Sensitivity Analysis Consider the CrossChek hockey stick production problem: Management believes...
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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.
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Transcript of Sensitivity Analysis Consider the CrossChek hockey stick production problem: Management believes...
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 = 1000Objective lineSlope = -3/4
CrossChek Manufacturing Problem
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -3/4
Extreme points
CrossChek Manufacturing Problem
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -3/4
CrossChek Manufacturing Problem
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -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 = 1000Objective lineSlope = -3/5
CrossChek Manufacturing Problem
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -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 = 1000Objective lineSlope = -3/4
Range of Optimality
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -1/2
Range of Optimality
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -1/2
Range of Optimality
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -1/2
Both points optimal
Range of Optimality
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope > -1/2
New optimal solution
Range of Optimality
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -3/4
Range of Optimality
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -1
Range of Optimality
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -1
Range of Optimality
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -1
Both points optimal
Range of Optimality
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope < -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 = 1000Objective lineSlope = -3/4
1. Find Optimal Point
1000
2000
1000 2000
x1 + 2/3x2 = 1000
4/5x1 + 4/5x2 = 1000
1/2x1 + x2 = 1000Objective lineSlope = -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 = -1Slope = -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 = -1Slope = -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 = -1Slope = -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 = -1Slope = -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
