Readings
Readings
Chapter 3
Linear Programming: Sensitivity Analysis and Interpretation
of Solution
BA 452 Lesson A.7 Sensitivity Analysis Applications
1
Overview
Overview
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Overview
Proportion Constraints restrict the size of one decision variable as a proportion
of another. For example, B > 0.1J or B - 0.1J > 0 says that variable B is at
least 10% of variable J.
Portfolio Selection Problems with Risk Indices use one specific measure of the
risk of a portfolio. When the risk index is part of a linear risk function, risk
minimization is a linear-programming problem.
100% Rules expand the range of optimality and the range of feasibility to
simultaneous changes in two or more objective function coefficients, or in two or
more right-hand-side constants.
Negative Dual Prices measure the rate of deterioration (negative improvement)
in the objective function value per unit increase in a right-hand side constant.
Blending Problems with Sensitivity Analysis help production managers decide
how cost is affected by changing product characteristics (shade tolerance,
calories, vitamin content, …).
BA 452 Lesson A.7 Sensitivity Analysis Applications
3
Proportion Constraints
Proportion Constraints
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Proportion Constraints
Overview
Proportion Constraints restrict the size of one decision
variable as a proportion of another. For example, B > 0.1J
or B - 0.1J > 0 says that variable B is at least 10% of
variable J.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Proportion Constraints
Question: Slater & Gordon Lawyers is ready to award
contracts for printing their annual reports. Past reports
were printed by Johnson Printing and Lakeside Litho. A
new firm, Benson Printing, is interested in doing some of
the printing. The quality of Lakeside Litho has been high,
with only 0.5% of their reports being unacceptable.
Johnson Printing has also been high, with only 1% of
reports unacceptable. Because Slater & Gordon has no
experience with Benson Printing, they estimate their
unacceptable rate at 10%. Slater & Gordon needs to
determine how many reports should be printed by each firm
to obtain 75,000 acceptable reports.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Proportion Constraints
To ensure Benson Printing will receive some of the
contract, management specifies that Benson must print at
least 10% of the number printed by Johnson.
To spread the risk of unacceptable reports, the reports
assigned to Benson, Johnson, and Lakeside must not
exceed 30000, 50000, and 50000, respectively.
To reward their long-standing relationship, Lakeside will be
assigned at least 30,000 reports.
The unit cost per report (not just acceptable reports) is
$2.45 for Benson, $2.50 for Johnson, and $2.75 for
Lakeside.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Proportion Constraints
Formulate and solve a linear program to determine how
many copies should be assigned to each printer to
minimize the cost of 75,000 acceptable reports.
Suppose management reconsiders the requirement that
Lakeside be awarded at last 30,000 reports. What affect
does that have?
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Proportion Constraints
Answer: First, formulate the problem:
 Define decision variables:
• B = number of copies done by Benson Printing
• 0.9 B is the number of acceptable-quality reports
(10% defective).
• J = number of copies done by Johnson Printing
• 0.99 L is the number of acceptable-quality reports
(1% defective).
• L = number of copies done by Lakeside Litho
• 0.995 B is the number of acceptable-quality reports
(0.5% defective).
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Proportion Constraints




Minimize (cost) 2.45B + 2.5J + 2.75L subject to these
constraints:
B < 30,000; J < 50,000; L < 50,000 (volume should not
exceed)
0.9B + 0.99J + 0.995L = 75,000 (useful reports, =
means no disposal)
B - 0.1J > 0 (Benson at least 10% of Johnson); L >
30,000 (Lakeside min.)
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Proportion Constraints




Minimize (cost) 2.45B + 2.5J + 2.75L subject to these constraints:
B < 30,000; J < 50,000; L < 50,000 (volume should not exceed)
0.9B + 0.99J + 0.995L = 75,000 (useful reports, = means no
disposal)
B - 0.1J > 0 (Benson at least 10% of Johnson); L > 30,000 (Lakeside
min.)
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Proportion Constraints
Hence, answer questions about Minimize (cost) 2.45B + 2.5J + 2.75L s.t.
 B < 30,000; J < 50,000; L < 50,000 (volume should not exceed)

0.9B + 0.99J + 0.995L = 75,000 (useful reports, = means no disposal)
 B - 0.1J > 0 (Benson at least 10% of Johnson); L > 30,000 (Lakeside min.)


How many copies should be assigned to
each printer? Rounding off to the nearest
integer, B = 4181, J = 41806, L = 30000
(we later see that rounding is not always
optimal)
Suppose management reconsiders the
requirement that Lakeside be awarded at
last 30,000 reports. What affect does that
have? For every unit increase in the
Lakeside minimum (up to 50,000), costs
increase $0.221. For every unit decrease
in the minimum (down to 21,106), costs
decrease $0.221.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Portfolio Selection with Risk Indices
Portfolio Selection with Risk Indices
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Portfolio Selection with Risk Indices
Overview
Portfolio Selection Problems with Risk Indices use one
specific measure of the risk of a portfolio. When the risk
index is part of a linear risk function, that measure makes
risk minimization a linear-programming problem. (But the
leading measures of risk are variance or standard
deviation, which are not linear functions.)
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Portfolio Selection with Risk Indices
Question: Fidelity Investments has been authorized to
invest up to $1,000,000 for a new client, in two investment
funds: a stock fund and a money-market fund.
 Each unit invested in the stock fund costs $100 and has
an expected return of 5%; each unit in the money-market
fund costs $50 and has an expected return of 2%.
 The client wants to minimize risk subject to the constraint
that the expected return on the investment be at least
$4,000. Fidelity computes a risk index that a) is a linear
function of the units invested in each fund, b) is 6 for a
unit invested in the stock fund, and c) is 5 for a unit
invested in the money market.
 At least 100 units need to be in the money-market fund.
How many units should be invested in each fund?
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Portfolio Selection with Risk Indices
Answer:
 Verbal Statement of the Objective Function
Minimize risk index.
 Verbal Statement of the Constraints
Total investment < $1,000,000.
Expected return > $4,000.
Money-market units invested > 2.
 Definition of the Decision Variables
x1 = number of units invested in the stock fund.
x2 = number of units invested in the money market fund.
Min
s.t.
6x1 + 5x2
(Risk index)
100x1 + 50x2 < 1,000,000 (Funds available)
5x1 +
x2 >
4,000 (Expected return)
x2 >
100 (Minimum money market)
x 1, x 2 > 0
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Portfolio Selection with Risk Indices
Min
s.t.
6x1 + 5x2
100x1 + 50x2 < 1,000,000
5x1 + x2 >
4,000
x2 >
100
x1 > 0 and x2 > 0
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Portfolio Selection with Risk Indices
Min
s.t.
6x1 + 5x2
100x1 + 50x2 < 1,000,000
5x1 + x2 >
4,000
x2 >
100
x1 > 0 and x2 > 0
Range of optimality
Conclusions:
 780 units should be
invested in the stock fund,
and 100 units in the moneymarket fund.
 Suppose the risk index on
the money-market fund
increased to 5.5. Is the
previous solution still
optimal? The output states
that the solution remains
optimal for any increase in
the objective function
coefficient of x2 (the
number of money-market
units).
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules
100% Rules
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules
Overview
100% Rules expand the range of optimality and the range
of feasibility to simultaneous changes in two or more
parameters. That is, the rule indicates when simultaneous
changes in two or more objective function coefficients will
not cause a change in the optimal solution. And the rule
indicates when simultaneous changes in two or more righthand-side constants will not cause a change in any of the
dual prices.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules
Question: Wild Oats Bicycles 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.
 Each week, a supplier delivers 100 pounds of the aluminum alloy
and 80 pounds of the steel alloy.
 The number of pounds of each alloy needed per frame is as follows:
Aluminum Alloy
Deluxe
2
Professional
4
Steel Alloy
3
2
How many Deluxe and how many Professional frames should
Wild Oats produce each week?
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules
Answer:
 Verbal Statement of the Objective Function
Maximize total weekly profit.
 Verbal Statement of the Constraints
Total weekly use of aluminum alloy < 100 pounds.
Total weekly use of steel alloy < 80 pounds.
 Definition of the Decision Variables
x1 = number of Deluxe frames produced weekly.
x2 = number of Professional frames produced weekly.
Max
s.t.
10x1 + 15x2
(Total Weekly Profit)
2x1 + 4x2 < 100 (Aluminum Available)
3x1 + 2x2 < 80 (Steel Available)
x 1, x 2 > 0
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules
Max 10x1 + 15x2
s.t.
2x1 + 4x2 < 100
3x1 + 2x2 < 80
x1 > 0 and x2 > 0
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules
Max 10x1 + 15x2
s.t.
2x1 + 4x2 < 100
3x1 + 2x2 < 80
x1 > 0 and x2 > 0
Conclusions:
 Wild Oats should produce
15.0 Deluxe and 17.5
Professional frames each
week, for a profit of $412.50.
 Suppose the unit profit on
deluxe frames increases to
$20. Is the previous solution
still optimal? The output
states that the solution
remains optimal as long as
the objective function
coefficient of x1 (number
of deluxe frames) is in the range of optimality, between 7.500 and 22.500.
Because 20 is within that range, the optimal solution will not change if the unit
profit on deluxe frames increases to $20. And what is the value of the
objective function when that unit profit increases to $20? The optimal profit
changes to $20x1 + $15x2 = $20(15) + $15(17.5) = $562.50.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules
One 100% rule is 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%.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules
Max 10x1 + 15x2
s.t.
2x1 + 4x2 < 100
3x1 + 2x2 < 80
x1 > 0 and x2 > 0
If simultaneously
the profit on Deluxe
frames were raised to $16 and
the profit on Professional
frames were raised to $17,
would the current solution
(15.0 Deluxe and 17.5
Professional frames ) remain
optimal? 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 that does not exceed
100%, the current solution remains optimal.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules
The other 100% rule is 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%.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules

Suppose Wild Oats has to pay for all 100 pounds of the
aluminum alloy regardless of whether they use each pound.
What is the maximum amount the company should pay for 50
extra pounds of aluminum?

Answer:
Because aluminum is a
sunk cost, the dual price
is the value of each unit
of extra aluminum. The
dual price for aluminum is
$3.125 per pound and the
maximum allowable
increase is 60 pounds (to
160 pounds). Because
Range of feasibility
50 is in this range, the
$3.125 is valid. Thus, the
value of 50 additional
pounds is = 50($3.125) =
$156.25.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Negative Dual Prices
Negative Dual Prices
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Negative Dual Prices
Overview
Negative Dual Prices measure the rate of deterioration
(negative improvement) in the objective function value per
unit increase in a right-hand side constant.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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100% Rules
Question: Use the Management Scientist to solve the
following cost-minimization problem:
Min
s.t.
6x1 + 9x2 ($ cost)
x1 + 2x2 < 8
10x1 + 7.5x2 > 30
x2 > 2
x1, x2 > 0
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Negative Dual Prices
Min
s.t.
6x1 + 9x2 ($ cost)
x1 + 2x2 < 8
10x1 + 7.5x2 > 30
x2 > 2
x1, x2 > 0
Solution: The optimal x1 = 1.5 and optimal x2 = 2.0, with the
minimized objective function value = $27.00.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Negative Dual Prices

Suppose the unit cost of x1 is decreased in the objective function
6x1 + 9x2 from $6 to $4. Is the current solution still optimal? What is
the value of the objective function when this unit cost is decreased
to $4?

Answer: That Management Scientist output states that the solution
remains optimal as long as the objective function coefficient of x1 is
between 0 and 12. Because 4 is within this range, the current
solution is still optimal. However, the optimal cost will be affected:
4x1 + 9x2 = 4(1.5) + 9(2.0) = $24.00.
How much can the unit cost of x2 be decreased while keeping the
current solution optimal?
Answer: The current solution remains optimal as long as the
objective function coefficient of x2 does not fall below 4.5.
BA 452 Lesson A.7 Sensitivity Analysis Applications


33
Negative Dual Prices

If simultaneously the cost of x1 were raised to $7.5 and
the cost of x2 were reduced to $6, would the current
solution remain optimal?

Answer: If c1 = 7.5, the amount c1 changed is 7.5 - 6 =
1.5. The maximum allowable increase is 12 - 6 = 6, so
that 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 that is a 3/4.5 = 66.7% change. The
sum of those percentages is 25% + 66.7% = 91.7%.
Because that does not exceed 100%, the current solution
remains optimal.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Negative Dual Prices
If the right-hand side of
constraint 3 is increased by
1, what will be the effect on
the optimal objectivefunction value?
 Answer: A dual price
represents the
improvement in the
objective function value per
unit increase in the righthand 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 (0 to 4), there is no change in the dual price. However, the
objective function value increases by $4.50.

BA 452 Lesson A.7 Sensitivity Analysis Applications
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Negative Dual Prices
If the right-hand side of
Constraint 2 increases by 5
and the right-hand side of
Constraint 3 decreases by
1, what will be the effect on
the optimal objectivefunction value?
 Answer: The increase of 5
in Constraint 2 is 20% of
the maximum allowable
increase of 25, and the
decrease of 1 in Constraint
3 is 50% of the maximum
allowable decrease of 2.
The sum of the change percentages is 20% + 50% = 70%. Because
that does not exceed 100%, the dual prices do not change. So the
optimal objective-function value changes by 0.600 (5) + 4.500 (-1) = 1.500, which is a decrease (improvement) of 1.500.

BA 452 Lesson A.7 Sensitivity Analysis Applications
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Blending with Sensitivity Analysis
Blending with Sensitivity Analysis
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Blending with Sensitivity Analysis
Overview
Blending Problems with Sensitivity Analysis helps
production managers decide how cost is affected by
changing product characteristics (shade tolerance, calories,
vitamin content, …).
BA 452 Lesson A.7 Sensitivity Analysis Applications
38
Blending with Sensitivity Analysis
Question: Bluegrass Farms, located in Lexington, Kentucky, has been
experimenting with a special diet for racehorses. The feed components
available for the diet are:
• A standard horse feed product
• An enriched oat product
• A new vitamin and mineral feed additive.
 Each feed component combines ingredients A, B, C.
 Nutritional value and cost data for feed components:
Feed Component
Standard
Enriched Oat
Additive
Ingredient A
0.8
0.2
0.0
Ingredient B
1.0
1.5
3.0
Ingredient C
0.1
0.6
2.0
Cost per pound
$0.25
$0.50
$3.00
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Blending with Sensitivity Analysis


The minimum daily diet requirements for each horse are
• 3 units of Ingredient A
• 6 units of Ingredient B
• 4 units of Ingredient C
To control the weight of the horses, the total daily feed
for a horse should not exceed 6 pounds.
What is the cheapest way to satisfy the daily diet
requirements?
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Blending with Sensitivity Analysis



S = number of pounds of the standard horse feed
product.
E = number of pounds of the enriched oat product.
A = number of pounds of the vitamin and mineral feed
additive.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Blending with Sensitivity Analysis
Objective: minimize cost from blending S standard
feed, E enriched oats, and A additive.
Min 0.25 S + 0.50 E + 3 A
Minimum requirement of A
s.t. 0.8 S + 0.2 E
>3
1.0 S + 1.5 E + 3.0 A > 6
0.1 S + 0.6 E + 2.0 A > 4
S+
E+
A<6
S, E, A > 0
Minimum requirement of B
Minimum requirement of C
Maximum weight constraint
Non-negativity constraints
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Blending with Sensitivity Analysis
Min 0.25 S + 0.50 E + 3A
s.t. 0.8 S + 0.2 E
> 3 (A)
1.0 S + 1.5 E + 3.0 A > 6 (B)
0.1 S + 0.6 E + 2.0 A > 4 (C)
S+
E+
A < 6 (weight)
S, E, A > 0

Minimize cost ($5.973 per horse) from blending S =
3.514 pounds of standard feed, E = 0.946 pounds of
enriched oats, and A = 1.541 pounds of additive.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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Blending with Sensitivity Analysis
Min 0.25 S + 0.50 E + 3A
 How much could be saved
s.t. 0.8 S + 0.2 E
> 3 (A)
by decreasing the Ingredient
1.0 S + 1.5 E + 3.0 A > 6 (B)
A minimum from 3 to 2?
0.1 S + 0.6 E + 2.0 A > 4 (C)
 The range of feasibility of the
S+
E+
A < 6 (weight)
first constraint is 1.143 to
S, E, A > 0
3.368, so the decrease from


3 to 2 is in that range.
The dual price of -1.216
means saving $1.216 when
the right-hand side of
constraint 1 decreases by 1.
The negative dual price
means the objective function
improves (lowers) with a
right-hand side decrease.
BA 452 Lesson A.7 Sensitivity Analysis Applications
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BA 452
Quantitative Analysis
End of Lesson A.7
BA 452 Lesson A.7 Sensitivity Analysis Applications
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