Chapter 6 The Revised Simplex Method
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).1 Chapter 6 Chapter 6 The Revised Simplex The Revised Simplex Method Method This method is a modified version This method is a modified version of the of the Primal Simplex Method that that we studied in Chapter 5. we studied in Chapter 5. It is designed to exploit the It is designed to exploit the fact that in many practical fact that in many practical applications the coefficient applications the coefficient matrix {a matrix {a ij ij } is very } is very sparse , , namely most of its elements are namely most of its elements are equal to zero. equal to zero.
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Chapter 6 The Revised Simplex Method. This method is a modified version of the Primal Simplex Method that we studied in Chapter 5. - PowerPoint PPT Presentation
Transcript of Chapter 6 The Revised Simplex Method
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Chapter 6Chapter 6
The Revised Simplex MethodThe Revised Simplex MethodChapter 6Chapter 6
The Revised Simplex MethodThe Revised Simplex Method This method is a modified version of the This method is a modified version of the
Primal Simplex Method that we studied in that we studied in Chapter 5.Chapter 5.
It is designed to exploit the fact that in It is designed to exploit the fact that in many practical applications the coefficient many practical applications the coefficient matrix {amatrix {aijij} is very } is very sparse, namely most of , namely most of
its elements are equal to zero.its elements are equal to zero.
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Bottom line:Bottom line:
– Don’t update all the columns of the Don’t update all the columns of the simplex tableau: update only those simplex tableau: update only those columns that you need!columns that you need!
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Standard FormStandard FormStandard FormStandard Form
• opt=maxopt=max
• ~ ~ • bbii ≥ 0 , ≥ 0 ,
for all i.for all i.
a11x1 a12x2 . .. a1n xn b1
a21x1 a22x2 . .. a2 nxn b2
. ... . .. .. . .. . ... .
. ... . .. .. . .. . ... .
am1x1 am 2x2 .. . amn xn bm
x j 0 , j 1,.. .,n
maxxZ cjx j
j1
n
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max
. .xZ
s t
Z cx
Ax b
x
0
0
More Convenient Form
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Canonical Form
a11x1 a12x2 . .. a1n xn xn1 b1
a21x1 a22x2 ... a2 nxn xn2 b2
. ... . .. ... .. . ... . .. ..
. ... . .. ... .. . ... . .. ..
am1 x1 am2 x2 .. . amn xn xnm bm
x j 0 , j 1, ...,n m
max z c jj1
n
x j
As in the standard format, bi≥0 for all i.
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System PSystem PSystem PSystem P
maxxZ
s.t .
Z cx 0
Sx bx 0
(6.4)
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2 40
30
15
0
1 2 3
1 2 4
1 5
1 2 3 4 5
x x x
x x x
x x
x x x x x
, , , ,
maxxZ x x 4 31 2
Example 6.1.1Example 6.1.1Example 6.1.1Example 6.1.1
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S 2 1 1
1 1 0
1 0 0
0
1
0
0
0
1
b( , , )40 30 15c( , , , , )4 3 0 0 0 .
System P
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max '
. .
' ' ' '
' ' '
'
xZ
s t
Z c x z
S x b
x
0
System P’
After a number of Pivot OperationsAfter a number of Pivot OperationsAfter a number of Pivot OperationsAfter a number of Pivot Operations
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ObservationObservationObservationObservation
After any iteration of the simplex method After any iteration of the simplex method the columns of the m basic variables the columns of the m basic variables comprise the columns of the mxm comprise the columns of the mxm identity matrix.
The The order in which these columns are in which these columns are arranged for this purpose is important.arranged for this purpose is important.
This order is specified in the This order is specified in the BV column of of the simplex tableau.the simplex tableau.
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BV Eq. # x1 x2 x3 x4 x5 RHSx3 1 0 1 1 0 - 2 10x4 2 0 1 0 1 - 1 15x1 3 1 0 0 0 1 15Z Z 0 - 3 0 0 4 60
Example 6.1.2
S' 0 1 1
0 1 0
1 0 0
0
1
0
2
1
1
b' ( , , ) 10 1515
c' ( , , , , ) 0 3 0 0 4
Z'=60.
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6.2 The Transformation6.2 The Transformation6.2 The Transformation6.2 The Transformation How can we compute S’ from S ?How can we compute S’ from S ? From Linear Algebra we know that any finite From Linear Algebra we know that any finite
sequence of pivot operations is equivalent to sequence of pivot operations is equivalent to (left) (left) multiplication by a matrix..
In other words,In other words,
S’ = TS S’ = TS The question is then:The question is then:
T = ??????
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What is T ???What is T ???What is T ???What is T ???
Observation 1Observation 1::
After any number of iterations of the simplex After any number of iterations of the simplex method, the columns of the coefficient matrix method, the columns of the coefficient matrix corresponding to the corresponding to the basic variables at that at that iteration, comprise the iteration, comprise the identity matrix..
Observation 2Observation 2::
Initially, the Initially, the last m columns of the coefficient of the coefficient matrix comprise the identity matrix.matrix comprise the identity matrix.
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AnalysisAnalysisAnalysisAnalysis
If we group the columns of the basic variables into If we group the columns of the basic variables into I and the nonbasic variables into D’, thenI and the nonbasic variables into D’, then
S’ = [I,D’]S’ = [I,D’] If we do the same for the initial matrix S, we haveIf we do the same for the initial matrix S, we have
S = [B,D]S = [B,D]
where B is the matrix constructed from the columns where B is the matrix constructed from the columns of the initial matrix corresponding to the current of the initial matrix corresponding to the current basic variables.basic variables.
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Since S’ = TS, it follows that Since S’ = TS, it follows that
S’ = [I,D’] = TS = [TB,TD]S’ = [I,D’] = TS = [TB,TD]
hencehence
I = TBI = TB
from which we conclude that from which we conclude that
T = B-1
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BV Eq. # x1 x2 x3 x4 x5 RHSx3 1 2 1 1 0 0 40x4 2 1 1 0 1 0 30x5 3 1 0 0 0 1 15Z Z - 4 - 3 0 0 0 0
Example 6.2.1
S 2 1 1
1 1 0
1 0 0
0
1
0
0
0
1
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BV Eq. # x1 x2 x3 x4 x5 RHSx2 1 0 1 1 0 - 2 10x4 2 0 0 - 1 1 1 5x1 3 1 0 0 0 1 15Z Z 0 0 3 0 - 2 90
S (2) 0 1 1
0 0 1
1 0 0
0
1
0
2
1
1
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I S.2(2), S.4
(2),S.1(2)
1 0 0
0 1 0
0 0 1
D' S.3(2),S.5
(2) 1
1
0
2
1
1
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BS.I B S.2,S.4,S1
1 0 2
1 1 1
0 0 1
DS.ID S.3,S.5
1
0
0
0
0
1
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B1 1 0 2
1 1 1
0 0 1
S B S( )2 1
B1S 1 0 2
1 1 1
0 0 1
2
1
1
1
1
0
1 0 0
0 1 0
0 0 1
0
0
1
1
0
0
1 0 2
1 1 1
0 0 1
S(2)
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Notation:Notation:Notation:Notation: IIBB = Indices of the basic elements (in = Indices of the basic elements (in
canonical form)canonical form) IIDD = indices of the nonbasic variables in = indices of the nonbasic variables in
increasing orderincreasing order ccBB = = Initial Initial cost vector of the basic variablescost vector of the basic variables
ccDD = = Initial cost vector of the nonbasic cost vector of the nonbasic
variablesvariables
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r = reduced costs vectorr = reduced costs vector D = columns of the coefficient matrix in the D = columns of the coefficient matrix in the
initial simplex tableau corresponding to the simplex tableau corresponding to the current current nonbasic variables.
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Behind the Formula (NILN)Behind the Formula (NILN)Behind the Formula (NILN)Behind the Formula (NILN)
Each column of the coefficient matrix in the new Each column of the coefficient matrix in the new tableau is equal to Btableau is equal to B-1-1 times the corresponding times the corresponding initial column, i.einitial column, i.e
new column = B-1 initial column This is also true for the right-hand-side vector, i.eThis is also true for the right-hand-side vector, i.e
new RHS = B-1 initial RHS Observe that the Observe that the z-row is is not included in this in this
formulation (why?)formulation (why?)
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D B D' 1
b B b' 1
c c c c c c cB D B I D IB D
' , ( : ; : ), , , , , ,
c cD' , , 0
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( cD, ) cBB
1D cD
z c B bB' 1
rD =
C’B = (0,0,...,0)
correction
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(NILN)(NILN)But how do we compute BBut how do we compute B-1 -1 ??
(NILN)(NILN)But how do we compute BBut how do we compute B-1 -1 ??
Bad news:: We have to compute it as we go alongWe have to compute it as we go along Good News:: We do not have to compute it from scratchWe do not have to compute it from scratch ObservationObservation::
S’ = BS’ = B-1-1S = BS = B-1-1 [M,I] = [B [M,I] = [B-1-1M,BM,B-1-1I] = [BI] = [B-1-1M,BM,B-1-1]] Hence, BHence, B-1-1 is equal to the matrix comprising the is equal to the matrix comprising the
last m columns of the of the LHS matrix.
