Chap3 (1) Introduction to Management Sciences

Slides Prepared by
JOHN S. LOUCKS
ST. EDWARD’S UNIVERSITY

© 2005 Thomson/South-Western 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


Standard Computer Output

Software packages such as The Management Scientist and
Microsoft Excel provide the following LP information:
 Information about the objective function:
• its optimal value
• coefficient ranges (ranges of optimality)
 Information about the decision variables:
• their optimal values
• their reduced costs
 Information about the constraints:
• the amount of slack or surplus
• the dual prices
• right-hand side ranges (ranges of feasibility)


Standard Computer Output

 In the previous chapter we discussed:
• objective function value
• values of the decision variables
• reduced costs
• slack/surplus
 In this chapter we will discuss:
• changes in the coefficients of the objective function
• changes in the right-hand side value of a
constraint


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 the manager who
must operate in a dynamic environment with
imprecise estimates of the coefficients.
 Sensitivity analysis allows him to ask certain what-if
questions about the problem.


Example 1

 LP Formulation

Max 5x1 + 7x2

s.t. x1 < 6
2x1 + 3x2 < 19
x1 + x2 < 8

x1, x2 > 0


Example 1

 Graphical Solution
x2
8
x1 + x2 < 8
Max 5x1 + 7x2
7

6
x1 < 6
5
Optimal:
4
x1 = 5, x2 = 3, z = 46
3

2
2x1 + 3x2 < 19
1

1 2 3 4 5 6 7 8 9 10
x1


Objective Function Coefficients

 Let us consider how changes in the objective function
coefficients might affect the optimal solution.
 The range of optimality for each coefficient provides
the range of values over which the current solution
will remain optimal.
 Managers should focus on those objective coefficients
that have a narrow range of optimality and
coefficients near the endpoints of the range.


Example 1

 Changing Slope of Objective Function
x2
8

7

6

5 5
4

3
Feasible 4
2
Region 3
1

1 2
1 2 3 4 5 6 7 8 9 10
x1


Range of Optimality

 Graphically, the limits of a range of optimality are
found by changing the slope of the objective function
line within the limits of the slopes of the binding
constraint lines.
 The slope of an objective function line, Max c1x1 +
c2x2, is -c1/c2, and the slope of a constraint, a1x1 + a2x2
= b, is -a1/a2.


Example 1

 Range of Optimality for c1
The slope of the objective function line is -c1/c2.
The slope of the first binding constraint, x1 + x2 = 8, is -
1 and the slope of the second binding constraint, x1 +
3x2 = 19, is -2/3.
Find the range of values for c1 (with c2 staying 7)
such that the objective function line slope lies between
that of the two binding constraints:
-1 < -c1/7 < -2/3
Multiplying through by -7 (and reversing the
inequalities):
14/3 < c1 < 7


Example 1

 Range of Optimality for c2
Find the range of values for c2 ( with c1 staying 5)
such that the objective function line slope lies between
that of the two binding constraints:
-1 < -5/c2 < -2/3

Multiplying by -1: 1 > 5/c2 > 2/3
Inverting, 1 < c2/5 < 3/2

Multiplying by 5: 5 < c2 < 15/2


Example 1

 Range of Optimality for c1 and c2
Adjustable Cells
Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease
$B$8 X1 5.0 0.0 5 2 0.33333333
$C$8 X2 3.0 0.0 7 0.5 2

Constraints
Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$B$13 #1 5 0 6 1E+30 1
$B$14 #2 19 2 19 5 1
$B$15 #3 8 1 8 0.33333333 1.66666667


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.


Dual Price

 Graphically, a dual price is determined by adding +1
to the right hand side value in question and then
resolving for the optimal solution in terms of the
same two binding constraints.
 The dual price is equal to the difference in the values
of the objective functions between the new and
original problems.
 The dual price for a nonbinding constraint is 0.
 A negative dual price indicates that the objective
function will not improve if the RHS is increased.


Relevant Cost and Sunk Cost

 A resource cost is a relevant cost if the amount paid for
it is dependent upon the amount of the resource used
by the decision variables.
 Relevant costs are reflected in the objective function
coefficients.
 A resource cost is a sunk cost if it must be paid
regardless of the amount of the resource actually used
by the decision variables.
 Sunk resource costs are not reflected in the objective
function coefficients.


A Cautionary Note
on the Interpretation of Dual Prices
 Resource cost is sunk
The dual price is the maximum amount you should
be willing to pay for one additional unit of the
resource.
 Resource cost is relevant
The dual price is the maximum premium over the
normal cost that you should be willing to pay for one
unit of the resource.


Example 1

 Dual Prices
Constraint 1: Since x1 < 6 is not a binding constraint,
its dual price is 0.
Constraint 2: Change the RHS value of the second
constraint to 20 and resolve for the optimal point
determined by the last two constraints:
2x1 + 3x2 = 20 and x1 + x2 = 8.
The solution is x1 = 4, x2 = 4, z = 48. Hence, the
dual price = znew - zold = 48 - 46 = 2.


Example 1

 Dual Prices
Constraint 3: Change the RHS value of the third
constraint to 9 and resolve for the optimal point
determined by the last two constraints: 2x1 + 3x2 = 19
and x1 + x2 = 9.
The solution is: x1 = 8, x2 = 1, z = 47.
The dual price is znew - zold = 47 - 46 = 1.


Example 1

 Dual Prices
Adjustable Cells
$B$8 X1 5.0 0.0 5 2 0.33333333
$C$8 X2 3.0 0.0 7 0.5 2

Constraints
$B$13 #1 5 0 6 1E+30 1
$B$14 #2 19 2 19 5 1
$B$15 #3 8 1 8 0.33333333 1.66666667


Range of Feasibility

 The range of feasibility for a change in the right hand
side value is the range of values for this coefficient in
which the original dual price remains constant.
 Graphically, the range of feasibility is determined by
finding the values of a right hand side coefficient
such that the same two lines that determined the
original optimal solution continue to determine the
optimal solution for the problem.


Example 1

 Range of Feasibility
Adjustable Cells
$B$8 X1 5.0 0.0 5 2 0.33333333
$C$8 X2 3.0 0.0 7 0.5 2

Constraints
$B$13 #1 5 0 6 1E+30 1
$B$14 #2 19 2 19 5 1
$B$15 #3 8 1 8 0.33333333 1.66666667


Example 2: Olympic Bike Co.

Olympic Bike is introducing two new lightweight
bicycle frames, the Deluxe and the Professional, to be
made from special aluminum and
steel alloys. The anticipated unit
profits are $10 for the Deluxe
and $15 for the Professional.
The number of pounds of
each alloy needed per
frame is summarized on the next slide.



A supplier delivers 100 pounds of the
aluminum alloy and 80 pounds of the steel
alloy weekly.

Aluminum Alloy Steel Alloy
Deluxe 2 3
Professional 4 2

How many Deluxe and Professional frames should
Olympic produce each week?



 Model Formulation
• Verbal Statement of the Objective Function
Maximize total weekly profit.
• Verbal Statement of the Constraints
Total weekly usage of aluminum alloy < 100 pounds.
Total weekly usage of steel alloy < 80 pounds.
• Definition of the Decision Variables
x1 = number of Deluxe frames produced weekly.
x2 = number of Professional frames produced weekly.



 Model Formulation (continued)

Max 10x1 + 15x2 (Total Weekly Profit)

s.t. 2x1 + 4x2 < 100 (Aluminum Available)
3x1 + 2x2 < 80 (Steel Available)

x1, x2 > 0



 Partial Spreadsheet: Problem Data
A B C D
1 Material Requirements Amount
2 Material Deluxe Profess. Available
3 Aluminum 2 4 100
4 Steel 3 2 80



 Partial Spreadsheet Showing Solution
A B C D
6 Decision Variables
7 Deluxe Professional
8 Bikes Made 15 17.500
9
10 Maximized Total Profit 412.500
11
12 Constraints Amount Used Amount Avail.
13 Aluminum 100 <= 100
14 Steel 80 <= 80



 Optimal Solution

According to the output:
x1 (Deluxe frames) = 15
x2 (Professional frames) = 17.5
Objective function value = $412.50



 Range of Optimality
Question
Suppose the profit on deluxe frames is increased
to $20. Is the above solution still optimal? What is
the value of the objective function when this unit
profit is increased to $20?



 Sensitivity Report

Adjustable Cells
$B$8 Deluxe 15 0 10 12.5 2.5
$C$8 Profess. 17.500 0.000 15 5 8.333333333

Constraints
$B$13 Aluminum 100 3.125 100 60 46.66666667
$B$14 Steel 80 1.25 80 70 30



Answer
The output states that the solution remains
optimal as long as the objective function coefficient of
x1 is between 7.5 and 22.5. Since 20 is within this
range, the optimal solution will not change. The
optimal profit will change: 20x1 + 15x2 = 20(15) +
15(17.5) = $562.50.



Question
If the unit profit on deluxe frames were $6
instead of $10, would the optimal solution change?




Adjustable Cells
$B$8 Deluxe 15 0 10 12.5 2.5
$C$8 Profess. 17.500 0.000 15 5 8.33333333

Constraints
$B$13 Aluminum 100 3.125 100 60 46.66666667
$B$14 Steel 80 1.25 80 70 30



Answer
x1 is between 7.5 and 22.5. Since 6 is outside this
range, the optimal solution would change.


Range of Optimality and 100% Rule

 The 100% rule states that simultaneous changes in
objective function coefficients will not change the
optimal solution as long as the sum of the
percentages of the change divided by the
corresponding maximum allowable change in the
range of optimality for each coefficient does not
exceed 100%.



 Range of Optimality and 100% Rule
Question
If simultaneously the profit on Deluxe frames
was raised to $16 and the profit on Professional
frames was raised to $17, would the current solution
be optimal?



Answer
If c1 = 16, the amount c1 changed is 16 - 10 = 6 .
The maximum allowable increase is 22.5 - 10 = 12.5,
so this is a 6/12.5 = 48% change. If c2 = 17, the
amount that c2 changed is 17 - 15 = 2. The maximum
allowable increase is 20 - 15 = 5 so this is a 2/5 = 40%
change. The sum of the change percentages is 88%.
Since this does not exceed 100%, the optimal solution
would not change.


Range of Feasibility and 100% Rule

 The 100% rule states that simultaneous changes in
right-hand sides will not change the dual prices as
long as the sum of the percentages of the changes
divided by the corresponding maximum allowable
change in the range of feasibility for each right-hand
side does not exceed 100%.



 Range of Feasibility and Sunk Costs
Question
Given that aluminum is a sunk cost, what is the
maximum amount the company should pay for 50
extra pounds of aluminum?




Adjustable Cells
$B$8 Deluxe 15 0 10 12.5 2.5
$C$8 Profess. 17.500 0.000 15 5 8.33333333

Constraints
$B$13 Aluminum 100 3.125 100 60 46.66666667
$B$14 Steel 80 1.25 80 70 30



Answer
Since the cost for aluminum is a sunk cost, the
shadow price provides the value of extra aluminum.
The shadow price for aluminum is the same as its
dual price (for a maximization problem). The shadow
price for aluminum is $3.125 per pound and the
maximum allowable increase is 60 pounds. Because
50 is in this range, the $3.125 is valid. Thus, the value
of 50 additional pounds is = 50($3.125) = $156.25.



 Range of Feasibility and Relevant Costs
Question
If aluminum were a relevant cost, what is the
maximum amount the company should pay for 50
extra pounds of aluminum?



 Range of Feasibility and Relevant Costs
Answer
If aluminum were a relevant cost, the shadow
price would be the amount above the normal price of
aluminum the company would be willing to pay.
Thus if initially aluminum cost $4 per pound, then
additional units in the range of feasibility would be
worth $4 + $3.125 = $7.125 per pound.


Example 3

 Consider the following linear program:

Min 6x1 + 9x2 ($ cost)

s.t. x1 + 2x2 < 8
10x1 + 7.5x2 > 30
x2 > 2

x1, x2 > 0


Example 3

 The Management Scientist Output

OBJECTIVE FUNCTION VALUE = 27.000
Variable Value Reduced Cost
x1 1.500 0.000
x2 2.000 0.000
Constraint Slack/Surplus Dual Price
1 2.500 0.000
2 0.000 -0.600
3 0.000 -4.500


Example 3

 The Management Scientist Output (continued)

OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value Upper Limit
x1 0.000 6.000 12.000
x2 4.500 9.000 No Limit

RIGHTHAND SIDE RANGES
Constraint Lower Limit Current Value Upper Limit
1 5.500 8.000 No Limit
2 15.000 30.000 55.000
3 0.000 2.000 4.000


Example 3

 Optimal Solution

According to the output:
x1 = 1.5
x2 = 2.0
Objective function value = 27.00


Example 3

Question
Suppose the unit cost of x1 is decreased to $4. Is
the current solution still optimal? What is the value
of the objective function when this unit cost is
decreased to $4?


Example 3


x1 0.000 6.000 12.000
x2 4.500 9.000 No Limit

1 5.500 8.000 No Limit
2 15.000 30.000 55.000
3 0.000 2.000 4.000


Example 3

Answer
x1 is between 0 and 12. Because 4 is within this range,
the optimal solution will not change. However, the
optimal total cost will be affected: 6x1 + 9x2 = 4(1.5) +
9(2.0) = $24.00.


Example 3

Question
How much can the unit cost of x2 be decreased
without concern for the optimal solution changing?


Example 3


x1 0.000 6.000 12.000
x2 4.500 9.000 No Limit

1 5.500 8.000 No Limit
2 15.000 30.000 55.000
3 0.000 2.000 4.000


Example 3

Answer
x2 does not fall below 4.5.


Example 3

Question
If simultaneously the cost of x1 was raised to $7.5
and the cost of x2 was reduced to $6, would the current
solution remain optimal?


Example 3

Answer
If c1 = 7.5, the amount c1 changed is 7.5 - 6 = 1.5.
The maximum allowable increase is 12 - 6 = 6, so this
is a 1.5/6 = 25% change. If c2 = 6, the amount that c2
changed is 9 - 6 = 3. The maximum allowable
decrease is 9 - 4.5 = 4.5, so this is a 3/4.5 = 66.7%
change. The sum of the change percentages is 25% +
66.7% = 91.7%. Since this does not exceed 100% the
optimal solution would not change.


Example 3

Question
If the right-hand side of constraint 3 is increased
by 1, what will be the effect on the optimal solution?


Example 3


x1 0.000 6.000 12.000
x2 4.500 9.000 No Limit

1 5.500 8.000 No Limit
2 15.000 30.000 55.000
3 0.000 2.000 4.000


Example 3

Answer
A dual price represents the improvement in the
objective function value per unit increase in the right-
hand side. A negative dual price indicates a
deterioration (negative improvement) in the
objective, which in this problem means an increase in
total cost because we're minimizing. Since the right-
hand side remains within the range of feasibility,
there is no change in the optimal solution. However,
the objective function value increases by $4.50.


End of Chapter 3


Chap3 (1) Introduction to Management Sciences

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Chap3 (1) Introduction to Management Sciences