單純形法求解max z 4X1 3X2 6X3 S T 3X1 X2 3X3 30 2X1 2X2 3X3 40 X1,X2,X

時間 2021-09-17 08:46:08

1樓:匿名使用者

3x1+x2+3x3≤30 (1')

2x1+2x2+3x3≤40 (2')

x1≥0 (3')

x2≥0 (4')

x3≥0 (5')

3x1+x2+3x3=30 (1)

2x1+2x2+3x3=40 (2)

x1=0 (3)

x2=0 (4)

x3=0 (5)

case 1:

from (1) ,(2), (3), x1=0

x2 +3x3=30

2x2+3x3=40

x2 =10, x3= 20/3

satisfy (4') and (5')

(x1,x2,x3) = (0,10, 20/3)

z =4x1+3x2+6x3

=30+40

=70case 2:

from (1) ,(2), (4), x2=0

3x1+3x3=30

2x1+3x3=40

x1 =-10, does not satisfy (3')

rejected case 2

case 3:

from (1) ,(2), (5), x3=0

3x1+x2=30

2x1+2x2=40

x1=5, x2=15

satisfy (3') and (4')

(x1,x2,x3) = (5,15, 0)

z =4x1+3x2+6x3

=20+45

=65case 4:

from (1) ,(3), (4), x1=x2=0

3x1+x2+3x3=30

x3=10

2x1+2x2+3x3≤40 (2')

satisfy (2') and (5')

(x1,x2,x3) = (0,0, 10)

z =4x1+3x2+6x3

=60case 5:

from (1) ,(3), (5), x1=x3=0

3x1+x2+3x3=30

x2=30

2x1+2x2+3x3≤40 (2')

does not satisfy (2')

rejected case 5

case 6:

from (1) ,(4), (5), x2=x3=0

3x1+x2+3x3=30

x1=10

2x1+2x2+3x3≤40 (2')

satisfy (2') and (3')

(x1,x2,x3) = (10,0, 0)

z =4x1+3x2+6x3

=40case 7:

from (2) ,(3), (4), x1=x2=0

2x1+2x2+3x3=40

x3=40/3

3x1+x2+3x3≤30 (1')

does not satisfy (1')

rejected case 7

case 8:

from (2) ,(3), (5), x1=x3=0

2x1+2x2+3x3=40

x2=20

3x1+x2+3x3≤30 (1')

satisfy (1') and (4')

(x1,x2,x3) = ( 0,20,0)

z =4x1+3x2+6x3

=60case 9:

from (2) ,(4), (5), x2=x3=0

2x1+2x2+3x3=40

x1=20

3x1+x2+3x3≤30 (1')

does not satisfy (1')

rejected case 9

ie max z = case 1=70

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