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sensitivity analysis in linear programming

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- 1. SENSITIVITY ANALYSIS IN LINEAR PROGRAMMING PRESENTED BY KIRAN C JADHAV GOVERNMENT COLLEGE OF ENGINEERING, AURANGABAD (An Autonomous Institute Of Government Of Maharashtra) DEPARTMENT OF CIVIL ENGINEERING 14/17/2017
- 2. CONTENT • Introduction • Basic parameter in Sensitivity Analysis • Duality and sensitivity analysis • Example of duality and sensitivity analysis • References 24/17/2017
- 3. Definition of Sensitivity Analysis/post optimality • Sensitivity analysis investigates the changes in the optimum solution resulting from changes in parameters of linear programming model . • In linear programming modal parameters are I ) objective function II ) constraint coefficients 34/17/2017
- 4. BASIC PARAMETER CHANGES THAT AFFECT THE OPTIMAL SOLUTION ARE : • Changes in right –hand – side constants • Changes in cost coefficients • Addition of new variables • Addition of new constraints 44/17/2017
- 5. DUALITY AND SENSITIVITY ANALYSIS • No learning of linear programming is complete unless we learn the concept of “duality” in linear programming • Associated with every linear programming problem called the primal, there is another linear programming called its dual . • These two problems possess very interesting and closely related properties 54/17/2017
- 6. • In most LP treatment, the dual is defined for various form of the primal depending on the sense of optimization (maximization or minimization) / types of constraints 64/17/2017
- 7. Source : NPTEL Linear programming lecture pdf74/17/2017
- 8. • we formulated and solved the LP problem to maximize the revenue for the bakery. The problem is to , Source : NPTEL Linear programming lecture pdf 84/17/2017
- 9. • Now considering the maximization problem (called P1) and the minimization problem (called P2) we make the following observations: 1 . P1 and P2 had the same value of the objective function at the optimum z = 374 and w =374 2. The objective function coefficient coefficients of P1 are the RHS values of P2 and vice versa. Source : NPTEL Linear programming lecture pdf 94/17/2017
- 10. 3. The number of variables in P1 and the number of constraints in P2 are equal and vice versa. Source : NPTEL Linear programming lecture pdf 104/17/2017
- 11. Is there a relationship between P1 and P2? • The relationship is established using the discussion that follows : • Consider P1 (the maximization problem). If there were no constraints the objective function value is ∞. • Let us try to get upper estimates for the value of Z. 114/17/2017
- 12. 1. Consider P1 (the maximization problem). If there were no constraints the objective function value is ∞. Let us try to get upper estimates for the value of Z. 2. We multiply the second constraint by 9 to get (4X1 + 3X2 ≤ 46 ) X ( 9 ) 36X1 + 27X2 ≤ 414. Since X1 and X2 are ≥ 0, 32X1 + 25X2 ≤ 36X1 + 27X2 ≤ 414. Therefore Z* ≤ 414. 124/17/2017 Source : NPTEL Linear programming lecture pdf
- 13. 3) We multiply the first constraint by 7 to get (5X1 + 4X2 ≤ 59 ) X ( 7 ) 35X1 + 28X2 ≤ 413 Since X1 and X2 are ≥ 0, 32X1 + 25X2 ≤ 35X1 + 28X2 ≤ 413. Therefore Z* ≤ 413 4) We add the two constraints to get , 4X1 + 3X2 ≤ 46 5X1 + 4X2 ≤ 59 9X1 + 7X2 ≤ 105 134/17/2017
- 14. • We add the two constraints to get 9X1 + 7X2 ≤ 105. • This inequality holds because X1 and X2 are ≥ 0, • (9X1 + 7X2 ≤ 105). x (4 ) • We multiply this constraint by 4 to get 36X1 + 28X2 ≤ 420. • Since X1 and X2 are ≥ 0, 32X1 + 25X2 ≤ 36X1 + 28X2 ≤ 420. • Therefore 420 is an upper estimate of Z* but we ignore this because our current best estimate is 413. 144/17/2017
- 15. 5) We multiply the constraint 4X1 + 3X2 ≤ 46 by 25/3 to get • 33.33X1 + 25X2 ≤ 383.33. • Based on the above discussions, 383.33 is a better upper estimate for Z*. • We multiply the constraint 5X1 + 4X2 ≤ 59 by 32/5 to get 32X1 + 25.6X2 ≤ 377.6. Based on the above discussions, 377.6 is a better upper estimate for Z* 6) We multiply the constraint 5X1 + 4X2 ≤ 59 by 32/5 to get • 32X1 + 25.6X2 ≤ 377.6. • Based on the above discussions, 377.6 is a better upper estimate for Z*. 154/17/2017
- 16. 6) We multiply the constraint 5X1 + 4X2 ≤ 59 by 32/5 to get • 32X1 + 25.6X2 ≤ 377.6. • Based on the above discussions, 377.6 is a better upper estimate for Z*. 7) We multiply the constraint 9X1 + 7X2 ≤ 105 by 25/7 to get • 32.14X1 + 25X2 ≤ 375. • Here we have added the two constraints and multiplied by 25/7. • Now 375 is a better upper estimate for Z*. 164/17/2017
- 17. • We can multiply the first constraint by a and the second by b and add them. If on addition the coefficients of X1 and X2 are ≥ 59 and 46 respectively, the RHS value (which is 59a + 46b) is an upper estimate of Z • If we want the best estimate of Z* (as small an upper estimate as possible) we need to define a and b such that 59a + 46b is as small as possible. • We therefore define a and b to Minimize 59a + 46b is minimized 174/17/2017
- 18. • The values Y1 = 4 and Y2 =3 represent the worth of the 59 units of the first resource and 46 units of the second resource at the optimum. • Therefore Problem P2 Is Born Out Of P1. The Problem P2 Is Called The Dual Of The Given Problem P1 (Called The Primal). 184/17/2017
- 19. References • Taha H.A., Operations Research – An Introduction, 9 th Edition, Pearson Education Inc. 2011 • Rao S.S., Engineering Optimization – Theory and Practice, Third Edition, New Age International Limited, New Delhi, 2000 • IIT Madras Operations Research Applications – Linear and Integer Programming (Web), by Prof. G. Srinivasan The National Programme on Technology Enhanced Learning (NPTEL) 194/17/2017
- 20. Thank You 4/17/2017 20

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