Optimal Product Mix for Wyndor Glass Co

McGraw-Hill/Irwin © The McGraw-Hill Companies, Inc., 2003
2.1
Three Classic Applications of LP
• Product Mix at Ponderosa Industrial
– Considered limited resources, and determined optimal mix of plywood
products.
– Increased overall profitability of company by 20%.
• Personnel Scheduling at United Airlines
– Designed work schedules for all employees at a location to meet service
requirements most efficiently.
– Saved $6 million annually.
• Planning Supply, Distribution, and Marketing at Citgo Petroleum
Corporation
– The SDM system uses LP to coordinate the supply, distribution, and marketing
of each of Citgo’s major products throughout the United States.
– The resulting reduction in inventory added $14 million annually to Citgo’s
profits.

2.2
Wyndor Glass Co. Product Mix Problem
• Wyndor has developed the following new products: An 8-foot glass door
with aluminum framing (unit profit = $300); a 4-foot by 6-foot double-
hung, wood-framed window (unit profit = $500).
• The company has three plants: Plant 1 (capacity: 4 hrs available)
produces aluminum frames and hardware; Plant 2 (capacity: 12 hrs
available) produces wood frames; Plant 3 (18 hrs available) produces glass
and assembles the windows and doors.
• The 8-foot glass door requires some of the production capacity in Plants 1
and 3, but not Plant 2; The 4-foot wood-framed window needs only Plants
2 and 3. Technological coefficients (hrs/unit): Plant 1: 1 hr/doors, 0
hrs/windows; Plant 2: 0 hrs/doors, 2 hrs/windows; Plant 3: 3 hrs/doors, 2
hrs/windows.
• Questions: What should be the product mix?

2.3
Algebraic Model for Wyndor Glass Co.
Let D = the number of doors to produce
W = the number of windows to produce
Maximize P = $300D + $500W
subject to
D ≤ 4
2W ≤ 12
3D + 2W ≤ 18
and
D ≥ 0, W ≥ 0.

2.4
Graph of Feasible Region
0 2 4 6 8
8
6
4
10
2
Feasible
region
Production rate for doors
D
W
2 W =12
D = 4
3 D + 2 W = 18
Production rate for windows

2.5
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)
Optimalsolution
10
W
D
P =3600= 300D + 500W
P =3000= 300D + 500W
P =1500= 300D + 500W

2.6
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.

2.7
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 Windows
Unit Profit $300 $500
Hours Hours
Used Available
Plant 1 1 0 1 <= 4
Plant 2 0 2 2 <= 12
Plant 3 3 2 5 <= 18
Doors Windows Total Profit
Hours Used Per Unit Produced

2.8
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 Windows
Hours Hours
Used Available
Plant 1 1 0 1 <= 4
Plant 2 0 2 2 <= 12
Plant 3 3 2 5 <= 18
Units Produced 1 1 $800

2.9
The Complete Solver Dialogue Box

2.10
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 non-negativity constraints to all the changing cells.

2.11
The Solver Results Dialogue Box

2.12
The Optimal Solution
Doors Windows
Hours Hours
Used Available
Plant 1 1 0 2 <= 4
Plant 2 0 2 12 <= 12
Plant 3 3 2 18 <= 18
Units Produced 2 6 $3,600

2.13
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?

2.14
Profit & Gambit Co. Spreadsheet Model
3
4
5
6
7
8
9
10
11
12
13
14
B C D E F G
Television Print Media
Unit Cost ($millions) 1 2
Increased Minimum
Sales Increase
Stain Remover 0% 1% 3% >= 3%
Liquid Detergent 3% 2% 18% >= 18%
Powder Detergent -1% 4% 8% >= 4%
Total Cost
Television Print Media ($millions)
Advertising Units 4 3 10
Increase in Sales per Unit of Advertising

2.15
Algebraic Model for Profit & Gambit
Let TV = the number of units of advertising on television
PM = 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 ≥ 3
Liquid detergent increased sales: 3TV + 2PM ≥ 18
Powder detergent increased sales: –TV + 4PM ≥ 4
and
TV ≥ 0, PM ≥ 0.

2.16
Applying the Graphical Method
8
6
4
2
0 2 4 6 8 1 0
-2
-4
Amount of TV advertising
Feasible
region
10
3 TV + 2 PM = 18
PM = 3
PM
TV
-TV + 4 PM = 4
Amount of print media advertising

2.17
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

2.18
A Production Problem
Weekly supply of raw materials:
8 Small Bricks 6 Large Bricks
Products:
Table
Profit = $20 / Table
Chair
Profit = $15 / Chair

2.19
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 produce
C = Number of chairs to produce
Maximize Profit = ($20)T + ($15)C
subject to
2T + C ≤ 6 large bricks
2T + 2C ≤ 8 small bricks
and
T ≥ 0, C ≥ 0.

2.20
Graphical Representation

2.21
Components of a Linear Program
• Data Cells
• Changing Cells (“Decision Variables”)
• Target Cell (“Objective Function”)
• Constraints

2.22
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.

2.23
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

2.24
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).
3
4
5
6
7
8
B C D E F G
Tables Chairs
Profit $20.00 $15.00
Available
Large Bricks 2 1 6
Small Bricks 2 2 8
Bill of Materials

2.25
• 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).
3
4
5
6
7
8
9
10
11
B C D E F G
Tables Chairs
Profit $20.00 $15.00
Available
Large Bricks 2 1 6
Small Bricks 2 2 8
Tables Chairs
Production Quantity: 0 0
Bill of Materials

2.26
• 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
G
Total Profit
=SUMPRODUCT(C4:D4,C11:D11)
3
4
5
6
7
8
9
10
11
B C D E F G
Tables Chairs
Profit $20.00 $15.00
Available
Large Bricks 2 1 6
Small Bricks 2 2 8
Tables Chairs Total Profit
Production Quantity: 1 0 $20.00
Bill of Materials

2.27
• 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.
3
4
5
6
7
8
9
10
11
B C D E F G
Tables Chairs
Profit $20.00 $15.00
Total Used Available
Large Bricks 2 1 3 <= 6
Small Bricks 2 2 4 <= 8
Bill of Materials
6
7
8
E
Total Used
=SUMPRODUCT(C7:D7,$C$11:$D$11)
=SUMPRODUCT(C8:D8,$C$11:$D$11)

2.28
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.
3
4
5
6
7
8
9
10
11
B C D E F G
Tables Chairs
Profit $20.00 $15.00
Bill of Materials

2.29
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.
3
4
5
6
7
8
9
10
11
B C D E F G
Tables Chairs
Profit $20.00 $15.00
Bill of Materials

2.30
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.
3
4
5
6
7
8
9
10
11
B C D E F G
Tables Chairs
Profit $20.00 $15.00
Bill of Materials

2.31
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.

2.32
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.”
3
4
5
6
7
8
9
10
11
B C D E F G
Tables Chairs
Profit $20.00 $15.00
Bill of Materials

2.33
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.

2.34
Example: (Multiple Optimal Solutions)
Minimize Z = 6x1 + 4x2
subject to
x1 ≤ 4
2x2 ≤ 12
3x1 + 2x2 ≤ 18
and
x1 ≥ 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

2.35
Example: (No Feasible Solution)
Maximize Z = 3x1 + 5x2
subject to
x1 ≥ 5
x2 ≥ 4
3x1 + 2x2 ≤ 18
and
x1 ≥ 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

2.36
Example: (Unbounded Solution)
Maximize Z = 5x1 + 12x2
subject to
x1 ≤ 5
2x1 –x2 ≤ 2
and
x1 ≥ 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

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