Sensitivity Analysis

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. Pet Food Co. – Linear Equations Linear Programming

3. Pet Food Co. – Graph Solution Line 3 Line 2 Linear Programming

4. Pet Food Co. – Optimal Solution • Extreme Point is optimal if: • Slope of Line 3 <= Slope of objective function <= Slope of Line 2 Linear Programming

5. Pet Food Co. – Calculate Slope of Line 3 1P1 + 1P2 >= 500 1P2 >= -1P1 + 500 P2 >= -P1 + 500 Slope of Intercept of Line 3 Line 3 on P2 axis Linear Programming

6. Pet Food Co. – Calculate Slope of Line 2 0P1 + 1P2 >= 200 1P2 >= -0P1 + 200 Slope of Intercept of Line 2 Line 2 on P2 axis Linear Programming

7. Pet Food Co. – Optimal Solution • Extreme Point 4 is optimal if: • -1 <= Slope of objective function <= 0 Linear Programming

8. Calculating Slope-Intercept • General form of objective function • Z = CP1P1 + CP2P2 • Slope-intercept for objective function • P2 = -(CP1/CP2) P1 + Z/CP2 Slope of Intercept of Obj. Function Obj. Function on x2 axis Linear Programming

9. Pet Food Co. – Optimal Solution • Extreme Point is optimal if: • -1 <= -(CP1/CP2) <= 0 Or • 0 <= (CP1/CP2) <= 1 Linear Programming

10. Pet Food Co. – Compute the Range of Optimality • Extreme Point is optimal if: • 0 <= (CP1/CP2) <= 1 • Compute range for CP1, hold CP2constant • 0 <= (CP1/8) <= 1 Linear Programming

11. Pet Food Co. – Compute the Range of Optimality • From the left-hand inequality, we have • 0 <= (CP1/8) • Thus, • 0 <= CP1 Linear Programming

12. Pet Food Co. – Compute the Range of Optimality • From the right-hand inequality, we have • (CP1/8) <= 1 • Thus, • CP1 <= 8 Linear Programming

13. Pet Food Co. – Compute the Range of Optimality • Summarizing these limits • 0 <= CP1 <= 8 Linear Programming

14. Pet Food Co. – Compute the Range of Optimality • Extreme Point is optimal if: • 0 <= (CP1/CP2) <= 1 • Compute range for CP2, hold CP1constant • 0 <= (5/CP2) <= 1 Linear Programming

15. Pet Food Co. – Compute the Range of Optimality • From the left-hand inequality, we have • 0 <= (5/CP2) • Thus, • (1/5) * 0 <= (1/CP2) • Invert • 5/0 >= CP2 • Division by zero is undefined (infinite). • This means the cost of P2 can increase to infinity without changing the optimal solution Linear Programming

16. Pet Food Co. – Compute the Range of Optimality • From the right-hand inequality, we have • (5/CP2) <= 1 • Thus, • CP2 >= 5 Linear Programming

17. Pet Food Co. – Compute the Range of Optimality • Summarizing these limits • 0 <= CP1 <= 8 • 5 <= CP2 <= Infinite 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. Pet Food Company – Graph Solution Line 3 Line 2 Linear Programming

20. Pet Food Co. – Range of Feasibility • Constraint 1 – is not binding • Therefore, the shadow price is zero • Slack is 100 • Range of Feasibility • 300 <= Constraint 1 RHS <= Infinite Linear Programming

21. Pet Food Co. – Change in the Right-hand Side • Constraint 2 – add 1 to right-hand side • 0P1 + 1P2 >= 201 • 1P1 + 1P2 >= 500 • Solve for P1 • -1(0P1 + 1P2 = 201) • 1P1 + 1P2 = 500 • P1 = 299 • Solve for P2 • 1(299) + 1P2 >= 500 • P2 =201 Linear Programming

22. Pet Food Co. – Change in the Right-hand Side • Solve objective function • z = 5(299) + 8(201) • z = 3103 • Shadow Price • 3103 – 3100 = 3 • Thus cost increases at $3.00 per lb. added of P2 per batch • Conversely, if we decrease lbs. of P2 per batch by 1 the objective function will decrease by $3.00 Linear Programming

23. Pet Food Co. – Range of Feasibility • Constraint 2 RHS = 200 • Allowable Increase = 300 • Allowable Decrease = 100 • Range of Feasibility • 100 <= Constraint 2 RHS <= 500 Linear Programming

24. Pet Food Co. – Change in the Right-hand Side • Constraint 3 – add 1 to right-hand side • 0P1 + 1P2 >= 200 • 1P1 + 1P2 >= 501 • Solve for P1 • -1(0P1 + 1P2 = 200) • 1P1 + 1P2 = 501 • P1 = 301 • Solve for P2 • 1(301) + 1P2 >= 501 • P2 = 200 Linear Programming

25. Pet Food Co. – Change in the Right-hand Side • Solve objective function • z = 5(301) + 8(199) • z = 3097 • Shadow Price • 3105 – 3100 = 5 • Thus cost increases at $5.00 per lb. added of P2 per batch • Conversely, if we decrease lbs. of P2 per batch by 1 the objective function will decrease by $5.00 Linear Programming

26. Pet Food Co. – Range of Feasibility • Constraint 3 RHS = 500 • Allowable Increase = 100 • Allowable Decrease = 300 • Range of Feasibility • 200 <= Constraint 3 RHS <= 600 Linear Programming

27. Pet Food Co. – Linear Equations Slack/ Surplus Variables Min 5P1 + 8P2 + 0S1 + 0S2 + 0S3 s.t. 1P1 + 1S1 = 400 1P2 - 1S2 = 200 1P1 + 1P2 - 1S3 = 500 P1, P2, S1 ,S2 ,S3 >= 0 Linear Programming

28. Pet Food Co. – 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 = 100 lbs. Linear Programming

29. Pet Food Co. – Surplus Variables • For each ≥ constraint the difference between the LHS and RHS (LHS-RHS). It is the amount bt which a minimum requirement is exceeded. • Constraint 2; S2 = 0 lbs. • Constraint 3; S3 = 0 lbs. Linear Programming

30. Pet Food Co. – Constraint Limits • Range of Feasibility • 300 <= Constraint 1 <= Infinite • 100 <= Constraint 2 <= 500 • 200 <= Constraint 3 <= 600 Linear Programming