1. Sensitivity Analysis
Consider the CrossChek hockey stick production problem:
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?
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.
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 total hours
phase 2 960
phase
How many of each product should be produced?
Maximize profit
Satisfy constraints.
Max 150x x2
s.t x1 + 2/3x2 <= 1000
4/5x1 + 4/5x2 <= 960
1/2x1 + x2 <= 1000
x1, x2 >= 0

2. 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 =
Original was , so these 4 hours translate into $500 more profit

3. 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 =
Original was , so these 4 hours translate into $1000 more profit
How much more profit if increased by 12 hours? $1500?

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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.

11. 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 = 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.

12. Example
The range of feasibility for the phase 2 constraint is [800, 1000]. What is the value of objective function if the right-hand side is changed to:
1000: 125*40=5000  =
860: 125*-100 =  =
700: don’t know

13. 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

14. 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.

15. Post-Optimality Analysis
What will happen to the optimal solution if the cost of each training program on teaming is decreased to $9000?
What will happen to the optimal solution if the cost of each training program on problem solving is increased to $9500?
What if both changes occur simultaneously?
What will be the new overall minimum cost if the minimum total number of courses is increased from 25 to 27?
What will be the new overall minimum cost if the total number of days available for training is decreased from 84 to 64? 54?
What will be the new overall minimum cost if the minimum total number of training courses on teaming is decreased from 8 to 5?
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?