Sensitivity Analysis
• Consider the CrossChek hockey stick production problem:
• Consider two high-end hockey sticks, A and B.
$150 and $200 profit are earned from each sale
of A and B, respectively. Each product goes
through 3 phases of production.
Max
150x1 + 200x2
s.t
x1 + 2/3x2 <= 1000
4/5x1 + 4/5x2 <= 960
1/2x1 + x2 <= 1000
x1, x2 >= 0
• A requires 1 hour of work in phase 1, 48 min in
phase 2, and 30 min in phase 3.
• B requires 40 min, 48 min and 1 hour, respectively.
• Limited manufacturing capacity:
• phase 1 1000 total hours
• phase 2 960
• phase 3 1000
• How many of each product should be
produced?
• Maximize profit
• Satisfy constraints.
• Management believes that CrossChek might only receive $120 profit
from the sale of each lower-profit hockey stick. Will that affect the
optimal solution? What about a (separate) change in the profit from
higher-profit sticks to $125?
Constraint Sensitivity
• What if 4 more hours were allocated to phase 2 of the
manufacturing stage? Thus the constraint:
4/5x1 + 4/5x2 <= 964
• New profit = 220500
• Original was 220000, so these 4 hours translate into $500
more profit
Constraint Sensitivity
• What if 8 more hours were allocated to phase 2 of the
manufacturing stage? Thus the constraint:
4/5x1 + 4/5x2 <= 968
• New profit = 221000
• Original was 220000, so these 4 hours translate into
$1000 more profit
• How much more profit if increased by 12 hours? $1500?
Example
2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1000
Solution: x1 = 400, x2 = 800
Objective line
Slope = -3/4
1/2x1 + x2 = 1000
1000
2000
Move a constraint
2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1000
Solution: x1 = 400, x2 = 800
Objective line
Slope = -3/4
1/2x1 + x2 = 1000
1000
2000
Move a constraint
2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1000
New solution
Objective line
Slope = -3/4
1/2x1 + x2 = 1000
1000
2000
Dual Prices
• The dual price of a constraint is the improvement in the
optimal solution, per unit increase in the right-hand side
value of the constraint. The dual price of the processor
configuration constraint is thus $500/4 = $125.
• A negative dual price indicates how much worse the
objective function value will get with each unit increase
Move a constraint
2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1000
New solution
Objective line
Slope = -3/4
1/2x1 + x2 = 1000
1000
2000
Move the constraint further
2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1000
Solution unchanged!
Objective line
Slope = -3/4
1/2x1 + x2 = 1000
1000
2000
Range of Feasibility
• The dual price may only be applicable for small increases.
Large increases may result in a change in the optimal
extreme point, and thus increasing this value further may
not have the same effect.
• The range of values the right-hand side can take without
affecting the dual price is called the range of feasibility.
This is similar to the concept of range of optimality for
objective function coefficients. There is no easy way to
manually calculate these ranges, but they can be found
under the RIGHT HAND SIDE RANGES heading in
LINDO.
Range of Feasibility
• For phase 2 of manufacturing, the allowable increase is
40 (LINDO)
• Thus the dual price of $125 is valid for any increase in the
allowable hours for phase 2, up to 960 + 40 = 1000
• i.e. any increase n (<= 40) added to 960 will increase the objective
value by $125n.
• Any increase x (> 40) added to 960 may or may not
increase the objective value by $125x.
Example
• The range of feasibility for the phase 2 constraint is [800,
1000]. What is the value of objective function if the righthand side is changed to:
• 1000: 125*40=5000  220000 + 5000 = 225000
• 860: 125*-100 = -12500 220000-12500 = 207500
• 700: don’t know
The 100% rule for constraint right-hand
sides
• To determine whether simultaneous changes will not
change dual prices
• For each constraint:
• Compute change as a percentage of the allowable change
• Sum all percentage changes
• If the sum is less than or equal to 100%, the dual prices
will not change
• If the sum exceeds 100%, the dual prices may change
Consolidated Electronics
• As part of a quality improvement initiative, Consolidated Electronics
employees complete a three-day training program on teaming and a
two-day training program on problem solving. The manager of quality
improvement has requested that at least 8 training programs on
teaming and at least 10 training programs on problem solving be
offered during the next six months. In addition, senior-level
management has specified that at least 25 training programs must be
offered during this period. Consolidated Electronics uses a consultant
to teach the training programs. During the next six months, the
consultant has up to 84 days of training time available. Each training
program on teaming costs $10,000 and each training program on
problem solving costs $8,000. Determine the number of training
programs on each teaming and problem solving that minimizes cost.
Post-Optimality Analysis
1.
What will happen to the optimal solution if the cost of each training program on
teaming is decreased to $9000?
2.
What will happen to the optimal solution if the cost of each training program on
problem solving is increased to $9500?
3.
What if both changes occur simultaneously?
4.
What will be the new overall minimum cost if the minimum total number of
courses is increased from 25 to 27?
5.
What will be the new overall minimum cost if the total number of days available
for training is decreased from 84 to 64? 54?
6.
What will be the new overall minimum cost if the minimum total number of
training courses on teaming is decreased from 8 to 5?
7.
What will be the new overall minimum cost if the minimum total number of
training courses on teaming is decreased from 8 to 2 but the minimum total
number of courses is increased from 25 to 26?