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15 / 63

50 0

x s2

2

20 60

1/2 5/2

1 0

1/2 -3/2

0 1

z

j

1,000

25

50

25

0

c j - z j

15

0

  • -

    25

0

The new tableau pivot row (x1) in the third simplex tableau is computed using the same formula used previously. Thus, all old pivot row values are divided through by 5/2, the pivot number. These values are shown in Table A-16. The values for the other row (x2) are computed as shown in Table A-15.

Basic

40

50

0

0

Variables

Quantity

x1

x2

s1

s2

Table A-14

The Simplex Method

A-15

The Pivot Row, Pivot Column, and Pivot Number

c j

Table A-15

C o m p u t a t i o n o f t h e x 2 R o w f o the Third Simplex Tableau r

Column

Quantity x x s s 1 2 1 2

20

-

(1/2

*

24)

=

8

1/2

-

(1/2

*

1)

=

0

1

-

(1/2

*

0)

=

1

1/2

-

(1/2

*

-3/5)

=

4/5

0

-

(1/2

*

2/5)

=

-1/5

Old Tableau Row Value

-

Coefficients in P Corresponding Pivot Column

*

New Tableau Pivot Row Value

Q

=

New Tableau Row Value

T h e s e n e w r o w v a l u e s , a s w e l l a s t h e n e w r o w a n d r o w , a r e s h o w n i n t h e c o m - c j - z pleted third simplex tableau in Table A-16. j z j

Table A-16

The Completed Third Simplex Tableau

c j

Basic

40

50

0

0

Variables

Quantity

x1

x2

s1

s2

50 40

x2 x 1

8

0

1

4/5

-1/5

24

1

0

-3/5

2/5

j z

1,360

40

50

16

6

j j c -z

0

0

-16

-6

v The solution is optimal when all a l u e s 0 . c j - z j

O b s e r v i n g t h e r o w t o d e t e r m i n e t h e e n t e r i n g v a r i a b l e , w e s e e t h a t a n o n b a s i c v a r i a b l e w o u l d n o t r e s u l t i n a p o s i t i v e n e t i n c r e a s e i n p r o f i t , a s a l l v a l u e s i n t h e r o w c j - z j c are zero or negative. This means that the optimal solution has been reached. The solution is j - z j

x 2 = 8 m u g s x 1 = 2 4 b o w Z = $1,360 profit l s

which corresponds to point B in Figure A-1.

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