sensitivity analysis

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Sensitivity Analysis or Post Optimal Analysis Lekshmi Krishna M.R(100609)

Renjini R(100611)MTECH – Technology ManagementDepartment of Futures StudiesUniversity of Kerala

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Thanks to Prof Sreenivasan(IIT Chennai) for his lecture series in Operations Research!!

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Introduction

Common theme of OR – search of optimal solution

An optimal solution for the original model may vary from ideal for the real problem

Additional analysis – Post optimal analysis

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Post Optimal Analysis

Post Optimal Analysis - Analysis done after finding an optimal solution

Some times called What-if analysis

Addressing questions about what would happen to the optimal solution if

different assumptions are made about future conditions

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Cont……

In part post optimality analysis involves conducting sensitivity analysis

To determine which parameter of the model are most critical in determining the solution - Sensitive parameter

Sensitive parametersThe parameters whose values cannot be changed

without changing the optimal solution

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Sensitivity Analysis

Involves the effect on the optimal solution making changes in values of model parameters aij, bi & cj

Changing parameter in primal also effect corresponding values in dual problem

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Procedure for sensitivity analysis

REVISION OF MODEL

REVISION OF FINALTABLEAU

CONVERSION TO PROPER FORM

FEASIBILITY TEST

OPTIMALITY TEST

REOPTIMIZATION

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Different case to do Sensitivity AnalysisCase 1: Changes in basic variable

of objective function

case 2: Changes in the non basic variable of objective function

Case 3 : Change in RHS

Case 4 : Change in the coefficients of constrains

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Consider a LPP

Objective function : Max 4X1 +3X2 + 5X3

Sub to X1+2X2+3X3 ≤ 9

2X1+3X2+X3 ≤ 12

X1,X2,X3 ≥ 0

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Simplex table4 3 5 0 0

B Cb X1 X2 X3 X4 X5 XB 0

X4 0 1 2 3 1 0 9 3

X5 0 2 3 1 0 1 12 12CJ-ZJ 4 3 5 0 0 0

X3 5 1/3 2/3 1 1/3 0 3 0

X5 0 5/3 7/3 0 -1/3 1 9 27/5CJ-ZJ 7/3 -1/3 0 -5/3 0 15

X3 5 0 1/5 1 2/5 -1/5 6/5

X1 4 1 7/5 0 -1/5 3/5 27/5CJ-ZJ 0 -18/5 0 -6/5 -7/5 138/5

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Solution : X1 = 27/5

X3 = 6/5

Z = 138/5

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CASE 1 : Change in objective functionCoefficient of non-basic variable Max Z = 4X1+ 3X2 + 5X3 + 0S1 + 0S2

Lets take X2

Effect in optimality 1. RHS = -ve

2. Cj-Zj = +ve

Here only change in C2 – Z2

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C2 – Z2 = C2 – ( 1+28/5)

= C2 – 33/5 ≤ 0 If C2 > 33/5 then C2-Z2 becomes +ve & C2

becomes incoming row again and need to do

iteration till we get the optimum solution

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4 7 5 0 0

B Cb X1 X2 X3 X4 X5 XB 0

X3 5 0 1/5 1 2/5 -1/5 6/5 6

X1 4 1 7/5 0 -1/5 3/5 27/5 27/7

CJ-ZJ 0 2/5 0 -6/5 -7/5 138/5

X3 5 -1/7 0 1 3/7 -2/7 3/7

X2 7 5/7 1 0 -1/7 3/7 27/7

Cj-Zj -2/7 0 0 -8/7 -11/7 204/7

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Solution : X2 = 27/7

X3 = 3/7

Z = 204/7

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Case 2 : Change in objective function coefficient of basic variable Max Z = 4X1+ 3X2 + 5X3 + 0S1 + 0S2

Lets take C1

Effect in optimality 1. RHS = -ve

2. Cj-Zj = +ve

Here only change in Cj-Zj of all the values

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C1 3 5 0 0

B Cb X1 X2 X3 X4 X5 RHS

5 X3 0 1/5 1 2/5 -1/5 6/5

C1 X1 1 7/5 0 -1/5 3/5 27/5

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Evaluation C2-Z2

C2 –Z2 =

3 - (1+7C1/3)

3 -7C1/5 ≤ 0

C1 ≥ 10/7

If C1 < 10/7

(X2 enters)

C4-Z4

C4 – Z4 =

0-(2 – C1/5)

C1/5 – 2 ≤ 0

C1 ≤ 10

If C1 > 10

(X4 enters)

C5-Z5

C5 - Z5 =

0 – (-1+3C1/5)

1-3C1/5 ≤ 0

C1 ≥ 5/3

If C1 < 5/3

(X5 enters)

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Suppose C1 = 12

b CB X1 X2 X3 X4 X5 XB 0

X3 5 0 1/5 1 2/5 -1/5 6/5 3

X1 12 1 7/5 0 -1/5 3/5 27/5

Cj-Zj 0 - 74/5 0 2/5 -31/5 354/5

X4 0 0 3/2 5/2 1 -1/2 3

X1 12 1 3/2 1/2 0 1/3 6

Cj-Zj 0 -15 -1 0 -6 72

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Solution : X1 = 6

X4 = 3

Z = 72

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Case:3 Right Hand Side ChangesObjective function : Max 4X1 +3X2 + 5X3

Sub to X1+2X2+3X3 ≤ 9

2X1+3X2+X3 ≤ 12

X1,X2,X3 ≥ 0

b = 9

12

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Effect of that change only in RHS RHS = B b

B = 2/5 - 1/5 b1

-1/5 3/5 12

= (2b1-12)/5

(36-b1)/5

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Evaluation

(2b1-12)/5 ≥ 0

2b1-12 ≥ 0

b1 ≥ 6

(36-b1)/5 ≥ 0

b1 ≤ 36

Hence the range of b1 should be

6 ≤ b1≤ 36

for the solution to be optimum

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Suppose b1 =40

RHS = (2b1-12)/5

(36-b1)/5

= 68/5

-4/5

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Substituting the new RHS values

4 3 5 0 0

b Cb X1 X2 X3 X4 X5 RHS

X3 5 0 1/5 1 2/5 -1/5 68/5

X1 4 1 7/5 0 -1/5 3/5 -4/5

Cj-Zj 0 -18/5 0 -6/5 -7/5 138/5

Primal infeasible

Here dual is feasible .Hence calculate 0 to do dual simplex iteration

0 - 6 -

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Evaluate by taking X4 comes in and X1 goes out

4 3 5 0 0

b Cb X1 X2 X3 X4 X5 Xb

X3 5 1 3 1 0 1 12

X4 0 -5 -7 0 1 -1 4

Cj-Zj -1 -9 0 0 -1 48

Solution X3=12 X4=4

Z =48

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Case 4 : Change in the constrain coefficient of a non basic variable Objective function : Max 4X1 +3X2 + 5X3

Sub to X1+2X2+3X3 ≤ 9

2X+ 3X2+X3 ≤ 12

X1,X2,X3 ≥ 0

P2 = 2 a

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Evaluation

P2 = B P2

= 2/5 -1/5 2

-1/5 3/5 a

= (4-a)/5

(3a-2)/5

Here C2-Z2 changes

C2-Z2

= C2 – y P2

= 3 – [6/5 7/5] 2

a

= (3 – 7a)/5

a= 0(3-7a)/5

C2-Z2 =3/5

P2 = 4/5 -2/5

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New simplex table4 3 5 0 0

b Cb X1 X2 X3 X4 X5 Xb 0

X3 5 0 4/5 1 2/5 -1/5 6/5 3/2

X1 4 1 -2/5 0 -1/5 3/5 27/5 -

Cj-Zj 0 3/5 0 -6/5 -7/5 138/5

X2 3 0 1 5/4 1/2 -1/4 3/2

X1 4 1 0 1/2 0 1/2 6

Cj-Zj 0 0 -3/4 -3/2 -3/4 57/2

X1 = 6 ; X2= 3/2 ; Z = 57/2

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Conclusion Sensitivity analysis carried out by

mathematical programming systems are called ranging (RANGE Procedure)

Sensitivity analysis needs to be performed to investigate what happens when the estimates are wrong

Helps to find out the range of likely values of the sensitive parameters

Hence an important part of most linear programming study

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Reference

Introduction to operations research

Frederick S Hillier & Gerald J Lieberman

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Thank you!!!!

lekshmi.uce05@gmail.com