[ppt]table of contents chapter 2 (linear programming:...

Table of ContentsChapter 2 (Linear Programming: Basic Concepts)

The Wyndor Glass Company Product Mix Problem (Section 2.1)

2.2Formulating the Wyndor Problem on a Spreadsheet (Section 2.2)

2.3–2.8The Algebraic Model for Wyndor (Section 2.3)

2.9The Graphical Method Applied to the Wyndor Problem (Section 2.4)

2.10–2.20Using the Excel Solver with the Wyndor Problem (Section 2.5)

2.21–2.28A Minimization Example—The Profit & Gambit Co. (Section 2.6)

2.29–2.34

Copyright © 2011 by the McGraw-Hill Companies, Inc. All rights reserved.McGraw-Hill/Irwin

Wyndor Glass Co. Product Mix Problem

• Wyndor has developed the following new products:– An 8-foot glass door with aluminum framing.– A 4-foot by 6-foot double-hung, wood-framed window.

• The company has three plants– Plant 1 produces aluminum frames and hardware.– Plant 2 produces wood frames.– Plant 3 produces glass and assembles the windows and doors.

Questions:1. Should they go ahead with launching these two new products?2. If so, what should be the product mix?

2-2

Developing a Spreadsheet Model

• Step #1: Data Cells– Enter all of the data for the problem on the spreadsheet.– Make consistent use of rows and columns.– It is a good idea to color code these “data cells” (e.g., light blue).

Wyndor Glass Co. Product-Mix Problem

Doors Windows

Unit Profit $300 $500

Hours

Hours Used Per Unit Produced Available

Plant 1 1 0 4

Plant 2 0 2 12

Plant 3 3 2 18

2-3


• Step #2: Changing Cells– Add a cell in the spreadsheet for every decision that needs to be made.– If you don’t have any particular initial values, just enter 0 in each.– It is a good idea to color code these “changing cells” (e.g., yellow with border).

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A B C D E F GWyndor Glass Co. Product-Mix Problem

Doors WindowsUnit Profit $300 $500

HoursAvailable

Plant 1 1 0 4Plant 2 0 2 12Plant 3 3 2 18

Hours Used Per Unit Produced

2-4


• Step #3: Target Cell– Develop an equation that defines the objective of the model.– Typically this equation involves the data cells and the changing cells in order to

determine a quantity of interest (e.g., total profit or total cost).– It is a good idea to color code this cell (e.g., orange with heavy border).

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B C D E F GDoors Windows

Unit Profit $300 $500Hours

AvailablePlant 1 1 0 1Plant 2 0 2 12Plant 3 3 2 18

Doors Windows Total ProfitUnits Produced 1 1 $800


1112

GTotal Profit

=SUMPRODUCT(UnitProfit,UnitsProduced)

2-5


• Step #4: Constraints– For any resource that is restricted, calculate the amount of that resource used in a

cell on the spreadsheet (an output cell).– Define the constraint in three consecutive cells. For example, if Quantity A <=

Quantity B, put these three items (Quantity A, <=, Quantity B) in consecutive cells.

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Hours HoursUsed Available

Plant 1 1 0 0 <= 4Plant 2 0 2 0 <= 12Plant 3 3 2 0 <= 18



2-6

Formulas for the Spreadsheet Model

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Hours HoursUsed Available




Range Name CellsHoursAvailable G7:G9HoursUsed E7:E9HoursUsedPerUnitProduced C7:D9TotalProfit G12UnitProfit C4:D4UnitsProduced C12:D12

56789

EHoursUsed

=SUMPRODUCT(C7:D7,UnitsProduced)=SUMPRODUCT(C8:D8,UnitsProduced)=SUMPRODUCT(C9:D9,UnitsProduced)

1112

GTotal Profit

=SUMPRODUCT(UnitProfit,UnitsProduced)

2-7

A Trial Solution

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Unit Profit $300 $500Hours HoursUsed Available


Doors Windows Total ProfitUnits Produced 4 3 $2,700


The spreadsheet for the Wyndor problem with a trial solution (4 doors and 3 windows) entered into the changing cells.

2-8

Algebraic Model for Wyndor Glass Co.

Let D = the number of doors to produceW = the number of windows to produce

Maximize P = $300D + $500Wsubject to

D ≤ 42W ≤ 123D + 2W ≤ 18

andD ≥ 0, W ≥ 0.

2-9

Graphing the Product Mix

Prod

uctio

n ra

te (u

nits

per

wee

k) fo

r win

dow

s

A product mix of

A product mix of

1

2

3

4

5

6

7

8

0

-1

-1-2 1 2 3 4 5 6 7 8

-2

Prod

uctio

n ra

te (u

nits

per

wee

k) fo

r win

dow

s

Production rate (units per week) for doors

(4, 6)

(2, 3)

D = 4 and W = 6

D = 2 and W = 3

Origin

D

W

2-10

Graph Showing Constraints: D ≥ 0 and W ≥ 0

Prod

uctio

n ra

te fo

r win

dow

s

8

6

4

2

2 4 6 80Production rate for doors

Prod

uctio

n ra

te fo

r win

dow

s

D

W

2-11

Nonnegative Solutions Permitted by D ≤ 4

Prod

uctio

n ra

te fo

r win

dow

s

D

W

8

6

4

2

2 4 6 80Production rate for doors

Prod

uctio

n ra

te fo

r win

dow

sD = 4

2-12

Nonnegative Solutions Permitted by 2W ≤ 12

Production rate for doors

8

6

4

2

2 4 6 80

2 W = 12

D

WProduction rate for windows

2-13

Boundary Line for Constraint 3D + 2W ≤ 18


8

6

4

2

2 4 6 80

10

(0, 9)

(2, 6)

(4, 3)

21_(1, 7 )

21_(3, 4 )

21_(5, 1 )

(6, 0)

3 D + 2 W = 18

D

WProduction rate for windows

2-14

Changing Right-Hand Side Creates Parallel Constraint Boundary Lines

12

10

8

6

4

2

0 2 4 6 8 10

Production rate for doorsD

W

3D + 2W = 24

3D + 2W = 18

3D + 2W = 12

Production rate for windows

2-15

Nonnegative Solutions Permitted by3D + 2W ≤ 18

8

6

4

0 2 4 6 8

10

2


W

3D + 2W = 18


2-16

Graph of Feasible Region

0 2 4 6 8

8

6

4

10

2

Feasible

region


W

2 W =12

D = 4

3 D + 2 W = 18


2-17

Objective Function (P = 1,500)

0 2 4 6 8

8

6

4

2

Production rate

for windows


Feasible

regionP = 1500 = 300D + 500W

D

W

2-18

Finding the Optimal Solution

0 2 4 6 8

8

6

4

2

Production rate

for windows


Feasible

region

(2, 6)

Optimal solution

10

W

D

P = 3600 = 300D + 500W

P = 3000 = 300D + 500W

P = 1500 = 300D + 500W

2-19

Summary of the Graphical Method

• Draw the constraint boundary line for each constraint. Use the origin (or any point not on the line) to determine which side of the line is permitted by the constraint.

• Find the feasible region by determining where all constraints are satisfied simultaneously.

• Determine the slope of one objective function line. All other objective function lines will have the same slope.

• Move a straight edge with this slope through the feasible region in the direction of improving values of the objective function. Stop at the last instant that the straight edge still passes through a point in the feasible region. This line given by the straight edge is the optimal objective function line.

• A feasible point on the optimal objective function line is an optimal solution.

2-20

Identifying the Target Cell and Changing Cells (Excel 2010)

• Choose the “Solver” from the Data tab.• Select the cell you wish to optimize in the “Set Target Cell” window.• Choose “Max” or “Min” depending on whether you want to maximize or minimize the

target cell.• Enter all the changing cells in the “By Changing Cells” window.

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101112






2-21

Identifying the Target Cell and Changing Cells (Excel 2007)

• Choose the “Solver” from the Data tab (Excel 2007) or Tools menu (earlier versions).• Select the cell you wish to optimize in the “Set Target Cell” window.• Choose “Max” or “Min” depending on whether you want to maximize or minimize the

target cell.• Enter all the changing cells in the “By Changing Cells” window.

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101112






2-22

Adding Constraints

• To begin entering constraints, click the “Add” button to the right of the constraints window.

• Fill in the entries in the resulting Add Constraint dialogue box.

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2-23

Some Important Options (Excel 2007)

• Click on the “Options” button, and click in both the “Assume Linear Model” and the “Assume Non-Negative” box.

– “Assume Linear Model” tells the Solver that this is a linear programming model.– “Assume Non-Negative” adds nonnegativity constraints to all the changing cells.

2-24

The Complete Solver Dialogue Box (Excel 2010)

2-25

The Complete Solver Dialogue Box (Excel 2007)

2-26

The Solver Results Dialogue Box

2-27

The Optimal Solution

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Doors Windows Total ProfitUnits Produced 2 6 $3,600


2-28

The Profit & Gambit Co.

• Management has decided to undertake a major advertising campaign that will focus on the following three key products:

– A spray prewash stain remover.– A liquid laundry detergent.– A powder laundry detergent.

• The campaign will use both television and print media

• The general goal is to increase sales of these products.

• Management has set the following goals for the campaign:– Sales of the stain remover should increase by at least 3%.– Sales of the liquid detergent should increase by at least 18%.– Sales of the powder detergent should increase by at least 4%.

Question: how much should they advertise in each medium to meet the sales goals at a minimum total cost?

2-29

Profit & Gambit Co. Spreadsheet Model

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1011121314

B C D E F GTelevision Print Media

Unit Cost ($millions) 1 2

Increased MinimumSales Increase

Stain Remover 0% 1% 3% >= 3%Liquid Detergent 3% 2% 18% >= 18%

Powder Detergent -1% 4% 8% >= 4%

Total CostTelevision Print Media ($millions)

Advertising Units 4 3 10

Increase in Sales per Unit of Advertising

2-30

Algebraic Model for Profit & Gambit

Let TV = the number of units of advertising on televisionPM = the number of units of advertising in the print media

Minimize Cost = TV + 2PM (in millions of dollars)subject to

Stain remover increased sales: PM ≥ 3Liquid detergent increased sales: 3TV + 2PM ≥ 18Powder detergent increased sales: –TV + 4PM ≥ 4

andTV ≥ 0, PM ≥ 0.

2-31

Applying the Graphical Method

8

6

4

2

0 2 4 6 8 1 0-2-4Amount of TV advertising

Feasible

region10

3 TV + 2 PM = 18

PM = 3

PM

TV

-TV + 4 PM = 4

Amount of print media advertising

2-32

The Optimal Solution

Amount of TV advertising

Feasible region

0 5 10 15

10

4

(4,3)

optimal solution

Cost = 15 = TV + 2 PM

Cost = 10 = TV + 2 PM

TV

PM

2-33

Summary of the Graphical Method

• Draw the constraint boundary line for each constraint. Use the origin (or any point not on the line) to determine which side of the line is permitted by the constraint.

• Find the feasible region by determining where all constraints are satisfied simultaneously.

• Determine the slope of one objective function line. All other objective function lines will have the same slope.

• Move a straight edge with this slope through the feasible region in the direction of improving values of the objective function. Stop at the last instant that the straight edge still passes through a point in the feasible region. This line given by the straight edge is the optimal objective function line.

• A feasible point on the optimal objective function line is an optimal solution.

2-34

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