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DSCI 3870 Chapter 3 SENSITIVITY ANALYSIS Additional Reading Material

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18 diapositives · 369 Views · Nov 11, 2014

DSCI 3870 Chapter 3 SENSITIVITY ANALYSIS Additional Reading Material. Chapter 3 – Sensitivity Analysis Additional Reading Material. Example 3. Example 3. Consider the following linear program:. Min 6 x 1 + 9 x 2 ($ cost) s.t. x 1 + 2 x 2 < 8

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  1. DSCI 3870 Chapter 3 SENSITIVITY ANALYSIS Additional Reading Material

  2. Chapter 3 – Sensitivity Analysis Additional Reading Material • Example 3

  3. Example 3 • Consider the following linear program: Min 6x1 + 9x2 ($ cost) s.t. x1 + 2x2 < 8 10x1 + 7.5x2 > 30 x2 > 2 x1, x2> 0

  4. Example 3 • The Management Scientist Output OBJECTIVE FUNCTION VALUE = 27.000 VariableValueReduced Cost x1 1.500 0.000 x2 2.000 0.000 Constraint Slack/SurplusDual Price 1 2.500 0.000 2 0.000 -0.600 3 0.000 -4.500

  5. Example 3 • The Management Scientist Output (continued) OBJECTIVE COEFFICIENT RANGES VariableLower Limit Current ValueUpper Limit x1 0.000 6.000 12.000 x2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES Constraint Lower LimitCurrent ValueUpper Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

  6. Example 3 • Optimal Solution According to the output: x1 = 1.5 x2 = 2.0 Objective function value = 27.00

  7. Example 3 • Range of Optimality Question: Suppose the unit cost of x1 is decreased to $4. Is the current solution still optimal? What is the value of the objective function when this unit cost is decreased to $4?

  8. Example 3 • The Management Scientist Output OBJECTIVE COEFFICIENT RANGES VariableLower Limit Current ValueUpper Limit x1 0.000 6.000 12.000 x2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES Constraint Lower LimitCurrent ValueUpper Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

  9. Example 3 • Range of Optimality Answer: The output states that the solution remains optimal as long as the objective function coefficient of x1 is between 0 and 12. Because 4 is within this range, the optimal solution will not change. However, the optimal total cost will be affected: 6x1 + 9x2 = 4(1.5) + 9(2.0) = $24.00.

  10. Example 3 • Range of Optimality Question: How much can the unit cost of x2 be decreased without concern for the optimal solution changing?

  11. Example 3 • The Management Scientist Output OBJECTIVE COEFFICIENT RANGES VariableLower Limit Current ValueUpper Limit x1 0.000 6.000 12.000 x2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES Constraint Lower LimitCurrent ValueUpper Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

  12. Example 3 • Range of Optimality Answer: The output states that the solution remains optimal as long as the objective function coefficient of x2 does not fall below 4.5.

  13. Example 3 • Range of Optimality and 100% Rule Question: If simultaneously the cost of x1 was raised to $7.5 and the cost of x2 was reduced to $6, would the current solution remain optimal?

  14. Example 3 • Range of Optimality and 100% Rule Answer: If c1 = 7.5, the amount c1 changed is 7.5 - 6 = 1.5. The maximum allowable increase is 12 - 6 = 6, so this is a 1.5/6 = 25% change. If c2 = 6, the amount that c2 changed is 9 - 6 = 3. The maximum allowable decrease is 9 - 4.5 = 4.5, so this is a 3/4.5 = 66.7% change. The sum of the change percentages is 25% + 66.7% = 91.7%. Because this does not exceed 100%, the optimal solution would not change.

  15. Example 3 • Range of Feasibility Question: If the right-hand side of constraint 3 is increased by 1, what will be the effect on the optimal solution?

  16. Example 3 • The Management Scientist Output OBJECTIVE COEFFICIENT RANGES VariableLower Limit Current ValueUpper Limit x1 0.000 6.000 12.000 x2 4.500 9.000 No Limit RIGHTHAND SIDE RANGES Constraint Lower LimitCurrent ValueUpper Limit 1 5.500 8.000 No Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

  17. Example 3 • Range of Feasibility Answer: A dual price represents the improvement in the objective function value per unit increase in the right-hand side. A negative dual price indicates a deterioration (negative improvement) in the objective, which in this problem means an increase in total cost because we're minimizing. Since the right-hand side remains within the range of feasibility, there is no change in the optimal solution. However, the objective function value increases by $4.50.

  18. End of Chapter 3