3.1 Lecture_remark.pdf

8/9/2019 3.1 Lecture_remark.pdf

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Remarks on the Simplex Method

Jeff Chak-Fu WONG

Department of Mathematics

Chinese University of Hong Kong

MAT581SS

Mathematics for Logistics

Produced by Jeff Chak-Fu WONG 1


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SIMPLEX METHOD

IMPLEXMETHOD 2


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Consider the LLP problem

Maximise z= 4x1+ 3x2

subject to x1+x2 82x1+x2 10

and x1, x2 0

(1)

IMPLEXMETHOD 3


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Basic Variables: x3 = 8, x4 = 10

Non-Basic Variables: x1 = 0, x2 = 0

Expressing the basic variables in terms of the non-basic variables,

we have

x3 = 8 x1 x2

x4 = 10 2x1 x2(2)

x3 = 8 (0) (0) = 8 0

x4 = 10 2(0) (0) = 10 0

Putting Eq. (2) into the objective function, we have

z(x1, x2, x3, x4) = 4x1+ 3x2+ 0x3+ 0x4

= 4x1+ 3x2+ 0

= 4(0) + 3(0)

= 0.

IMPLEXMETHOD 4


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Clearly,X

(0) = (x1, x2, x3, x4) = (0, 0, 8, 10)

is not an optimal solution since from the economical point of view,

x1 = x2 = 0indicated that no production was made, i.e.,

z(X(0)) = 0.

x

x

2 x + x = 10

x + x = 8

O

2

2

2

1

1

P

Q

R

(0,0)z = 0

1

Figure 1: z(0, 0) = 0

IMPLEXMETHOD 5


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Consider the objective function:

z = 4x1+ 3x2

The main idea is

x1 (0 ?), x2 (0 ?) z(0 ?).

We look for a new B.F.S. from X(0). Then, the value of zwill improve.

Observe that when increasing the values of x1and x2by1unit,

4x1 >3x2.

Choosex1as the basic variable, known asan entering variable.

Known that two basic variables are(x3, x4). Hence one of these

basic variables will become a non-basic variable, known asa

leaving variable.

IMPLEXMETHOD 6


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z = 4x1+ 3x2

The main idea is

x1 (0 ?), x2 (0 ?) z(0 ?).



4x1 >3x2.




leaving variable.

IMPLEXMETHOD 7


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z = 4x1+ 3x2

The main idea is

x1 (0 ?), x2 (0 ?) z(0 ?).



4x1 >3x2.




leaving variable.

IMPLEXMETHOD 8


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z = 4x1+ 3x2

The main idea is

x1 (0 ?), x2 (0 ?) z(0 ?).



4x1 >3x2.




leaving variable.

IMPLEXMETHOD 9


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Or we consider

4x1 3x2+z = 0.

IMPLEXMETHOD 10


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NBVs: ( x1 an entering variable

, x2) BVs: (x3, x4)

a leaving variable

Consider

x3 = 8 x1 x2

x4 = 10 2x1 x2

Then

x3 = 8 x1 0

x4 = 10 2x1 0

Looking for a leaving variable that will quickly reach to a zero value (i.e., it

will become a non-basic variable):

x1 = min

8

1,10

2

=

10

2= 5

IMPLEXMETHOD 11


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NBVs : ( x1 an entering variable

, x2) BVs : (x3, x4)

a leaving variable

Consider

x3 = 8 x1 x2

x4 = 10 2x1 x2

Then

x3 = 8 x1 0

x4 = 10 2x1 0

Looking for a leaving variable that will quickly reach to a zero value (i.e., itwill become a non-basic variable):

x1 = min

8

1,10

2

=

10

2 = 5

Thereforex4will change a basic variable to a non-basic variable.

IMPLEXMETHOD 12


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x3 = 8 x1 x2

x4 = 10 2x1 x2(3)

We expect to get:

x3 = ?

x1 = ?(4)

(x1, x3)will become basic variables.

IMPLEXMETHOD 13


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x3 = 8 x1 x2

x4 = 10 2x1 x2

x3 + x1 = 8 x2

2x1 = 10 x2 x4(5)

x3 + x1 = 8 x2

x1 = 5 1

2x2

12

x4

12

Eq.(5)[2]

(6)

x3 = 3 1

2x2 +

12

x4 Eq.(6)[1] Eq.(6)[2]

x1 = 5 1

2x2

12

x4

(7)

IMPLEXMETHOD 14


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x3 = 3

1

2

x2 + 1

2

x4

x1 = 5 1

2x2

12

x4

Tableau 3(cf. Lecture Note1d)

cj : (4 3 0 0)

y1 y2 y3 y4

CB XB x1 x2 x3 x4 Min. Ratio

0 x3 3 0 1/2 1 -1/2

4 x1 5 1 1/2 0 1/2zj cj z(XB) = 20 0 -1 0 2

IMPLEXMETHOD 15


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x3 = 3 1

2x2 +

12

x4 = 3 0

x1 = 5 1

2x2

12

x4 = 5 0

Note that




z(x1, x2, x3, x4) = 4x1+ 3x2+ 0x3+ 0x4

= 4

5 12

x2+12

x4

+ 3x2+ 0

= 20 2x2+ 2x4+ 3x2

= 20 +x2+ 2x4

IMPLEXMETHOD 16


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Clearly,X

(1) = (x1, x2, x3, x4) = (5, 0, 3, 0)

is a new B.F.S with

z(X

(1)

) = 20 > z(X

(0)

) = 0.

x

x

2 x + x = 10

x + x = 8

O

2

2

2

1

1

P

Q

R

(0,0)z = 0

(5,0)

z = 20 1

Figure 2: z(5, 0) = 20

IMPLEXMETHOD 17


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Canzbe increased further?

IMPLEXMETHOD 18


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, x4) BVs: (x1, x3)

a leaving variable

x3 = 3 1

2 x2 + 1

2 x4x1 = 5

12

x2 1

2x4

x3 = 3

12

x2 + 1

2(0)

x1 = 5 1

2x2

12

(0)

x3 = 3 1

2x2

x1 = 5 1

2x2


x2 = min

3

1/2,

5

1/2

=

3

1/2 = 6


IMPLEXMETHOD 19


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, x4) BVs: (x1, x3)

a leaving variable

x3 = 3 1

2 x2 + 1

2 x4x1 = 5

12

x2 1

2x4

x3 = 3

12

x2 + 1

2(0)

x1 = 5 1

2x2

12

(0)

x3 = 3 1

2x2

x1 = 5 1

2x2


x2 = min

3

1/2,

5

1/2

=

3

1/2 = 6


IMPLEXMETHOD 20


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x3 = 3 1

2x2 +

12

x4

x1 = 5 1

2x2

12

x4

We expect to get:

x2 = ?

x1 = ?(8)

(x1, x2)will become basic variables.

IMPLEXMETHOD 21


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12

x2 = 3 x3 + 1

2x4

12

x2 + x1 = 5 1

2x4

(9)

x2 = 6 2x3 + x4 2 Eq.(9)[1]

x2 + 2x1 = 10 x4 2 Eq.(9)[2](10)

x2 = 6 2x3 + x4

2x1 = 4 + 2x3 2x4 Eq.(10)[2] Eq.(10)[1]

(11)

x2 = 6 2x3 + x4

x1 = 2 + x3 x4

12

Eq.(11)[2](12)

IMPLEXMETHOD 22


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x2 = 6 2x3 + x4

x1 = 2 + x3 x4

Tableau 5(cf. Lecture Note1d)

cj : (4 3 0 0)

y1 y2 y3 y4

CB XB x1 x2 x3 x4 Min. Ratio

3 x2 6 0 1 2 -1

4 x1 2 1 0 -1 0zj cj z(XB) = 26 0 0 2 1

IMPLEXMETHOD 23


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x2 = 6 2x3 + x4 = 6 0

x1 = 2 + x3 x4 = 2 0

Note that




z(x1, x2, x3, x4) = 4x1+ 3x2+ 0x3+ 0x4

= 4(2 +x3 x4)+ 3(6 2x3+x4)+ 0

= 8 + 4x3 4x4+ 18 6x3+ 3x4

= 26 2x3 x4

Clearly,

X(2) = (x1, x2, x3, x4) = (2, 6, 0, 0)

is a new B.F.S with

z(X(2)

) = 26 > z(X(1)

) = 20.

IMPLEXMETHOD 24


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x

x

2 x + x = 10

x + x = 8

O

2

2

2

1

1

P

Q

R

(0,0)z = 0

(2,6)

z = 26

(5,0)

z = 20 1

Figure 3: z(2, 6) = 26

The optimal solution isx1 = 2,x2 = 6with maximum ofz

= 26.

IMPLEXMETHOD 25

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