341-l06 revised simplex

8/2/2019 341-l06 Revised Simplex

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Revised simplex method

Matrix representation of the simplex tableSuppose that we have a LP model in the standard form:

Max z= c1x1 + c2x2 + ` + cnxn

Subject to

a11x1 + a12x2 + ` + a1nxn b1

a21x1 + a22x2 + ` + a2nxn b2

_

am1x1 + am2x2 + ` + amnxn bm

O

and

x1,x2,...,xn 0

Its augmented form is

Max z= c1x1 + c2x2 + ` + cnxn

Subject to

a11x1 + a12x2 + ` + a1nxn + xn + 1 = b1

a21x1 + a22x2 + ` + a2nxn + xn + 2 = b2

_

am1x1 + am2x2 + ` + amnxn + xn+ m = bm

A

and

x1,x2,...,xn,

slack variables

xn + 1,...,xn + m 0

Let


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c = c1 c2 ` cn 0 ` 0

A =

a11 a12 ` a1n 1 0 0

a21 a22 ` a2n 0 1 0

_am1 am2 amn 0 0 1

x =

x1

x2

_

xn

, xs =

xn + 1

xn + 2

_

xn + m

, b =

b1

b2

_

bm

.

Then the system (O) can be written as

Max z= cx

s.t. Ax b

x 0

The system (A) can be written as

Max z= cx

s.t. A Ix

xs

= b

x

xs

0

The first row of the initial table of the simplex method can be written in the matrix form

z ? cx = 0

or

1 ?c 0

z

x

xs

= 0

The initial table of the simplex method is


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z x1 ` xn xn + 1 ` xn+ m RHS

1 ?c1 ` ?cn 0 ` 0 0

xn+ 1 0 a11 ` a1n 1 ` 0 b1

_ 0 _ _ _ _ _

xn+ m 0 am1 ` amn 0 ` 1 bm

can be represented by

1 ?c 0 0

0 A I b

which is equivalent to the matrix equation:

1 ?c 0

0 A I

z

x

xs

=

0

b . M

Example Consider the system

Max z= 2x1 + 3x2

s.t.

x1 + 2x2 6

2x1 + x2 9

x1,x2 0

Its augmented form is

Max z= 2x1 + 3x2

s.t.

x1 + 2x2 + x3 = 6

2x1 + x2 + x4 = 9

x1,x2,x3,x4 0

We have

c = 2 3 ,A =1 2

2 1, b =

6

9, x =

x1

x2, xs =

x3

x4.


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1 ?c 0

z

x

xs

= 1 ?2 ?3 0 0

z

x1

x2

x3

x4

= z ? 2x1 ? 3x2

0 A I

z

x

xs

=

0 1 2 1 0

0 2 1 0 1

z

x1

x2

x3

x4

=

x1 + 2x2 + x3

2x1 + x2 + x4

Revised Simplex MethodIn the iterations of simplex method, the most important information is the choice of basicvariables.. Once we know what are the basic variables, all the other information can be figured outfrom the initial table (M). For example, suppose we chose x2,x4 to be our basic variables after some

iteration.

A Ix

xs

=

1 2 1 0

2 1 0 1

x1

x2

x3

x4

(Let x1,x3 be 0) =1 2 1 0

2 1 0 1

0

x2

0

x4

= x22

1+ x4

0

1

=

2 0

1 1

x2

x4

We know that the effect of the row operation can be represented by multiplying a matrix from the

left. If after one iteration, x2,x4 are the current basic variable, then all the row operations up to thatpoint will change the columns ofx2,x4 into an identity matrix. Therefore all those operation is

equivalent to multiplying the inverse ofB =2 0

1 1.


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B ? 1 =2 0

1 1

? 1

=

12

0

?12

1

Therefore the lower part of the new simplex table should be

B ? 1 0 A I =

12

0

?12

1

0 1 2 1 0

0 2 1 0 1=

0 12

1 12

0

0 32

0 ? 12

1

Because the same operation is applied to the right hand side, we have the new lower part of the righthand side

B ? 1b =

12

0

?12

1

6

9=

3

6=

x2

x4= xB

That gives us x2 = 3 and x4 = 9, so z =

cB

3 0

1

2 0?

12

169

= 3 3 + 0 6 = 9.

In general, suppose that B is formed by the columns ofA that correspond to the basic variables,and we define xB and cB similarly. Then initially, the simplex table in the matrix form is

1 ?c 0

0 A I

z

x

xs

=

0

b

After some iteration, we get

z

xB

=

cBB? 1b

B ? 1b=

1 cBB? 1

0 B ? 1

0

b

Therefore,1 cBB

? 1

0 B ? 1is the matrix representing all the row operations and the matrix for the

simplex table after the iteration is

1 cBB? 1

0 B? 1

1 ?c 0

0 A I

z

x

xs

=

1 cBB? 1

0 B? 1

0

b

1 cBB? 1A ? c cBB

? 1

0 B ? 1A B ? 1

z

x

xs

=

cBB? 1b

B ? 1b.

In our example, we have


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cBB? 1= 3 0

12

0

?12

1=

32

0

cBB? 1A ? c = 3

20

1 2

2 1?

2 3= ?

1

20

So the complete table would be

1 ? 12

0 32

0 9

0 1/2 1 1/2 0 3

0 3/2 0 ?1/2 1 6

Next step: x1 would be the entering basic variable and x4 would be the leaving basic variable.

xB =x2

x1, B =

2 1

1 2

. We have

B ? 1 =

23

?13

?13

23

, cB = 3 2 , cBB? 1=

43

13

cBB? 1A ? c = 3 2

23

?13

?13

23

1 2

2 1? 2 3 = 0 0

The calculation in the second line is not necessary.. That tells us that this is an optimal solution. Thesolution is

xB =x2

x1= B ? 1b =

23

?13

?13

23

6

9=

1

4

The steps of the Revised Simplex Method:

1. Initialization. Same as before.

2. Optimality test: Calculate cBB ? 1 and cBB ? 1A ? c for the nonbasic variables. If the solution is notoptimal, find the entering basic variable.

3. Determine the leaving basic variable: calculate only the column of the entering basic variable,say it is xj, and then the ratios bi/aij for the positive entries aij.

The column ofxj can be calculated by multiplying B? 1 to that column in A.

Bnew? 1

= E6 Bold? 1

where Eis the matrix corresponding to the row operations used. For example:

Bold? 1

=

12

0

?12

1

The row operations we are going to use are:

a. Multiply row 2 with 2/3;


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341-l06 revised simplex

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