lectures 18-19 linear programming. preparation linear algebra
TRANSCRIPT
Lectures 18-19
Linear Programming
Preparation
Linear Algebra
Linearly Independent
.0 such that
0, toequal allnot ,,...,, scalersexist there
ifdependent linearly are ,...,, Vectors
.0 0
ift independenlinearly are ,...,, Vectors
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Maximal Independent Subset
.
i.e., set,t independen maximal
ain vectorsofn combinatiolinearly a is or Every vect
y.cardinalit same thehave subsetst independen maximal All
subset.t independenlinearly another of
subset proper anot isit andt independenlinearly isit if
subsett independen maximal a called is },...,,{
subset A }.,...,,{ vectors,ofset aConsider
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Rank of Matrix
| vectorsrow ofsubset t independen maximal|
|torscolumn vec ofsubset t independen maximal|)(rank
.matrix aConsider
A
A
Linear Programming
LP examples
• A post office requires different numbers of full-time employees on different days of the week. The number of full-time employees required on each day is given in the table. Union rules state that each full-time employee must work five consecutive days and then receive two days off. The post office wants to meet its daily requirements using only full-time employees. Formulate an LP that the post office can use to minimize the number of full-time employees that must be hired.
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yxz 54
Feasible domain
Optimal occurs at a vertex!!!
.0,0,0
6032 s.t.
54 max
wyx
wyx
yxz
Slack Form
.)(rank
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s.t.
max
mA
x
bAx
cxz
What’s a vertex?
. ,),(2
1
if
vertexa called is polyhadren ain point A
zyxzyzyx
x
. of sin vertice found becan it then
solution, optimalan has over max If
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xcx
Ax = b, xx =
Fundamental Theorem
.constraint oneleast at violates' is, that ,in not
'point a havemust line theThus, line.any contain not
does However, solutions. optimal are *)( line
on points feasible all that followsIt solutions. optimal
also are and that means This .
havemust we2,)( and ,
,* Since distinct. are ,*, and 2/)(*
such that ,exist thereis, that not, is * suppose
on,contraditiBy . of vertex a is * that show willWe
solutions. optimal all among components zero of
number maximum with *solution optimalan Consider
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/cy+czcx* = czcx*
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x
x
Proof.
ion.contradict a
,*than component -zero more one has which ,* and
'between solution optimalan findeasily can weNow,
.0 with somefor 0 constraint a violate
must ' Hence, 0. constraint latecannot vio '
that means This .any for 0 toequal is *)( of
component th theTherefore, .0* havemust
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Moreover, . constraint latecannot vio ' Thus,
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Proof (cont’s).
t.independenlinearly
are 0 with 1for all if
only and if vertex a is point feasible a
Then . ofcolumn th thedenote Let
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Characterization of Vertex
Proof
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and ,
0, smallly sufficientfor and 0Then
.0 if 0
0 if
settingby )( Define zero. are
allnot and 0such that 0 with for
exists e then thert,independenlinearly not are 0for If
}.0|{set index by determineduniquely are
then t,independenlinearly are 0for If
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Basic Feasible Solution
0. and
|| = m = )(rank ifonly and if basis feasible
a is subset index an Then .)( Denote
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feasible basic a exists thereif feasible is basisA
basis. a called torscolumn vec of
subsett independen maximum a ofset index The
solution. feasible basic a called also isA vertex
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Optimality Condition
condition.acy nondegenerunder
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ifonly and if optimal is Moreover,
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basic a with associated is basis feasibleEach
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Degeneracy Condition
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Sufficiency
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reaches ,0 and 0 Since
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Necessary
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optimal.not is 0
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solution. optimal
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(pivoting)
Simplex Method
method.
simplex called g,propromminlinear thesolve
tomethod a givesy necessarit of proof The
Simplex Table
1 2 1 4 36
1 5 2 2 24
1 3 1 1 30
2 1 3
0,,,,,
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303 s.t.
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6321
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1 5 2 2 24
1 3 1 1 30
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1 5 2 2 24
1 3 1 1 30
2 1 3
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1/4 1/2 1/4 1 9
1/2- 1 4 3/2 0 6
1/4- 1 5/2 3/4 0 21
3/4- 1/2 1/4 0 27-
}5,4,1{ 1
z
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1/4 1/2 1/4 1 9
1/3- 3/2 8/3 1 0 4
1/4- 1 5/2 3/4 0 21
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}2{})5{\}5,4,1({ 2
z
I
1/3 3/8- 1/6- 0 1 8
1/3- 3/2 8/3 1 0 4
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}2,4,1{ 2
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Puzzle 1
? basis feasible
1st or thesolution feasible basic1st thefind wedo How
Puzzle 2
LP? solve
tohow hold,not does assumptionacy nondegenerWhen
lexicographical ordering
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A vector .1 somefor ...
if , as written ,n larger tha
hicallylexicograp be tosaid is .)...(
and )...,( vectorswoConsider t
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Method
positive.hically lexicograp is row topthe
except blesimplex ta initial in the rowevery that makes This
columns. first at the placed is basis feasible initial the
such that columns of ordering therearrange Initially,
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Method(cont’)
pivoting.
under preserved be willrow top theother than rows all of
spositveneshically lexicograp that theguarantees choice This
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Method (cont’)
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such that basis feasibleith solution w basic feasible optimalan
findor solution optimal of cenonexisten findseither algorithm theTherefore,
ing.nonincreas is aluefunction v objective theand oncemost at basis feasible
each visit algorithm that theguarantee rules additional above theTherefore,
ordering. hicallexicograpin strictly decreasse top themake pivot will
each positive,hically lexicograp are row top theother than rows all Since
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Theorem
optimal. is with associatedsolution feasible
basic then the0, satisfies basis feasible a if Moreove,
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feasible with associatedsolution feasible basic optimalan hasit then
solution, optimalan has gprogramminlinear theIf
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then solution, optimal no has gprogramminlinear theIf
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