Week 4 - Feb 3
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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? CrossChek Manufacturing Problem 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -3/4 1/2x1 + x2 = 1000 1000 2000 CrossChek Manufacturing Problem Extreme points 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -3/4 1/2x1 + x2 = 1000 1000 2000 CrossChek Manufacturing Problem 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -3/4 1/2x1 + x2 = 1000 1000 2000 CrossChek Manufacturing Problem 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 CrossChek Manufacturing Problem • Consider change in profit of lower-end stick from $150 to $120 • Change in objective function: 150x1 + 200x2 -> 120x1 + 200x2 • Change in slope of objective function: -3/4 -> -3/5 CrossChek Manufacturing Problem 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -3/4 1/2x1 + x2 = 1000 1000 2000 CrossChek Manufacturing Problem 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -3/5 1/2x1 + x2 = 1000 1000 2000 CrossChek Manufacturing Problem Same solution! 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Solution: x1 = 400, x2 = 800 Objective line Slope = -3/5 1/2x1 + x2 = 1000 1000 2000 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 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -3/4 1/2x1 + x2 = 1000 1000 2000 Range of Optimality 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -1/2 1/2x1 + x2 = 1000 1000 2000 Range of Optimality 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -1/2 1/2x1 + x2 = 1000 1000 2000 Range of Optimality Both points optimal 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -1/2 1/2x1 + x2 = 1000 1000 2000 Range of Optimality New optimal solution 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope > -1/2 1/2x1 + x2 = 1000 1000 2000 Range of Optimality 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -3/4 1/2x1 + x2 = 1000 1000 2000 Range of Optimality 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -1 1/2x1 + x2 = 1000 1000 2000 Range of Optimality 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -1 1/2x1 + x2 = 1000 1000 2000 Range of Optimality Both points optimal 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -1 1/2x1 + x2 = 1000 1000 2000 Range of Optimality New optimal solution 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope < -1 1/2x1 + x2 = 1000 1000 2000 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 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 1000 Objective line Slope = -3/4 1/2x1 + x2 = 1000 1000 2000 1. Find Optimal Point 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 2. Determine slope of the 2 lines that meet at optimal point 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 Slope = -1 1000 Slope = -1/2 1/2x1 + x2 = 1000 1000 2000 3. For objective function c1x1 + c2x2 with slope = – c1/c2 Let sL be the lower slope (-1) Let sH be the higher slope (-1/2) 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 Slope = -1 1000 Slope = -1/2 1/2x1 + x2 = 1000 1000 2000 3. For objective function c1x1 + c2x2 with slope = – c1/c2 Let sL be the lower slope (-1) Let sH be the higher slope (-1/2) 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 Slope = -1 1000 a) Range for c1. Fix c2 and solve for c1 -c1/c2 >= sL -c1/c2 <= sH Slope = -1/2 1/2x1 + x2 = 1000 1000 2000 3. For objective function c1x1 + c2x2 with slope = – c1/c2 Let sL be the lower slope (-1) Let sH be the higher slope (-1/2) 2000 x1 + 2/3x2 = 1000 4/5x1 + 4/5x2 = 1000 Slope = -1 1000 b) Range for c2. Fix c1 and solve for c2 -c1/c2 >= sL -c1/c2 <= sH Slope = -1/2 1/2x1 + x2 = 1000 1000 2000 Sensitivity Analysis - LINDO • 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? 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 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 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 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 8 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 = 960 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 optimal solution will get with each unit increase in the right-hand side value of the constraint 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 • LINDO Example Max 150x1 + 200x2 st x1 + 0.667x2 < 1000 0.8x1 + 0.8x2 < 960 0.5x1 + x2 < 1000 x1 > 0 x2 > 0
