INTRODUCTION TO
MANAGEMENT
SCIENCE, 13e
Anderson
Sweeney
Williams
Martin
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2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slides by
JOHN
LOUCKS
St. Edward’s
University
Slide 1
Chapter 3
Linear Programming: Sensitivity Analysis
and Interpretation of Solution




Introduction to Sensitivity Analysis
Graphical Sensitivity Analysis
Sensitivity Analysis: Computer Solution
Simultaneous Changes
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2008 Thomson South-Western. All Rights Reserved
publicly accessible website, in whole or in part.
Slide 2
Introduction to Sensitivity Analysis



Sensitivity analysis (or post-optimality analysis) is
used to determine how the optimal solution is
affected by changes, within specified ranges, in:
• the objective function coefficients
• the right-hand side (RHS) values
Sensitivity analysis is important to a manager who
must operate in a dynamic environment with
imprecise estimates of the coefficients.
Sensitivity analysis allows a manager to ask certain
what-if questions about the problem.
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2008 Thomson South-Western. All Rights Reserved
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Slide 3
Example 1

LP Formulation
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Slide 4
Example 1

Graphical Solution (objective function coefficient)
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Slide 5
Example 1

Graphical Solution (objective function coefficient)
―3/2 <= slope of objective function <= ―7/10
CS
3
7
 

2
9
10
6.3  CS  13.5
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2008 Thomson South-Western. All Rights Reserved
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Slide 6
Objective Function Coefficients



The range of optimality for each coefficient provides
the range of values over which the current solution
will remain optimal.
Objective function coefficient’s range (range of
optimality) is just for one variable
given that all others are not changed
What if the coefficients are 13 and 8 for S and D
respectively.
6.3  CS  13.5
but
6.667  CD  14.286

CS
13
   1.625
CD
8
which is out of range of
―3/2 <= slope of objective function <= ―7/10
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Slide 7
Right-Hand Sides




Let us consider how a change in the right-hand side
for a constraint might affect the feasible region and
perhaps cause a change in the optimal solution.
The improvement in the value of the optimal solution
per unit increase in the right-hand side is called the
dual price.
The range of feasibility is the range over which the
dual price is applicable.
As the RHS increases, other constraints will become
binding and limit the change in the value of the
objective function.
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Slide 8
Example 1

Graphical Solution (Right Hand Side)
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Slide 9
Example 1

Graphical Solution (Right Hand Side of Constraint 1)
Intersection of constraints (3) & (4) : (474.545, 350.182)
(1) 7/10*474.545 + 1*350.182 = 682.364
Intersection of S-axis & (3) : (708, 0)
(1) 7/10*708 + 1*0 = 495.6
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Slide 10
Computer Solutions

Management Scientist
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Slide 11
Example 1

Graphical Solution (Right Hand Side of Constraint 2)
No upper limit
Intersection of constraints (1) & (3) : (540, 252)
(1) 1/2*540 + 5/6*252 = 480
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Slide 12
Dual Price




The improvement in the value of the optimal solution
per unit increase in the right-hand side is called the
dual price.
The dual price for a nonbinding constraint is 0.
For >= constraints, dual price of 0 surplus is ―
For <= constraints, dual price of 0 slack is +
A negative dual price indicates that the objective
function will not improve if the RHS is increased.
The range of feasibility (range of RHS) is the range
over which the dual price is applicable (not changed).
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Slide 13
Sensitivity Analysis: Computer Solution


Simultaneous Changes
• Until now, the sensitivity analysis information is
based on the assumption that only one coefficient
changes
100% rule
• More than 2 objective coefficients
or more than 2 RHS
• Optimal solution basis (positive valued decision
variables) are not changed
if sum of all the (changes / allowable changes)
ratios is less than 1.
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2008 Thomson South-Western. All Rights Reserved
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Slide 14
Sensitivity Analysis: Computer Solution

Objective function coefficients
• Ex1 : CS  13, CD  8  13  1.625   3
8
2
3
1
100 
 100  128.6  100
3.5
2.3333
• Ex2 :
CS  11.5, CD  8.25
3
11.5
7
 
 1.394  
2
8.25
10
1.5
0.75
100 
 100  75  100
3.5
2.3333
• Ex3 :
CS  13, CD  14
3
13
7
    0.9286  
2
14
10
3
5
100 
 100  180.3  100
3.5
5.28571
(not simultaneously binding)
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Slide 15
Sensitivity Analysis: Computer Solution

RHS values
• Ex1 : RHS1  670, RHS3  650
40
58
100 
100  121.7  100
52.36316
128
• Ex2 :
RHS1  650, RHS3  650
20
58
100 
100  83.5  100
52.36316
128
• Ex3 :
RHS1  670, RHS3  808
40
100
100 
 100  128.5  100
52.36316
192
(not simultaneously binding)
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Slide 16
Sensitivity Analysis: Computer Solution

RHS values
RHS1  670, RHS3  650
Global optimal solution found.
Objective value:
Infeasibilities:
Total solver iterations:
7390.898
0.000000
3
Variable
Value
S
395.4528
D
381.8189
Reduced Cost
0.000000
0.000000
Row Slack or Surplus
Dual Price
1
7390.898
1.000000
2
11.36416
0.000000
3
84.09247
0.000000
4
0.000000
8.727289
5
0.000000
12.72711
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2008 Thomson South-Western. All Rights Reserved
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Slide 17
Sensitivity Analysis: Computer Solution

RHS values
RHS1  650, RHS3  650
Global optimal solution found.
Objective value:
Infeasibilities:
Total solver iterations:
7353.117
0.000000
2
Variable
Value
S
406.2477
D
365.6266
Reduced Cost
0.000000
0.000000
Row Slack or Surplus
Dual Price
1
7353.117
1.000000
2
0.000000
4.374957
3
92.18853
0.000000
4
0.000000
6.937530
5
2.968579
0.000000
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2008 Thomson South-Western. All Rights Reserved
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Slide 18
Sensitivity Analysis: Computer Solution

RHS values
RHS1  670, RHS3  808
Global optimal solution found.
Objective value:
Infeasibilities:
Total solver iterations:
8536.745
0.000000
2
Variable
Value
S
677.4988
D
195.7509
Reduced Cost
0.000000
0.000000
Row Slack or Surplus
Dual Price
1
8536.745
1.000000
2
0.000000
4.374957
3
98.12555
0.000000
4
0.000000
6.937530
5
18.31241
0.000000
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Slide 19
Sensitivity Analysis: Second Example (p.110)
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Slide 20
Sensitivity Analysis: Second Example (p.110)
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Slide 21
Sensitivity Analysis: Second Example (p.110)


Dual price
• The improvement of the objective function value
per 1 unit increase of the RHS.
• Total production requirement and Processing time
are binding
• Dual price of processing time is 1
• Dual price of total minimum (350) is -4
Notes
• Dual price is an extra cost. If the profit
contribution is calculated considering the
purchasing cost of the resource, the price we are
willing to pay for that resource is
purchasing cost + dual price for 1 unit.
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Slide 22
Sensitivity Analysis: Note and Comments

Degeneracy
• Consider the available Sewing time is 480
which is calculated with 1/2*540 + 5/6*252
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Slide 23
Sensitivity Analysis: Note and Comments

Degeneracy
• Consider the available Sewing time is 480
which is calculated with 1/2*540 + 5/6*252
Global optimal solution found.
Objective value:
7668.000
Infeasibilities:
0.000000
Total solver iterations:
2
Variable
Value
S
540.0000
D
252.0000
Row
1
2
3
4
5
Objective Coefficient Ranges:
Current
Allowable
Variable Coefficient
Increase
S
10.00000
3.500000
9.000000
5.285714
Reduced Cost D
0.000000
Righthand Side Ranges:
0.000000
Slack or Surplus
Dual Price
7668.000
1.000000
0.000000
4.375000
0.000000
0.000000
0.000000
6.937500
18.00000
0.000000
Row
2
3
4
5
Current
RHS
630.0000
480.0000
708.0000
135.0000
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Allowable
Increase
0.000000
INFINITY
192.0000
INFINITY
Allowable
Decrease
3.700000
2.333333
Allowable
Decrease
134.4000
0.000000
0.000000
18.00000
Slide 24
Sensitivity Analysis: Note and Comments


Degeneracy
• Consider the available Sewing time is 480
which is calculated with 1/2*540 + 5/6*252
• In the standard form number of variables (2+3=5),
number of constraints 3.
Thus, basic solution has (set 2 variables to 0, and
solve simultaneous equations (연립방정식).
Now, at the optimal solution 3 variables are 0.
• Dual price of binding constraints is 0.
• Constraint 2 (Sewing) has 0 slack, but dual price is 0
• Range of Feasibility (range of RHS) for constraints 2,
3 and 4 are only one direction.
100% rule works only when sum of ratios are less than
100.
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Slide 25
Example 3 (more than 2 variables)

Consider the following linear program:
Max
10S + 9D + 12.85L
s.t.
0.7S + 1D + 0.8L <
0.5S + 5/6D + 1L <
1S + 2/3D + 1L <
0.1S + 0.25D + 0.25L <
630
600
708
135
x1, x2 > 0
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Slide 26
Example 3
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Slide 27
Example 3 (more than 2 variables)

Interpretation
• Deluxe model is not produced
• Finishing (Constraint 3) and Inspection and
Packaging (Constraint 4) are binding
• Range of objective function for Deluxe is
― infinity < current 9 < 10.15
• Reduced cost : the amount that an objective function
coefficient would have to improve in order for the
corresponding decision variable becomes positive.
Reduced cost of Deluxe is 1.15
= 1 * 0 + 5/6*0 + 2/3*8.1 + 0.25*19 – 9
(sum of dual prices consumed to produce 1 unit of
Deluxe)
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Slide 28
Example 3 (more than 2 variables)

Max
Primal problem vs. Dual problem
10S + 9D + 12.85L
0.7S + 1D + 0.8L <
0.5S + 5/6D + 1L <
1S + 2/3D + 1L <
0.1S + 0.25D + 0.25L <
S, D, L > 0
630
600
708
135
Min
630C + 600W + 708F + 135I
0.7C + 0.5W + 1F + 0.1I –R1 = 10
1C + 5/6W + 2/3F + 0.25I –R2 = 9
0.8C + 1W + 1F + 0.25I
–R3 = 12.85
C, W, F, I, R1, R2, R3 > 0
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Slide 29
Example 3 (more than 2 variables)

Primal problem vs. Dual problem
• Primal problem
maximize the total profit contribution with the
constraints of limited available resources
• Dual problem
minimize the total cost allocation to resources with
the constraints of guaranteeing the minimum
profitability.
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Slide 30
Example 3 (more than 2 variables)

Alternative optimal solution (p.116 Fig. 3.7)
• Profit contribution of Deluxe is 10.15
• Slack of Constraint 1 is 0, but the dual price is also 0.
• Range of optimality (range of objective function
coefficient) has one direction
• If the primal problem has an alternative optima, the

dual is degenerate and vice versa.
Extra constraint (p.117 Fig. 3.8)
• Deluxe should be produces at least 30% of standard bag.
D > 0.3S  –0.3S + D > 0
• Dual price –1.38 means that the total profit will decrease
if Deluxe is produce 1 more than 30% of standard bag.
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Slide 31
Example 4 (Bluegrass Farms Problem, p.118)

Decision variables
S = pounds of standard horse feed product to feed
E = pounds of vitamin-enriched oat product to feed
A = pounds of new vitamin and mineral feed additive
Min
0.25S + 0.5E + 3A
0.8S + 0.2E + 0.0A
1S + 1.5E + 3.0A
0.1S + 0.6E + 2.0A
1S + 1E + 1A
>
>
>
<
3
6
4
6
S, E, A > 0
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Slide 32
Example 4 (Bluegrass Farms Problem, p.118)
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Slide 33
Example 4 (Bluegrass Farms Problem, p.118)

Interpretation
• What’s the optimal decision?
• What’s the optimal cost?
• Which constraint has slack/surplus?
• What the dual prices for binding constraints?
0.919 of maximum weight means
if maximum weight requirement is increased, some
cheaper product will be feed to meet the requirements
of ingredients by allowing more weights
• Explain with the ranges of objective function
What will happen if the standard horse feed product is
free
• Explain with the ranges of RHS
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Slide 34
Example 5
(Electronic Communication Problem, p.123)



Maximize or minimize
What are the constraints
How many?
Decision variables
M = number of unit to produce for the marine
equipment distribution channel
B = number of units to produce for the business
equipment distribution channel
R = number of units to produce for the national retail
chain distribution channel
D = number of units to produce for the direct mail
distribution channel
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Slide 35
Example 5
(Electronic Communication Problem, p.123)

Model Formulation
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Slide 36
Example 5
(Electronic Communication Problem, p.123)
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Slide 37
Example 5
(Electronic Communication Problem, p.123)
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Slide 38
Example 5
(Electronic Communication Problem, p.123)

Interpretation
• What’s the optimal decision?
• What’s the optimal cost?
• What should be the profit for the direct mail channel
in order to produce some for the direct model?
• Which constraint has slack/surplus?
• What the dual prices for binding constraints?
• Explain with the ranges of objective function
• Explain with the ranges of RHS
What if the production requirement of 600 is changed to 601?
How much of the advertising budget is allocated to business
distributors?
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Slide 39
Ch.3 Homework

Q29 on p.149
• Formulate the model
• Solve with Excel
In Excel, you choose all options of 보고서 after
해찾기 to get the output of sensitivity analysis
• Solve with LINGO
• Answer all questions on p.149 Q29.
• Put all output answers in one file except Excel file and
upload through mis3nt.gnu.ac.kr
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Slide 40
End of Chapter 3
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2008 Thomson South-Western. All Rights Reserved
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Slide 41