30 0.571428571$E$4 Soakers 0 -2.5 22 2.5 1E+30$F$4 O/T Hours 32 0 -180 1E+30 90

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$C$4 Zappers 200 0 0 1E+30 252$D$4 Big Squirts 365 0 0 1E+30 87$E$4 Soakers 0 0 0 1E+30 452$B$4 Space Rays 565 5 0 486.6666667 91.57894737$G$7 Platistic Total 2425 0 3000 1E+30 575$G$8 Prod. Min. Total 2400 4.5 2400 920 520$G$9 Overtime Hours Total 32 90 32 15.33333333 8.666666667$G$10 Contract Total 200 -0.5 200 280 89.23076923$G$4 Total Production 1130 0 1000 130 1E+30

Galaxy Industries Expansion Plan

D E F G H

7

89

10

11

16

19

21

22

2324

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Galaxy Industries Expansion Plan

Modelo Produção (Dúzias) Lucro Bruto % Total

Space Rays 565 9,040 50

Zappers 200 3,000 17.7

Big Squirts 365 7,300 32.3

Soakers 0 0 0

Total 1130 19,340

$ Horas-extras 5,760

Lucro Líquido 13,580

•Foram utilizados todos os minutos de hora regular (2,400) e hora extra (32) – Restrições “binding”;

•2,425 das 3,000 libras disponíveis de plástico foram utilizadas – Restrição “nonbinding”;

•Total produzido (1130) excedeu o mínimo em 130 dúzias – Restrição “nonbinding”.

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Galaxy Industries Expansion PlanIntervalos de Otimalidade

Modelo Lucro/Dúzia Lucro Mínimo Lucro Máximo

Space Rays 16 4 = 16 - 12 20 = 16 + 4

Zappers 15 Não há mínimo 15.5 = 15 + 0.5

Big Squirts 20 19.43 = 20 - 0.57 Não há Máximo

Soakers 22 Não há mínimo 24.5 = 22 + 2.5

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Galaxy Industries Expansion Plan - Outras constatações• Solução permanece ótima enquanto custo H-E < $270 (H11);

• Para produzir Soakers seu lucro deverá aumentar de $2.5 (E10);

• Horas Regulares adicionais melhoram o lucro total em $4.5/min ou $270/h (E21), sem passar de 920 minutos ou 15 1/3 h (G21);

• Horas-Extras adicionais (ou a menos) acima ou abaixo de 32 melhoram (ou pioram) o lucro total de $90 (E22), se total de H-E ficar no intervalo de 23 1/3 = 32 – 8 2/3 (H22) e 47 1/3 = 32 + 15 1/3 (G22);

• Cada dúzia de Zapper adicional ao contrato, até 280 dúzias (G23), subtrai $0.5 (E23) do lucro total. Reduções no contrato melhoram o lucro total em $0.5/dúzia desde que não excedam 89.23 dúzias (H23);

• Cada Dúzia de Space Rays que seja permitido produzir acima de 50% do total produzido melhora o lucro total em $5 (E19), até 486 2/3 dúzias a mais que 50% (G19).

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Galaxy Industries Expansion Plan• Recomendações ao Gerente

– Autorizar mais H-E;

– Aumentar a % de Space Rays produzidos;

– Reduzir o contrato dos Soakers ou melhorar o lucro unitário

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3.4.2 Portfolio Models

• Portfolio models are usually designed to: – Maximized return on investment,– Minimize risk.

• Factors considered include: – Liquidity requirements,– Long and short term investment goals,– Funds available.

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Jones Investment

Potential Expected Jones's Liquidity RiskInvestment Return Rating Analysis FactorSavings Account 4.0% A Immediate 0Certificate of Deposite 5.2% A 5-year 0Atlantic Lighting 7.1% B + immediate 25Arkansas REIT 10.0% B immediate 30Bedrock Insurance Annuity 8.2% A 1-year 20Nocal Mining Bond 6.5% B+ 1-year 15Minicomp Systems 20.0% A immediate 65Antony Hotel 12.5% C mediate 40

Potential Expected Jones's Liquidity RiskInvestment Return Rating Analysis Factor

Savings Account 4.0% A Immediate 0Certificate of Deposite 5.2% A 5-year 0Atlantic Lighting 7.1% B + immediate 25Arkansas REIT 10.0% B immediate 30Bedrock Insurance Annuity 8.2% A 1-year 20Nocal Mining Bond 6.5% B+ 1-year 15Minicomp Systems 20.0% A immediate 65Antony Hotel 12.5% C mediate 40

• Charles’ Evaluation

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Jones Investment• Portfolio goals

– Expected annual return of at least 7.5%.– At least 50% invested in “A-Rated” investments.– At least 40% invested in immediately liquid investments.– No more than $30,000 in savings accounts and

certificates of deposit.

• Problem summary– Determine the amount to be placed in each investment.– Minimize total overall risk.– Invest all $100,000.– Meet the investor goals (diversify).

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• VariablesXi = the amount allotted to each investment;

• The Mathematical Model

Minimize 25X3+30X4+20X5+15X6+65X7 + 40X8

ST: X1+ X2+ X3+ X4+ X5+ X6 + X7+ X8 = 100,000.04X1+.052X2+.071X3+.10X4+.082X5+.056X6+.27X7+.125X8 7500

X1+ X2 + X5 + X7 50,000X1+ X3+ X4 + X7 40,000X1+ X2 30,000

All the variables are non-negative

Risk function

Total investment

Return

A - Rate Liquid

Savings/Certificate

Jones Investment – Solution

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=SUM(B5:B12)

=SUMPRODUCT(B5,B12,C5:C12)

=SUMIF(E5:E12,"A",B5:B12)=SUMIF(F5:F12,"Immediate",B5:B12)

=SUMPRODUCT(B5,B12,D5:D12)

=B5+B6

Jones Investment - Spreadsheet

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Jones Investment - Spreadsheet

100,000

INVESTMENT AMOUNT Risk Factor Expected Return Rating LiquiditySavings Account 17333.33 0 0.040 A ImmediateCertificate of Deposit 12666.67 0 0.052 A 5-YearAtlantic Lighting 0 25 0.071 B+ ImmediateArkansas REIT 22666.67 30 0.100 B+ ImmediateBedrock Insurance Annuity 47333.33 20 0.082 A 1-YearNocal Mining Bond 0 15 0.065 B+ 1-YearMinicomp Systems 0 65 0.200 A ImmediateAntony Hotels 0 40 0.125 C Immediate

TOTAL INVESTMENT 1000001626666.667

7500 MIN 750077333.33333 MIN 50000

40000 MIN 4000030000 MAX 30000

Total in Liquid Investments Total in Savings and Certificates of Deposit

Recommend Portfolio for Frank BaklarzPortfolio Amount =

TOTAL RISK

Total Expected ReturnREQUIREMENTS

Total in A-rated Investments

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Jones Investment - SpreadsheetMicrosoft Excel Sensitivity ReportWorksheet: [Jones.xls]Jones Investment Input

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$5 Savings Account AMOUNT 17333.33333 0 0 1.176470588 0.5$B$6 Certificate of Deposit AMOUNT 12666.66667 0 0 0.5 1.176470588$B$7 Atlantic Lighting AMOUNT 0 4.666666667 25 1E+30 4.666666667$B$8 Arkansas REIT AMOUNT 22666.66667 0 30 0.384615385 1.176470588$B$9 Bedrock Insurance Annuity AMOUNT 47333.33333 0 20 0.425531915 0.5$B$10 Nocal Mining Bond AMOUNT 0 0.666666667 15 1E+30 0.666666667$B$11 Minicomp Systems AMOUNT 0 1.666666667 65 1E+30 1.666666667$B$12 Antony Hotels AMOUNT 0 1.666666667 40 1E+30 1.666666667

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$B$13 TOTAL INVESTMENT AMOUNT 100000 -7.333333333 100000 4634.146341 6341.463415$D$16 Total Expected Return Expected Return 7500 333.3333333 7500 520 380$D$17 Total in A-rated Investments Expected Return77333.33333 0 50000 27333.33333 1E+30$D$18 Total in Liquid Investments Expected Return 40000 4 40000 21111.11111 28888.88889$D$19 Total in Savings and Certificates of Deposit Expected Return30000 -10 30000 17333.33333 6333.333333

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Jones Investment• Restrições Binding:

– Retorno médio anual de $7,500– Aplicar mínimo de $40,000 em Investimentos com

liquidez– Aplicar no máximo $30,000 em poupança e

certificado de depósito bancário• Restrição Nonbinding:

– Aplicar no mínimo $50,000 em investimentos com ranking A

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Jones Investment

• Recomendações:– Aplicar $17,333 em poupança– Aplicar $12,667 em Certificados de Depósito– Aplicar $22,667 em Arkansas REIT– Aplicar $47,333 em Bedrock Insurance Annuity– Risco total = 1,626,667 ou seja fator médio de risco

de 16.27 por dolar aplicado

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Jones Investment• Custos reduzidos

– Para Atlantic ser incluída seu fator de risco deve baixar para 20.33 = 25 – 4.67

– Para Nocal Mining ser incluída seu fator de risco deve baixar de 0.67– Para Minicomp ser incluída seu fator de risco deve baixar de 1.67– Para Antony Hotels ser incluída seu fator de risco deve baixar de 1.67

• Intervalos de Otimalidade– Bedrock: 19.5 = 20 – 0.5 e 20.43 = 20 + 0.43– Como fator de risco negativo não tem sentido, os limitantes mínimos

para poupança e certificado de depósitos devem ser zero!

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Jones Investment• Preços Sombra

– Para cada dolar extra aplicado acima de $100,000 o risco total melhora (cai) em 7.33;

– Para cada dolar extra que aumente o retorno médio esperado mínimo o risco total aumenta em 333.33;

– Para cada dolar extra que tenha ter maior liquidez o risco total aumenta em 4;

– Para cada dolar extra que se permita investir em poupança e certificado de depósito o risco total cai em 10;

– Não há mudança no risco total se dolar adicional for aplicado em investimentos de ranking A .

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Jones Investment

• Intervalos de Otimalidade– Indicam os intervalos para os quais os preços

sombra não se alteram• Para o retorno mínimo de $7,500: 7120 = 7500 – 380 e

8020 = 7500 + 520• Allowable Decrease de 1E+30 = infinito ()

– Limitante inferior para o investimento em ranking A é -

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• Governments in the public sector are charged with distributing resources for the public good.

• The public good can be measured by – Traditional objectives (i.e. cost minimization), – Specific functions developed to measure the public

satisfaction or preference.• The constraints represent (among others)

– Resource availability, – Social issues (diversity, equality)

3.4.3 Public Sector Models

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St. Joseph Inspection Problem

• St. Joseph Public Utility Commission needs to inspect and report utility problems in a flood area.

Experts are assigned to inspect:

• Homes• Offices• Plants

Three types of inspection

will be conducted: • Electrical• Gas• Insulation

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• Problem Summary

– St. Joseph Public Utility Commission needs to determine the number of homes, office complexes, and plants to be inspected.

– The objective is to maximize the total number of structures inspected, under certain requirements.

St. Joseph Inspection Problem

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• Data– Requirements

• At least eight offices and eight plants must be inspected.• At least 60% of the inspections should involve private

homes.– Resources

• At most, 120 hours can be allocated for electrical inspections.

• At most 80 hours can be allocated for gas inspection.• At most 100 consulting hours can be allocated for

insulation inspection.

St. Joseph Inspection Problem

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• Variables– X1, X2, X3 = number of homes, office complexes, and industrial plants to be

inspected, respectively.– X4 = a summation variable: The total number of structures to be inspected.

• The Complete Mathematical ModelMaximize X4

ST:X1 + X2 + X3 - X4 = 0 (Summation)X2 8 (Min. Office) X3 8 (Min. Plants)

X1 -.6X4 0 (At least 60% Homes)2X1 + 4X2 + 6X3 120 (Electrical)1X1 + 3X2 + 3X3 80 (Gas)3X1 + 2X2 + 1X3 100 (Insulation)

X1, X2, X3 0

St. Joseph Inspection Problem – Solution

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St. Joseph Inspection Problem – Solution

Estruturas inspecionadas

Horas usadas

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The problem is properly formulated, but there is no feasible solution to the problem.

Here is why...

Infeasible Solution

St. Joseph Inspection Problem – Solution

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Even when the smallest number allowed of offices and plants are inspected, the number of houses that can be inspected with the given resources is too small.

Here is the reasoning.

St. Joseph Inspection Problem spreadsheet

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St. Joseph Inspection Problem – the infeasibility

• Recall that X2 8 and X3 8.

• Let X2 = 8 (offices) and X3 = 8 (plants). That is, take their minimum feasible value.

• Hours used on electrical inspection for offices and plants are 4(8) + 6(8) = 80.

• Hours left over to inspect houses are 120 – 80 = 40,thus at most 40/2 = 20 houses can be inspected.

• Total number of structures inspected is 8+8+20=36, of which houses are only 20/36 = 55.55%

• This is less than the 60% required.

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The revised problem

INSPECTIONS MINIMUM ELECTRICAL GAS INSULATIONHOUSE 27 24 2 1 3OFFICE 6 6 4 3 2PLANT 7 6 6 3 1

TOTAL 40 HOURS USED 120 66 100AVAILABLE 120 80 100

HOURS PER UNIT

St. Joseph's Public Utility Commission

St. Joseph Inspection Problem – Solution Suppose the commission would accept 6 offices and 6 plants.

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3.4.4 Purchasing Modeling

• These models can consider: – Demand– Budget– Cash flow– Advertising– Inventory restrictions.

• In solving purchasing problems, we attempt to balance customer satisfaction with resource utilization by the business enterprise.

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Euromerica Liquor Purchasing Problem

• Euromerica Liquors purchases and distributes a number of wines to retailers.

• There are four different wines to be ordered. • Requirements

– Order at least 800 of each type.– Order at least twice as many domestic bottles as

imported bottles.

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• Data:Wine Country Cost Selling Price

Napa Gold U.S. $2.50 $4.25Cayuga Lake U.S. $3.00 $4.50

Seine Soir France $5.00 $8.00Bella Bella Italy $4.00 $6.00

Wine Country Cost Selling Price

Napa Gold U.S. $2.50 $4.25Cayuga Lake U.S. $3.00 $4.50

Seine Soir France $5.00 $8.00Bella Bella Italy $4.00 $6.00

• Management wishes to determine how many bottles of each type to order.

• The objective is to maximize the total profit from purchasing and distributing the wine bottles.

Euromerica Liquor Purchasing Problem

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• VariablesX1 = bottles of Napa Gold purchased

X2 = bottles of Cayuga Lake purchased

X3 = bottles of Seine Soir purchased

X4 = bottles of Bella Bella purchased.

Euromerica Liquor Purchasing Problem – Solution

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• The Mathematical Model

Maximize 1.75X1 + 1.50X2 + 3X3 + 2X4

ST:X1 800

X2 800X3 800

X4 800X1 + X2 - 2X3 - 2X4 0

X1, X2, X3, X4 0

$4.25 - $2.50 = $1.75

[Domestic wines] [ Are at least] [ Twice the imported wines]

Euromerica Liquor Purchasing Problem – Solution

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Euromerica Liquor Purchasing Problem – Solution

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An Unbounded solution

Euromerica Liquor Purchasing Problem – Solution

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The revised model: Given budget, limited supply, and limited demand

Maximize 1.75X1+ 1.50X2+ 3X3+ 2X4

ST:X1 800

X2 800

X3 800

X4 800

X1 + X2 - 2X3 - 2X4 0 2.50X1+ 3.00X2+ 5X3+ 4X4 28000 (Budget)

X1 3600 (Napa) X3 2400 (Seine)

X1+ X2+ X3+ X4 10000 (Total)

X1, X2, X3, X4 0

Euromerica Liquor Purchasing Problem – Revised Solution

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Euromerica Liquor Purchasing Problem – Solution

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Solution Summary

Wine Bottles Cases Cost Profit

Napa Gold 3600 300 9000 6300Cayuga Lake 1968 164 5904 2952Seine Soir 1980 165 9900 5940Bella Bella 804 67 3216 1608Total 8352 696 28020 16800

Wine Bottles Cases Cost Profit

Napa Gold 3600 300 9000 6300Cayuga Lake 1968 164 5904 2952Seine Soir 1980 165 9900 5940Bella Bella 804 67 3216 1608Total 8352 696 28020 16800

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3.4.5 Blending Models

• Blending models were successfully implemented first by the oil industry.

• Blending problems have the following characteristics:– Each of several products have certain specifications that

must be met.– The products can be produced by blending various

components, each with different properties of its own.– The problem is to find the least costly (most profitable) blends

that meet the requirements and specifications of all the products.

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United Oil Company

• United Oil blends two input streams of crude oil– Alkylate – Catalytic Cracked.

• The outputs of the blending process are – Regular gasoline.– Mid-Grade gasoline.– Premium gasoline.

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• Restrictions

– Weekly supply of Crude oil is limited.

– Contracted weekly demand for commercial gasoline has to be met.

– To classify gasoline as Regular, Mid-Grade, or Premium, certain levels (specifications) of octane and vapor pressure must be met.

• Profit per barrel of each type of commercial gasoline depends on the blend it was made of.

United Oil Company

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• DataCrude Oil Product Data

Blended Gasoline Data

Octane Vapor Cost WeeklyProduct Rating Pressure per Barrel Supply (barrels)Alkylate 98 5 $19 15000Catalytic- Cracked 86 9 $16 15000

Minimum Maximum Selling PriceGasoline Octane Rating Vapor Pressure Per Barrel Demand Regular 87 9 $18 12000

Mid-Grade 89 7 $20 7500 Premium 92 6 $23 4500

United Oil Company

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• Problem Summary– Determine how many barrels of alkylate and catalytic cracked to

blend into regular, mid – grade, and premium each week.

– Maximize total weekly profit.

– Remain within raw gas availability.

– Meet contract requirements.

– Produce gasoline blends that meet the octane and vapor pressure requirements.

United Oil Company

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• Decision Variables

X1, X2, X3 = number of barrels of Alkylate blended each week into Regular, Mid-Grade, and Premium gas

respectively. Y1, Y2, Y3 = number of barrels of Catalytic Cracked blended each

week into Regular Mid-Grade, and Premium respectively.

R, M, P = barrels of Regular, Mid-Grade, Premium respectively, produced weekly (summation variables).

United Oil Company – Solution

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• The Mathematical ModelMax –1X1 + 1X2 + 4X3 +2Y1 +4Y2 + 7Y3

1X1 +1X2 +1X3 150001Y1 +1Y2 +1Y3 15000

1X1 +1Y1 -R = 0

1X2 +1Y2 -M = 0

1X1 +1Y1 -P = 0

R 12000M 7500P4500

98X1 + 86Y1 87R 0

98X2 + 86Y2 89M 0

98X3 + 86Y3 92P 0

Click for more constraints

5x1 + 9Y1 –9R 0

5X2 + 9Y2 – 7M 0

5X3 + 9Y3 –6P 0

All the variables are non-negative

5x1 + 9Y1 –9R 0

5X2 + 9Y2 – 7M 0

5X3 + 9Y3 –6P 0

All the variables are non-negative

United Oil Company – Solution

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Availible Octane Vap. Pr. Cost/bbl Regular Midgrade Premium TotalAlkylate 15000 98 5 $19 Alkylate 1000 3750 10250 15000

Catalytic Cracked 15000 86 9 $16 Catalytic Cracked 11000 3750 250 15000TOTAL 12000 7500 10500 30000

Required Min Oct.MAX V.P. Price/bbl Regular Midgrade Premium TotalRegular 12000 87 9 $18 Alkylate ($1,000) $3,750 $41,000 $43,750

Midgrade 7500 89 7 $20 Catalytic Cracked $22,000 $15,000 $1,750 $38,750Premium 4000 92 6 $23 TOTAL $21,000 $18,750 $42,750 $82,500

Regular Midgrade PremiumOctane 87 92 97.71429

Vapor Pressure 8.666667 7 5.095238

Profits

Performance Measures

CRUDES

GASOLINES

OUTPUTS

RECOMMENDED BLENDED AMOUNTS

United Oil Company

INPUTS

United Oil Company – Solution

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Availible Octane Vap. Pr. Cost/bbl Regular Midgrade Premium TotalAlkylate 15000 98 5 $19 Alkylate 1000 3750 10250 15000

Catalytic Cracked 15000 86 9 $16 Catalytic Cracked 11000 3750 250 15000TOTAL 12000 7500 10500 30000

Required Min Oct.MAX V.P. Price/bbl Regular Midgrade Premium TotalRegular 12000 87 9 $18 Alkylate ($1,000) $3,750 $41,000 $43,750

Midgrade 7500 89 7 $20 Catalytic Cracked $22,000 $15,000 $1,750 $38,750Premium 4000 92 6 $23 TOTAL $21,000 $18,750 $42,750 $82,500

Regular Midgrade PremiumOctane 87 92 97.71429

Vapor Pressure 8.666667 7 5.095238

Profits

Performance Measures

CRUDES

GASOLINES

OUTPUTS

RECOMMENDED BLENDED AMOUNTS

United Oil Company

INPUTS

Crude Availability

Blended Requirements

Total ProfitVapor Pr. Constraints (Hidden)

Octane Constraints (Hidden)

Decision Variables

United Oil Company – Solution

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• These models cover a planning horizon of several periods.

• Linking constraints secure the proper transfer of quantities from one period to the next one. The form of these constraints is:

Amount this period = Amount last period +

Inflow for the period – Outflow for the period

• These models are useful for accounting analysis.

Appendix 3.2 (CD): Cash Flow Models

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Appendix 3.2 (CD): Cash Flow Models

• For example:Casht = Casht-1 + [Interest paid]t – [Loan repaid]t

t-1 t

100

-20

+30 110

110 = 100 + 30 – 20

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• Data– There are $9 million available for short-term investments over a

period of five months (it is now January 1).– There are three possible investments.– Interest earned on each investment is:

• 0.7% over two months for two month term account.• 1.5% over three months for three months construction loan.• 0.2% per one-month period for passbook saving account.

– Funds invested in term account are not liquid before the term ends.– Interest earned on investment before it is matured is calculated proportionately to the term rate.

The Powers Group Cash Flow Problem

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• Objective function– Maximize the book value at the end of May.

• Constraints– No more than $4 million should be invested in any one of the

three short term investments.– Total investment each month in the liquid passbook account

should be at least $2 million.– Cash available at the end of each month should be at least

$3.5 million.– Cash available at the end of May should be at least $5 million.

The Powers Group Cash Flow Problem

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• Decision variables– Tj = the amount of funds invested in the 2 month term

account at the beginning of month j = 1, 2,…, 6 (j=1, Jan.)– Cj = the amount of funds invested in construction loan at the

beginning of month j = 1, 2,…, 6 (j=1, Jan.)– Pj = the amount of funds invested in passbook saving account

at the beginning of month j = 1, 2,…, 6 (j=1, Jan.)– Summation variables: see next slide

The Powers Group Cash Flow Problem – Solution

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Summation variablesTTj = TTj-1 –Tj-2 + Tj

TCj = TCj-1 – Cj-3 + Cj

TPj = Pj

The Powers Group Cash Flow Problem – Solution

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• Objective functionBook value consists of cash, and proportionate interest paid on investments before maturity.

1.007T4 + 1.0035T5 + 1.015C3 + 1.010C4 + 1.005C5 +1.002P5

Half of the full two months interest is considered at the end of Mayfor a 2-month term investment made at thebeginning of May.

.0035 .0035May 1 June 1

The Powers Group Cash Flow Problem – Solution

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• Constraints– Not more than $4,000,000 can be invested in any investment

TTj 4,000,000;TCj 4,000,000TPj 4,000,000

– Total in Passbook Saving is at least $2,000,000TPj 2,000,000

– Total investment at the beginning of each month = cash available for investment at the end of the previous month.Tj + Cj + Pj = Lj-1 Then, Lj 3,500,000for j=Feb, March, …) and L1 = 9,000,000. Also L5 5,000,000

The Powers Group Cash Flow Problem – Solution

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The Powers Group Cash Flow Problem - spreadsheet

=1.007*H6+1.015*G7+1.002*I8Drag back to E9:H9

=H13+I6-G6=H14+I&7-G7=I8Drag back to column E:H

=1.007*H6+1.0035*I6+1.015*G7+1.01*H7+1.005*I7+1.002*I8

Drag back to E16:H16=(I16 – E18)/E18*(12/5)

=Sum(I6:I8)Drag back to E5:H5

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Appendix 3.3 (CD): Data Envelopment Analysis Models

Relative efficiency =

• In these models the relative efficiency of facilities with similar goals and objectives is studied.

• The relative efficiency is calculated as a ratioof outputs to inputs.

Weighted sum of outputsWeighted sum of inputs

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Sir Loin Restaurants - DEA

• KATTLECORP Inc. owns and operates four restaurants located in different states.

• The restaurants are of different size, personnel, and traffic density.

• KATTLECORP wishes to determine which restaurant operates efficiently.

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Sir Loin Restaurants - DEA

• To calculate efficiency KATTLECORP needs to compare input to output in each restaurant

• Input– Capacity – # of employees– Population around

• Output– Gross revenue– % of returning customers– Food rating

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• Data

Sir Loin Restaurants - DEA

INPUTS TAMPA ATLANTA MOBILE COLUMBIACapacity 213 265 157 152

# Employees 52 65 40 48Service Area 650000 900000 200000 275000

OUTPUTSTAMPA ATLANTA MOBILE COLUMBIA

Gross Revenue 604000 663000 375000 354000% Repeat 89 85 94 88

Food Rating 7.3 6.8 9.1 7.5

Population within 10 miles around the restaurant location

Customer survey: 1. Would you return to this restaurant?2. Rate the food on a scale of 0 – 10.

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• Decision variables– X1, X2, X3 = relative input weight for capacity, number

of employee, and service area respectively

– Y1, Y2, Y3 = relative input weight for gross revenue, repeat business, and food rating respectively

Sir Loin Restaurants – Solution

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• Objective functionMaximize the efficiency for the Columbia restaurant

Max , which becomes

Sir Loin Restaurants – Solution

354000Y1+ 88Y2+ 7.5Y3

152X1+ 48X2+ 275000X3

Since we maximize theratio, the denominator canbe selected as equal to 1,without loss of generality.

Max 354000Y1+ 88Y2+ 7.5Y3

S.T. 152X1+ 48X2+ 275000X3 = 1,

and other constraints shown next…

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• Constraints– We have already established that

152X1+ 48X2+ 275000X3 = 1 – In general, we require that the efficiency of each restaurant is

not greater than 1.Output/input 1, or, Output Input

– 604000Y1+89Y2+7.3Y3 213X1+52X2+650000X3 (Tampa)663000Y1+85Y2+6.8Y3 265X1+ 5X2+900000X3 (Atlanta)

375000Y1+94Y2+9.1Y3 157X1+40X2+200000X3 (Mobile)

354000Y1+88Y2+7.5Y3 152X2+48X2+275000X3 (Columbia)

Sir Loin Restaurants – Solution

All the variables are non-negative

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Sir Loin Restaurants - Spreadsheet

=SUMPRODUCT($F$4:$F$6,E4:E6) Drag back to B7:D7

=SUMPRODUCT($F$10:$F$12,E10:E12) Drag back to B13:D13

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