Introduction to Management Science Chapter Two 1.

Introduction to Management Science

Chapter Two

1

2-2

A Production ProblemWeekly supply of raw materials:

8 Small Bricks 6 Large Bricks

Products:

TableProfit = $20 / Table

ChairProfit = $15 / Chair

2-3

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.

2-4

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

2-5

Developing a Spreadsheet Model• 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

B C D E F GTables Chairs

Profit $20.00 $15.00

AvailableLarge Bricks 2 1 6Small Bricks 2 2 8

Tables ChairsProduction Quantity: 0 0

Bill of Materials

2-6

Developing a Spreadsheet Model• 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

B C D E F GTables Chairs

Profit $20.00 $15.00

AvailableLarge Bricks 2 1 6Small Bricks 2 2 8

Tables Chairs Total Profit

Production Quantity: 1 0 $20.00

Bill of Materials

2-7

Developing a Spreadsheet Model• 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

B C D E F GTables Chairs

Profit $20.00 $15.00

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

Tables Chairs Total Profit

Production Quantity: 1 1 $35.00

Bill of Materials

678

ETotal Used

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

2-8

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

B C D E F GTables Chairs

Profit $20.00 $15.00

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

Tables Chairs Total Profit

Production Quantity: 1 1 $35.00

Bill of Materials

2-9

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

B C D E F GTables Chairs

Profit $20.00 $15.00

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

Tables Chairs Total Profit

Production Quantity: 1 1 $35.00

Bill of Materials

2-10

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

B C D E F GTables Chairs

Profit $20.00 $15.00

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

Tables Chairs Total Profit

Production Quantity: 1 1 $35.00

Bill of Materials

2-11

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.

2-12

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

B C D E F GTables Chairs

Profit $20.00 $15.00

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

Tables Chairs Total Profit

Production Quantity: 2 2 $70.00

Bill of Materials

2-13

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

2-14

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?

2-15

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.

2-16

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).

Doors WindowsUnit Profit $300 $500

HoursAvailable

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

Hours Used Per Unit Produced

2-17

Developing a Spreadsheet Model• 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).

Doors WindowsUnit Profit $300 $500

HoursAvailable

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

Doors WindowsUnits Produced 0 0

Hours Used Per Unit Produced

2-18

Developing a Spreadsheet Model• 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).

Doors WindowsUnit Profit $300 $500

HoursAvailable

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

Doors Windows Total ProfitUnits Produced 1 1 $800

Hours Used Per Unit Produced

1112

GTotal Profit

=SUMPRODUCT(UnitProfit,UnitsProduced)

2-19

Developing a Spreadsheet Model• 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.

Doors WindowsUnit Profit $300 $500

Hours HoursUsed Available

Plant 1 1 0 1 <= 4Plant 2 0 2 2 <= 12Plant 3 3 2 5 <= 18

Doors Windows Total ProfitUnits Produced 1 1 $800

Hours Used Per Unit Produced

56789

EHoursUsed

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

2-20

A Trial SolutionDoors Windows

Unit Profit $300 $500Hours HoursUsed Available

Plant 1 1 0 4 <= 4Plant 2 0 2 6 <= 12Plant 3 3 2 18 <= 18

Doors Windows Total ProfitUnits Produced 4 3 $2,700

Hours Used Per Unit Produced

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

2-21

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.

Doors WindowsUnit Profit $300 $500

Hours HoursUsed Available

Plant 1 1 0 1 <= 4Plant 2 0 2 2 <= 12Plant 3 3 2 5 <= 18

Doors Windows Total ProfitUnits Produced 1 1 $800

Hours Used Per Unit Produced

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.

Doors WindowsUnit Profit $300 $500

Hours HoursUsed Available

Plant 1 1 0 1 <= 4Plant 2 0 2 2 <= 12Plant 3 3 2 5 <= 18

Doors Windows Total ProfitUnits Produced 1 1 $800

Hours Used Per Unit Produced

2-23

The Complete Solver Dialogue Box

2-24

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.

2-25

The Optimal Solution

Doors WindowsUnit Profit $300 $500

Hours HoursUsed Available

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

Doors Windows Total ProfitUnits Produced 2 6 $3,600

Hours Used Per Unit Produced

2-26

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

2-27

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.

2-28

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

2-29

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

2-30

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

2-31

Boundary Line for Constraint 3D + 2W ≤ 18

Production rate for doors

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

2-32

Changing Right-Hand Side Creates Parallel Constraint Boundary Lines

12

10

8

6

4

2

0 2 4 6 8 10

Production rate for doors

D

W

3D + 2W = 24

3D + 2W = 18

3D + 2W = 12

Production rate for windows

2-33

Nonnegative Solutions Permitted by3D + 2W ≤ 18

8

6

4

0 2 4 6 8

10

2

Production rate for doorsD

W

3D + 2W = 18

Production rate for windows

2-34

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

Production rate for windows

2-35

Objective Function (P = 1,500)

0 2 4 6 8

8

6

4

2

Production rate

for windows

Production rate for doors

Feasible

regionP = 1500 = 300D + 500W

D

W

2-36

Finding the Optimal Solution

0 2 4 6 8

8

6

4

2

Production rate

for windows

Production rate for doors

Feasible

region

(2, 6)

Optimal solution

10

W

D

P = 3600 = 300D + 500W

P = 3000 = 300D + 500W

P = 1500 = 300D + 500W

2-37

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?

2-38

Profit & Gambit Co. Spreadsheet Model

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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

2-39

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.

2-40

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

2-41

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

2-42

Components of a Linear Program

• Data Cells

• Changing Cells (“Decision Variables”)

• Target Cell (“Objective Function”)

• Constraints

2-43

Four Assumptions of Linear Programming

• Linearity

• Divisibility

• Certainty

• Nonnegativity

2-44

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

2-45

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.

2-46

(Multiple Optimal Solutions)Maximize 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

2-47

(No Feasible Solution)Maximize Z = 3x1 + 5x2

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

2-48

(Unbounded Solution)Maximize Z = 5x1 + 12x2

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