Final Simplex Tableau for the Flair Furniture Problem C j Solution

Chapter 9
Linear
Programming:
The Simplex
Method
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
9-1
© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Learning Objectives
Students will be able to
• Convert LP constraints to
equalities with slack, surplus, and
artificial variables.
• Set up and solve both
maximization and minimization
LP problems with simplex
tableaus.
• Interpret the meaning of every
number in a simplex tableau.
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
9-2
© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Learning Objectives continued
Students will be able to
• Recognize cases of infeasibility,
unboundedness, degeneracy, and
multiple optimal solutions in a
simplex output.
• Understand the relationship
between the primal and dual and
when to formulate and use the
dual.
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by Render/Stair/Hanna
9-3
© 2003 by Prentice Hall, Inc.
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Chapter Outline
9.1 Introduction
9.2 How to Set Up the Initial
Solution
9.3 Simplex Solution Procedures
9.4 The Second Simplex Tableau
9.5 Developing the Third
Simplex Tableau
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by Render/Stair/Hanna
9-4
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Chapter Outline continued
9.6 Review of Procedures for
Solving LP Maximization
Problems
9.7 Surplus and Artificial
Variables
9.8 Solving Minimization
Problems
9.9 Review of Procedures for
Solving LP Minimization
Problems
9.10 Special Cases
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by Render/Stair/Hanna
9-5
© 2003 by Prentice Hall, Inc.
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Flair Furniture
Company
Hours Required to Produce One Unit
X2
Chairs
Available
Hours This
Week
Carpentry
4
Painting/Varnishing 2
3
1
240
100
Profit/unit
$5
X1
Tables
Department
Constraints:
$7
4 X1 + 3 X 2  240 (carpentry )
2 X 1 + 1 X 2  100
(painting & varnishing )
Objective:
Maximize: 7 X1 +5 X2
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by Render/Stair/Hanna
9-6
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Flair Furniture Company's
Feasible Region & Corner
Points
X2
Number of Chairs
100
80 B = (0,80)
60
4X1 +3X2 240
40
C = (30,40)
20
Feasible
Region
2X1+1X1100
D = (50,0)
0
20
40
60
80
100 X
Number of Tables
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by Render/Stair/Hanna
9-7
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Flair Furniture Adding Slack Variables
Constraints:
4 X 1 + 3 X 2  240 (carpentry )
2 X 1 + 1 X 2  100 (painting & varnishing )
Constraints with Slack Variables
4 X 1 + 3 X 2 + S1
2 X1 + 1 X 2
= 240 (carpentry )
+ S 2 = 100 (painting &
varnishing )
Objective Function
7 X1 +5 X2
Objective Function with Slack
Variables
7 X1 +5 X2 +0S1 +0S2
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by Render/Stair/Hanna
9-8
© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Flair Furniture’s Initial
Simplex Tableau
Real
Profit
Variables
per
Unit Prod. Columns Slack
Constant
Column Mix
Variables
Column
Columns
Column
Cj
$7
$5
$0
$0
Solution
Mix
X1
X2
S1
S2
$0
S1
2
1
1
$0
S2
4
3
0
Zj
$0
$0
$0
Cj $7 $5
Z
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$0
for Management, 8e
by Render/Stair/Hanna
9-9
Profit
per
unit row
Quantity
Constraint
100 equation
rows
1
240
Gross
Profit
$0
$0
row
Net
$0
$0
Profit
row
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0
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Pivot Row, Pivot Number
Identified in the Initial
Simplex Tableau
Cj
$7
$5
$0
$0
Solution
Mix
X1
X2
S1
S2
Quantity
1
1
0
100
3
0
1
Pivot number
240
$0
S1
2
$0
S2
4
Zj
$0
$0
$0
$0
$0
Cj Zj
$7
$5
$0
$0
$0
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by Render/Stair/Hanna
Pivot column
9-10
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Pivot
row
Completed Second Simplex
Tableau for Flair Furniture
Cj
$7
$5
$0
$0
Solution
Mix
X1
X2
S1
S2
Quantity
$7
X1
1
1/2
1/2
0
50
$0
S2
0
1
-2
1
40
Zj
$7 $7/2 $7/2 $0
Cj Zj
$0 $3/2 -$7/2 $0
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by Render/Stair/Hanna
9-11
$350
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Pivot Row, Column, and
Number Identified in Second
Simplex Tableau
Cj
$7
$0
$7
$5
$0
$0
Solution
Mix
X1
X2
S1
S2
Quantity
1/2
1/2
0
50
X1
1
S2
0
Zj
$7 $7/2 $7/2 $0
Cj Zj
$0 $3/2 -$7/2 $0
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by Render/Stair/Hanna
1
-2 1
40
Pivot number
$350
(Total
Profit)
Pivot column
9-12
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Pivot
row
Calculating the New X1
Row for Flair’s Third
Tableau
 Number   Correspond ing 

 
 Number   Number  
number

 
  above  

in
New
=
in
old



 
 

 
in
new
pivot
 X Row   X row  

 i
  i
  number  
X row 

 
1
0
3/2
-1/2
30
=
=
=
=
1
1/2
1/2
0
50
=
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by Render/Stair/Hanna
-
(1/2)
(1/2)
(1/2)
(1/2)
(1/2)
-
x
x
x
x
(0)
(1)
(-2)
(1)
(40)
x
9-13
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Upper Saddle River, NJ 07458
Final Simplex Tableau for
the Flair Furniture
Problem
Cj
$7
$5
$0
$0
Solution
Mix
X1
X2
S1
S2
Quantity
$7
X1
1
0
3/2 -1/2
30
$5
X2
0
1
-2
40
Zj
$7
5
Cj Zj
$0
$0 -$1/2-$3/2
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
9-14
1
$1/2 $3/2
$410
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Simplex Steps for
Maximization
1. Choose the variable with the greatest
positive
Cj - Zj to enter the
solution.
2. Determine the row to be replaced by
selecting that one with the smallest
(non-negative) quantity-to-pivotcolumn ratio.
3. Calculate the new values for the
pivot row.
4. Calculate the new values for the
other row(s).
5. Calculate the Cj and Cj - Zj values for
this tableau. If there are any Cj - Zj
values greater than zero, return to
Step 1.
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
9-15
© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Surplus & Artificial
Variables
Constraints
5 X1 +10X2 +8 X3  210
25X1 +30X2
= 900
Constraints-Surplus & Artificial
Variables
5 X1 +10X2 +8X3 S1 +A1 =210
25X1 +30X2
+A2 =900
Objective Function
Min: 5 X1 +9 X2 +7 X3
Objective Function-Surplus & Artificial
Variables
Min: 5 X1 +9 X2 +7 X3 +0 S1 + MA1 + MA2
To accompany Quantitative Analysis
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by Render/Stair/Hanna
9-16
© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Simplex Steps for
Minimization
1. Choose the variable with the greatest
negative Cj - Zj to enter the solution.
2. Determine the row to be replaced by
selecting that one with the smallest
(non-negative) quantity-to-pivotcolumn ratio.
3. Calculate the new values for the pivot
row.
4. Calculate the new values for the other
row(s).
5. Calculate the Cj and Cj - Zj values for
this tableau. If there are any Cj - Zj
values less than zero, return to Step 1.
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
9-17
© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Special Cases
Infeasibility
Cj
5
8
0
0
Sol X1 X2 S1 S2
Mix
5 X1 1 0 -2 3
8 X2 0 1 1 2
M A2 0 0 0 -1
Zj
5 8 -2 31
M
Cj - 0 0 2 M
Zj
31
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9-18
M
M
A1 A2 Qty
-1
-1
-1
21M
2M
+21
0
0
1
0
200
100
20
180
0+2
M
0
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Special Cases
Unboundedness
Cj
6
9
0
0
Sol X1
Mix
X1
-1
S1
-2
Zj
-9
Cj - Zj 15
X2
S1
S2 Qty
1
0
9
0
2
-1
18
-18
0
1
0
0
30
10
270
Pivot
Column
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by Render/Stair/Hanna
9-19
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Upper Saddle River, NJ 07458
Special Cases
Degeneracy
Cj
5
Solution X1
Mix
8
X2
2
X3
0
S1
0
S2
0
S3 Qty
8
X2
1/4
1
1
-2
0
0
10
0
S2
4
0
1/3
-1
1
0
20
0
S3
2
0
2
2/5
0
1
10
Zj
2
8
8
16
0
0
80
3
0
6
16
0
0
C j-Z j
Pivot Column
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by Render/Stair/Hanna
9-20
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Upper Saddle River, NJ 07458
Special Cases
Multiple Optima
Cj
3
2
0
0
Sol X1 X2 S1 S2 Qty
Mix
2
X1 3/2 1
1 0
6
0
S2
1
0 1/2 1
3
Zj
3
2
2 0 12
Cj - Zj 0
0 -2 0
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for Management, 8e
by Render/Stair/Hanna
9-21
© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Sensitivity Analysis
High Note Sound Company
Max: 50 X1 +120X 2
Subject to :
2 X1 + 4 X 2  80
3 X1 +1 X 2  60
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by Render/Stair/Hanna
9-22
© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Sensitivity Analysis
High Note Sound Company
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by Render/Stair/Hanna
9-23
© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Simplex Solution
High Note Sound Company
Cj
Sol
Mix
120 X2
0
S2
Zj
Cj Zj
50
120
0
0
X1
X2
S1
S2
Qty
0
1
0
0
20
40
2400
1/2 1 1/4
5/2 0 -1/4
60 120 30
-10 0 -30
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Nonbasic Objective
Function Coefficients
Cj
120
0
Sol
Mix
X2
S2
Zj
Cj – Zj
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by Render/Stair/Hanna
50
120
0
0
X1
X2
S1
S2 Qty
1/2
5/2
60
-10
1
1/4
0 -1/4
120 30
0
-30
9-25
0 20
1 40
0 2400
0
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Basic Objective Function
Coefficients
Cj
120
+
0
Sol
Mix
X1
S2
Zj
Cj - Zj
50
120
0
0
X1
X2
S1
S2
Qty
1/2
1
1/4
0
20
-1/4
30+
/4
-30/4
1
0
40
2400
+20 
5/2
0
60+ 120
/2 + 
-10- 0
/2
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9-26
0
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Simplex Solution
High Note Sound Company
Cj
Sol
Mix
X1
S2
Zj
Cj Zj
50
120
0
0
X1
X2
S1
S2
Qty
½
5/2
1
0
1/4
1/4
120 30
0 -30
0
1
20
40
0
0
40
2400
60
0
Objective increases by 30 if 1
additional hour of electricians time
is available.
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9-27
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Shadow Prices
• Shadow Price: Value of One
Additional Unit of a Scarce
Resource
• Found in Final Simplex Tableau
in C-Z Row
• Negatives of Numbers in Slack
Variable Column
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Steps to Form the Dual
To form the Dual:
• If the primal is max., the dual is min.,
and vice versa.
• The right-hand-side values of the primal
constraints become the objective
coefficients of the dual.
• The primal objective function
coefficients become the right-hand-side
of the dual constraints.
• The transpose of the primal constraint
coefficients become the dual constraint
coefficients.
• Constraint inequality signs are reversed.
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9-29
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Primal & Dual
Primal:
Dual
Max: 50X1 +120X2
Min : 80U1 + 60U2
Subject to:
Subject to:
2 X1 +4 X2 80
2U1 + 3U2  50
3 X1 +1X2 60
4U1 + 1U2 120
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9-30
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Primal’s Optimal Solution
Comparison of the Primal
and Dual Optimal Tableaus
Cj
$50 $120
Solution Quantity
Mix
X1
$0
X2
S1
S2
$7
X2
20
1/2
1
1/4
0
$5
S2
40
5/2
0
-1/4
1
Zj
$2,400
60
120
30
0
-10
0
-30
0
80
60
$0
$0
M
M
X2
S1
S2
A1
A2
Cj - Zj
Dual’s Optimal Solution
$0
Cj
Solution Quantity
Mix
X1
$7
U1
30
1
1/4
0 -1/4
0
1/2
$5
S1
10
0
-5/2
1 -1/2
-1
1/2
Zj
$2,400
80
20
0
-20
0
40
$0
40
0
20
M M-40
Cj - Zj
To accompany Quantitative Analysis
for Management, 8e
by Render/Stair/Hanna
9-31
© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458