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? Max 150x x2 s.t x1 + 2/3x2 <= /5x1 + 4/5x2 <= 960 1/2x1 + x2 <= 1000 x1, x2 >= 0 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 phase How many of each product should be produced? Maximize profit Satisfy constraints.

CrossChek Manufacturing Problem x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4

CrossChek Manufacturing Problem x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4 Extreme points

CrossChek Manufacturing Problem x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4

CrossChek Manufacturing Problem x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4 Solution: x1 = 400, x2 = 800

CrossChek Manufacturing Problem Consider change in profit of lower-end stick from $150 to $120 Change in objective function: 150x x2 -> 120x x2 Change in slope of objective function: -3/4 -> -3/5

CrossChek Manufacturing Problem x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4

CrossChek Manufacturing Problem x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/5

CrossChek Manufacturing Problem x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/5 Solution: x1 = 400, x2 = 800 Same solution!

Range of Optimality We know that changing the coefficient of x1 in the objective function 150x x2 from 150 to 120 does not change the solution What is the range (i.e. lowest and highest) of values that will not change the solution?

Range of Optimality x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4

Range of Optimality x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -1/2

Range of Optimality x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -1/2

Range of Optimality x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -1/2 Both points optimal

Range of Optimality x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope > -1/2 New optimal solution

Range of Optimality x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4

Range of Optimality x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -1

Range of Optimality x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -1

Range of Optimality x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -1 Both points optimal

Range of Optimality x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope < -1 New optimal solution

Conclusion: Any objective function with slope less than or equal to -½ and greater than or equal to -1 will not change the optimal solution!

Recap x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4

1. Find Optimal Point x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4 Solution: x1 = 400, x2 = 800

2. Determine slope of the 2 lines that meet at optimal point x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Slope = -1 Slope = -1/2

3. For objective function c1x1 + c2x2 with slope = – c1/c x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Slope = -1 Slope = -1/2 Let sL be the lower slope (-1) Let sH be the higher slope (-1/2)

3. For objective function c1x1 + c2x2 with slope = – c1/c x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Slope = -1 Slope = -1/2 Let sL be the lower slope (-1) Let sH be the higher slope (-1/2) a) Range for c1. Fix c2 and solve for c1 -c1/c2 >= sL -c1/c2 <= sH

3. For objective function c1x1 + c2x2 with slope = – c1/c x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Slope = -1 Slope = -1/2 Let sL be the lower slope (-1) Let sH be the higher slope (-1/2) b) Range for c2. Fix c1 and solve for c2 -c1/c2 >= sL -c1/c2 <= sH

Sensitivity Analysis - LINDO 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? Max 150x x2 s.t x1 + 2/3x2 <= /5x1 + 4/5x2 <= 960 1/2x1 + x2 <= 1000 x1, x2 >= 0 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 phase How many of each product should be produced? Maximize profit Satisfy constraints.

Caveat We know that changing the coefficient of x1 in the objective function 150x x2 from 150 to 120 does not change the solution We know that changing the coefficient of x2 in the objective function 150x x2 from, say 200 to 175 does not change the solution However, these ranges are only valid when the other coefficient remains fixed Changing both simultaneously may or may not change the optimal solution

100% Rule for Objective Function Coefficients To determine whether simultaneous changes will not change solution For each coefficient: Compute change as a percentage of the allowable change Sum all percentage changes If the sum is less than or equal to 100%, the optimal solution will not change If the sum exceeds 100%, the solution may change

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? Max 150x x2 s.t x1 + 2/3x2 <= /5x1 + 4/5x2 <= 960 1/2x1 + x2 <= 1000 x1, x2 >= 0 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 phase How many of each product should be produced? Maximize profit Satisfy constraints.

Constraint Sensitivity What if 4 more hours were allocated to phase 2 of the manufacturing stage? Thus the constraint: 4/5x1 + 4/5x2 <= 964

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

Constraint Sensitivity What if 8 more hours were allocated to phase 2 of the manufacturing stage? Thus the constraint: 4/5x1 + 4/5x2 <= 968

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 8 hours translate into $1000 more profit How much more profit if increased by 12 hours? $1500?

Example x1 + 2/3x2 = /5x1 + 4/5x2 = 960 1/2x1 + x2 = 1000 Objective line Slope = -3/4 Solution: x1 = 400, x2 = 800

Move a constraint x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4 Solution: x1 = 400, x2 = 800

Move a constraint x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4 New solution

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 optimal solution will get with each unit increase in the right-hand side value of the constraint

Move a constraint x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4 New solution

Move the constraint further x1 + 2/3x2 = /5x1 + 4/5x2 = /2x1 + x2 = 1000 Objective line Slope = -3/4 Solution unchanged!

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 LINDO Example Max 150x x2 st x x2 < x x2 < x1 + x2 < 1000 x1 > 0 x2 > 0