Sensitivity Analysis Consider the Cross Chek hockey stick

Sensitivity Analysis • Consider the Cross. Chek 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. • 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. Max s. t 150 x 1 + 200 x 2 x 1 + 2/3 x 2 <= 1000 4/5 x 1 + 4/5 x 2 <= 960 1/2 x 1 + x 2 <= 1000 x 1, x 2 >= 0 • Management believes that Cross. Chek 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: 4/5 x 1 + 4/5 x 2 <= 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: 4/5 x 1 + 4/5 x 2 <= 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 x 1 + 2/3 x 2 = 1000 4/5 x 1 + 4/5 x 2 = 1000 Solution: x 1 = 400, x 2 = 800 Objective line Slope = -3/4 1/2 x 1 + x 2 = 1000 2000

Move a constraint 2000 x 1 + 2/3 x 2 = 1000 4/5 x 1 + 4/5 x 2 = 1000 Solution: x 1 = 400, x 2 = 800 Objective line Slope = -3/4 1/2 x 1 + x 2 = 1000 2000

Move a constraint 2000 x 1 + 2/3 x 2 = 1000 4/5 x 1 + 4/5 x 2 = 1000 New solution Objective line Slope = -3/4 1/2 x 1 + x 2 = 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 x 1 + 2/3 x 2 = 1000 4/5 x 1 + 4/5 x 2 = 1000 New solution Objective line Slope = -3/4 1/2 x 1 + x 2 = 1000 2000

Move the constraint further 2000 x 1 + 2/3 x 2 = 1000 4/5 x 1 + 4/5 x 2 = 1000 Solution unchanged! Objective line Slope = -3/4 1/2 x 1 + x 2 = 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 $125 n. • Any increase x (> 40) added to 960 may or may not increase the objective value by $125 x.

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?