Chapter 8 PD-Method and Local Ratio

(1) Duality of LP

Ding-Zhu Du

Dual LP

.0 s.t. min

xbAx

cxz

s.t.

maxcyA

ybz

Primal Dual

Lemma

cxyAxyb

ybcxyx

. ,point feasible dualany and point feasible primalany For

Proof.

.** Moreover,ly.respective ,* and *say solutions, optimal haveboth they

then solution, feasible have dual theand primal both the Ifsolution.

feasible no has dual theiff valueminimum has primal The. value

maximum has dual theiffsolution feasible no has primal The

bycxyx

Theorem

.),(Then

.0 and i.e., , with associatedsolution optimalan is Suppose

feasible. dual is S,

.),(),()(

Then .Set

.0

satisfying basis feasible optimalan obtaincan wemethod, hicallexicagrapby theat recall statement, 3rd theshow To

.from follows statements First two

11

1

1,

1

1

1

cxxAAcxcxAAccyAxyb

xbAxIx

y

cccAAccAAAcyA

Acy

AAcc

I

ybcx

IIIIIIIIII

III

IIIIIIIIII

II

IIII

Proof.

Complementary Slackness Condition

.)(

optimal. are and both case, asuch In

.0)(such that solution feasible dual a exists there

iff optimal is solution feasible primalA

xyAcybcx

yx

xyAcy

x

Remark

Chapter 8 PD-Method and Local Ratio

(2) Primal-Dual Method

Ding-Zhu Du

Primal and Dual

0 s.t.

min

xbAx

cx

0

s.t. max

ycyA

yb

Complementary Slackness Condition

.optimal. are and both case, asuch In

.0)( ,0)(

such that solution feasible dual a exists there iff optimal is solution feasible primalA

ybyAxcxyx

bAxyxyAc

yx

Remark

Primal c-s ConditionDual c-s Condition

General Cover

.0 ,0 where

}1,0{

1 s.t. min

Ac

x

Axcx

n

M × n

Primal and Dual

.0 ,0

where0

1 s.t. min

Ac

xAxcx

.0 ,0

where,0

s.t.1 max

Ac

ycyA

y

Idea0)()1( xyAcAxy

.01 if increaseonly ,much too)1( increase not toorder in

0)( with choose

0)( keep

;1 to0 some change ; feasible primalnot is

xayAxy

yacj

xyac

xxx

ii

jj

jj

j

jj

iteration?each in allat increased benot would)1(you think Do AxyQuestion:

Idea0)()1( xyAcAxy

.}1,0{

and ,0)( feasible dual - :keep

nx

xyAcy

.Output

while.-end;1

)(for and )(for set

;1 and if set

min

such that )( choose

}0|{)(

and 0}|{)(set

begin do feasible primenot is while

.0 0, 0,set Initially,

11

11

)(

1)(

)(

1

1

00

kA

ki

ki

ki

ki

kr

kj

kj

m

kIiij

m

i

kiijj

kJjm

kIiir

m

i

kiirr

n

j

kjij

kj

k

xx

kkkIiyykIiyy

xrjxx

a

yac

a

yac

kJr

xaikI

xjkJ

x

kyx

))(( nmnO

.feasible-dual

still is Hence, . Thus, .0 ),(

),(for that Note ).(for , of choiceby and

0 that note we,1for (a) see To .1consider weNow,

.0for hold they Suppose . trivially trueare (b) and (a) all Initially,on induction by it prove We

and 0, (b)

feasible,-dual is (a)

:following thehave wek,any For

1

11

1

1

1

1

1

kj

m

i

kiij

m

i

kiijij

j

m

i

kiij

ki

ki

kjj

m

i

kiij

k

ycyayaakJj

kIikJjcya

yyk+k +

kk.

xcya

y

Lemma

Proof

0. = Therefore, . know we, Since

.0 ,hypothesisinduction By . then , If

. allfor that note we,1for (b) see To

and 0, (b)

1

1

1

11

1

1

1

1

1

kj

kjr

m

i

kiir

kjj

m

i

kiijj

m

i

kiij

m

i

kiij

m

i

kiij

kjj

m

i

kiij

xxrjcya

xcyacya

jyayak+

xcya

optf

yfyf

Ax = ycx

x

xzAyc

xxh

afoptfcx

hm

i

hi

AhA

Aj

Ahh

hA

n

jijmi

A

)1(

Thus, .10 that Note

0)(

and Then .iterations at stops Algorithm Suppose

.max = where

1

11

Lemma

Proof

Two observations

Primal and Dual

.0 ,0

where01

1 s.t. min

Ac

xAxcx

.0 ,0

where0,0

s.t.11 max

Ac

zyczyA

zy

1. Is redundant?z Yes!!!

Complementary-Slackness

LP. dualfor optimal is and LP, primalfor optimal is

ifonly and if0)()1()1(

0)()1()1(1

yx

xzyAcxzAxyxzyAcxzAxyycx

.Output

while.-end;1

,0);(max set

)(for and )(for set

;1 and if set

min

such that )( choose

}0|{)(

and 0}|{)(set

begin do feasible primenot is while

.0 0,),( 0,set Initially,

1

11

11

11

)(

1)(

)(

1

1

000

k

m

ij

kiij

kj

ki

ki

ki

ki

kr

kj

kj

m

kIiij

m

iiijj

kJjm

kIiir

m

iiirr

n

j

kjij

kj

k

x

kk

cyaz

kIiyykIiyy

xrjxx

a

yac

a

yac

kJr

xaikI

xjkJ

x

kzyx

))(( nmnO

2. Could we explain the approximation without using LP? Yes!!!

.Output

while.-end;1

)(for and )(for set

;1 and if set

min

such that )( choose

}1|{)(

and 0}|{)(set

begin do feasible primenot is while

.0 0, 0,set Initially,

11

11

)(

1)(

)(

1

1

00

kA

ki

ki

ki

ki

kr

kj

kj

m

kIiij

m

i

kiijj

kJjm

kIiir

m

i

kiirr

n

ji

kjij

kj

k

xx

kkkIiyykIiyy

xrjxx

a

yac

a

yac

kJr

bxaikI

xjkJ

x

kyx

do covered,not edgean exists thereiftices}chosen vernot {)( kJ

edge} coverednot {)( kI

of degree edge uncovered of weight remaining

jj

ight vertex weupdate

.Output

while.-end;1

)(for

and )(for set

;1 and if set

min

such that )( choose

}1|{)(

and 0}|{)(set

begin do feasible primenot is while

.0 0, 0,set Initially,

1

1

11

)(

1)(

)(

1

1

00

kA

ki

ki

ki

ki

kr

kj

kj

m

kIiij

m

i

kiijj

kJjm

kIiir

m

i

kiirr

n

ji

kjij

kj

k

xx

kkkIiyy

kIiyy

xrjxx

a

yac

a

yac

kJr

bxaikI

xjkJ

x

kyx

.while.-end

)at edges uncovered of (# xeach verteat weight update

};{

)(at edge uncovered of # of weight remaining

minimize to vertex choose edges}; uncovered{

};in not vertices{ exists edge uncovered while

. Initially,

jcc

rCCr

rCr

ICJ

C

jj

A Special Case

.2||2and || weight totalhascover ex every vert because is This

ion.approximat-2 iscover ex every vert ),deg( that caseIn

1 optEccE

jc

n

j

Weight Decomposition

.)(deg))(deg(

Hence,

).(deg))(deg(:parts in two decomposed is weight iteration,each In

1cov

1cov

1

covcov

n

jjeredun

n

jjeredunj

n

jjj

ereduneredunjj

xjxjcxc

jjcc

Covering Type

0 ,0 ,0 where

}1,0{

s.t. min

bAc

x

bAxcx

n

M × n

Packing Type

0 ,0 ,0 where

}1,0{

s.t. max

bAc

y

cyAyb

n

m × n

Thanks, End