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PPT – Chapter 8 Sensitivity Analysis PowerPoint presentation | free to download - id: 1e509d-ZDc1Z

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Chapter 8Sensitivity Analysis

- Bottom line
- How does the optimal solution change as some of

the elements of the model change? - For obvious reasons we shall focus on Linear

Programming Models.

Ingredients of LP Models

- Linear objective function
- A system of linear constraints
- RHS values
- Coefficient matrix (LHS)
- Signs (, lt, gt)
- How does the optimal solution change as these

elements change?

8.1 Introduction

- What kind of changes are to be considered?
- How do we handle such changes?
- To motivate the discussion it is instructive to

classify the changes into two categories - Structural changes
- Parametric changes

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Structural Changes

- eg.
- New decision variable
- New constraint
- Loss of a decision variable
- Relaxation of a constraint

Parametric Changes

- Changes in one or more of the coefficients of the

objective function (cj) - Changes in the RHS values (bi)
- Changes in the coefficients of the LHS matrix

aij.

our emphasis will be on

- Basic principles and ideas rather than

investigation of many specific cases (as done in

many textbooks ....) - Thus, in the exam you may have to demonstrate

that you know how to apply the principles to

solve new problems.

it is time for a concrete example!

Question?

- Suppose that the RHS value of the second

constraint is changing. - What kind of changes should be expect in the

optimal solution?

Geometric Analysis

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Back to algebra ....

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- How do we conduct such an analysis in higher

dimensions?

8.2 The Basic Principles

- Impact of changes
- Feasibility
- Optimality
- Terminology
- Old before the change
- New after the change

Two Cases

- The old optimal solution remains optimal
- The old optimal solution remains feasible but is

no longer optimal - The second case has two sub-cases depending on

the impact of the changes on the basis - All the old basic variables remain in the basis
- There is a change in the basis

- The latter means that a number of pivot

operations will be required to construct the new

optimal solution from the old optimal solution. - In the former case, the new optimal solution can

be easily computed from the new RHS values

(why?).

General Observation

- Changes in the objective function coefficients

(c) are manifested in changes in the reduced

costs of the non-basic variables of the (old)

optimal solution.

r cBB-1D - c

(optimality criterion!)

Observation

- Changes in vector b are manifested in changes in

the RHS values of the final simplex tableau.

b B-1 b

(feasibility)

- Structural changes, as well as changes in the

coefficient matrix (A), may require restoration

of the canonical form of the simplex tableau.

In short, ... the tasks are

- Checking whether the new RHS values are

non-negative - Checking the signs of the new reduced costs
- Pivot operations to restore the canonical form of

the simplex tableau

Recipe

- Given the final tableau of the old solution and

the changes to the model, compute the new

tableau. - If the new tableau is not in a canonical form,

restore the canonical form by appropriate pivot

operations. - Check, if necessary, that the RHS values are

non-negative. - Check, if necessary, that the reduced costs are

of the appropriate sign (non-negative for

optmax, non-positive for optmin).

Implications ....

- to accomplish these tasks we need to be able to

quickly compute new RHS and/or reduced costs

resulting from the changes in the problem.

8.3 Overview

- The formulas used by the Revised Simplex Method

suggest what factors should be under

consideration when parameters of the LP model

change - These formulas are constructive in developing

recipes for sensitivity analysis of a number of

important cases. - Thus,
- read again Chapter 6 ....... ???

Case 1 All the components of the new RHS are

non-negative.

- In this case the old optimal solution is still

feasible, namely the changes in the problem do

not have any impact on the feasibility of the

"old" optimal solution. We do not have to take

any "corrective action" as far as feasibility is

concerned. - The new values of the basic variables are simply

equal to the new values of the RHS.

Case 2. At least one of the components of the

new RHS is negative.

- Clearly in this case the "old" optimal solution

is no longer feasible, as the non-negativity

constraint is violated by at least one decision

variable. - Thus, in this case the changes in the parameters

of the model caused the "old" solution to become

infeasible and therefore corrective actions

involving changes in the basis, must be taken.

Case 3. All the components of r satisfy the

optimality condition.

- Since the optimality conditions are satisfied,

(by) the basic variables remain basic, thus no

corrective actions are required. - Observe, however, that the value of x may have

changed due to changes in the vector b.

Case 4. At least one of the components of r

violates the optimality conditions

- Obviously, if we exclude degeneracy (why?), there

is a need to changes the basis itself. - Thus, the changes in the problem results in a new

basis.

8.4 Common Cases

- 8.4.1 Changes in the RHS values, b.
- Suppose that we change one of the elements of b,

say bk, by ?. so that the new b is equal to the

old one except that the new value of bk is equal

to bk?. - In short,
- b(new) b ?ek
- where ek is the kth column of the identity matrix

(i.e. ek(0,0,...0,1,0,...,0) where the 1 is in

the kth position).

- In this case the new RHS value is given by
- b B-1b(new) B-1(b?ek)
- This yields
- b B-1b ?B-1ek B-1b (?B-1).k
- Hence,
- b old RHS (?B-1).k

Recipe

- If the old basis is to remain optimal after the

change occurs, the new RHS value must be

non-negative, i.e bgt0. - This will be the case if we require
- ?(B-1).k gt - old RHS
- or equivalently
- ?(B-1)i,k gt - old RHSi
- for i1,2,...,m.

This produces ....

correction

min

- Dont take it seriously .....

8.4.2 Example

k2

final

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- We therefore conclude that the old basis

(whatever it is), will remain optimal if the

value of b2 is in the interval 20-4,20416,24

. - Comment
- You do not have to use the recipe Compute the

RHS after the change and determine the critical

values of ?.

Direct Approach (NILN)

k2

New final RHS B-1 b

- Thus, the non-negativity constraint requires
- 484? gt 0 , hence ? gt -12
- 164??gt 0 , hence ? gt - 4
- 4 - ? gt 0 , hence ? lt 4
- In short,
- -4 lt ? lt 4
- (16 lt b2 lt 24)

8.4.3 Changes in the elements of the cost

vector, c.

- Suppose that the value of ck changes for some k.

How will this affect the optimal solution to the

LP problem? - We can distinguish between two cases
- (1) xk is not in the old basis
- (2) xk is in the old basis

Case 1 xk is not in the old basis

- Thus
- Recipe
- rk gt ? , if optmax
- rk lt ? , if opt min

Case 2 xk is in the old basis

Observations

- rj 0 for basic variables xj.
- (ek)j 0 for all nonbasic variables xj.
- if optmax all the old reduced costs are

non-negative - if optmin all the old reduced costs are

non-positive.

Recipe

if optmax

if optmin

Remark

- It is very unfortunate that sometime (often?)

mathematical notation tends to obscure the

essential elements of the situation under

investigation. This is a typical example! - As an exercise (on your own) try to translate

this to the language of the simplex tableau

8.4.3 Example

- Suppose that the reduced costs in the final

simplex tableau are as follows - r (0,0,0,2 3 4)
- with IB(2,3,1), namely with x2,x3 and x1

comprising the basis. - What would happen if we change the value of c4 ?
- First we observe that x4 is not in the basis and

that the optmax (why?)

- The recipe for this case, namely (8.20) is that

the old optimal solution remains optimal as long

as r4 ?, or in our case, 2 ?. - Note that we do not need to know the current

(old) value of c4 to reach this conclusion. - Next, suppose that consider changes in c1,

recalling that x1 is in the basis.

recipe for this case

- However, it will be instructive to apply the

basic ideas directly! Let's do it.

Preliminary Analysis

- We see that in order to analyze this case we have

to know the entries in the row of the final

tableau (tp. ) that is represented by x1 in the

basis. - What is the value of p?
- Since IB(2,3,1), this is row p3.
- Suppose that this row is as follows
- t3. (0,0,1,3,-4,0)

We can display this in a "tableau" form as

follows

If we add ? to the old c1, we would have instead

correction

So we now have to restore the canonical form of

the x1 column.

end result ....

- To ensure that the current basis remains optimal

we have to make sure that all the reduced costs

are non-negative (optmax). Hence, - 23? 0 and 3-4? 0
- Thus,
- 3/4 ? -2/3

in words, ....

- the old optimal solution will remain optimal if

we keep the increase in c1 in the interval -2/3,

3/4. If ? is too small it will be better to

enter x4 into the basis, if ? is too large it

will better to put x5 into the basis.

- It is strongly recommended that you consider the

following structural change on your own - A new decision variable is introduced. Its cost

coefficient is given and its constraint

coefficients are given. - How does this affect the given final tableau of

the old problem?

- Warning
- If you are asked about the range of admissible

changes in say b1 then it is not sufficient to

report on the admissible changes in d1 . That is,

you have to translate the range of admissible

changes in d1 into the range of admissible

changes in b1. - Example Suppose that b1 12, and -2 lt d1 lt 3.

Then the admissible range of b1 is 12-2 , 123

10,15.

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