1

Optimization – Part I

Applications of Optimization

To Operations Management

For this session, the learning objectives are:

Learn what a Linear Program is.

Learn how to formulate a Linear Program and solve it using Excel’s Solver.

Using Solver to solve a Make-or-Buy Problem.

Using Solver to solve a Transshipment Problem (Product Distribution).

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Optimization involves the maximization or minimization of an objective subject to a set of constraints.

Every copy of Microsoft Excel includes Solver, which enables you to solve the following types of optimization problems:

a Linear Program,

an Integer Linear Program,

a Nonlinear Program.

The next page summarizes the use of Excel’s Solver.

Excel’s Solver

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Using Excel’s Built-in Solver 1. Launch Excel in the usual manner. 2. Construct the appropriate spreadsheet model. 3. Choose the Tools, Solver menu selection, and wait for the opening of the Solver

Parameters dialog box. (NOTE: If Solver does not appear under the Tools menu, then choose the Tools, Add-Ins menu selection, and check the Solver box. If a check-box for Solver does not appear under the Tools, Add-Ins menu, then you must install Solver from Excel’s original installation disk or CD-ROM.)

4. In the Set Target Cell box, identify the cell to optimize (the Target Cell in the jargon

of Solver). Then select either the Max or Min option. (Max is the default.) 5. In the By Changing Cells box, identify the cells that represent the decision

variables (the Changing Cells in the jargon of Solver). 6. By clicking the Add button as often as needed, specify the constraints. 7. Click the Options button. If appropriate (as will be true for most of this course),

check the Assume Non-Negative box to specify that all changing cells must have non-negative values, and check the Assume Linear Model box to specify that the optimization problem is a linear program. (NOTE: If you forget to do the latter when solving a linear program, you will most likely obtain the correct optimal solution, but your computer will work harder and longer. Also, for advanced users, your Sensitivity Report will be incomplete.)

8. To save the model for future use, temporarily leave the Solver Parameters dialog

box by clicking Close, and then save your spreadsheet in the usual manner. 9. Return to the Solver Parameters dialog box by once again choosing the Tools,

Solver menu selection. (NOTE: To retain a copy of the Solver Parameters dialog box, simultaneously depress the Alt and Print Screen keys, thereby placing a copy of the Solver Parameters dialog box into the clipboard. Then, after closing the dialog box, you can paste the copy into any Excel spreadsheet or Word document.)

10. Optimize the model by clicking the Solve button. 11. In the Solver Results dialog box, click OK to keep Solver’s solution. 12. Print the optimized spreadsheet. (NOTE: To have the row and column headings

printed with your spreadsheet, choose the File, Page Setup menu selection, then click the Sheet tab, and finally check the Row and Column Headings box.

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DuPunt, Inc. manufactures three types of chemicals. For the upcoming month, DuPunt has contracted to supply its customers with the following amounts of the three chemicals:

DuPunt’s production is limited by the availability of processing time in two chemical reactors. Each chemical must be processed first in Reactor 1 and then in Reactor 2. The following table provides the hours of processing time available next month for each reactor and the processing time required in each reactor by each chemical:

Because of the limited availability of reactor processing time, DuPunt has insufficient capacity to meet its demand with in-house production. Consequently, DuPunt must purchase some chemicals from vendors having excess capacity and resell them to its own customers. The following table provides each chemical’s in-house production cost and outside purchase cost:

Chemical 1

Chemical 2

Chemical 3

Contracted Sales (lbs) 2000 3500 1800

Reactor Processing Times (hrs per lb)

Chemical 1

Chemical

2

Chemical

3

Reactor Availabilities

(hrs) Reactor 1 0.05 0.04 0.01 200 Reactor 2 0.02 0.06 0.03 150

Chemical 1

Chemical 2

Chemical 3

In-house Production Cost $2.50 $1.75 $2.90 Outside Purchase Cost $2.80 $2.50 $3.25

DuPunt’s objective is to fill its customers’ orders with the cheapest combination of in-house production and outside purchases. In short, DuPunt must decide how much of each chemical to produce in-house (i.e., “make”) and how much of each chemical to purchase outside (i.e.,”buy”).

A Make-or-Buy Problem

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Nonnegativity Constraints

Formulation of the Make-or-Buy Problem as a Linear Program

Minimize Total Costs

Contracted Sales

Reactor Availabilities

Define the following 6 decision variables:

CHEMICAL 1 CHEMICAL 2 CHEMICAL 3MAKE M1 = Pounds of Chemical 1 to "Make" M2 = Pounds of Chemical 2 to "Make" M3 = Pounds of Chemical 3 to "Make"

BUY B1 = Pounds of Chemical 1 to "Buy" B2 = Pounds of Chemical 2 to "Buy" B3 = Pounds of Chemical 3 to "Buy"

6

7

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A Transshipment ProblemConsider a firm that for simplicity produces a single product. The firm has 3 plants (Tokyo, Hong Kong, and Bangkok), 2 warehouses (Seattle and Los Angeles), and 4 customers (Chicago, New York, Atlanta, and Dallas) geographically dispersed as diagrammed below. The firm ships its product from a plant to a warehouse and then on to a customer. In the diagram below:

The number to the left of each plant represents the plant’s supply.

The number to the right of each customer represents the customer’s demand.

The number appearing along an arrow from a plant to a warehouse or from a warehouse to a customer represents the corresponding unit shipping cost. For example, the unit shipping cost from Bangkok to Seattle is $25 per unit.

The firm wants to distribute its product at minimum cost.

185

237

14

19 11

1221

1525 10

816

10

25

29

18

19

24

17

Chicago

New York

Atlanta

Dallas

Tokyo

Hong Kong

Bangkok

Seattle

Los Angeles

72 = Total Supply

Total Demand = 70

9Nonnegativity Constraints

Formulation of the Transshipment Problem as an LP

Min Total Shipping Costs

Supply Constraints

Transshipment Constraints

Demand Constraints

Let AZ denote the amount shipped from location A to location Z. As examples, TL denotes the amount shipped from Tokyo to Los Angeles, and SN denotes the amount shipped from Seattle to New York.

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Solving the Transshipment LP Using Excel’s Solver

The Spreadsheet Before Optimization

123456789101112131415161718192021222324252627282930313233343536373839

A B C D E F G H I J K L M N

TRANSSHIPMENT PROBLEM

TOTAL SHIPPING COST = $0

UNIT SHIPPING COSTS ($) FROM/TO UNITS SHIPPED FROM/TO

TO WAREHOUSE IN … TO WAREHOUSE IN …SHIPPED

S L S L OUT SUPPLY

T 18 23 T 0 0 0 25FROM PLANT IN … H 19 21 H 0 0 0 29

B 25 16 B 0 0 0 18SHIPPED IN 0 0

SHIPPEDC N A D C N A D OUT

S 5 7 14 11 S 0 0 0 0 0L 12 15 10 8 L 0 0 0 0 0

SHIPPED IN 0 0 0 0DEMAND 19 24 17 10

FROM WAREHOUSE IN …

TO CUSTOMER IN … TO CUSTOMER IN …

Changing Cells

Target Cell=SUMPRODUCT(C12:D14,J 12:K14)+SUMPRODUCT(C21:F22,J 21:M22)

=SUM(J 12:J 14) ; copied to K15

=SUM(J 21:J 22) ; copied to K23:M23

=SUM(J 12:K12) ; copied to L13:L14

=SUM(J 21:M21) ; copied to N22

Changing Cells

Transshipment

Demand

Supply

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Solving the Transshipment LP Using Excel’s SolverThe Spreadsheet After Optimization

1234567891011121314151617181920212223242526

A B C D E F G H I J K L M N

TRANSSHIPMENT PROBLEM

TOTAL SHIPPING COST = $1,782

UNIT SHIPPING COSTS ($) FROM/TO UNITS SHIPPED FROM/TO

TO WAREHOUSE IN … TO WAREHOUSE IN …SHIPPED

S L S L OUT SUPPLY

T 18 23 T 25 0 25 25FROM PLANT IN … H 19 21 H 18 9 27 29

B 25 16 B 0 18 18 18SHIPPED IN 43 27

SHIPPEDC N A D C N A D OUT

S 5 7 14 11 S 19 24 0 0 43L 12 15 10 8 L 0 0 17 10 27

SHIPPED IN 19 24 17 10DEMAND 19 24 17 10

FROM WAREHOUSE IN …

TO CUSTOMER IN … TO CUSTOMER IN …

Changing Cells

Target Cell=SUMPRODUCT(C12:D14,J 12:K14)+SUMPRODUCT(C21:F22,J 21:M22)

=SUM(J 12:J 14) ; copied to K15

=SUM(J 21:J 22) ; copied to K23:M23

=SUM(J 12:K12) ; copied to L13:L14

=SUM(J 21:M21) ; copied to N22

Changing Cells

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INTRODUCTION TO THE BLENDING PROBLEM

In many businesses and industrial environments, the goal is to find the optimal “recipe” for blending a variety of “ingredients” to obtain a product that meets lower and/or upper limits on a variety of characteristics. The table below summarizes several applications.

Final Product Partial List of Ingredients Lower and/or Upper Limits

on Such Things as …

Gasoline or Heating Oil

“Stock” Crude Oils with differing characteristics, Butane, & Catalytic Reformate

Octane, Lead Content, & Volatility

Steel “Stock” Steels with differing characteristics, Pig Iron, & Carbon

Strength, Melting Point, & Resistance to Rust

Fertilizer “Stock” Fertilizers with differing chemical compositions

Nitrogen Content, Phosphorus Content, & Potassium Content

Recycled Paper “Stock” Pulps such as Newspaper, Office Paper, & Cardboard

Color, Strength, & Texture

Alcoholic Drinks (such as wine &

whiskey)

“Stock” Wines/Whiskeys with differing levels of Quality and differing Ages

Quality, Average Age, Color

Animal Feed (for cattle, pigs,

dogs, etc.)

Grains such as Corn & Wheat, Soybean Meal, Fishbones, Hay, & Sawdust

Fat Content, Protein Content, Carbohydrate Content, Calcium Content, Iron Content, Fiber, & Calories

Sausage

Various Grades of Beef & Pork, Water, and “Filler” that consists of some things you really don’t want to know

Content of various types of meats, Coloring, & Texture

Mixed Nuts Peanuts, Almonds, Pecans, Cashews, & Hazelnuts,

Content of various types of nuts

Power Generation Using Coal

“Stock” Coals with differing characteristics

Emissions of pollutants as measured by Sulfur Dioxide & Coal Dust

Portfolio of Investments

Stocks, Bonds, Gold, Loans Risk, Diversity, Beta, Duration, & Convexity

Portfolio of Advertisements

Television Ads, Radio Ads, Newspaper Ads, Magazine Ads, Billboard Ads

Exposure, Impact, & Diversity

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A Blending Problem

Harrus Feeding Company’s Blending Problem

The Harrus Feeding Company (HFC) operates a feedlot to which cattle are brought for the final fattening process. Since HFC’s cattle population averages about 100,000, it is important for HFC to feed the cattle in a way that meets their nutritional requirements at minimum cost.

The mixture HFC feeds the cattle is blend of four feedstuffs: corn, wheat, barley, and hay. The table below provides the relevant dietary and cost data per pound of each feedstuff, along with a steer’s daily nutritional requirement. For example, for each pound of corn a steer consumes, it receives 2 grams of fat, 20 grams of protein, 4 milligrams of iron, and 200 calories.

a) Assuming a steer’s daily consumption of feedstuffs must be exactly 24 pounds, formulate and solve a linear program for determining the dietary blend that satisfies HFC’s daily requirements at minimum cost.

b) How would you modify your formulation if a steer’s daily consumption of feedstuffs must be in the range of 23-25 pounds?

c) How would you modify your formulation if there were no daily limit on the pounds of feedstuffs that a steer must consume?

d) Can the formulations in part (a) and part (c) result in distinct optimal solutions? Can you anticipate a potential problem with the optimal solution to the linear program in part (c)?

The exercise below is designed to review the “basics” of formulating a linear program and solving it using Solver.

Units of Nutrient per Pound of Feedstuff

Nutrient Corn Wheat Barley Hay

Minimum Daily

Requirement

Maximum Daily

Requirement

Fat (g) 2 1 3 4 25 g 100 g Protein (g) 20 15 15 10 400 g no limit

Iron (mg) 4 7 6 5 125 mg no limit Calories 200 400 300 500 6000 no limit

Cost per Pound

60¢ 40¢ 35¢ 5¢