1

Chapter 8Chapter 8

Sensitivity AnalysisSensitivity AnalysisChapter 8Chapter 8

Sensitivity AnalysisSensitivity Analysis

Bottom line:Bottom line: How does the optimal solution change as

some of the elements of the model change?

For obvious reasons we shall focus on For obvious reasons we shall focus on Linear Programming Models.Models.

2

Ingredients of LP ModelsIngredients of LP ModelsIngredients of LP ModelsIngredients of LP Models

Linear objective functionLinear objective function A system of linear constraintsA system of linear constraints

– RHS valuesRHS values

– Coefficient matrix (LHS)Coefficient matrix (LHS)

– Signs (=, <=, >=)Signs (=, <=, >=) How does the optimal solution change as

these elements change?

3

Parametric ChangesParametric ChangesParametric ChangesParametric Changes

Changes in one or more of the coefficients Changes in one or more of the coefficients of the objective function (cof the objective function (c jj))

4

8.4.3 Changes in the 8.4.3 Changes in the elements of the cost vector, elements of the cost vector,

c.c.

8.4.3 Changes in the 8.4.3 Changes in the elements of the cost vector, elements of the cost vector,

c.c. Suppose that the value of cSuppose that the value of ckk changes for some k. changes for some k.

How will this affect the optimal solution to the How will this affect the optimal solution to the LP problem?LP problem?

We therefore can distinguish between two cases: We therefore can distinguish between two cases:

(1) x(1) xkk is not in the old basis is not in the old basis

(2) x(2) xkk is in the old basis is in the old basisr

j= c

BB

− 1

D, j

− cj

, j = 1 , 2 , . . . , n

5

Case 1: xCase 1: xkk is not in the old is not in the old

basisbasis

Case 1: xCase 1: xkk is not in the old is not in the old

basisbasis

ThusThus

Recipe::

rrkk >= >= , if opt=max , if opt=max

rrkk <= <= , if opt = min , if opt = min

r 'k

= cBB

− 1

D, k

− ( ck

+ ) = ( cB

B− 1

D, k

− c ) − = rk

−

6

Case 2: xCase 2: xkk is in the old basis is in the old basisCase 2: xCase 2: xkk is in the old basis is in the old basis

c 'B

= cB

+ ep

r ' = c 'B

B

− 1

D − c ' = ( cB

+ δ ep

) B

− 1

D − ( c + δ ek

)

= cBB

− 1

D − c + δ ( epB

− 1

D − ek

)

= r + δ ( epB

− 1

D − ek

)

r ' = r + δ ( t

p •− e

k)

t

p •: = e

pB

− 1

D

7

ObservationsObservationsObservationsObservations

r’r’jj = 0 for basic variables x = 0 for basic variables x jj..

(e(ekk))jj = 0 for all nonbasic variables x = 0 for all nonbasic variables x jj..

if opt=max all the old reduced costs are if opt=max all the old reduced costs are non-negativenon-negative

if opt=min all the old reduced costs are non-if opt=min all the old reduced costs are non-positive.positive.

8

8.4.3 Example8.4.3 Example8.4.3 Example8.4.3 Example

Suppose that the reduced costs in the final simplex Suppose that the reduced costs in the final simplex tableau are as follows:tableau are as follows:

r = (0,0,0,2 3 4)r = (0,0,0,2 3 4)

with Iwith IBB=(2,3,1), namely with x=(2,3,1), namely with x2,x,x3 and x and x1 comprising comprising

the basis.the basis.

What would happen if we change What would happen if we change the value of ce value of c44 ? ?

First we observe that xFirst we observe that x44 is not in the basis (why?) is not in the basis (why?)

and that the opt=maxand that the opt=max

9

The recipe for this case, namely (8.20) is The recipe for this case, namely (8.20) is that the old optimal solution remains that the old optimal solution remains optimal as long as roptimal as long as r44 ≥ ≥ , or in our case, 2 ≥ , or in our case, 2 ≥

. . Note that we do not need to know the Note that we do not need to know the

current (old) value of ccurrent (old) value of c4 4 to reach this to reach this

conclusion.conclusion. Next, suppose that consider changes in cNext, suppose that consider changes in c11, ,

recalling that xrecalling that x11 is in the basis. is in the basis.

10

Preliminary AnalysisPreliminary AnalysisPreliminary AnalysisPreliminary Analysis

We see that in order to analyse this case we have We see that in order to analyse this case we have to know the entries in the row of the final to know the entries in the row of the final tableau which is represented by xtableau which is represented by x11 in the basis in the basis

(t(tp.p.).).

What is the value of p?What is the value of p? Since ISince IBB=(2,3,1), this is row p=3.=(2,3,1), this is row p=3.

Suppose that this row is as follows:Suppose that this row is as follows: tt3.3. = (0,0,1,3,-4,0) = (0,0,1,3,-4,0)

11

We can display this in a We can display this in a "tableau" form as follows:"tableau" form as follows:We can display this in a We can display this in a

"tableau" form as follows:"tableau" form as follows:B V E q # x 1 x 2 x 3 x 4 x 5 x 6

x 1 3 1 0 0 3 - 4 0

Z 0 0 0 2 3 4

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

B V E q # x 1 x 2 x 3 x 4 x 5 x 6

x 1 3 1 0 0 3 - 4 0

Z − 0 0 2 3 4

So we now have to restore the canonical form of the x1 column.

correction

12

end result ....end result ....end result ....end result ....

To ensure that the current basis remains optimal we To ensure that the current basis remains optimal we have to make sure that all the reduced costs are non-have to make sure that all the reduced costs are non-negative (opt=max). Hence,negative (opt=max). Hence,

2+32+3 ≥ 0 and 3-4 ≥ 0 and 3-4 ≥ 0 ≥ 0 Thus,Thus,

3/4 ≥ 3/4 ≥ ≥ -2/3 ≥ -2/3

B V E q # x 1 x 2 x 3 x 4 x 5 x 6

x 1 3 1 0 0 3 - 4 0

Z 0 0 0 2 + 3 3 - 4 4

13

in words, ....in words, ....in words, ....in words, ....

the old optimal solution will remain optimal the old optimal solution will remain optimal if we keep the increase in cif we keep the increase in c11 in the interval in the interval

[-2/3, 3/4]. If [-2/3, 3/4]. If is too small it will be better is too small it will be better to enter xto enter x44 into the basis, if into the basis, if is too large it is too large it

will better to put xwill better to put x55 into the basis. into the basis.