Engineering Optimization: Theory and Practice, Fourth Edition

4.5 Sensitivity or Postoptimality Analysis 219

Thus Eq. (E 1 ) can be expressed as

2 x 1 + 5 (^403 − 301 x 1 + 301 x 4 + 1501 x 5 − 752 x 6 )

+ 3 (^8003 −^53 x 1 −^73 x 4 − 154 x 5 + 151 x 6 )+ 4 x 4 = 006

that is,

−^196 x 1 −^176 x 4 −^2330 x 5 + 151 x 6 = −^8003 (E 2 )

Step 2Transform this constraint such that the right-hand side becomes positive,
that is,
19
6 x^1 +

17

6 x^4 +

23

30 x^5 −

1

15 x^6 =

800

3 (E^3 )

Step 3Add an artifical variable, say,xk, the new constraint given by Eq. (E 3 ) and the
infeasibility formw=xkinto the original optimum tableau to obtain the new
canonical system as follows:

Basic Variables

variables x 1 x 2 x 3 x 4 x 5 x 6 xk −f −w bi (bi/ais)

x 3 53 0 1^7345 − 151 0 0 0 8003 160

x 2 301 1 0 − 301 − 1501 752 0 0 0 403 400

xk^196 0 0^1762330 − 151 1 0 (^08003160019)
Pivot element
−f^253 0 0^50322323 0 1 0 28 , 3000
−w −^196 0 0 −^176 −^2330151 0 0 1 −^8003
↑
Step 4Eliminate the artificial variable by applying the phase I procedure:
Basic Variables
variables x 1 x 2 x 3 x 4 x 5 x 6 xk −f −w bi
x 3 0 0 1 1916 285113 − 953 −^1019002 , 19400
x 2 0 1 0 − 956 − 4757 47513 − 951 0 0 20019
x 1 1 0 0 1917 9523 − 952 196 0 0 1 , 19600
−f 0 0 0 17519 10119 1619 −^50191016419 ,^000
−w 0 0 0 0 0 0 0 0 1 0
Thus the new optimum solution is given by
x 1 =^160019 , x 2 =^20019 , x 3 =^240019 ( asic variablesb )
x 4 =x 5 =x 6 = 0 (nonbasic variables)
fmin= −

164 , 000

19

and maximum profit=

$164, 000

19