1

A polynomial relaxation-type algorithm for linear programming

Sergei ChubanovUniversity of Siegen, Germanysergei.chubanov@uni-siegen.de

2

Relaxation method

x

Project the current point onto the half-spacegenerated by a constraint which is not satisfied:

Agmon, and Motzkin and Schoenberg (1954)

3

Relaxation method

is exponential

z

*x

4

Relaxation method

*xz

5

Outline

A strongly polynomial algorithm

which either finds a solution or

proves that there are no 0,1-solutions

A polynomial algorithm

for linear programming

10 x

bAx

uxu

b Ax -

6

dCx

bAx

Linear system

hx

k

1lll

m

1iii

k

1lll

m

1iii dbcah 0l

is induced by the system if and only if

)A,,A(a in1ii )C,,C(c nl1ll

7

1 dCx,bAx **

)r,z(Bxhx

:x*

an induced inequality(ii)

(i)

Given , construct one of the two objects:),r,z(

z r

Task

hxsuch that

8

bAx

zr

maxc2r

Elementary case

)z(px* 1

dCx

bAx*

*

irc

dzc

i

ii

r)z(pz

zproj)z(p }bAx|x{

maxc Cis a row of of max. length

9

bAx

z

T)z(pz:h )z(hp:

)z(p

r

Elementary case

ic:h id:

zr

r)z(pz rc

dzci

i

ii

r,zBxdxc ii

r,zBx

z)z(pzx)z(pz TT

ii dxc

10

Divide-and-conquer algorithm

1. If , then the elementary case. maxc2

r

,r

2

1,z2. D&C returns or 11xh *x

3. Calculate :z

r

2

1,zBxh|xr,zB 11

,r

2

1,z D&C returns or 22xh *x

4. Calculate withhx r,zBxhx

11

Recursion

rz

12

Recursion

Recursive call for the same center and a smaller radius

rz

r2

1

13

11xh

Recursion

Recursive call either produces an approximate solution or a valid inequality

rz

r2

1

14

11xh

Recursion

Recursive call either produces an approximate solution or a valid inequality

rz z

r2

1

15

11xh

Recursion

Recursive call for the same radiusand another center which isthe projection of the currentcenter onto the half-space.

rz z

r2

1

16

11xh

22xh

Recursion

The second recursive call either produces an approximate solution or an induced inequality

rz z

r2

1

17

11xh

22xh

21 h)1(hh

21 )1(

Recursion

Y

2211 xh,xh|xY

hx

rz z

r2

1

18

.0,hh 21

11xh 22xh

Recursion

The algorithm may failto construct an inducedinequality

19

max

2

crlog

…

……

maxc2

2

r

r

Depth of recursion

20

2

maxcrOAt most recursive calls

Running time

21

2

max2223 crNnmnnmmO

CN nonzero components of

bAx m equations

n variables

Running time

Azb)AA(Az)z(p 1TT

22

D&C algorithm

• Not faster than the relaxation method• Can solve the task, but not always

23

1t

dtCx

btAx

0

0

dCx

bAx

x

t

1 )t,x( )1,x( x

Parameterized system

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11t

dtCx

btAx

1

0

Strengthened parameterized system

D&C is applied to

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(I) 1t,dtCx,btAx 00

:),h( )r),t,x((B)t,x()t,x(h 00

:)t,x(

is induced by the strengthened parameterized system

)t,x(h

(II)

Task

),r),t,x(( 00 Given

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If D&C finds an approximate solution to the strengthened parameterized system

is an exact solution to the parameterized system

)t,x(

)t,x(

If D&C finds a solution

is a solution to the system in questionxt

1

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)1(ttdxc)t,x(ak

1ii

k

1iiii

bA

The two recursive calls at the iteration where it fails produce the inequalities

where a are linear combinations of the rows ofa

)1(ttdxc)t,x(ak

1ii

k

1iiii

and

0txctd)t,x(a

txctd)t,x(a

k

1iiii

k

1iiii

If D&C fails

= 0

= 021 hh

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Px*

0xcdxcdk

1i

*iii

k

1i

*iii

0 0 contradiction infeasible

0i or i*

i dxc 0i .dCx,bAx,x

dxc ii

or

0 0

)1,x( x

dCx,bAx|xP

If D&C fails

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)1(ttdxc)t,x(ak

1ii

k

1iiii

a is a linear combination of the rows of bA

)1()t,x(hk

1ii

)t,x(h is induced by the original parameterized system

If D&C returns an inequality

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)1()t,x(h:)t,x(Hk

1ii

)0,(0

)t,x(h:)t,x(H

The smaller balldoes not contain any solution of the original parameterized system

If D&C returns an inequality

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H),0,(distH),0,(dist 00

)1,x( **

k

1i

*iii

k

1ii

*

xcd

k

1ii

k

1i

*iii

* xcdh

If D&C returns an inequality

32

h

)1(H),0,(dist

k

1ii

0

H),0,(disth

hH),0,(dist

h

hH),0,(dist

hH),0,(dist

k

1ii

00

00

H),0,(distH),0,(dist 00

If D&C returns an inequality

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1rrH),0,(distH),0,(dist ** 00

1r)1,x( ****

** rx

If D&C returns an inequality

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1k1k

xcd

k

1ii

k

1i

*iii

k

1ii

*

i

1r3k2r *

1r1r1kr1rrH),0,(dist **** 0

1

Case 1.

If D&C returns an inequality

0

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1k2

xcd

1k

xcd*

lll

lk

1i

*iii

k

1ii

*

,max iil

1r1r1k2rH),0,(dist ** 0

2

1xcd *

ll ,Px* Case 2.

1r3k2r * 1

If D&C returns an inequality

36

i

1rH),0,(dist * 0

,il

2

1xcd,rx,Px *

ll***

or

*** rx,Px

Px1r1,x * Pxrx *

1r3k2r * 1

If D&C returns an inequality

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Pxrx * i

,il *ll rx,Px

2

1xcd

If D&C returns an inequality

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Algorithm

If D&C fails, then either no solutions or

If D&C generates an induced inequality, either no solutions or

*ll rx,Px

2

1xcd

Pxdxc ll

nll ZPxdxc

)1r(3k2r * 1

Repeated application of the following argument:

Pxrx *

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Algorithm

The algorithm either finds a solution or decides that there are no 0,1- solutions in strongly polynomial time

10 x

bAx

2

max2223 crNnnmnnmmO

1)1n(3n4r 2

1

2

1* )1n(r

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Algorithm

If the system is feasible and the bounds are tight, a solution can be found in strongly polynomial time

uxu

b Ax - 10

x

bxA

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n

n

n

K0

uK1x0

uK1x0

uuxx

bbAx

ux

bAx

0 (1) (2)

(1) is feasible if and only if (2) has an integer solution

jiij ubA!nK

General case

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ux

bxA

0

10

z

bzz2a i

n

1jjssjsl

1s

0l

l1ulog

0sjsij

j

(2) (3)

(2) has an integer solution if and only if (3) has an integer solution

122u1s

0l

l1ulog

0sjs

s1ulog

0sjsj

jj

0u j

By solving (3) we also solve (1) a polynomial algorithm for linear programming

Polynomial algorithm