the theory of the simplex method
DESCRIPTION
The Theory of the Simplex Method. The Essence. Simplex method is an algebraic procedure However, its underlying concepts are geometric Understanding these geometric concepts helps before going into their algebraic equivalents - PowerPoint PPT PresentationTRANSCRIPT
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Foundations-1
The Theory of the Simplex MethodThe Theory of the Simplex Method
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Foundations-2
The EssenceThe Essence
• Simplex method is an algebraic procedure• However, its underlying concepts are geometric• Understanding these geometric concepts helps before
going into their algebraic equivalents• With two decision variables, the geometric concepts are
easy to visualize• Should understand the generalization of these concepts
to higher dimensions
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Foundations-3
Standard (Canonical) Form of an LP ModelStandard (Canonical) Form of an LP Model
Maximize Z = c1x1 + c2x2 + … + cnxn
subject to a11x1 + a12x2 + … + a1nxn ≤ b1
a21x1 + a22x2 + … + a2nxn ≤ b2
…am1x1 + am2x2 + … + amnxn ≤ bm
x1 ≥ 0, x2 ≥ 0, …, xn ≥ 0
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Foundations-4
Extended TerminologyExtended Terminology
max cx n variabless.to Ax b m inequalities
x 0 n inequalities
max cx n+m variabless.to Ax + Ixs = b m equations
x, xs 0 n+m inequalities
• Constraint boundary equation– For any constraint, obtain by replacing its , =, by =– It defines a “flat” geometric shape: hyperplane
• Corner-point solution– Simultaneous solution of n constraint boundary equations
( )
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Foundations-5
Extended TerminologyExtended Terminology
max cx + 0xs n+m variabless.to Ax + Ixs = b m equations
x, xs 0 n+m inequalities
• Indicating variables (in the augmented form)
Original Constraint in Augmented Form
Constraint Boundary Equation Indicating Variable
xj 0 (j=1,2,…,n)
xj = 0 xj
(i=1,2,…,m)
xsn+iisin
n
jjij bxxa
1i
n
jjij bxa
1
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Foundations-6
' '1 2 3 3
1 11
2 22
3 ' ' 33
Max 0 0 variables
s.to = equations
= equations
- + equations
, ,
s a s a
s
a
s a
s a
cx x Mx x Mx n m m m m
A x Ix b m
A x Ix b m
A x Ix Ix b m
x x x ' '1 2 3 3, , 0 inequalities s ax x n m m m m
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Foundations-7
Original Constraint
Original Constraint in Augmented Form
Constraint Boundary Equation
Indicating Variable
xj 0 xj 0 xj = 0 xj
(i=1,2,…,m1)
(i=1,2,…,m2)
(i=1,2,…,m3)
Indicating variables, overallIndicating variables, overall
' ' s an i n ix x
sn ix1 1
1
n
sij j n i i
j
a x x b 1 1
1
n
ij j ij
a x b1 1
1
n
ij j ij
a x b
2 2
1
n
ij j ij
a x b
3 3
1
n
ij j ij
a x b
2 2
1
n
aij j n i i
j
a x x b
3 ' ' 3
1
n
s aij j n i n i i
j
a x x x b
2 2
1
n
ij j ij
a x b
3 3
1
n
ij j ij
a x b
an ix
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Foundations-8
Properties of CPF (BF) SolutionsProperties of CPF (BF) Solutions
Property 1:a. If there is exactly one optimal solution, then it must be a
corner-point feasible solutionb. If there are multiple optimal solutions (and a bounded
feasible region), then at least two must be adjacent corner-point feasible solutions
Proof of (a) by contradiction
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Foundations-9
Properties of CPF (BF) SolutionsProperties of CPF (BF) Solutions
Property 2:There are only a finite number of corner-point feasible solutions
max cx n variabless.to Ax b m inequalities
x 0 n inequalities
CP solution is defined as the intersection of n constraint boundary equations(n+m) choose n:
!m!n)!mn(
nmn
Upper bound on # of CPF solutions
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Foundations-10
Properties of CPF (BF) SolutionsProperties of CPF (BF) Solutions
Property 3:a. If a corner-point feasible solution has no adjacent CPF
solution that are better, then there are no better CPF solutions elsewhere
b. Hence, such a CPF solution is an optimal solution (assuming LP is feasible and bounded)
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Foundations-11
Simplex in Matrix (Product) Form Simplex in Matrix (Product) Form
max cx n variabless.to Ax b m inequalities
x 0 n inequalities
max cx m+n variabless.to Ax + Ixs = b m equations
x, xs 0 m+n inequalities
Initial Tableau:
( )B.V.
Original Variables Slack Variablesr.h.s.
x1 x2 … xn xsn+1 … xs
n+m
Z -c 0 0xs
n+1
A I b…
xsn+m
XB=
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Foundations-12
Simplex in Matrix (Product) Form Simplex in Matrix (Product) Form
Intermediate iterations:• Let B denote the square matrix that contains the columns from
[ A | I ] that correspond to the current set of basic variables, xB (in order)
• Similarly, let cB be the vector of elements in c that correspond to xB
• Then at any intermediate step, the simplex tableau is given by
B.V.Original Variables Slack Variables
r.h.s.x1 x2 … xn xs
n+1 … xsn+m
Z cBB-1A-c cBB-1 cBB-1b
xB B-1A B-1 B-1b
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Foundations-13
Revised Simplex MethodRevised Simplex Method
• InitializationFind an initial BFS (if one not immediately available, do Phase 1. If Phase 1 terminates with z1>0, LP is infeasible)
• Optimality testCalculate cBB-1A-c for basic, and cBB-1 for nonbasic variables. If all ≥ 0, stop with optimality
• Iterative step– Determine the entering variable as before (steepest ascent)– Let xk be the entering variable– Determine the leaving variable as before (minimum ratio test), but now only need to
calculate the coefficients of the pivot column (from B-1A or B-1) and the updated rhs (B-1b)– Let xl : the leaving variable
r : the equation number of the leaving variablea’rk : coefficient of xk in rth equation (the pivot element)
– Determine B-1new=E B-1
old where E is I expect for rth column replaced with
– Can calculate the new BFS using xB=B-1b and z=cBB-1b – Back to optimality test
ri if'a1
ri if'a'a
,
rk
rk
ik
i
m
1
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Foundations-14
Revised Simplex ExampleRevised Simplex Example
Original data:
Maximize z = 2x1+ 3x2
subject to x1 + 2x2 +s1 = 103x1+ x2 +s2 = 15
x2 +s3 = 4
x1,x2, s1, s2, s3 ≥ 0
41510
b 101321
A
32c
Maximize z = 2x1+ 3x2
subject to x1 + 2x2 ≤ 10
3x1+ x2 ≤ 15
x2 ≤ 4
x1,x2 ≥0
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Foundations-15
Iteration 0Iteration 0
bBcz
bBx
B B x
1-B
1-B
1-B
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Foundations-16
Iteration 0Iteration 0
• Calculate row-0 coefficients:cBB-1A-c =
cBB-1 =
Optimal?• Entering variable is• Calculate coefficients in pivot column using B-1A or B-1
• Calculate updated rhs using B-1b, do the ratio test
• Leaving variable is• Update B-1
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Foundations-17
Iteration 1Iteration 1
bBcz
bBx
B B x
1-B
1-B
1-B
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Foundations-18
Iteration 1Iteration 1
• Calculate row-0 coefficients:cBB-1A-c =
cBB-1 =
Optimal?• Entering variable is• Calculate coefficients in pivot column using B-1A or B-1
• Calculate updated rhs using B-1b, do the ratio test
• Leaving variable is• Update B-1
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Foundations-19
Iteration 2Iteration 2
bBcz
bBx
B B x
1-B
1-B
1-B
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Foundations-20
Iteration 2Iteration 2
• Calculate row-0 coefficients:cBB-1A-c =
cBB-1 =
Optimal?• Entering variable is• Calculate coefficients in pivot column using B-1A or B-1
• Calculate updated rhs using B-1b, do the ratio test
• Leaving variable is• Update B-1
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Foundations-21
Iteration 3Iteration 3
bBcz
bBx
B B x
1-B
1-B
1-B
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Foundations-22
Iteration 3Iteration 3
• Calculate row-0 coefficients:cBB-1A-c =
cBB-1 =
Optimal?• Entering variable is• Calculate coefficients in pivot column using B-1A or B-1
• Calculate updated rhs using B-1b, do the ratio test
• Leaving variable is• Update B-1