Engineering Optimization: Theory and Practice, Fourth Edition

4.2 Revised Simplex Method 191

Table 4.10 Tableau at the Beginning of Cycle 2

Columns of the original canonical form

Basic variables x 4 x 5 x 6 −f

Value of the basic

variable xsa

x 2 54 14 0 0 4

x 3 14 14 0 0 2

x 6 −^64 −^24110

−f^114340110

aThis column is blank at the beginning of cycle 2.

Here

b 5

a 53

=

8

4

= 2

b 6

a 63

=

4

2

= 2

Since there is a tie betweenx 5 andx 6 , we selectxr=x 5 arbitrarily.
Step 7To bringx 3 into the basic set in place ofx 5 , pivot onars=a 53 in Table 4.9.
Enterthe result as shown in Table 4.10, keeping its last column blank. Since a
new cycle has to be started, we go to step 3.
Step 3The simplex multipliers are given by the negative values of the numbers appear-
ing in the row of−fin Table 4.10. Therefore,π 1 = −^114 , π 2 = −^34 , andπ 3 =. 0
The relative cost factors are given by
cj=cj= −πTAj

Then

c 1 =c 1 −π 1 a 11 −π 2 a 21 = − 1 −(−^114 )( 2 )−(−^34 )( 2 )= 6

c 2 =c 2 −π 1 a 12 −π 2 a 22 = − 2 −(−^114 )( 1 )−(−^34 )(− 1 )= 0

c 3 =c 3 −π 1 a 13 −π 2 a 23 = − 1 −(−^114 )(− 1 )−(−^34 )( 5 )= 0

c 4 =c 4 −π 1 a 14 −π 2 a 24 = 0 −(−^114 )( 1 )−(−^34 )( 0 )=^114

c 5 =c 5 −π 1 a 15 −π 2 a 25 = 0 −(−^114 )( 0 )−(−^34 )( 1 )=^34

c 6 =c 6 −π 1 a 16 −π 2 a 26 = 0 −(−^114 )( 0 )−(−^34 )( 0 )= 0

These values are entered as third row in Table 4.8.
Step 4Since allcjare ≥ 0 , the present solution will be optimum. Hence the optimum
solution is given by
x 2 = 4 , x 3 = 2 ,x 6 = 0 (basic variables)
x 1 =x 4 =x 5 = (nonbasic variables) 0

fmin= − 10