Matrix Simplex Method

8/18/2019 Matrix Simplex Method

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Matrix simplex method LP standard model in matrix form Basic solutions and basis The simplex tableau in matrix form Reviewed primal simplex method Product form for inverse matrix Steps of the reviewed primal

simplex method


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LP standard model in

matrix form The LP standard problem can be

expressed in matrix form as follows:

Maximize o Minimize z = CX

s. t.

(A,I) X = b

X ≥ 0


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matrix form Where I is the mxm identit matrix:

=

=

==

−

−

−

mmnmmm

mn

mn

n

T

n

b

b

b

b

aaa

aaa

aaa

A

cccC x x x X

...

...

............

...),...,,( ,),...,,(

2

1

,21

,22221

,11211

2121


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matrix form The identit matrix I can alwas be

written in the form it appears in

the constraint e!uations" This canbe done au#mentin# or arran#in#the defect$ excess and%or arti&cial

variables$ as necessar"


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matrix form This means that the n elements of

vector X include an au#mented

variables 'defect$ excess andarti&cial($ where the m elementslocated to the ri#ht ed#e represent

the variables correspondin# to theinitial solution"

Example:


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Basic solutions and basis Since (A,I) X = b has m e!uations and

n un)nowns *+,'x-$x.$/$xn( T0$ a basic

solution can be obtained settin# n-m variables to zero$ and then solvin# theremainin# m e!uations with m un)nowns$ whenever a uni!ue solution

does exist" 1n extreme point is associated to one

'or several( basic solution"


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Basic solutions and basis Let

Where P 2 is the 2th column vector of 'A,I("Whatever m linearly inepenentvectors ta)en from P-$ P.$/ Pncorrespond to a basic solution of

(A,I)X=b and so$ to an extreme point inthe space of solutions" Such vectors forma basis$ which associated matrix isnonsin#ular "

Example:

∑=

=n

j

j j x1

PX)I,A(


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The simplex tableau in

matrix form Maximize

3=CX

s.t. (A,I)X= b, X≥0

4ector X is divided into X5 and X55

where X55 corresponds to the elements

of X associated to the initial basis !=I

5n the same wa$ C is also divided intoC5 and C55 correspondin# to X5 and X55"


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matrix form The LP standard problem can be written as:

=

−−

b

0

X

XIA0

CC1

II

I

II I

z


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matrix form 6or each iteration$ let XB be the

representation of the current basic

variables and ! its associated basis" This means that XB represents m elements of X and ! represents thevectors of (A,I) associated to XB"

1ccordin#l$ let CB be therepresentation of the elements of C associated to XB"


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matrix form !X!=b and z=C!X!

5n other words we have:

=

− b

0

XB0

C1

B

B z


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matrix form 5t is possible to obtain the current

values of 3 and XB invertin# the sub7

divided matrix$ #ettin#:

=

=

−

−

−

−

bB

bBC

b

0

B0

BC1

X 1

1

1

1

B B

B

z


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matrix form The #eneral simplex tableau

correspondin# to XB is determined

ta)in# into consideration that:

=

−−

−

−

−

−

b

0

B0

BC1

X

XIA0

CC1

B0

BC1

1

1

1

1

B

II

I

II I B

z


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matrix form 8omputin# the indicated operations between

matrixes$ it is obtained the followin# #eneralsimplex iteration$ expressed in matrix form:

Basic X5 X55

z CB!7-

A7C5 CB!7-

7C55 CB!7-

b

XB !7-A !7- !7-b


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matrix form The complete tableau$ at an

iteration$ can be computed once it

is )nown the basis ! associated toXB 'and therefore its inverse !7-("

9ach element in the tableau is a

function of !7-

and the original dataof the problem

Example:


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Reviewed primal simplex

method The simplex methods$ primal or

dual$ expressed in matrix form$

dier onl on the selection of theincomin# and the out#oin# vector"9ver compute else are the same"


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Reviewed primal simplexmethod

The reviewed simplex method oersan advanta#eous procedure fromthe point of view of the computin#precision$ due to the wa tocompute the inverse matrix B7-


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Product form for inversematrix

;iven the current basis B$ the nextbasis Bnxt in the followin# iteration

will dier from B onl in onecolumn"

B7-nxt is obtained pre7multiplin#

the current inverse B7- b anspeciall constructed matrix 9"


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Let us de&ne the identitmatrix Im as

Where ei is a unitar column

vector with a number - at theith place and < elsewhere

),...,,( 21 mm eee I =


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Suppose that ! and !7- are #iven andthat vector Pr is replaced b vector P 2 in

matrix ! 'P 2 and Pr are the incomin# andthe out#oin# vectors$ respectivel(

j

j

PB

1−

=α


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So that = 2) is the )th element of = 2"

Then the new inverse B7-nxt can be

computed as follows:

Bnxt7-,9B7-


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Where:

),...,,,,...,( 111 mr r eeee E +−= ξ


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−

+

−

−

=

j

r

j

m

j

r

jr j

j

r

j

α α

α

α α

α α

ξ

/

.../1

...

/

/

2

1

rth

place


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Whenever =r 2>


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Steps of the reviewedprimal simplex method

The main idea of the reviewedmethod is to use the inverse of the

current base !7-$ and the ori#inaldata of the problem to donecessar computin# in order to

determine the incomin# and theout#oin# variables


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Steps of the reviewedprimal simplex method

;iven the initial basis I$ it isdetermined its ob2ective

coeAcients vector CB associated$dependin# whether the initial basicvariables are of defect$ excess

and%or arti&cial tpe


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Step 1

Incoming vector P j 8ompute , 8BB7-" Then for each

non basic vector P 2$ compute

j j j j cYP c z −=−


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Step 1

5n maximization 'minimization(problems$ choose the incomin# vector P 2

with the more ne#ative 'positive( z 27c 2Cties are bro)en arbitraril"

5f ever z 27c 2 are #reater or e!ual than


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Step 2

Outgoing vector Pr ;iven the incomin# vector P 2$

compute: -" The values of the current basic

variables$ that is: X!=!#$b

." The constraint coeAcients of theincomin# variables$ that is: %=!#$P %


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Step 2 The out#oin# vector Pr 'for maximization

and minimization( must be associated to:

Where 'B7-b() and = 2) are the )th elementsof B7-b and α 2

5f ever = 2) D=

−0,

)( 1

min j

k j

k

k

k

b Bα

α θ


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Step 3

Next basis

;iven the current inverse basis B7-

$ itis )nown that the next inverse basisis #iven b:

!nxt#$="!#$

@ow set B7- , B7-nxt and #o bac) to step-"


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Step 3

Steps - and . are exactl the same as

those in the ori#inal simplex tableau$as it is shown in the followin# table:


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Step 3

Basic x- x. / x 2 / xn Solution

z z-7c- z.7c. / z 27c 2 / zn7

cn

+B !7-P 2 !7-b

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Matrix Simplex Method