Review A.6

A.6 Sensitivity Analysis Theory Review Questions

1

Lesson Topics

Sensitivity Analysis (or post-optimality analysis)

determines how an optimal solution is affected by changes in the objective function coefficients or in the right-hand side constants.

Sensitivity to Coefficients (1) measures how

changes in the objective function coefficients change optimal solutions. The range of optimality are those values keeping the current solution optimal.

Sensitivity to Constants (7) measures how

changes in the right-hand-side constants of constraints change optimal solutions. The dual price is the rate of improvement in the objective function. Computer Analysis computes the range of optimality of each decision variable, the dual price of each constraint, and range of feasibility for each dual price.

Resource Allocation Problems with Sensitivity

Analysis help production managers compute the value to them of buying additional resources to produce goods. That value guides mutually-beneficial trade.


2

Sensitivity to Coefficients

Question. The following problem that maximizes

profit from the production of two products has an

optimum where the first and third constraints are

binding.

Min x1 + 2x2

s.t. x1 + x2 30

2x1 + x2 40

2x1 + 5x2 > 75

x1 , x2 > 0

If the objective function becomes Min 2x1 + 2x2, what will be the optimal

values of x1, x2, and the objective function? Be sure to show your work and

your reasoning.


3

Answer to Question: For partial credit, find an optimal solution to the

new problem:

The solution to the original problem remains optimal if the slope of the

objective function remains between the slopes of the binding forms of the

first and third constraints.

Slope of the objective function c1x1 + c2x2 is -c1/c2.

Slope of the binding first constraint x1 + x2 = 30 is -1.

Slope of the binding third constraint 2x1 + 5x2 = 75 is -2/5.

Therefore, the solution to the original problem remains optimal if

-1 < -c1/c2 < -2/5, which means

1 > c1/c2 > 2/5, which is satisfied by the new objective function

Min 2x1 + 2x2.

Finally, find the solution to the original and the new problems by solving the

binding forms of the first the third constraints,

x1 + x2 = 30 and 2x1 + 5x2 = 75. And insert the solutions x1 = 25 and x2 =

5 into the objective function, Z = 2x1 + 2x2 = 50 + 10 = 60.

For full credit, find any alternative optimal solutions to the new

problem:

A constraint and iso-value-line graph reveals the iso-value-line passing

through the optimum (25,5) above lies along an edge of the feasible set. In

particular, (25,5) is not the only optimum. One other optimum is where the

first and second constraints bind, at (10,20). And there are an infinite

number of other optima. They are those points (x1 , x2) on the line segment

joining (25,5) to (10,20).


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Sensitivity to Constants

Question. Blue Ridge Hot Tubs produces two types

of hot tubs: Aqua-Spas and Hydro-Luxes.

There are 5 pumps, 18 hours of labor, and 16 feet of

tubing available to make the tubs. Here are the input

requirements, and unit profits:

Aqua-Spa Hydro-Lux

Pumps 1 1

Labor 3 hours 6 hours

Tubing 6 feet 4 feet

Unit Profit $8 $20

Formulate Blue Ridge’s problem as a linear program.

Solve graphically for the optimum.

How much should Blue Ridge be willing to pay for one more pump? for two

more pumps? for one more foot of tubing? for two more feet of tubing?

Tip: Your written answer should define the decision variables, formulate the

objective and constraints, and solve the problem.


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Answer to Question:

Let A = number of Aqua-Spa’s and H = number of Hydro-Lux’s produced.

Max 8 A + 20 H

s.t.

A + H < 5 (pump constraint)

3A + 6H < 18 (labor constraint)

6A + 4H < 16 (tubing constraint)

Graphing the constraints and isovalue lines indicates the non-negativity of

A and the labor constraint are binding at the optimum (so A = 0 and 3A +

6H = 18), with the pump constraint and the tubing constraint both slack.

Therefore, the solution is to produce no Aqua-Spa’s and 3 Hydro-Lux’s.

Since both the pump and tubing constraints are slack at the optimum,

Blue Ridge should only be willing to pay whatever input prices are

embedded into the unit profits for one more pump, or for two more pumps,

or for one more foot of tubing, or for two more feet of tubing,

6

5

4

3

2

1

0

0 1 2 3 4 5 6 7

Aqua Spa’s on horizontal axis

Hydro-Lux’s on vertical axis

Pump-constraint intercepts A=5 and H=5

Labor-constraint intercepts A=6 and H=3

Tubing-constraint intercepts A=2.7 and H=4

Sample isovalue line (dashed)


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To make the tubs, there are 5 pumps available at a

cost of $1 each, 16 hours of labor available at a cost

of $1 per hour, and 10 feet of tubing available at a cost of $1 per foot. Here

are the input requirements, sales prices, and unit profits:

Aqua-Spa Hydro-Lux

Pumps 1 1



Sales Price $20 $32

Unit Profit $10 $20


Solve for the optimum.

How much should Blue Ridge be willing to pay for one more pump? for one

more hour of labor? for one more foot of tubing?

You may use computer software to compute your answers, but be sure to

define the computer input, write down the essential computer output, and

explain how the computer output solves the problem.




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Answer to Question:


Max 10 A + 20 H

s.t.




Management Scientist output below reveals the optimum A = .2, H = 2.2,

with the pump constraint slack and the labor and tubing constraints binding,

with dual prices 2.667 and 0.333. Hence, Blue Ridge should be willing to

pay $1 for one more pump (the input price embedded into the unit profits),

$1 + 2.667 = $3.667 for one more hour of labor, and $1 + 0.333 = $1.333

for one more foot of tubing.


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9




To make the tubs, there are 4 pumps available at the

cost of $4 each, 18 hours of labor available at a cost

of $1 per hour, and 16 feet of tubing available at a cost of $0.50 per foot.

Here are the input requirements, sales prices, and unit profits:

Aqua-Spa Hydro-Lux

Pumps 1 1



Sales Price $13 $19

Unit Profit $3 $7


Solve graphically for the optimum.

How much should Blue Ridge be willing to pay for one more pump? for

one more hour of labor? for one more foot of tubing?




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Answer to Question:


Max 3 A + 7 H

s.t.




Graphing the constraints and isovalue lines indicates the non-negativity of

A and the labor constraint are binding at the optimum (so A = 0 and 3A +

6H = 18), with the pump constraint and the tubing constraint both slack.

Therefore, the solution is to produce 0 Aqua-Spa’s and 3 Hydro-Lux’s.

Blue Ridge should be willing to pay just the embedded input prices of $4 for

one more pump, $0.50 for one more foot of tubing?

To find the dual price of labor, recompute the optimum for 19 hours

available: A = 0 and 3A + 6H = 19. So, 0 Aqua-Spa’s and 3 1/6 Hydro-

Lux’s. That increases profit $7x(1/6) = $1.17. Hence, Blue Ridge should

be willing to pay $1 + 1.17 = $2.17 for one more hour of labor.

6

5

4

3

2

1

0

0 1 2 3 4 5 6 7

Aqua Spa’s on horizontal axis

Hydro-Lux’s on vertical axis

Pump-constraint intercepts A=5 and H=5

Labor-constraint intercepts A=6 and H=3

Tubing-constraint intercepts A=2.7 and H=4



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Question. The Boeing Company seeks to maximize

profit by making Model 747 (4-engine widebody) and

Model 787 (2-engine widebody) commercial aircraft

from aluminum, titanium and labor.

It just received this month's allocation of 18 tons of aluminum and 5 tons of

titanium, and there is 16 thousand hours of labor available.

It takes 3 tons of aluminum to make each Model 747 aircraft, and 2 tons of

aluminum to make each Model 787 aircraft.

And it takes 1 ton of titanium to make each Model 747 aircraft .

And it takes 2 thousand hours of labor to make each Model 747 aircraft,

and 2 thousand hours of labor to make each Model 787 aircraft.

Model 747 aircraft have unit profit of 2 million dollars, and Model 787

aircraft have unit profit of 1 million dollars.

Graphically determine how much The Boeing Company should produce of

each aircraft.

How much should The Boeing Company be willing to pay to expand its

labor force from 16 thousand hours to 17 thousand hours? That is, how

much more profit can be made if The Boeing Company can use 17

thousand hours of labor?


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Answer to Question:

Let X = units of Model 747 produced.

Let Y = units of Model 787 produced.

Max 2X + Y

s.t. 3X + 2Y < 18 (tons of aluminum)

X < 5 (tons of titanium)

2X + 2Y < 16 (thousand hours of labor)

X, Y 0

Graphing the constraints and isovalue lines indicates the third constraint

(labor) is slack when the amount of labor is either 16 thousand hours or 17

thousand hours. Solving the binding form of the first two constraints yields

the optimal solution: X = 5, Y = 1.5.

How much should The Boeing Company produce of each aircraft? 5 units

of Model 747 and 1.5 units of Model 787.

How much should The Boeing Company be willing to pay to expand its

labor force from 16 thousand hours to 17 thousand hours? Since the labor

constraint is slack, they should not pay anything to expand that constraint.

6

5

4

3

2

1

0

0 1 2 3 4 5 6 7

Model 747 on horizontal axis

Model 787 on vertical axis

Aluminum-constraint intercepts X=6 and Y=9

Titanium-constraint intercept X=5

Labor-constraint intercepts X=8 and Y=8 (when

there are 16 units of labor)



13


Question. Davison Electronics manufactures two

LCD television monitors, identified as model A and

model B. Each model has its lowest possible

production cost when produced on Davison’s new

production line. However, the new production line does not have the

capacity to handle the total production of both models. As a result, as least

some of the production must be routed to a higher-cost, old production line.

The following table shows the minimum production requirements for next

month, the production line table shows the minimum production

requirements for next month, the production line capacities in units per

month, and the production cost per unit for each production line:

Let: AN= Units of model A produced on the new production line AO= Units of model A produced on the old production line BN = Units of model B produced on the new production line BO= Units of model B produced on the old production line

Production Cost per

Unit on

Production Cost per

Unit on

Model New Line Old Line Minimum Production

Requirements

A $30 $50 50,000

B $25 $40 70,000

Production Line

Capacity

80,000 60,000


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Davison’s objective is to determine the minimum cost production plan. The

computer solution is shown below.

a. Formulate the linear programming model for this problem using the

following four constraints: Constraint 1: Minimum production for model A Constraint 2: Minimum production for model B Constraint 3: Capacity of the new production line Constraint 4: Capacity of the old production line

b. Using computer solution below, what is the optimal solution, and what is

the total production cost associated with this solution?

c. Which constraints are binding? Explain.

d. The production manager noted that the only constraint with a positive dual price is the constraint on the capacity of the new production line. The manager’s interpretation of the dual value was that a one-unit increase in the right-hand side of this constraint would actually increase the total production cost by $15 per unit. Do you agree with this interpretation? Would an increase in capacity for the new production line be desirable? Explain.

e. Would you recommend increasing the capacity of the old production line? Explain.

f. Suppose that the minimum production requirement for model B is reduced from 70,000 units to 60,000 units. What effect would this change have on the total production cost? Explain.


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Optimal Objective Value = 3850000.00000

Variable Value Reduced Cost

AN 50000.00000 0.00000

AO 0.0000 5.00000

BN 30000.00000 0.00000

BO 40000.00000 0.00000

Constraint Slack/Surplus Dual Price

1 0.00000 -45.00000

2 0.00000 -40.00000

3 0.00000 15.00000

4 20000.00000 0.00000

OBJECTIVE COEFFICIENT RANGES

Variable Objective Coefficient Allowable Increase Allowable Decrease

AN 30.00000 5.00000 Infinite

AO 50.00000 Infinite 5.00000

BN 25.00000 15.00000 5.00000

BO 40.00000 5.00000 15.00000

RIGHT HAND SIDE RANGES

Constraint RHS Value Allowable Increase Allowable Decrease

1 50000.00000 20000.00000 40000.00000

2 70000.00000 20000.00000 40000.00000

3 80000.00000 40000.00000 20000.00000

4 60000.00000 Infinite 20000.00000


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Answer to Question:

a. The linear programming model is as follows:

Min 30AN + 50AO + 25BN + 40BO

s.t.

AN + AO 50,000

BN + BO 70,000

AN + BN 80,000

AO + BO 60,000

b. Optimal solution:

New Line Old Line

Model A 50,000 0

Model B 30,000 40,000

Total Cost $3,850,000

c. The first three constraints are binding because the values in the

Slack/Surplus column for these constraints are zero. The fourth

constraint, with a slack of 0 is nonbinding.

d. The dual price for the new production line capacity constraint is 15.

Because the dual price is positive, increasing the right-hand side of

constraint 3 will cause the objective function value to decrease

(improve). Thus, every one unit increase in the right hand side of this

constraint will reduce the total production cost by $15. In other words,

an increase in capacity for the new production line is desirable.


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e. Because constraint 4 is not a binding constraint, any increase in the

production line capacity of the old production line will have no effect on

the optimal solution. Thus, there is no benefit in increasing the capacity

of the old production line.

f. The right hand side range for constraint 2 shows an allowable

decrease of 20,000. Thus, if the minimum production requirement is

reduced 10,000 units to 60,000, the dual value of 40 is applicable.

Thus, total cost would decrease by 10,000(40) = $400,000.


18


Question: Vollmer Manufacturing makes three

components for sale to refrigeration companies.

The components are processed on two machines: a

sharper and a grinder. The times (in minutes)

required on each machine are as follows:

The sharper is available for 120 hours, and the grinder is available for 110

hours. No more than 200 units of component 3 can be sold, but up to 1000

units of each of the other components can be sold. In fact, the company

already has orders for 600 units of component 1 that must be satisfied. The

profit contributions for components 1,2, and 3 are $8, $6, and $9,

respectively.

a. Formulate and use the Management Scientist to solve for the

recommended production quantities.

b. What are the objective coefficient ranges for the three components?

Interpret these ranges for company management.

c. What are the right-hand-side ranges? Interpret these ranges for

company management.

d. If more time could be made available on the grinder, how much would

it be worth?

e. If more units of component 3 can be sold by reducing the sales price

by $4, should the company reduce the price?

Machine

Component Sharper Grinder

1 6 4

2 4 5

3 4 2


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Answer to Question:

a. LetC1 = units of component 1 manufactured C2 = units of component 2 manufactured

C3 = units of component 3 manufactured

Max 8C1 + 6C2 + 9C3

s.t. 6C1 + 4C2 + 4C3 < 7200

4C1 + 5C2 + 2C3 < 6600

C3 < 200

C1 < 1000

C2 < 1000

C1 > 600

C1, C2, C3 > 0

The optimal solution is

C1 = 600

C2 = 700

C3 = 200

b.

Variable Objective Coefficient Range

C1 No Lower Limit to 9.0

C2 5.33 to 9.0

C3 6.00 to No Lower Limit

Individual changes in the profit coefficients within these ranges will not cause a change in the optimal number of components to produce.

c. Constraint Right-Hand-Side Range

1 4400 to 7440 2 6300 to No Upper Limit

3 100 to 900 4 600 to No Upper Limit

5 700 to No Upper Limit 6 514.29 to 1000


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These are the ranges over which the dual values for the associated

constraints are applicable.

d. Nothing, since there are 300 minutes of slack time on the grinder at

the optimal solution.

e. No, since at that price it would not be profitable to produce any of

component 3. To see that, the proposed reduction of $4 in the sales

price means a $4 reduction in the unit profit, from $9 to $5. That is

outside the range of optimality, so you need to re-compute an answer.

That answer is that the output of component 3 falls to 0. So, it is not

worth lowering the price by $4 to sell more.


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Question: The Ford Motor Company makes cars

and trucks on an assembly line. Each car and truck

requires labor and needs to visit three stations:

engine installation, hood installation, and wheel

installation. The hours of labor and hours required at each station are as

follows:

There are available 12 hours of labor, 48 hours of engine installation, 20

hours of hood installation, and 6 hours of wheel installation. Labor costs

$20 per hour, and each station costs $10 per hour to operate. Cars sell for

$700 each, and trucks sell for $540 each.

a. Formulate and graphically solve for the recommended production

quantities. Do not require production units to be integers.

b. How much should Ford be willing to pay to another potential worker

(Joe) to supply one more unit of labor. (Joe is not part of the 12

hours of labor mentioned above.)

Engine Hood Wheel

Vehicle Labor Installation Installation Installation

Car 4 6 4 3

Truck 3 8 5 4


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Answer to Question:

a. The profit contributions are $500 for each car, and $300 for each truck.

Let C = cars produced.

Let T = trucks produced.

Max 500C + 300T

s.t. 4C + 3T < 12 (labor)

6C + 8T < 48 (engine)

4C + 5T < 20 (hood)

3C + 4T < 12 (wheel)

C, T 0

A graph of the feasible set shows the engine and hood constraints are

redundant, and the iso-value lines show the optimal solution occurs where

the labor constraint and the non-negativity of T bind (C = 3 and T = 0).

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5 6 7 8 9 10


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b. How much should Ford be willing to pay to change the problem by

having Joe supply one more unit of labor.

Repeat analysis after changing the right hand side of the first constraint

from 12 to 13:

Max 500C + 300T

s.t. 4C + 3T < 13 (labor)

6C + 8T < 48 (engine)

4C + 5T < 20 (hood)

3C + 4T < 12 (wheel)

C, T 0

The new graph of the feasible set shows the engine and hood constraints

are redundant, and the iso-value lines show the optimal solution occurs

where the labor constraint and the non-negativity of T bind (C = 13/4 = 3.25

and T = 0). That new solution yields profit 500C + 300T = 1625, which is

125 higher than the previous profit of 1500. So, Ford should be willing to

pay up to 145 = 20+125 dollars to Joe.

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5 6 7 8 9 10


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Question. The Iron Works, Inc. seeks to maximize

profit by making two products from steel and labor.

It just received this month's allocation of 18 pounds of

steel, and there is 160 hours of labor available.

It takes 3 pounds of steel to make a unit of Product 1, and 2 pounds of

steel to make a unit of Product 2. It also takes 40 hours of labor to make a

unit of Product 1, and 20 hours of labor to make a unit of Product 2.

The physical plant has the capacity to make up to 6 units of total product

(Product 1 plus Product 2).

Product 1 has unit profit 6 dollars, and Product 2 has 5 dollars.

Graphically determine how much Iron Works should produce of each

product.

How much should Iron Works be willing to pay to expand its plant capacity

from 6 to 7? That is, how much more profit can be made if Iron Works can

make up to 7 units of total product?


25

Answer to Question:

Let X = units of Product 1 produced.

Let Y = units of Product 1 produced.

Max 6X + 5Y

s.t. 3X + 2Y < 18 (steel)

40X + 20Y < 160 (labor)

X + Y < 6 (capacity)

X, Y 0

A graph of the feasible set reveals that the first costraint is redundant --- it

does not affect the feasible set. Adding a graph of isovalue lines reveals

the optimum occurs where the second and third constraints bind. Solving

the binding form of those two constraints yields the optimal solution: X = 2,

Y = 4.

How much should Iron Works produce of each product? 2 units of Product

1, and 4 units of Product 2.

How much should Iron Works be willing to pay to expand its plant capacity

from 6 to 7? That is, how much more profit can be made if Iron Works can

make up to 7 units of total product? After changing the third constraint from

X+Y < 6 to X+Y < 7, a graph of the feasible set and isovalue lines reveals

the first constraint is still redundant, and the optimum still occurs where the

second and third constraints bind. Solving the binding form of those two

constraints yields the new optimal solution: X = 1, Y = 6. That new

solution yields profit 6X + 5Y = 36, which is higher than the profit 6X + 5Y

= 32. Precisely, 4 dollars more profit can be made if Iron Works can make

up to 7 units of total product, so Iron Works should be willing to pay up to 4

dollars to expand its plant capacity from 6 to 7

Review A.6

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