4.5 Sensitivity or Postoptimality Analysis 211Result of pivot operation:
x 3 53 0 1 73 154 − 151 0 8003
x 2 301 1 0 − 301 − 1501 752 0 403−f^2530050322323128 , 3000
The optimum solution is given by
x 2 =40
3
, x 3 =800
3
(basic variables)x 1 =x 4 =x 5 =x 6 = (nonbasic variables) 0fmin=− 28 , 000
3
or maximum profit=$28, 000
3
From the final tableau, one can find that
XB=
{
x 3
x 2}
=
{ 800
3
40
3}
=
vector of basic variables in
the optimum solution(E 1 )
cB={
c 3
c 2}
=
{
− 30
− 100
}
=
vector of original cost
coefficients corresponding
to the basic variables(E 2 )
B=
[
4 10
1 40
]
=
matrix of original coefficients
corresponding to the basic variables(E 3 )
B−^1 =
[
β 33 β 32
β 23 β 22]
=
[ 4
15 −
1
15
− 1501 752]
=
inverse of the coefficient
matrixB, which appears
in the final tableau also(E 4 )
π=cTBB−^1 = (− 30 − 100 )[ 4
15 −
1
15
− 1501 752]
=
{
−^223
−^23
}
=
simplex multipliers, the
negatives of which appear
in the final tableau also(E 5 )
Example 4.6 Find the effect of changing the total time available per day on the two
machines from 1200 and 800 min to 1500 and 1000 min in Example 4.5.
SOLUTION Equation (4.36) gives
xi+∑mj= 1βijbj≥ 0 , i= 1 , 2 ,... , m (4.36)wherexiis the optimum value of theithbasic variable. (This equation assumes that
the variables are renumbered such thatx 1 toxmrepresent the basic variables.)