sensitivity analysis in linear programming

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
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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
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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
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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
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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
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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
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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.
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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.
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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
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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.
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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*.
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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*.
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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
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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).
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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)
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20. Thank You
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