# Sensitivity Analysis Consider the CrossChek hockey stick production problem: Management believes that CrossChek might only receive \$120 profit from the.

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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 150x1 + 200x2 s.t x1 + 2/3x2 <= 1000 4/5x1 + 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 1 1000 total hours phase 2 960 phase 3 1000 How many of each product should be produced? Maximize profit Satisfy constraints.

CrossChek Manufacturing Problem 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -3/4

CrossChek Manufacturing Problem 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -3/4 Extreme points

CrossChek Manufacturing Problem 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -3/4

CrossChek Manufacturing Problem 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -3/4 Solution: x1 = 400, x2 = 800

CrossChek Manufacturing Problem 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -3/5

CrossChek Manufacturing Problem 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + 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 150x1 + 200x2 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 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -3/4

Range of Optimality 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -1/2

Range of Optimality 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -1/2

Range of Optimality 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -1/2 Both points optimal

Range of Optimality 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope > -1/2 New optimal solution

Range of Optimality 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -3/4

Range of Optimality 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -1

Range of Optimality 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -1

Range of Optimality 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -1 Both points optimal

Range of Optimality 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + 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 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -3/4

1. Find Optimal Point 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Objective line Slope = -3/4 Solution: x1 = 400, x2 = 800

2. Determine slope of the 2 lines that meet at optimal point 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Slope = -1 Slope = -1/2

3. For objective function c1x1 + c2x2 with slope = – c1/c2 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + 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/c2 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Slope = -1 Slope = -1/2 Let sL be the lower slope (-1) Let sH be the higher slope (-1/2) a) Fix c2 and solve for c1 -c1/c2 >= sL -c1/c2 <= sH

3. For objective function c1x1 + c2x2 with slope = – c1/c2 1000 2000 1000 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1/2x1 + x2 = 1000 Slope = -1 Slope = -1/2 Let sL be the lower slope (-1) Let sH be the higher slope (-1/2) b) 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 150x1 + 200x2 s.t x1 + 2/3x2 <= 1000 4/5x1 + 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 1 1000 total hours phase 2 960 phase 3 1000 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 150x1 + 200x2 from 150 to 120 does not change the solution We know that changing the coefficient of x2 in the objective function 150x1 + 200x2 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

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