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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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LP standard model in
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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LP standard model in
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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LP standard model in
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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The simplex tableau in
matrix form The LP standard problem can be written as:
=
−−
b
0
X
XIA0
CC1
II
I
II I
z
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The simplex tableau in
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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The simplex tableau in
matrix form !X!=b and z=C!X!
5n other words we have:
=
− b
0
XB0
C1
B
B z
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The simplex tableau in
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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The simplex tableau in
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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The simplex tableau in
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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The simplex tableau in
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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Product form for inversematrix
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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Product form for inversematrix
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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Product form for inversematrix
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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Product form for inversematrix
Where:
),...,,,,...,( 111 mr r eeee E +−= ξ
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Product form for inversematrix
−
+
−
−
=
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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Product form for inversematrix
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