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Sensitivity Analysis

Sensitivity Analysis. How will a change in a coefficient of the objective function affect the optimal solution? How will a change in the right-hand side value for a constraint affect the optimal solution?. Solving Linear Equations.

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Sensitivity Analysis

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  1. Sensitivity Analysis • How will a change in a coefficient of the objective function affect the optimal solution? • How will a change in the right-hand side value for a constraint affect the optimal solution? Linear Programming

  2. Solving Linear Equations • All operations that apply to linear equations also apply to linear inequalities with the following exceptions: • If you multiply or divide by a negative number it will switch the direction of the inequality. • If you invert an inequality it will also switch the direction of the inequality Linear Programming

  3. Sherwood – Linear Equations Linear Programming

  4. Sherwood – Graph Solution Line 1 5 4 3 Line 2 1 2 Linear Programming

  5. Sherwood – Optimal Solution • Extreme Point 3 is optimal if: • Slope of Line 1 <= Slope of objective function <= Slope of Line 2 Linear Programming

  6. Sherwood – Calculate Slope of Line 1 8x1 + 2x2 <= 160 2x2 = -8x1 + 160 x2 = -4x1 + 80 Slope of Intercept of Line 1 Line 1 on x2 axis Linear Programming

  7. Sherwood – Calculate Slope of Line 2 4x1 + 3 x2 <= 120 3x2 = -4x1 + 120 x2 = -4/3x1 + 40 Slope of Intercept of Line 2 Line 2 on x2 axis Linear Programming

  8. Sherwood – Optimal Solution • Extreme Point 3 is optimal if: • -4 <= Slope of objective function <= -4/3 Linear Programming

  9. Calculating Slope-Intercept • General form of objective function • P = Cx1x1 + Cx2x2 • Slope-intercept for objective function • x2 = -(Cx1/Cx2) x1 + P/Cx2 Slope of Intercept of Obj. Function Obj. Function on x2 axis Linear Programming

  10. Sherwood – Optimal Solution • Extreme Point 3 is optimal if: • -4 <= -(Cx1/Cx2) <= -4/3 Or • 4/3 <= (Cx1/Cx2) <= 4 Linear Programming

  11. Sherwood – Compute the Range of Optimality • Extreme Point 3 is optimal if: • 4/3 <= (Cx1/Cx2) <= 4 • Compute range for Cx1, hold Cx2constant • 4/3 <= (Cx1/10) <= 4 Linear Programming

  12. Sherwood – Compute the Range of Optimality • From the left-hand inequality, we have • 4/3 <= (Cx1/10) • Thus, • 40/3 <= Cx1 Linear Programming

  13. Sherwood – Compute the Range of Optimality • From the right-hand inequality, we have • (Cx1/10) <= 4 • Thus, • Cx1 <= 40 Linear Programming

  14. Sherwood – Compute the Range of Optimality • Summarizing these limits • 40/3 <= Cx1 <= 40 Linear Programming

  15. Sherwood – Compute the Range of Optimality • Extreme Point 3 is optimal if: • 4/3 <= (Cx1/Cx2) <= 4 • Compute range for Cx2, hold Cx1constant • 4/3 <= (20/Cx2) <= 4 Linear Programming

  16. Sherwood – Compute the Range of Optimality • From the inequality, we have • 4/3 <= (20/Cx2) <= 4 • Thus, • 4/60 <= (1/Cx2) <= 4/20 • 5 <= Cx2 <= 15 Linear Programming

  17. Sherwood – Compute the Range of Optimality • Summarizing these limits • 40/3 <= Cx1 <= 40 • 5 <= Cx2 <= 15 Linear Programming

  18. Sensitivity Analysis • How will a change in a coefficient of the objective function affect the optimal solution? • How will a change in the right-hand side value for a constraint affect the optimal solution? Linear Programming

  19. Sherwood – Graph Solution Line 1 5 4 3 Line 2 1 2 Linear Programming

  20. Sherwood – Change in the Right-hand Side • Constraint 1 – add 1 to right-hand side • 4x1 + 3x2 <= 121 • 8x1 + 2x2 <= 160 • Solve for x2 • 2(4x1 + 3x2 = 121) • -1(8x1 + 2x2 = 160) • 4x2 = 82 • x2 = 20.5 • Solve for x1 • 8x1 + 2(20.5)= 160 • x1 =14.875 Linear Programming

  21. Sherwood – Change in the Right-hand Side • Solve objective function • z = 20(14.875) + 10(20.5) • z = 502.5 • Shadow Price • 502.5 – 500 = 2.5 • Thus profit increases at $2.50 per hour of labor added to assembly • Conversely, if we decrease labor for assembly by 1 hour the objective function will decrease by $2.50 Linear Programming

  22. Sherwood – Range of Feasibility • Constraint 1 RHS = 120 • Allowable Increase = 24 • Allowable Decrease = 40 • Range of Feasibility • 80 <= Constraint 1 RHS <= 144 Linear Programming

  23. Sherwood – Change in the Right-hand Side • Constraint 2 – add 1 to right-hand side • 4x1 + 3x2 <= 120 • 8x1 + 2x2 <= 161 • Solve for x2 • 2(4x1 + 3x2 = 120) • -1(8x1 + 2x2 = 161) • 4x2 = 79 • x2 = 19.75 • Solve for x1 • 4x1 + 3(19.75)= 120 • x1 =15.1875 Linear Programming

  24. Sherwood – Change in the Right-hand Side • Solve objective function • z = 20(15.1875) + 10(19.75) • z = 501.25 • Shadow Price • 501.25 – 500 = 1.25 • Thus profit increases at $1.25 per hour of labor added to finishing • Conversely, if we decrease labor for finishing by 1 hour the objective function will decrease by $1.25 Linear Programming

  25. Sherwood – Range of Feasibility • Constraint 2 RHS = 160 • Allowable Increase = 80 • Allowable Decrease = 48 • Range of Feasibility • 112 <= Constraint 2 RHS <= 240 Linear Programming

  26. Sherwood – Range of Feasibility • Constraint 3 RHS • Slack = 12 • Shadow Price = 0 • Range of Feasibility • 20 <= Constraint 3 RHS <= Infinite Linear Programming

  27. Non-Binding Constraints • There is more resource then needed (i.e. there is slack). • When you have a non-binding constraint the shadow price is zero • Also, the allowable increase will be 1E+30 (infinite) represents that no upper limit exists for the range of feasibility • The lower limit allowable decrease equals the amount of slack Linear Programming

  28. Reduced Costs • For each decision variable, the absolute value of the reduced costs indicates how much the objective coefficient would have to improve before that variable could assume a positive value in the optimal solution. • If the decision variable is already positive in the optimal solution, its reduced costs variable is zero. Linear Programming

  29. Sherwood - Slack Variables Max 20x1 + 10x2 + 0S1 + 0S2 + 0S3 s.t. 4x1 + 3x2 + 1S1 = 120 8x1 + 2x2 + 1S2 = 160 x2 + 1S3 = 32 x1, x2, S1 ,S2 ,S3 >= 0 Linear Programming

  30. Sherwood – Slack Variables • For each ≤ constraint the difference between the RHS and LHS (RHS-LHS). It is the amount of resource left over. • Constraint 1; S1 = 0 hrs. • Constraint 2; S2 = 0 hrs. • Constraint 3; S3 = 12 Custom Linear Programming

  31. Binding vs. Non-Binding Constraints • Constraints that have zero slack are considered binding constraints • Constraints that have slack or unused capacity available are non-binding. They have a shadow price of zero. This shows that additional units of this resource will not increase the value of the objective function Linear Programming

  32. Summary • In summary, the right-hand-side ranges provide limits within which the shadow prices are applicable. For changes outsides the range, the problem must be resolved to find the new optimal solution and the new shadow price. The ranges of feasibility for the Sherwood problem are: • 80 <= Constraint 1 <= 144 • 112 <= Constraint 2 <= 240 • 20 <= Constraint 3 <= Infinite Linear Programming

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