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POSTOPTIMALITY ANALYSIS PowerPoint Presentation
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POSTOPTIMALITY ANALYSIS

POSTOPTIMALITY ANALYSIS

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POSTOPTIMALITY ANALYSIS

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  1. LINEAR PROGRAMMING PROBLEM POSTOPTIMALITY ANALYSIS Minggu 5

  2. DUALITY • For every LP problem, there is another related LP that incorporate the same input parameters. • As a pair of LP problems, one problem is called the primal problem, and the other is the dual problem. • A fundamental relationship between the primal and dual problem is that if the primal problem has a finite optimal (bounded) solutions, then the dual problem has it too, and both optimal objective values are the same.

  3. Example… The dual problem Minimize Z = 1200y1 + 3000y2 + 3600y3 subject to y1 + 2y2 + y3 3 y1 + 3y2 + 4y3 4 with y1 , y2 , y3 0 The primal problem Maximize Z = 3x1 + 4x2 subject to x1 + x2 1200 2x1 + 3x2 3000 x1 + 4x2 3600 with x1, x2 0

  4. specific relationship… • The variables in the dual problem (y1, y2, and y3) are associated with the individual constraint in the primal problem. • The RHS value of the constraint in the primal become the objective function coefficients in the dual. • The objective function coefficients in the primal problem become the RHS value for the constraints in the dual problem. • The constraints coefficients for a single variable in the primal become the constraint coefficients for the corresponding constraint in the dual problem. • It is assumed that all constraints in the primal problem are [] and [] for the dual problem. All the variables in both problem are nonnegative, and that the objective is to maximize, and minimize in primal and dual problem, respectively.

  5. The primal problem Maximize The dual problem Minimize

  6. The primal and dual solutions

  7. Additional Primal/Dual Relationship…(mixed constraint) Maximize Z = 20x1 + 30x2 subject to x1 + 2x2 400 2x1 + 15x2  1000 3x1 + x2 = 500 with x1, x2  0 Mathematically the 2nd and 3rd constraints can be rewritten in an equivalent form. Initially 2nd constraint is modified by multiplying both sides of the constraint by -1 - 2x1 - 15x2  -1000

  8. The 3rd constraint can be replaced by two inequalities 3x1 + x2  500 3x1 + x2  500 The second condition can be modified by multiplying by -1 as before. -3x1 - x2  -500 Combining those modification, the primal problem can be rewritten as follows: Maximize Z = 20x1 + 30x2 subject to x1 + 2x2 400 - 2x1 - 15x2  -1000 3x1 + x2  500 -3x1 - x2  -500 with x1, x2  0

  9. and The Dual Problem Minimize Z = 400y1 – 1000y2 + 500(y3-y4) subject to y1 - 2y2 + 3(y3 - y4)  20 2 y1 - 15y2 + (y3 - y4)  30 with y1, y2, y3, y4  0 Since it is found that the difference (y3 - y4) appears in both constraints and the objective function as well, then it can be represented with a single variable that is unconstrained in sign.

  10. Here, Minimize Z = 400y1 – 1000y2 + 500y3 subject to y1 - 2y2 + 3y3  20 2 y1 - 15y2 + y3  30 with y1, y2  0, y3 unconstrained in sign. The general relationship is that if the primal problem has an equality constraint, the corresponding variable in the dual is unconstrained in sign

  11. Manfaat dual • Informasi yang dihasilkan, yaitu tentang sumber-sumber model. Seringkali manajer tidak terlalu menaruh perhatian pada laba bersih tetapi lebih pada penggunaan sumber-sumber daripada akumulasi laba

  12. SENSITIVITY ANALYSIS • The initial coefficient matrix x1 x2 S1 S2 S3 • The basis matrix S3x1 x2

  13. The inverse of matrix B S1S2 S2 RHS Optimal RHS B-1B = I x1 x2 S1 S2 S3

  14. Changes in the RHS values Note : if the change is the first constraint RHS from 1200 to 1200 + 1, the only impact on the optimal tableau is in the solution column.The basis will not change as long as the basic variable remain greater than or equal to :S3 = 600 + 5  0x1 = 600 + 3  0 x2 = 600 - 2  0

  15. -120    300 • Or   -120   -200   300 The range for b1 is then 1200 – 120  b1  1200 + 300 or 1080  b1  1500 Suppose the actual number of resource b1 available is 1050, then S3 = 600 + 5(-150) = -150 x1 = 600 + 3(-150) = 150 x2 = 600 - 2(-150) = 900 This solution is not feasible, need an operation with the simplex method

  16. Important to note that the availability of 1st resource decreased by 150 while the objective decreased by 200. The shadow price of 1st resource is 1 only if the availability of that resource is restricted to be between 1080 unit and 1500 unit. For 2nd resource, the changing in its value will result as -600    200 or 2400  b2  3200, for and 3rd resource will be   -600 or the availability of this resource can be reduced by 600 unit without changing the optimal basis. Variable S3 = 600 implies that we already have 600 unit (minutes) of precision adjustment time unused.

  17. Changes in coefficients of the objective function • For the change of the objective function coefficient c1to c1+, the changes in the optimal simplex tableau are as follows:

  18. The solution in the above table is remain optimal as long as -1 - 3  0 -1 +   0 or -1/3    1 or 8/3  c1  4 . Here, the objective coefficient could be underestimated by as much as $1 and overestimated by as much as $1/3 without changing the optimal basis. However, is c1 = 2 ( = -1), c1-Z1 = +2 for variable S1an there will be a basis change. It can be verified by performing one additional pivot with the simplex method, and the new optimal solution would be x1 = 240, x2 = 840, while the objective value is 3840. Similarly if c2 is replaced by c2 +  the condition is satisfied by -1    ½ or 3  c2  9/2. For example if c2 = 7/2 the only change in the solution will be Z = 4200 + 600 (-½ ) = 3900.

  19. Summary of sensitivity analysis for all objective function coefficients

  20. The addition of a constraint • For example, a constraint is added x1 + 1.75x21400 Look at previous optimal simplex tableau 5S1– 3S2 + S3 = 600 x1 + 3S1 - S2 = 600 x2 - 2S1 + S2 = 600 (add the new constraint) x1 + 1.75x2 + S4 = 1400

  21. When a constraint must be added after the optimal solution is computed, that constraint cannot generally be added to the optimal simplex tableau without some addition calculations. -1(x1 + 3S1 - S2 = 600) x1 + 1.75x2+ S4 = 1400 1.75x2 - 3S1 + S2 + S4 = 800 and -1.75( x2– 2S1 + S2 = 600) 1.75x2 - 3S1 + S2 + S4 = 800 0.5S1– 0.75 S2 + S4 = -250 The result of this calculations is that S3is basic in the first equation; x1 is basic in the second equation, x2 is basic in the third equation, and S4 is basic in the fourth equation. However, this solution is not feasible S4 = -250, thus need more iteration within simplex method.

  22. Adding a variable For example: Maximize Z = 3x1 + 4x2+ 2.5x3 subject to x1 + x2+ x3 1200 2x1 + 3x2+ x3 3000 x1 + 4x2+ x3 3600 with x1, x2, x3 0 The solution is first incorporating decision variable x3into the optimal tableau as a non basic variable.

  23. The constraint coefficient for x3in optimal tableau also can be calculated by