Adv. Notes 1

© The McGraw-Hill Companies, Inc., 2008

2.1McGraw-Hill/Irwin

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) 2.3–2.7The Algebraic Model for Wyndor (Section 2.3) 2.8The Graphical Method Applied to the Wyndor Problem (Section 2.4) 2.9–2.19Using the Excel Solver with the Wyndor Problem (Section 2.5) 2.20–2.25A Minimization Example—The Profit & Gambit Co. (Section 2.6) 2.26–2.31

Introduction to Linear Programming (UW Lecture) 2.32–2.47These slides are based upon a lecture introducing the basic concepts of linear programming and the Solver to first-year MBA students at the University of Washington (as taught by one of the authors). The lecture is largely based upon a production problem using lego building blocks.

The Graphical Method and Properties of LP Solutions (UW Lecture) 2.48–2.56These slides are based upon a lecture introducing the graphical method and other concepts about linear programming solutions to first-year MBA students at the University of Washington (as taught by one of the authors).



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?



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).

345678

B C D E F GTables Chairs

Profit $20.00 $15.00

AvailableLarge Bricks 2 1 6Small Bricks 2 2 8

Bill of Materials




• 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).

3456789101112

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 WindowsUnits Produced 0 0

Hours Used Per Unit Produced




• 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).

3456789101112


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)




• 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.

3456789

101112


Unit Profit $300 $500Hours HoursUsed Available

Plant 1 1 0 1 <= 1Plant 2 0 2 2 <= 12Plant 3 3 2 5 <= 18



56789

EHoursUsed

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



A Trial Solution

3456789101112




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.



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.



Graphing the Product Mix

Pro

duct

ion

rate

(un

its

per

wee

k) f

or w

indo

ws

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

Pro

duct

ion

rate

(un

its

per

wee

k) f

or w

indo

ws

Production rate (units per week) for doors

(4, 6)

(2, 3)

D = 4 and W = 6

D = 2 and W = 3

Origin

D

W



Graph Showing Constraints: D ≥ 0 and W ≥ 0

Pro

duct

ion

rate

for

win

dow

s

8

6

4

2

2 4 6 80

Production rate for doors

Pro

duct

ion

rate

for

win

dow

s

D

W



Nonnegative Solutions Permitted by D ≤ 4

Prod

uctio

n ra

te f

or w

indo

ws

D

W

8

6

4

2

2 4 6 80Production rate for doors

Prod

uctio

n ra

te f

or w

indo

ws

D = 4



Nonnegative Solutions Permitted by 2W ≤ 12


8

6

4

2

2 4 6 80

2 W = 12

D

WProduction rate for windows



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



Changing Right-Hand Side Creates Parallel Constraint Boundary Lines

12

10

8

6

4

2

0 2 4 6 8 10


D

W

3D + 2W = 24

3D + 2W = 18

3D + 2W = 12

Production rate for windows



Nonnegative Solutions Permitted by3D + 2W ≤ 18

8

6

4

0 2 4 6 8

10

2

Production rate for doorsD

W

3D + 2W = 18




Graph of Feasible Region

0 2 4 6 8

8

6

4

10

2

Feasible

region

Production rate for doorsD

W

2 W =12

D = 4

3 D + 2 W = 18




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



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



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.



Identifying the Target Cell and Changing Cells

• Choose the “Solver” from the Tools menu.• 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.

3456789101112








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.

3456789101112








The Complete Solver Dialogue Box



Some Important Options

• 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.



The Solver Results Dialogue Box



The Optimal Solution

3456789101112




Doors Windows Total ProfitUnits Produced 2 6 $3,600




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?



Profit & Gambit Co. Spreadsheet Model

34567891011121314

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



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.



Applying the Graphical Method

8

6

4

2

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

Feasible

region

10

3 TV + 2 PM = 18

PM = 3

PM

TV

-TV + 4 PM = 4

Amount of print media advertising



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



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.



A Production Problem

Weekly supply of raw materials:

8 Small Bricks 6 Large Bricks

Products:

TableProfit = $20 / Table

ChairProfit = $15 / Chair



Linear Programming

• Linear programming uses a mathematical model to find the best allocation of scarce resources to various activities so as to maximize profit or minimize cost.

Let T = Number of tables to produceC = Number of chairs to produce

Maximize Profit = ($20)T + ($15)Csubject to

2T + C ≤ 6 large bricks2T + 2C ≤ 8 small bricks

andT ≥ 0, C ≥ 0.



Graphical Representation

1 2 3 4 5 6

1

2

3

4

5

Chairs

Tables

Chairs + 2 Tables = 6 Large Bricks

2 Chairs + 2 Tables = 8 Small Bricks



Components of a Linear Program

• Data Cells

• Changing Cells (“Decision Variables”)

• Target Cell (“Objective Function”)

• Constraints



Four Assumptions of Linear Programming

• Linearity

• Divisibility

• Certainty

• Nonnegativity



When is a Spreadsheet Model Linear?

• All equations (output cells) must be of the form= ax + by + cz + …

where a, b, c are constants (data cells) and x, y, z are changing cells.

• Suppose C1:C6 are changing cells and D1:D6 are data cells.Which of the following can be part of an LP?

– SUMPRODUCT(D1:D6, C1:C6)

– SUM(C1:C6)

– C1 * SUM(C4:C6)

– SUMPRODUCT(C1:C3, C4:C6)

– IF(C1 > 3, 2*C3 + C4, 3*C3 + C5)

– IF(D1 > 3, C1, C2)

– MIN(C1, C2)

– MIN(D1, D2) * C1

– ROUND(C1)



Why Use Linear Programming?

• Linear programs are easy (efficient) to solve

• The best (optimal) solution is guaranteed to be found (if it exists)

• Useful sensitivity analysis information is generated

• Many problems are essentially linear




• 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).

345678


Profit $20.00 $15.00


Bill of Materials




• 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).

34567891011


Profit $20.00 $15.00


Tables ChairsProduction Quantity: 0 0

Bill of Materials




• 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).

10

11

GTotal Profit

=SUMPRODUCT(C4:D4,C11:D11)

345678910

11


Profit $20.00 $15.00


Tables Chairs Total Profit

Production Quantity: 1 0 $20.00

Bill of Materials




• 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.

– Note the use of relative and absolute addressing to make it easy to copy formulas in column E.

345678910

11


Profit $20.00 $15.00

Total Used AvailableLarge Bricks 2 1 3 <= 6Small Bricks 2 2 4 <= 8



Bill of Materials

678

ETotal Used

=SUMPRODUCT(C7:D7,$C$11:$D$11)=SUMPRODUCT(C8:D8,$C$11:$D$11)



Defining the Target Cell

• Choose the “Solver” from the Tools menu.• 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.

345678910

11


Profit $20.00 $15.00




Bill of Materials



Identifying the Changing Cells

• Enter all the changing cells in the “By Changing Cells” window.– You may either drag the cursor across the cells or type the addresses.

– If there are multiple sets of changing cells, separate them by typing a comma.

345678910

11


Profit $20.00 $15.00




Bill of Materials



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.

345678910

11


Profit $20.00 $15.00




Bill of Materials



Some Important Options

• 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.



The Solution

• After clicking “Solve”, you will receive one of four messages:

– “Solver found a solution. All constraints and optimality conditions are satisfied.”

– “Set cell values did not converge.”

– “Solver could not find a feasible solution.”

– “Conditions for Assume Linear Model are not satisfied.”

345678910

11


Profit $20.00 $15.00




Bill of Materials



The Graphical Method for Solving LP’s

• Formulate the problem as a linear program

• Plot the constraints

• Identify the feasible region

• Draw an imaginary line parallel to the objective function (Z = a)

• Find the optimal solution



Example #1

Maximize Z = 3x1 + 5x2

subject tox1 ≤ 42x2 ≤ 123x1 + 2x2 ≤ 18

andx1 ≥ 0, x2 ≥ 0.

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

x2

x1



Example #2

Minimize Z = 15x1 + 20x2

subject tox1 +2x2 ≥ 102x1 – 3x2 ≤ 6x1 + x2 ≥ 6

andx1 ≥ 0, x2 ≥ 0.

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

x2

x1



Example #3

Maximize Z = x1 + x2

subject tox1 +2x2 = 8x1 – x2 ≤ 0

andx1 ≥ 0, x2 ≥ 0.

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

x2

x1



Properties of Linear Programming Solutions

• An optimal solution must lie on the boundary of the feasible region.

• There are exactly four possible outcomes of linear programming:– A unique optimal solution is found.

– An infinite number of optimal solutions exist.

– No feasible solutions exist.

– The objective function is unbounded (there is no optimal solution).

• If an LP model has one optimal solution, it must be at a corner point.

• If an LP model has many optimal solutions, at least two of these optimal solutions are at corner points.



Example #4 (Multiple Optimal Solutions)

Minimize Z = 6x1 + 4x2

subject tox1 ≤ 42x2 ≤ 123x1 + 2x2 ≤ 18

andx1 ≥ 0, x2 ≥ 0.

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

x2

x1



Example #5 (No Feasible Solution)


subject tox1 ≥ 5x2 ≥ 43x1 + 2x2 ≤ 18

andx1 ≥ 0, x2 ≥ 0.

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

x2

x1



Example #6 (Unbounded Solution)


subject tox1 ≤ 52x1 –x2 ≤ 2

andx1 ≥ 0, x2 ≥ 0.

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

x2

x1



The Simplex Method Algorithm

1. Start at a feasible corner point (often the origin).

2. Check if adjacent corner points improve the objective function:a) If so, move to adjacent corner and repeat step 2.

b) If not, current corner point is optimal. Stop.

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

x2

x1

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Adv. Notes 1