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A-36

Module A

The Simplex Solution Method

b e c o m e s a m u l t i p l e o f t h e c o l u m n v a l u e s w h e n t h e n e w r o w v a l u e s a n d t h e s u b s e q u e n t r o w v a l u e s , a l s o s h o w n i n T a b l e A - 3 4 , a r e c o m p u t e d . c j - z j z j

c

j

Basic Variables

Quantity

160 + ¢ x1

200 x2

200 160 + ¢ 0

x x s

2 1 3

8 4 48

0 1 0

1 0 0

z c -z

j

j

j

2,240 + 4¢

160 + ¢ 0

200 0

1/2 -1>2 6

-1>18 1/9 -2

0 0 1

20 - ¢/2 -20 + ¢/2

20/3 + ¢/9 -20>3 - ¢/9

0 0

0

0

0

s1

s2

s3

Table A-34

The Optimal Simplex Tableau with c1 = 160 + ¢

For the solution to remain optimal all values in the cj - zj row must be 0.

T h e s o l u t i o n s h o w n i n T a b l e A - 3 4 w i l l r e m a i n o p t i m a l a s l o n g a s t h e r o w v a l u e s r e m a i n n e g a t i v e . ( I f b e c o m e s p o s i t i v e , t h e p r o d u c t m i x w i l l c h a n g e , a n d i f i t b e c o m e s c j - z j c zero, there will be an alternative solution.) Thus, for the solution to remain optimal, j - z j

  • -

    20 + ¢>2 0

and

  • -

    20>3 - ¢>9 0

Both of these inequalities must be solved for ¢:

  • -

    20 + ¢>2 0 ¢>2 20 ¢ … 40

and

  • -

    20>3 - ¢>9 0

    • -

      ¢>9 20>3

      • -

        ¢ … 60 ¢ Ú -60

Thus, ¢ 40 and ¢ Ú - 60. Now recall that c1 = 160 + ¢; therefore, ¢ = c1 - 160. Substituting the amount c1 - 160 for ¢ in these inequalities,

¢ … 40 c1 - 160 40 c1 200

and

¢ Ú -60 c1 - 160 Ú -60 c1 Ú 100

Therefore, the range of values for c1 over which the solution basis will remain optimal (although the value of the objective function may change) is

100 c1 200

Next, consider a ¢ change in c2 so that c2 = 200 + ¢. The effect of this change in the final simplex tableau is shown in Table A-35.

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