WORKSHEET 4: SENSITIVITY ANALYSIS I:
SENSITIVITY ANALYSIS OF OBJECTIVE FUNCTION COEFFICIENTS
 Analyzing Changes in the Objective Function Coefficients
“The optimal solution may change depending on the values of the objective function coefficients”
Once the optimal solution to a linear programming model has been found, the decision
maker may be concerned about how changes to any one of the objective function coefficients
affect the optimal solution. We may ask the question: “Keeping all other factors the same,
how much can an objective function coefficient change without changing the optimal
solution?”
Sensitivity analysis of an objective function coefficient focuses on answering this
question.
Consider the example:
The Beaver Creek Pottery Company is a small crafts operation that employs skilled
artisans to produce clay bowls and mugs. The two primary resources used by the company are
special pottery clay and skilled labor. Given these limited resources, the company desires to
know how many bowls and mugs to produce each day in order to maximize profit. The two
products have the following resource requirements for production and profit per item produced.
Labor
Clay
Product (hour/unit) (lb./unit)
Bowl
1
4
Mug
2
3
Profit
($/unit)
40
50
There are 40 hours of labor and 120 pounds of clay available each day for production.
The LP model for his problem is:
Let: x1: number of bowls to be produced
x2: number of mugs to be produced
Max.Z= 40 x1+ 50 x2
s.t.
x1+ 2x2 ≤ 40
4 x1+ 3x2 ≤ 120
x1, x2 ≥ 0
Now answer these questions:

Will the optimal solution change if the objective function coefficient per bowl changes
from its current value of $40 to $30; ie. the objective function becomes
Max.Z= 30 x1+ 50 x2 ?

Will the optimal solution change if the objective function coefficient per bowl changes
from its current value of $40 to $25; ie. the objective function becomes
Max.Z= 25 x1+ 50 x2 ?

Will the optimal solution change if the objective function coefficient per bowl changes
from its current value of $40 to $20; ie. the objective function becomes
Max.Z= 20 x1+ 50 x2 ?

Will the optimal solution change if the objective function coefficient per bowl changes
from its current value of $40 to $60; ie. the objective function becomes
Max.Z= 60 x1+ 50 x2 ?

Will the optimal solution change if the objective function coefficient per bowl changes
from its current value of $40 to $80; ie. the objective function becomes
Max.Z= 80 x1+ 50 x2 ?

Will the optimal solution change if the objective function coefficient per mug changes
from its current value of $50 to $40; ie. the objective function becomes
Max.Z= 40 x1+ 40 x2 ?

Will the optimal solution change if the objective function coefficient per mug changes
from its current value of $50 to $30; ie. the objective function becomes
Max.Z= 40 x1+ 30 x2 ?

Will the optimal solution change if the objective function coefficient per mug changes
from its current value of $50 to $25; ie. the objective function becomes
Max.Z= 40 x1+ 25 x2 ?

Will the optimal solution change if the objective function coefficient per mug changes
from its current value of $50 to $60; ie. the objective function becomes
Max.Z= 40 x1+ 60 x2 ?

Will the optimal solution change if the objective function coefficient per mug changes
from its current value of $50 to $80; ie. the objective function becomes
Max.Z= 40 x1+ 80 x2 ?

Will the optimal solution change if the objective function coefficient per mug changes
from its current value of $50 to $120; ie. the objective function becomes
Max.Z= 40 x1+ 120 x2
If you answer all the questions above, you will see that there is a range for each objective
function coefficient within which the optimal solution will not change. You will also see
that, beyond this range, the optimal solution will shift to some other corner point.
 Ranges of Optimality
For problems involving two variables, we can find the range of optimality for an
objective function coefficient by determining the range of values that give a slope of the
objective function line between the slopes the binding constraints.
The binding constraints of our example are:
x1+ 2x2 ≤ 40
4 x1+ 3x2 ≤ 120
The slopes of the binding constraints are: -1/2 and – 4/3 respectively.
The objective function of our example is:
Z= 40 x1+ 50 x2
Thus, the slope of the objective function line can be expressed as - c1/50; and the range
of optimality for c1 is found by solving the following for c1:
- c1
50
=
-4
3
c1 = 66,67
- c1
50
=
-1
2
c1 = 25
Thus:
25 ≤ c1 ≤ 66,67
Similarly, to calculate the range of optimality for c2 , the slope of the objective function
line, given that c1 is its original value of 40, can be expressed as -40/ c2
Then, the range of optimality for c2 is found by solving the following for c2
- 40 = -4
c2
3
c2 = 30
- 40 = -1
c2
2
c2 = 80
The range of optimality for c2 is therefore 30≤ c2 ≤ 80

Now change the profit per bowl from $40 to $20 so that the objective function
becomes:
Max.Z= 20 x1+ 50 x2
Calculate the optimality ranges for the objective function coefficients (ie.
Coefficients over which the optimal solution will not change.)

Now change the profit per bowl from $40 to $80 so that the objective function
becomes:
Max.Z= 80 x1+ 50 x2
Calculate the optimality ranges for the objective function coefficients (ie.
Coefficients over which the optimal solution will not change.)

Consider the example:
Puck and Pawn Campany manufactures hockey sticks and chess sets. Each hockey
stick yields an incremental profit of $2, and each chess set, $4. A hockey stick requires four
hours of processing at machine center A and two hours at machine center B. A chess set
requires six hours at machine center A, six hours at machine center B, and one hour at
machine center C. Machine center A has a maxumim of 120 hours of available capacity per
day, machine center B has 72 hours and machine center C has 10 hours. If the company
wishes to maximize profit, how many hockey sticks and chess sets should be produced per
day?
Machine Center
A
4
6
120 hours
Product
Hockey stick
Chess Set
Capacity
Machine Center
B
2
6
72 hours
Machine Center
C
1
10 hours
Profit
$2
$4
The LP model for his problem is:
Let: x1: number of hockey sticks to be produced
x2: number of chess sets to be produced
Max.Z= 2 x1+ 4 x2
s.t.
4 x1+ 6x2 ≤120 (machine center A constraint)
2 x1+ 6x2 ≤ 72 (machine center B constraint)
x2 ≤ 10 (machine center C constraint)
x1, x2 ≥ 0

Consider the example:
The United Aluminum Company of Cincinnati produces three grades (high, medium, and low)
of aluminum at two mills. Each mill has a different production capacity (in tons per day) for
each grade, as follows:
Mill
Aluminum Grade
1
2
High
6
2
Medium
2
2
Low
4
10
The company has contracted with a manufacturing firm to supply at least 12 tons of highgrade aluminum, 8 tons of medium-grade aluminum, and 5 tons of low-grade aluminum. It
costs United $6,000 per day to operate mill 1 and $7,000 per day to operate mill 2. The
company wants to know the number of days to operate each mill in order to meet the contract
at minimum cost. Solve the problem and find the ranges of optimality of the objective
function coefficients.
 Multiple Changes: Analyzing Simultaneous Changes Using the 100%
Rule
Until now, we have analyzed the impact of a change in just a single objective function
coefficient. The ranges of optimality are valid only when a change is made to a single
objective function coefficient. When multiple changes are made, it may be necessary to
resolve the problem to determine the new optimal solution.
Is it possible to analyze the impact of changes in more than one objective function coefficient?
The answer is yes, but only under a specific condition, ie., there is one case in which we can
be sure that the optimal solution does not change. It is expressed by the 100% rule.
100% Rule
1. For each increase in an objective function coefficient, calculate (and express as a
percentage) the ratio of the change in the coefficient to the maximum possible increase
as determined by the limits of the range of optimality.
2. For each decrease in an objective function coefficient, calculate (and express as a
percentage) the ratio of the change in the coefficient to the maximum possible
decrease as determined by the limits of the range of optimality.
3. Sum all these percent changes. If the total is less than 100%, the optimal solution will
not change If the total is greater than or equal to 100%, the optimal solution may
change.
To illustrate, suppose:

The profit per bowl is changed from $40 to $44, a $4 increase Since the upper
limit for its range of optimality is $66,67, a $4 increase from $40 is possible;
the percentage change is $4 / $26,67 = 0.149, or 15%

At the same time the profit per mug is decreased from $50 to $40, since the
lower limit for its range of optimality is $30, a $10 decrease from its original is
possible; the percentage change is $10 / $20 = 0.50, or 50%

Sum these percentages 15.0% + 50% =65%.
As this sum is less than 100% the opimal solution will not change.
 Reduced Cost
The optimal solution to the problem has positive values for both of the decision
variables. Now suppose that the objective function had been:
Z= 20 x1+ 50 x2
Then the optimal solution with this objective function will be:
x1= 0; x2=20, which gives an objective function value of $1000.
The reason why x1=0 (no bowls are to be produced) is because the $20 profit
coefficient of x is not large enough to justify the production of bowls.
It is then reasonable to ask: “How much will the profit coefficient of x1 have to
increase before x1 can be positive (i.e., to justify the production of some bowls) in the optimal
solution?’’ The answer is expressed by the reduced cost for this profit coefficient.
The reduced cost for a variable yields:
1. the amount the profit coefficient must change before the variable can take on a value
other zero.
2. the amount the optimal profit will change per unit increase in the variable from zero.
Reduced costs apply only to variables at zero value in the optimal solution. The reduced
cost corresonding to a variable with a positive value is 0. ie.,at the optimal solution either
the variable is 0, or the reduced cost for the variable is 0.
We see that x1=0; x2=20 is the optimal solution, until the profit coefficient for bowls
(c1) inreases to more than 25. At this point, the optimal solution changes to x1= 24; x2 = 8.
Thus c1 must increase by $5 (from $20 to $25) before it would be economically feasible to
produce bowls. Increasing the profit by $5 is equivalent to reducing the cost component of the
coefficient by $5, hence the reduced cost is -$5.
Another question that may arise is: If x1 were made positive (bowls were to be
produced) how much would the profit change per bowl produced? Again the answer is
expressed by the reduced cost.
If x1 were required to be at least equal to 1, we would add the constraint x1 ≥1. The
optimal solution is now x1=1; x2=19,5. This gives us an objective function value of $995,
which represents a $5 reduction to the optimal value of the objective function. (Scetch the
graph) Thus the reduced cost is $-5.