Chapter 8 Sensitivity Analysis

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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.

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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?

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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

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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.

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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.

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it is time for a concrete example!
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Question?

• Suppose that the RHS value of the second
  constraint is changing.
• What kind of changes should be expect in the
  optimal solution?

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Geometric Analysis
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Back to algebra ....
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• How do we conduct such an analysis in higher
  dimensions?

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8.2 The Basic Principles

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

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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

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• 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?).

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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!)
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Observation

• Changes in vector b are manifested in changes in
  the RHS values of the final simplex tableau.

b B-1 b
(feasibility)
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• Structural changes, as well as changes in the
  coefficient matrix (A), may require restoration
  of the canonical form of the simplex tableau.

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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

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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).

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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.

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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 ....... ???

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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.

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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.

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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.

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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.

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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).

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• 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

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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.

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This produces ....
correction
min

• Dont take it seriously .....

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8.4.2 Example
k2
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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 ?.

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Direct Approach (NILN)
k2
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New final RHS B-1 b
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• 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)

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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

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Case 1 xk is not in the old basis

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

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Case 2 xk is in the old basis
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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.

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Recipe
if optmax
if optmin
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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

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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?)

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• 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.

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recipe for this case

• However, it will be instructive to apply the
  basic ideas directly! Let's do it.

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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)

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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.
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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

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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.

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• 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?

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• 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.