1
Chapter 3
Applications of Linear and Applications of Linear and Integer ProgrammingInteger Programming
Models - 1Models - 1
2
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
3
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
4
• 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
5
• Three important factors may affect the successful process of building good models:– Familiarity– Simplicity– Clarity
3.2 Building Good Linear and Integer Programming Models
6
• 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
7
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.
8
• 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
9
Summation variables/constraints – TV production spreadsheet
=SUM(B2:D2) Total production
Decision Variables
Percentage
ConstraintsPlastic
Constraint
10
– 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)
11
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
12
• 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
13
• 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
14
• 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
15
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.
16
• 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
17
• 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
18
• 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
19
• 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 $180 more than one hour of regular time.
Maximize 16X1 +15X2 +20X3+22X4 - 180X5
Galaxy Industries Expansion Plan – Solution
20
• Constraints
200X2 :200produced] [Zappers
32X5 :32used] hours [Overtime
240060X5- 6X4+5X3+4X2+3x1
or 60X5,+24006X4+5X3+4X2+3X1
:available] minutes overtime) +meregular ti ( of[Number
minutes] production of[Number
30004X4+3X3+X2+2X
:3000used plastic ofAmount [
1
Galaxy Industries Expansion Plan – Solution
21
Introduce the summation Variable X6, that helps in setting up the production mix constraints
X6 = total weekly production (in dozens ),X6 = X1+X2+X3+X4, or X1+X2+X3+X4 -X6 =0
Galaxy Industries Expansion Plan – Solution
221000X61000least at is production Total
0.4X6X4production total of 40%Soakers of productionWeekly
0.4X6X3production total of 40%Squirts Big of productionWeekly
0.4X6X2production total of 40%Zappers of productionWeekly
0.5X6X1production total of 50%Rays Space of productionWeekly
Galaxy Industries Expansion Plan
23
The Complete Mathematical Model
Max 16X1 + 15X2 + 20X3 + 22X4 – 180X5
S.T. 2X1 + 1X2 + 3X3 + 4X4 3X1+ 4X2 + 5X3 + 6X4 – 60X5 2400
X5 321X2 200
X1 + X2 + X3 + X4 - - X6 = 0X1 -.5X6 = 0
X2 -.4X6 = 0X3 -.4X6 = 0
X4 -.4X6 = 0X6 1000
Xj are non-negative
Galaxy Industries Expansion Plan
24
=SUM(B4:E4)
PercentageConstraints
SUMPRODUCT($B$4:$F$4,B6,F6) Drag to G7:G10
Galaxy Industries Expansion Plan
25
Microsoft Excel Sensitivity ReportWorksheet: [Galaxy Expansion.xls]Galaxy Expansion Input
Adjustable CellsFinal Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease$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
26
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”.
27
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
28
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).
29
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
30
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.
31
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
32
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 $100,000.– Meet the investor goals (diversify).
33
• VariablesXi = the amount allotted to each investment;
• The Mathematical Model
Minimize 25X3+30X4+20X5+15X6+65X7 + 40X8
ST: X1+ X2+ X3+ X4+ X5+ X6 + X7+ X8 = 100,000.04X1+.052X2+.071X3+.10X4+.082X5+.056X6+.27X7+.125X8 7500
X1+ X2 + X5 + X7 50,000X1+ X3+ X4 + X7 40,000X1+ X2 30,000
All the variables are non-negative
Risk function
Total investment
Return
A - Rate Liquid
Savings/Certificate
Jones Investment – Solution
34
=SUM(B5:B12)
=SUMPRODUCT(B5,B12,C5:C12)
=SUMIF(E5:E12,"A",B5:B12)=SUMIF(F5:F12,"Immediate",B5:B12)
=SUMPRODUCT(B5,B12,D5:D12)
=B5+B6
Jones Investment - Spreadsheet
35
Jones Investment - Spreadsheet
100,000
INVESTMENT AMOUNT Risk Factor Expected Return Rating LiquiditySavings Account 17333.33 0 0.040 A ImmediateCertificate of Deposit 12666.67 0 0.052 A 5-YearAtlantic Lighting 0 25 0.071 B+ ImmediateArkansas REIT 22666.67 30 0.100 B+ ImmediateBedrock Insurance Annuity 47333.33 20 0.082 A 1-YearNocal Mining Bond 0 15 0.065 B+ 1-YearMinicomp Systems 0 65 0.200 A ImmediateAntony Hotels 0 40 0.125 C Immediate
TOTAL INVESTMENT 1000001626666.667
7500 MIN 750077333.33333 MIN 50000
40000 MIN 4000030000 MAX 30000
Total in Liquid Investments Total in Savings and Certificates of Deposit
Recommend Portfolio for Frank BaklarzPortfolio Amount =
TOTAL RISK
Total Expected ReturnREQUIREMENTS
Total in A-rated Investments
36
Jones Investment - SpreadsheetMicrosoft Excel Sensitivity ReportWorksheet: [Jones.xls]Jones Investment Input
Adjustable CellsFinal Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease$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
37
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
38
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
39
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!
40
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 .
41
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 -
42
• 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
43
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
44
• 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
45
• 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
46
• 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
47
St. Joseph Inspection Problem – Solution
Estruturas inspecionadas
Horas usadas
48
The problem is properly formulated, but there is no feasible solution to the problem.
Here is why...
Infeasible Solution
St. Joseph Inspection Problem – Solution
49
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
50
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.
51
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.
52
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.
53
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.
54
• 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
55
• 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
56
• 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
57
Euromerica Liquor Purchasing Problem – Solution
58
An Unbounded solution
Euromerica Liquor Purchasing Problem – Solution
59
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
60
Euromerica Liquor Purchasing Problem – Solution
61
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
62
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.
63
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.
64
• 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
65
• 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
66
• 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
67
• 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
68
• 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
69
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
70
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
71
• 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
72
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
73
• 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
74
• 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
75
• 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
76
Summation variablesTTj = TTj-1 –Tj-2 + Tj
TCj = TCj-1 – Cj-3 + Cj
TPj = Pj
The Powers Group Cash Flow Problem – Solution
77
• 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
78
• 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
79
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
80
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
81
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.
82
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
83
• 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.
84
• 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
85
• 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…
86
• 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
87
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
88
Copyright 2002 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that named in Section 117 of the United States Copyright Act without the express written consent of the copyright owner is unlawful. Requests for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. Adopters of the textbook are granted permission to make back-up copies for their own use only, to make copies for distribution to students of the course the textbook is used in, and to modify this material to best suit their instructional needs. Under no circumstances can copies be made for resale. The Publisher assumes no responsibility for errors, omissions, or damages, caused by the use of these programs or from the use of the information contained herein.