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Sensitivity Analysis

Sensitivity Analysis. CONTENTS Geometrical interpretation of sensitivity analysis Shadow prices Changing the cost of a nonbasic variable Changing the cost of a basic variable Changing the right-hand side vector Changing the column of a variable Adding a new activity

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Sensitivity Analysis

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  1. Sensitivity Analysis CONTENTS • Geometrical interpretation of sensitivity analysis • Shadow prices • Changing the cost of a nonbasic variable • Changing the cost of a basic variable • Changing the right-hand side vector • Changing the column of a variable • Adding a new activity Reference: Chapter 6 in Bazaraa, Jarvis, and Sherali.

  2. Sensitivity Analysis: Introduction Sensitivity analysis is concerned with how changes in a LP’s parameters affect the optimal solution. Max. z = 3x1 + 2x2(Obj. Func.) subject to 2x1 + x2 100 (Finishing constraint) x1 + x2 80 (Carpentry constraint) x1 40 (Bound on soldiers) x1 0 (Sign restriction) x2 0 (Sign restriction) x1, x2, x3 , …., xn 0 Optimal Solution: z = 180, x1 = 20, x2 = 60 How the optimal solution would change if some cost efficient changes or some right-hand side term changes, or some aij changes.

  3. Changing Cost Coefficient x2 Obj. Function: z = 3x1 + 2x2 How change in c1 will affect the optimal solution and optimal objective function value? How change in c2 will affect the optimal solution and optimal objective function value? Finishing constraint (slope = -2) Isoprofit line (slope = -1.5) Carpentry constraint (slope = -1) x1

  4. Changing Right-Hand Side Coefficients x2 Obj. Function: z = 3x1 + 2x2 How change in b1 will affect the optimal solution and optimal objective function value? How change in b2 will affect the optimal solution and optimal objective function value? Shadow prices tell us how objective function value changes with change in bi. Finishing constraint (b1 = 100) Isoprofit line (slope = -1.5) Carpentry constraint (b2 = 80) x1

  5. Shadow Prices • The shadow prices are associated with constraints of an LP. • The shadow price for the ith constraint of an LP is the amount by which the optimal z-value (that is, the objective function value) improves if the right-hand side of the ith constraint is increased by one unit. • This definition assumes that changing the right-hand side of the ith constraint leaves the current basis optimal. • What is the shadow price for the finishing constraint: • Solve 2x1 + x2 = 100 +  and x1 + x2 = 80. • x1 = 20 +  and x2 = 60 - . • z = 3x1 + 2x2 = 3(20 + ) + 2(60 - ) = 120 +  • The shadow price of finishing constraint is 1.

  6. Importance of Sensitivity Analysis • In many applications, values of LP parameters change. If a parameter changes, the sensitivity analysis allows us to determine the new optimal solution without solving the problem afresh. • It allows us to determine what changes do not change the solution. • The knowledge of sensitivity analysis enables an analyst to determine the impact of changes on the optimal objective function value. • Managerial use of shadow prices.

  7. Types of Sensitivity Analysis Considered • We will study six types of changes in LP parameters: • Changing the objective function coefficient of a nonbasic variable • Changing the objective function coefficient of a basic variable • Changing the right-hand side of a constraint • Changing the column of a nonbasic variable • Adding a new variable or activity

  8. The Example • The LP Problem: Max. z = 60x1 + 30x2 + 20x3 (Obj. Func.) subject to 8x1 + 6x2 + x3 48 (Lumber constraint) 4x1 + 2x2 + 1.5x3 20 (Finishing constraint) 2x1 + 1.5x2 + 0.5x3 8 (Bound on soldiers) x1, x2, x3 0 • Initial simplex tableau: z - 60x1 - 30x2 - 20x3 = 0 8x1 + 6x2 + x3 + s1 = 48 4x1 + 2x2 + 1.5x3 + s2 = 20 2x1 + 1.5x2 + 0.5x3 + s3 = 8

  9. The Example (contd.) • Optimal tableau: z + 5x2 + 10s2 + 10s3= 280 - 2x2 + s1 + 2s2 - 8s3 = 24 - 2x2 + x3 + 2s2 - 4s3 = 8 x1 + 1.25x2 - 0.5s2 + 1.5s3 = 2 B = {s1, x3, x1}, N = {x2, s2, s3}, Optimal BFS: z = 280, s1 = 24, x3 = 8, x1 = 2, x2 = 0, s2 = 0, s3 = 0 Optimal simplex multipliers p = {0, 10, 10} B = B-1 =

  10. Changing the Cost of a Nonbasic Variable • For what values of c2 would B = {s1, x3, x1} remain optimal? • If we change c2, the reduced cost coefficient changes. As long as remain nonnegative, the current basis will be optimal. = (a12p1 + a22p2 + a32p3) – c2 = (0, 10, 10)(6, 2, 1.5)T - (30 + D)  0 Alternatively, 5 - D 0 or D 5. Hence we can increase c2 by at most 5 units and the current basis remains optimal. We can decrease c2 by any amount and the current basis will remain optimal.

  11. Changing the Cost of a Basic Variable • For what values of c1 would B = {s1, x3, x1} remain optimal? • If we change the cost of a basic variable, then it also changes the simplex multipliers, which in turn changes several reduced cost coefficients, all of which must remain nonnegative. p = cBB-1 = (0, 20, 60+D) = (0, 10-.5D, 10+1.5D)

  12. Changing the Cost of a Basic Variable (contd.) of x2 = (pa2 – c2)= = [0, 10 - .5D, 10 + 1.5D] [6, 2, 1.5]T – 30 = 5 + 1.25D Reduced cost coefficient of s2 = 10 - .5D Reduced cost coefficient of s3 = 10 + 1.5D The current basis remains optimal as long as: 5 + 1.25D  0 or D  -4 10 - .5D  0 or D  20 10 + 1.5D  0 or D  -(20/3) Hence, for all -4  D  20, the current basis remains optimal. Alternatively, c1 must satisfy, -56 c1 80.

  13. When the Current Basis is Not Optimal Suppose that c1 = 100. Clearly the current basis is not optimal. How do we then obtain the new optimal solution? z + 55x2 - 10s2 + 70s3= 360 - 2x2 + s1 + 2s2 - 8s3 = 24 - 2x2 + x3 + 2s2 - 4s3 = 8 x1 + 1.25x2 - 0.5s2 + 1.5s3 = 2 Performing the pivot operation, gives us the following solution: z + 45x2 + 5x3 + 50s3= 400 - x3 + s1 - 4s3 = 16 - x2 + 0.5x3 + s2 - 2s3 = 4 x1 + 0.75x2 + 0.25x3 + 0.5s3 = 4 which is an optimal solution of the modified problem.

  14. Changing the Right Hand Side of a Constraint • Changing the right-hand side vector does not affect the optimality conditions of the solution, but it affects the feasibility of the solution. • As long as the right-hand side of each of each constrain in the optimal tableau remains nonnegative, the current basis remains feasible and optimal. • Suppose we want to know by what amount we can change b2 without changing the optimal basis. • Suppose that we change b2 to b2 + D. Determine the range of D for which the current basis remains optimal.

  15. Changing the Right Hand Side of a Constraint (contd.) = B-1 b = = The current basis remains optimal as long as: 24 + 2D  0 or D  -12 8 + 2D  0 or D  -4 2 - 0.5D  0 or D 4 Hence, for all -4  D  20, the current basis remains optimal. Alternatively, b2 must satisfy, 16 b2 24.

  16. Effect on Decision Variables and z = = New z-value = cBB-1(new b) = [0, 10, 10] = 300

  17. When the Current Basis is No Longer Optimal • If we change a right-hand side enough that the current basis is no longer optimal, how can we determine the optimal basis? • Suppose that we change to 30. In this case, the right-hand side vector becomes: = = • This vector is no longer feasible or optimal. We use dual simplex method to make this solution feasible and optimal.

  18. Changing the Column of a Nonbasic Variable • The original LP Problem: Max. z = 60x1 + 30x2 + 20x3 (Obj. Func.) subject to 8x1 + 6x2 + x3 48 (Lumber constraint) 4x1 + 2x2 + 1.5x3 20 (Finishing constraint) 2x1 + 1.5x2 + 0.5x3 8 (Bound on soldiers) x1, x2, x3 0 • The modified LP Problem: Max. z = 60x1 + 43x2+ 20x3 (Obj. Func.) subject to 8x1 + 5x2 + x3 48 (Lumber constraint) 4x1 + 2x2 + 1.5x3 20 (Finishing constraint) 2x1 + 2x2 + 0.5x3 8 (Bound on soldiers) x1, x2, x3 0

  19. Changing the Column of a Nonbasic Variable (contd.) • Changing the nonbasic column does not change the simplex multipliers p. • The new value of x2 is: = pa2 – c2 = [0, 10, 10] [5, 2, 2] – 43 = -3 < 0 • This operation is called a pricing out operation. • It now becomes profitable to increase x2. We will now perform a pivot operation to enter x2 in the basis. B-1a2 = =

  20. Changing the Column of a Nonbasic Variable (contd.) B-1b = = • What is the leaving variable? • What is the new B-1? • What are the new simplex multipliers?

  21. Changing the Column of a Basic Variable • Changing the column of a basic variable changes B-1. • It also changes the simplex multipliers p and the optimality of the solution. • It also changes the right-hand vector which may impact the feasibility of the solution. • We need to compute B-1 and p, and price out all nonbasic variables and check whether they satisfy the optimality conditions. We also need to compute B-1b and ensure that the new solution is feasible.

  22. Adding a New Activity • Suppose that the company is considering making footstools. • A footstool sells for $15 and requires 1 board foot of lumber, 1 finishing hour, and 1 carpentry hour. Should the company manufacture any stools? • New formulation: z - 60x1 - 30x2 - 20x3 - 15x4 = 0 8x1 + 6x2 + x3 + x4 + s1 = 48 4x1 + 2x2 + 1.5x3 + x4 + s2 = 20 2x1 + 1.5x2 + 0.5x3 + x4 + s3 = 8 • How to solve the modified problem?

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