chapter 2 linear programming models: graphical and computer methods jason c. h. chen, ph.d....
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Chapter 2Linear Programming Models:Graphical and Computer Methods
Jason C. H. Chen, Ph.D.Professor of MIS
School of Business AdministrationGonzaga UniversitySpokane, WA 99223
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Steps in Developing a Linear Programming (LP) Model
1) Formulation
2) Solution
3) Interpretation and Sensitivity Analysis
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Properties of LP Models
1) Seek to minimize or maximize
2) Include “constraints” or limitations
3) There must be alternatives available
4) All equations are linear
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Example LP Model Formulation:The Product Mix Problem
Decision: How much to make of > 2 products?
Objective: Maximize profit
Constraints: Limited resources
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Example: Flair Furniture Co.
Two products: Chairs and Tables
Decision: How many of each to make this month?
Objective: Maximize profit
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Flair Furniture Co. DataTables
(per table)
Chairs(per chair)
Hours Available
Profit Contribution
$7 $5
Carpentry 3 hrs 4 hrs 2400
Painting 2 hrs 1 hr 1000
Other Limitations:• Make no more than 450 chairs• Make at least 100 tables
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Decision Variables:
T = Num. of tables to make
C = Num. of chairs to make
Objective Function: Maximize Profit
Maximize $7 T + $5 C
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Constraints:
• Have 2400 hours of carpentry time available
3 T + 4 C < 2400 (hours)
• Have 1000 hours of painting time available
2 T + 1 C < 1000 (hours)
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More Constraints:• Make no more than 450 chairs
C < 450 (num. chairs)
• Make at least 100 tables T > 100 (num. tables)
Nonnegativity:Cannot make a negative number of chairs or tables
T > 0C > 0
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Model Summary
Max 7T + 5C (profit)
Subject to the constraints:
3T + 4C < 2400 (carpentry hrs)
2T + 1C < 1000 (painting hrs)
C < 450 (max # chairs)
T > 100 (min # tables)
T, C > 0 (nonnegativity)
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Using Excel’s Solver for LPRecall the Flair Furniture Example:
Max 7T + 5C (profit)Subject to the constraints:
3T + 4C < 2400 (carpentry hrs)
2T + 1C < 1000 (painting hrs)
C < 450 (max # chairs)
T > 100 (min # tables)
T, C > 0 (nonnegativity)
Go to file 2-1.xls
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Flair Furniture
T C
Tables Chairs
Number of units
Profit 7 5 =SUMPRODUCT(B6:C6,$B$5:$C$5)
Constraints:
Carpentry hours 3 4 =SUMPRODUCT(B8:C8,$B$5:$C$5) <= 2400
Painting hours 2 1 =SUMPRODUCT(B9:C9,$B$5:$C$5) <= 1000
Maximum chairs 1 =SUMPRODUCT(B10:C10,$B$5:$C$5) <= 450
Minimum tables 1 =SUMPRODUCT(B11:C11,$B$5:$C$5) >= 100
LHS Sign RHS
Flair Furniture
T C
Tables Chairs
Number of units 320.0 360.0
Profit $7 $5 $4,040.00
Constraints:
Carpentry hours 3 4 2400.0 <= 2400
Painting hours 2 1 1000.0 <= 1000
Maximum chairs 1 360.0 <= 450
Minimum tables 1 320.0 >= 100
LHS Sign RHS
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Microsoft Excel 10.0 Answer Report
Worksheet: [2-1.xls]Flair Furniture
Target Cell (Max)
Cell Name Original Value Final Value
$D$6 Profit $0.00 $4,040.00
Adjustable Cells
Cell Name Original Value Final Value
$B$5 Number of units Tables 0.0 320.0
$C$5 Number of units Chairs 0.0 360.0
Constraints
Cell Name Cell Value Formula Status Slack
$D$8 Carpentry hours 2400.0 $D$8<=$F$8 Binding 0.0
$D$9 Painting hours 1000.0 $D$9<=$F$9 Binding 0.0
$D$10 Maximum chairs 360.0 $D$10<=$F$10 Not Binding 90.0
$D$11 Minimum tables 320.0 $D$11>=$F$11 Not Binding 220.0
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Add a new constraint
• A new constraint specified by the marketing department.
• Specifically, they needed to ensure theat the number of chairs made this month is at least 75 more than the number of tables made. The constraint is expressed as:
C - T > 75
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Revised Model for Flair FurnitureMax 7T + 5C (profit)
Subject to the constraints:
3T + 4C < 2400 (carpentry hrs)
2T + 1C < 1000 (painting hrs)
C < 450 (max # chairs)
T > 100 (min # tables)
- 1T + 1C > 75
T, C > 0 (nonnegativity)
Go to file 2-2.xls
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Flair Furniture - Modified Problem
T C
Tables Chairs
Number of units 300.0 375.0
Profit $7 $5 $3,975.00
Constraints:
Carpentry hours 3 4 2400.0 <= 2400
Painting hours 2 1 975.0 <= 1000
Maximum chairs 0 1 375.0 <= 450
Minimum tables 1 0 300.0 >= 100
Tables vs Chairs -1 1 75.0 >= 75
LHS Sign RHSFlair Furniture
T C
Tables Chairs
Number of units 320.0 360.0
Profit $7 $5 $4,040.00
Constraints:
Carpentry hours 3 4 2400.0 <= 2400
Painting hours 2 1 1000.0 <= 1000
Maximum chairs 1 360.0 <= 450
Minimum tables 1 320.0 >= 100
LHS Sign RHS
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End of Chapter 2
• No Graphical Solution will be discussed
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Graphical Solution
• Graphing an LP model helps provide insight into LP models and their solutions.
• While this can only be done in two dimensions, the same properties apply to all LP models and solutions.
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Carpentry
Constraint Line
3T + 4C = 2400
Intercepts
(T = 0, C = 600)
(T = 800, C = 0)
0 800 T
C
600
0
Feasible
< 2400 hrs
Infeasible
> 2400 hrs
3T + 4C = 2400
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Painting
Constraint Line
2T + 1C = 1000
Intercepts
(T = 0, C = 1000)
(T = 500, C = 0)
0 500 800 T
C1000
600
0
2T + 1C = 1000
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0 100 500 800 T
C1000
600
450
0
Max Chair Line
C = 450
Min Table Line
T = 100
Feasible
Region
Dr. Chen, Decision Support Systems 220 100 200 300 400 500 T
C
500
400
300
200
100
0
Objective Function Line
7T + 5C = Profit
7T + 5C = $2,100
7T + 5C = $4,040
Optimal Point(T = 320, C = 360)7T + 5C
= $2,800
Dr. Chen, Decision Support Systems 230 100 200 300 400 500 T
C
500
400
300
200
100
0
Additional Constraint
Need at least 75 more chairs than tables
C > T + 75
Or
C – T > 75
T = 320C = 360
No longer feasible
New optimal pointT = 300, C = 375
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LP Characteristics
• Feasible Region: The set of points that satisfies all constraints
• Corner Point Property: An optimal solution must lie at one or more corner points
• Optimal Solution: The corner point with the best objective function value is optimal
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Special Situation in LP
1. Redundant Constraints - do not affect the feasible region
Example: x < 10
x < 12
The second constraint is redundant because it is less restrictive.
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Special Situation in LP
2. Infeasibility – when no feasible solution exists (there is no feasible region)
Example: x < 10
x > 15
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Special Situation in LP
3. Alternate Optimal Solutions – when there is more than one optimal solution
Max 2T + 2CSubject to:
T + C < 10T < 5 C < 6
T, C > 0
0 5 10 T
C
10
6
0
2T + 2C = 20All points on Red segment are optimal
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Special Situation in LP
4. Unbounded Solutions – when nothing prevents the solution from becoming infinitely large
Max 2T + 2CSubject to: 2T + 3C > 6
T, C > 0
0 1 2 3 T
C
2
1
0
Directi
on
of solutio
n