Principles of Mathematics in Operations Research

254 Solutions

0

0

0

<

Si

«2

S3

=

20

12

11

1

1

0

a < 6 — 1 = 5(bounds of x\)

=>a = min{20,12,5} = 5.

xi leaves immediately at its upper bound, xi = 6.

2. B = I,cB = T (2,3,1,4), xTB = (15,7,11) z = 12 + 0 + 3 + 0 = 15,
   CN ~ CgB N = (2, 3,1,4). Then, Bland's rule marks the second variable.
   Since the reduced cost of x 2 is positive and x 2 is at its lower bound; as x 2
   is increased, so is z. Hence, x 2 enters.

0

0

0

<

Si

Si

S3

=

15

7

11

2

1

0

a < 10 - 0 = 10(bounds of x 2 )

=> a = mini —-,7,10 > = 7.

Thus, S2 leaves.

3.

B= {si,x 2 ,s 3 } =4> B -

120

0 10

001

B'

"1

0

0

-2 0"

10

01

XB =

Si

X2

.S3.

=

"1 -2 0"

0 10

0 0 1

"301

13

20

"1 -20"

0 10

0 01

"1035"

1 100

0034

"6"

0

3

0

4

13

20

2 = (0,3,0)

-1 -2 3 5'

1 100

0 034

CO

O CO O

4

13

20

—

3

6

9

=

1

7

11

1

7

11

+ (2,0,1,4) = 21 + (12 + 3) = 36.

cTN-cTBB~lN = (2,0,1,4) -(0,3,0)

-2 3 5

100

034

= (2,0,1,4)-(3,3,0,0) = (-1,-3,1,4),