1 OM2, Supplementary Ch. C Modeling Using Linear Programming ©2010 Cengage Learning. All Rights...

1OM2, Supplementary Ch. C Modeling Using Linear Programming

©2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

MODELING USING LINEAR PROGRAMMING

SUPPLEMENTARY CHAPTER C

DAVID A. COLLIERAND

JAMES R. EVANS

OM2

2

OM2, Supplementary Ch. C Modeling Using Linear Programming


LO1 Explain how to recognize decision variables, the objective function, and constraints in formulating linear optimization models. LO2 Describe how to use linear optimization models for OM applications. LO3 Explain how to use Excel Solver to solve linear optimization models on spreadsheets.

Supplementary Chapter C. Modeling Using Linear Programming

l e a r n i n g o u t c o m e s

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aller’s Pub & Brewery is a small restaurant and microbrewery that makes six types of special beers, each having a unique taste and color. Jeremy Haller, one of the family owners who oversees the brewery operations, has become worried about increasing costs of grains and hops that are the principal ingredients and the difficulty they seem to be having in making the right product mix to meet demand and using the ingredients that are purchased under contract in the commodities market. Haller’s buys six different types of grains and four different types of hops.


h

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Each of the beers needs different amounts of brewing time and is produced in 30-keg (4,350-pint) batches. While the average customer demand is 55 kegs per week, the demand varies by type. In a meeting with the other owners, Jeremy stated that Haller’s has not been able to plan effectively to meet the expected demand. “I know there must be a better way of making our brewing decisions to improve our profitability.”


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Can you identify any examples when you needed to find a better way of planning, designing, or operating some system or process?

What do you think?


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Quantitative models that seek to maximize or minimize some objective function while satisfying a set of constraints are called optimization models.

Linear programming (LP) models are used widely for many types of operations design and planning problems that involve allocating limited resources among competing alternatives, and for supply chain management design and operations.


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Softwater Production Planning Problem

• Pellets are produced in 40- and 80-pound bags.

• Company has orders for 20,000 pounds

• 4,000 pounds are currently in inventory

• Limited amounts of packaging materials and packaging line time

• Determine how many bags of each size to produce to maximize profit.


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Decision Variables

A decision variable is a controllable input variable that represents the key decisions a manager must make to achieve an objective.

x1 = number of 40-pound bags produced

x2 = number of 80-pound bags produced


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Objective Function

Suppose that Softwater makes $2 for every 40-lb. bag and $4 for every 80-lb. bag produced and sold.

Max total profit = z = 2x1 + 4x2 [C.1]

The constant terms in the objective function are called objective function coefficients.


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SolutionsAny particular combination of

decision variables is referred to as a solution.

Solutions that satisfy all constraints are referred to as feasible solutions.

Any feasible solution that optimizes the objective function is called an optimal solution.


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A Solution for the Softwater Problem


Suppose that Softwater decided to produce 200 40-pound bags and 300 80-pound bags. The profit would be

z = 2(200) + 4(300) = 400 + 1,200 = $1,600

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ConstraintsA constraint is some limitation or

requirement that must be satisfied by the solution.

Suppose that each 40-pound bag requires 1.2 minutes of packaging time per bag and 80-pound bags require 3 minutes per bag. The total packaging time required is

1.2x1 + 3x2

Only 1,500 minutes of packaging time are available, so we have the constraint:

1.2x1 + 3x2 ≤ 1,500


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Packaging Material Constraint

Softwater has 6,000 square feet of packaging materials available; each 40-pound bag requires 6 square feet and each 80-pound bag requires 10 square feet. Since the amount of packaging materials used cannot exceed what is available, we have the constraint:

6x1 + 10x2 ≤ 6,000

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Aggregate Production ConstraintWe need to produce a net amount of

16,000 pounds. Because the small bags contain 40 pounds of pellets and the large bags contain 80 pounds, we must impose this aggregate-demand constraint:

40x1 + 80x2 ≥ 16,000


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Nonnegativity Constraints

We must prevent the decision variables from having negative values. Thus, we need the constraints:

x1 and x2 ≥ 0


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Softwater Optimization ModelMax z = 2x1 + 4x2 (profit) subject to

1.2x1 + 3x2 ≤ 1,500 (packaging line) 6x1 + 10x2 ≤ 6,000 (materials availability) 40x1 + 80x2 ≥16,000 (aggregate production) x1, x2 ≥ 0 (nonnegativity)

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Linear Functions

A function in which each variable appears in a separate term and is raised to the first power is called a linear function.

The objective function and all constraints of the Softwater problem consist of linear functions. This is a requirement for a linear program and its solution procedure.


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Production Scheduling

Bollinger Electronics Company produces two electronic components for an airplane engine manufacturer. Demand for the next three months is:

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Decision Variables

xim denotes the production volume in units for product i in month m. Here i =1, 2, and m = 1, 2, 3; i = 1 refers to component 322A, i = 2 to component 802B, m = 1 to April, m = 2 to May, and m = 3 to June.

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Objective Function

Component 322A costs $20 per unit to produce and component 802B costs $10 per unit to produce. The production-cost part of the objective function is:

20x11 + 20x12 + 20x13 + 10x21 + 10x22 + 10x23

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Objective Function

To incorporate the relevant inventory costs into the model, let Iim denote the inventory level for product i at the end of month m. Inventory-holding costs are 1.5 percent of the cost of the product; that is, (.015)($20) = $0.30 per unit for component 322A, and (.015)($10) = $0.15 per unit for component 802B. The inventory-holding cost portion of the objective function can be written as:

0.30I11 + 0.30I12 + 0.30I13 + 0.15I21 + 0.15I22 + 0.15I23

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Objective FunctionTo incorporate the costs due to fluctuations in

production levels from month to month, we need to define additional decision variables:

Rm = increase in the total production level during month m compared with month m – 1

Dm = decrease in the total production level during month m compared with month m – 1

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Complete Objective Function

Min 20x11 + 20x12 + 20x13 + 10x21 + 10x22 + 10x23

+

0.30I11 + 0.30I12 + 0.30I13 + 0.15I21 + 0.15I22 +

0.15I23

+ 0.50R1 + 0.50R2 + 0.50R3 + 0.20D1 + 0.20D2 +

0.20D3

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ConstraintsFirst we must guarantee that the schedule

meets customer demand. We have the basic equation:

Ending inventory from previous month + Current production – Ending inventory for this month = This month’s demand

Assume inventories at the beginning of the three-month scheduling period are 500 units for component 322A and 200 units for component 802B.

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ConstraintsMonth 1: 500 + x11 – I11 = 1000

200 + x21 – I21 = 1000

Month 2: I11 + x12 – I12 = 3,000 I21 + x22 – I22 = 500

Month 3: I12 + x13 – I13 = 5,000 I22 + x23 – I23 = 3,000

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Constraints

Minimum Inventory Level:

At least 400 units of component 322A and at least 200 units of component 802B:

I13 ≥ 400 and I23 ≥ 200

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Additional Constraint Data

Additional Constraint Data

Exhibits C.1 and C.2

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Constraints

Machine capacity: 0.10x11 + 0.08x21 ≤ 400 (month 1)

0.10x12 + 0.08x22 ≤ 500 (month 2)

0.10x13 +1 0.08x23 ≤ 600 (month 3)

Labor capacity: 0.05x11 + 0.07x21 ≤ 300 (month

1) 0.05x12 + 0.07x22 ≤ 300 (month

2) 0.05x13 + 0.07x23 ≤ 300 (month

3) Storage capacity: 2I11 + 3I21 ≤ 10,000 (month 1) 2I12 + 3I22 ≤ 10,000 (month 2)

2I13 + 3I23 ≤ 10,000 (month 3)

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ConstraintsWe must also guarantee that Rm and Dm will

reflect the increase or decrease in the total production level for month m. Suppose the production levels for March were 1,500 units of component 322A and 1,000 units of component 802B. Then

April production – March production = Change x11 + x21 – 2,500 = Change x11 + x21 – 2,500 = R1 – D1

Similar constraints for May and June.

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Constraints

Production Smoothing Constraints:

x11 + x21 – R1 + D1 = 2,500 – x11 – x21 + x12 + x22 – R2 + D2 = 0 – x12 – x22 + x13 + x23 – R3 + D3 = 0

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Blending Problems Grand Strand Oil Company produces

regular-grade and premium-grade gasoline products by blending three petroleum components. The gasolines are sold at different prices, and the petroleum components have different costs. The firm wants to determine how to blend the three components into the two products in such a way as to maximize profits.

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Petroleum Component Cost and Supply

DataRegular-grade gasoline can be sold for $2.20 per

gallon and the premium-grade gasoline for $2.40 per gallon. Current commitments to distributors require Grand Strand to produce at least 10,000 gallons of regular-grade gasoline.

Exhibit C.4

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Component Specifications for Grand Strand’s Products

Data

Exhibit C.5

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Supplementary Chapter C Modeling Using Linear Programming

Decision Variablesx1r = gallons of component 1 in regular

gasoline x2r = gallons of component 2 in regular

gasoline x3r = gallons of component 3 in regular

gasoline x1p = gallons of component 1 in premium



gasoline

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Objective FunctionMax 2.20(x1r + x2r + x3r) + 2.40(x1p + x2p +

x3p) – 1.00(x1r + x1p) - 1.20(x2r + x2p) - 1.64(x3r + x3p)

By combining terms, we can then write the objective function as:

Max 1.20x1r + 1.00x2r + 0.56x3r + 1.40x1p + 1.20x2p + 0.76x3p

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ConstraintsComponent availability:

x1r + x1p ≤ 5,000 (component 1) x2r + x2p ≤ 10,000 (component 2) x3r + x3p ≤ 10,000 (component 3)

Regular grade gasoline requirement:x1r + x2r +x3r ≥ 10,000

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ConstraintsComponent 1 must account for at most 30

percent of the total gallons of regular gasoline produced:x1r /(x1r + x2r + x3r) ≤ 0.30 or x1r ≤ 0.30(x1r + x2r + x3r)

Rewrite this as:0.70x1r - 0.30x2r - 0.30x3r ≤ 0

Other specification constraints:– 0.40x1r + 0.60x2r – 0.40x3r ≤0 – 0.20x1r – 0.20x2r + 0.80x3r ≤ 0 – 0.75x1p – 0.25x2p – 0.25x3p ≤ 0 – 0.40x1p + 0.60x2p – 0.40x3p ≤ 0 – 0.30x1p – 0.30x2p + 0.70x3p ≤ 0

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Transportation ProblemThe transportation problem is a

special type of linear program that arises in planning the distribution of goods and services from several supply points to several demand locations.

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Foster Generators Supply/Demand Data

Foster Generators Supply/Demand Data

Exhibits C.6 and C.7

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Foster Generators Transportation Cost per Unit

Foster Generators Cost Data

Exhibit C.8

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Transportation Table

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Transportation LP Model

Min total cost = 3x11 + 2x12 + 7x13 + 6x14 + 7x21 + 5x22 + 2x23 + 3x24 + 2x31 + 5x32 + 4x33 + 5x34

Subject toCleveland: x11 + x12 + x13 + x14 = 5,000. Bedford: x21 + x22 + x23 + x24 = 6,000. York: x31 + x32 + x33 + x34 = 2,500. Boston: x11 + x21 + x31 = 6,000. Chicago: x12 + x22 + x32 = 4,000 St. Louis: x13 + x23 + x33 = 2,000 Lexington: x14 + x24 + x34 = 1,500

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LP Model for Crashing Decisions

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Data

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Decision Variables and Objective Function

xi = start time of activity i yi = amount of crash time used for activity I

Min 2,000yA + 1,000yB + 2,500yC + 1,500yD + 500yE

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Constraints

For each arc from activity i to activity j in the network, the start time for the following activity must be at least as great as the finish time for each immediate predecessor with crashing applied

xj ≥ xi + normal time for activity i - yi

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Precedence Constraints

xB ≥ xA + 10 - yA xD ≥ xB + 14 - yB xC ≥ xB + 14 - yB xE ≥ xD + 11 - yD xE ≥ xC + 6 - yC xF ≥ xE + 8 - yE

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Other ConstraintsMaximum Crash Times:yA ≤ 3 yB ≤ 4 yC ≤ 2 yD ≤ 2 yE ≤ 4

Project Completion Time:xF = 35

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Using Excel Solver – Softwater Spreadsheet Model

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Solver Model

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Solver Results Dialog Box

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Solver Solution

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Solver Answer Report

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Solver Sensitivity Report

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Solver Limits Report

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Holcomb Candle Case Study


Formulate an LP model, solve it, and explain what the solution means for the company.

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