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Con

stra

int P

rogr

amm

ing

Sho

rt C

ours

e fo

r Air

Pro

duct

s &

Che

mic

als

Ap

ril 2

00

3

John

Hoo

ker

Ca

rneg

ie M

ello

n U

niv

ers

ity

I.O

verv

iew

and

Suc

cess

Sto

ries

(1 h

our)

II.B

asic

Con

cept

s an

d P

robl

em F

orm

ulat

ion

(1.5

hou

rs)

III.

Alg

orith

mic

Idea

s (1

hou

r)

I. O

verv

iew

and

Suc

cess

Sto

ries

Wha

t is

Con

stra

int P

rogr

am

min

g?

•It

is a

rel

ativ

ely

new

tec

hnol

ogy

deve

lope

d in

the

com

pute

r sc

ienc

e an

d ar

tific

ial i

ntel

ligen

ce c

omm

uniti

es.

•It

has

foun

d an

impo

rtan

t rol

e in

sc

hedu

ling,

logi

stic

s an

d su

pply

cha

in

man

agem

ent.

Ea

rly C

omm

erc

ial S

ucce

sse

s

•C

onta

iner

por

t sch

edul

ing

(Hon

g K

ong

and

Sin

gapo

re)

•C

ircui

t des

ign

(Sie

men

s)

•R

eal-t

ime

cont

rol

(Sie

men

s, X

erox

)

App

lica

tions

•Jo

b sh

op s

ched

ulin

g

•Ass

embl

y lin

e sm

ooth

ing

and

bala

ncin

g

•C

ellu

lar

freq

uenc

y as

sign

men

t

•N

urse

sch

edul

ing

•S

hift

plan

ning

•M

aint

enan

ce p

lann

ing

•Airl

ine

crew

ros

terin

g an

d sc

hedu

ling

•Airp

ort

gate

allo

catio

n an

d st

and

plan

ning

•P

rodu

ctio

n sc

hedu

ling

chem

ical

sav

iatio

no

il re

finin

gst

eel

lum

ber

pho

togr

aphi

c p

late

stir

es

•T

rans

port

sch

edul

ing

(foo

d,

nucl

ear

fuel

)

•W

areh

ouse

man

agem

ent

•C

ours

e tim

etab

ling

App

lica

tions

•T

radi

tiona

l mat

hem

atic

al p

rogr

amm

ing

met

hods

ar

e ve

ry g

ood

at m

any

supp

ly c

hain

pro

blem

s bu

t of

ten

have

diff

icul

ty w

ith sc

hedu

ling

and

sequ

enci

ng.

Con

stra

int P

rogr

am

min

g vs

. M

ath

em

atic

al P

rogr

am

min

g

•C

onst

rain

t pro

gram

min

g ca

n ex

cel a

t sche

dulin

g an

d se

quen

cing

.

•T

he tw

o na

tura

lly c

an w

ork

toge

ther

in a

sup

ply

chai

n co

ntex

t.

•M

athe

mat

ical

pro

gram

min

g m

etho

ds r

ely

heav

ily

on n

umer

ical

cal

cula

tion.

The

y in

clud

e

•lin

ear

prog

ram

min

g (L

P)

•m

ixed

inte

ger

prog

ram

min

g (M

IP)

•no

nlin

ear

prog

ram

min

g (N

LP)

Con

stra

int P

rogr

am

min

g vs

. M

ath

em

atic

al P

rogr

am

min

g

•C

onst

rain

t pro

gram

min

g re

lies

heav

ily o

n co

nstr

aint

pro

paga

tion

(a f

orm

of l

ogic

al in

fere

nce)

.

•In

cons

trai

ntpr

ogra

mm

ing

:

•p

rog

ram

min

g = a

form

of c

ompu

ter

prog

ram

min

g

Pro

gra

mm

ing ≠

Pro

gra

mm

ing

•In

mat

hem

atic

alpr

ogra

mm

ing:

•p

rog

ram

min

g= p

lann

ing

•In

mat

hem

atic

al p

rogr

amm

ing,

equ

atio

ns

(con

stra

ints

) de

scrib

e th

e pr

oble

m b

ut d

on’t

tell

how

to

sol

ve it

.

Con

stra

int P

rogr

am

min

g vs

. M

ath

em

atic

al P

rogr

am

min

g

•In

com

pute

r pr

ogra

mm

ing,

a p

roce

dure

tel

ls h

ow

to s

olve

the

prob

lem

.

•In

con

stra

int p

rogr

amm

ing

, ea

ch c

onst

rain

t in

voke

s a

proc

edur

e th

at s

cree

ns o

ut u

nacc

epta

ble

solu

tions

.

Ma

jor

Vend

ors

•C

HIP

–G

ener

al s

tate

-of-

the-

art

cons

trai

nt p

rogr

amm

ing

•W

AR

EP

LAN

–W

areh

ouse

m

anag

emen

t

Ma

jor

Vend

ors

•C

PLE

X–

Line

ar p

rogr

amm

ing

(LP

) an

d m

ixed

inte

ger/

linea

r p

rogr

amm

ing

(MIL

P)

•S

olve

r–G

ener

al s

tate

-of-

the-

art

cons

trai

nt p

rogr

amm

ing

(CP

)

•S

ched

uler

–C

P w

ith s

pec

ial-p

urp

ose

sch

edul

ing

cons

trai

nts

•O

PL

Stu

dio

–M

od

elin

g fr

amew

ork

fo

r b

oth

CP

and

MIL

P

•O

DF

(Op

timiz

atio

n D

evel

op

men

t F

ram

ewo

rk)

–To

ol f

or

dev

elo

pin

g o

ptim

izat

ion

exte

nsio

ns f

or

SA

P’s

AP

O.

Ma

jor

Vend

ors

•X

PR

ES

S-M

P–

Line

ar,

nonl

inea

r an

d m

ixed

inte

ger

prog

ram

min

g

•X

PR

ES

S-X

L –

Spr

eads

heet

-bas

ed o

ptim

izat

ion

•M

osel

–F

ram

ewor

k fo

r in

tegr

atin

g so

lver

s

•D

ash

has

a co

oper

ativ

e ag

reem

ent

with

Cos

ytec

to

use

CH

IP.

Ma

jor

Vend

ors

•E

CLi

PS

e –

Hig

h-le

vel m

odel

ing

lang

uage

for

MIL

P a

nd C

P

•IC

Par

c is

an

exte

nsio

n of

Im

peria

l Col

lege

, Lon

don

Som

e S

ucce

ss S

torie

s

•T

hese

are

cho

sen

beca

use:

•T

hey

are

very

rec

ent

(rel

ease

d 2

wee

ks a

go).

•T

hey

illus

trat

e ho

w s

ched

ulin

g in

tera

cts

with

oth

er

aspe

cts

of s

uppl

y ch

ain.

•A

nd t

hus

how

CP

can

inte

ract

with

oth

er m

etho

ds.

•S

ince

they

are

par

t of a

gov

ernm

ent

(EU

) su

ppor

ted

proj

ect (

LIS

CO

S),

a fa

ir am

ount

of d

etai

l w

as r

elea

sed

to p

ublic

.

•All

are

solv

ed w

ith h

elp

of D

ash’

s M

osel

sys

tem

.

Pro

cess

Sch

edu

ling

and

L

ot S

izin

g a

t BA

SF

Man

ufac

ture

of

poly

prop

ylen

es in

3 s

tage

s

poly

mer

izat

ion

inte

rmed

iate

st

orag

e

extr

usio

n

Pro

cess

Sch

edu

ling

and

Lo

t Siz

ing

at B

AS

F

•M

anua

l pla

nnin

g (o

ld m

etho

d)

•R

equi

red

3 da

ys

•Li

mite

d fle

xibi

lity

and

qual

ity c

ontr

ol

•24

/7 c

ontin

uous

pro

duct

ion

•Va

riabl

e ba

tch

size

.

•S

eque

nce-

depe

nden

t ch

ange

over

tim

es.

Pro

cess

Sch

edu

ling

and

Lo

t Siz

ing

at B

AS

F

•In

term

edia

te s

tora

ge

•Li

mite

d ca

paci

ty

•O

ne p

rodu

ct p

er s

ilo

•E

xtru

sion

•P

rodu

ctio

n ra

te d

epen

ds o

n pr

oduc

t and

m

achi

ne

Pro

cess

Sch

edu

ling

and

Lo

t Siz

ing

at B

AS

F

•T

hree

pro

blem

s in

one

•Lo

t si

zing

–ba

sed

on c

usto

mer

dem

and

fore

cast

s

•Ass

ignm

ent

–pu

t eac

h ba

tch

on a

par

ticul

ar

mac

hine

•S

eque

ncin

g –

deci

de t

he o

rder

in w

hich

eac

h m

achi

ne p

roce

sses

bat

ches

ass

igne

d to

it

Pro

cess

Sch

edu

ling

and

Lo

t Siz

ing

at B

AS

F

•T

he p

robl

ems

are

inte

rdep

ende

nt

•Lo

t si

zing

dep

ends

on

assi

gnm

ent,

sinc

e m

achi

nes

run

at d

iffer

ent

spee

ds

•Ass

ignm

ent

depe

nds

on s

eque

ncin

g, d

ue to

re

stric

tions

on

chan

geov

ers

•S

eque

ncin

g de

pend

s on

lot s

izin

g, d

ue to

lim

ited

inte

rmed

iate

sto

rage

Pro

cess

Sch

edu

ling

and

Lo

t Siz

ing

at B

AS

F

•S

olve

the

prob

lem

s si

mul

tane

ousl

y

•L

ot s

izin

g:s

olve

with

MIP

(us

ing

XP

RE

SS

-MP

)

•A

ssig

nm

en

t:sol

ve w

ith M

IP

•S

eq

uen

cin

g:so

lve

with

CP

(us

ing

CH

IP)

•T

he M

IP a

nd C

P a

re li

nked

mat

hem

atic

ally

.

•U

se lo

gic-

base

d B

ende

rs d

ecom

posi

tion,

de

velo

ped

only

in th

e la

st fe

w y

ears

.

Sam

ple

sche

dule

, ill

ustr

ated

with

Vis

ual S

ched

uler

(A

viS

/3)

Sou

rce

: B

AS

F

Pro

cess

Sch

edu

ling

and

Lo

t Siz

ing

at B

AS

F

•B

enef

its

•O

ptim

al s

olut

ion

obta

ined

in 1

0 m

ins.

•E

ntire

pla

nnin

g pr

oces

s (d

ata

gath

erin

g, e

tc.)

re

quire

s a

few

hou

rs.

•M

ore

flexi

bilit

y

•F

aste

r re

spon

se t

o cu

stom

ers

•B

ette

r qu

ality

con

trol

Pa

int P

rodu

ctio

n a

t Ba

rdot

•Tw

o pr

oble

ms

to s

olve

sim

ulta

neou

sly

•Lo

t si

zing

•M

achi

ne s

ched

ulin

g

•F

ocus

on

solv

ent-

base

d pa

ints

, for

w

hich

ther

e ar

e fe

wer

sta

ges.

•B

arbo

t is

a P

ortu

gues

e pa

int

man

ufac

ture

r.

Sev

eral

mac

hine

s o

f ea

ch ty

pe

Pai

nt P

rod

uct

ion

at B

arb

ot

•S

olut

ion

met

hod

sim

ilar

to B

AS

F c

ase

(MIP

+ C

P).

•B

enef

its

•O

ptim

al s

olut

ion

obta

ined

in a

few

min

utes

for

20 m

achi

nes

and

80 p

rodu

cts.

•P

rodu

ct s

hort

ages

elim

inat

ed.

•10

% in

crea

se in

out

put.

•F

ewer

cle

anup

mat

eria

ls.

•C

usto

mer

lead

tim

e re

duce

d.

Dis

cre

te L

ot S

izin

g a

t Pro

cte

r a

nd G

am

ble

•C

ontin

uous

ly r

unni

ng p

rodu

ctio

n lin

e.

•M

anuf

actu

res

50 s

nack

foo

ds in

dis

cret

e ba

tche

s.

•A b

atch

may

con

sum

e on

e or

mor

e 8-

hr p

erio

ds.

•At m

ost o

ne b

atch

per

per

iod

(90-

150

perio

ds).

•S

eque

ncin

g co

nstr

aint

s.

Sim

plifi

ed p

rodu

ctio

n pr

oces

s

Sou

rce

: P

&G

Dis

cret

e Lo

t Siz

ing

at P

&G

•H

ighl

y va

riabl

e de

man

d

Sou

rce

: P

&G

Dis

cret

e Lo

t Siz

ing

at P

&G

•O

bjec

tive:

Cho

ose

batc

h se

quen

cing

and

siz

es s

o as

to:

•Avo

id e

xces

sive

inve

ntor

y ac

cum

ulat

ion.

•Avo

id la

rge

batc

h si

zes.

•S

olut

ion

met

hod

did n

otu

se C

P

•U

sed

a sp

ecia

lized

MIP

app

roac

h.

•H

owev

er,

this

is a

nat

ural

pro

blem

for

MIP

+ C

P.

Dis

cret

e Lo

t Siz

ing

at P

&G

•B

enef

its

•Im

prov

ed c

usto

mer

ser

vice

.

•C

ompl

ete

plan

ning

cyc

le f

or 1

50 p

erio

ds

redu

ced

from

4-5

hou

rs to

2 h

ours

.

•M

etho

d is

bei

ng e

xten

ded

to 1

3 lin

es w

ith

inte

rmed

iate

sto

rage

and

pac

king

.

Pro

duct

ion

Lin

e S

equ

enc

ing

at P

eug

eot

/Citr

oën

•T

he P

euge

ot 2

06 c

an b

e m

anuf

actu

red

with

12,

000

optio

n co

mbi

natio

ns.

•P

lann

ing

horiz

on is

5 d

ays

Pro

du

ctio

n Li

ne

Seq

uen

cing

at P

eug

eot/C

itroë

n

•E

ach

car

pass

es t

hrou

gh 3

sho

ps.

•O

bjec

tives

•G

roup

sim

ilar

cars

(e.

g. in

pai

nt s

hop)

.

•R

educ

e se

tups

.

•B

alan

ce w

ork

stat

ion

load

s.

Pro

du

ctio

n Li

ne

Seq

uen

cing

at P

eug

eot/C

itroë

n

•S

peci

al c

onst

rain

ts

•C

ars

with

a s

un r

oof

shou

ld b

e gr

oupe

d to

geth

er in

ass

embl

y.

•Air-

cond

ition

ed c

ars

shou

ld n

ot b

e as

sem

bled

con

secu

tivel

y.

•E

tc.

Pro

du

ctio

n Li

ne

Seq

uen

cing

at P

eug

eot/C

itroë

n

•P

robl

em h

as t

wo

part

s

•D

eter

min

e nu

mbe

r of

car

s of

eac

h ty

pe

assi

gned

to

each

line

on

each

day

.

•D

eter

min

e se

quen

cing

for

eac

h lin

e on

ea

ch d

ay.

•P

robl

ems

are

solv

ed s

imul

tane

ousl

y.

•Aga

in b

y M

IP +

CP

.

Sam

ple

sche

dule

Sou

rce

: P

eug

eot

/Citr

oën

Pro

du

ctio

n Li

ne

Seq

uen

cing

at P

eug

eot/C

itroë

n

•B

enef

its

•G

reat

er a

bilit

y to

bal

ance

suc

h in

com

patib

le b

enef

its a

s fe

wer

set

ups

and

fast

er c

usto

mer

ser

vice

.

•B

ette

r sc

hedu

les.

Lin

e B

ala

ncin

g a

t Pe

uge

ot/C

itroë

n

A c

lass

ic p

rodu

ctio

n se

quen

cing

pro

blem

Sou

rce

: P

eug

eot

/Citr

oën

Lin

e B

alan

cing

at P

eug

eot/C

itroë

n

•O

bjec

tive

•E

qual

ize

load

at w

ork

stat

ions

.

•K

eep

each

wor

ker

on o

ne s

ide

of t

he c

ar

•C

onst

rain

ts

•P

rece

denc

e co

nstr

aint

s be

twee

n so

me

oper

atio

ns.

•E

rgon

omic

req

uire

men

ts.

•R

ight

equ

ipm

ent a

t sta

tions

(e.

g. a

ir so

cket

)

Lin

e B

alan

cing

at P

eug

eot/C

itroë

n

•S

olut

ion

agai

n ob

tain

ed b

y a

hybr

id m

etho

d.

•M

IP:

obta

in s

olut

ion

with

out r

egar

d to

pr

eced

ence

con

stra

ints

.

•C

P: R

esch

edul

e to

enf

orce

pre

cede

nce

cons

trai

nts.

•T

he tw

o m

etho

ds in

tera

ct.

Sou

rce

: P

eug

eot

/Citr

oën

Lin

e B

alan

cing

at P

eug

eot/C

itroë

n

•B

enef

its

•B

ette

r eq

ualiz

atio

n of

load

.

•S

ome

stat

ions

cou

ld b

e cl

osed

, red

ucin

g la

bor.

•Im

prov

emen

ts n

eede

d

•R

educ

e tr

acks

ide

clut

ter.

•E

qual

ize

spac

e re

quire

men

ts.

•K

eep

wor

kers

on

one

side

of c

ar.

Why

Se

que

ncin

g is

Ha

rd

•T

here

are

man

y w

ays

to s

eque

nce

only

a fe

w jo

bs.

•5

jobs

can

be

sequ

ence

d 12

0 w

ays.

•10

jobs

can

be

sequ

ence

d 3,

628,

800

way

s.

•20

jobs

can

be

sequ

ence

d 2,

432,

902,

008,

176,

640,

000

way

s.

•F

ast

com

pute

rs a

re u

sele

ss a

gain

st t

his

expo

nent

ial

expl

osio

n.

•C

leve

r m

etho

ds c

an r

educ

e th

e w

ork,

but

onl

y up

to a

po

int.

Adv

ant

age

s of

Con

stra

int P

rogr

am

min

g

•U

sual

ly b

ette

r at

seq

uenc

ing

than

oth

er m

etho

ds.

•Add

ing

mes

sy c

onst

rain

ts m

akes

the

pro

blem

eas

ier.

•M

ore

pow

erfu

l mod

elin

g la

ngua

ge m

akes

form

ulat

ion

and

debu

ggin

g ea

sier

.

Dis

adv

ant

age

s

•C

anno

t dea

l with

cos

t/pr

ofit

optim

izat

ion.

•N

ot s

o go

od a

t res

ourc

e al

loca

tion,

task

ass

ignm

ent,

inve

ntor

y co

ntro

l.

Tre

nds

•C

onst

rain

t pro

gram

min

g w

ill b

ecom

e be

tter

kno

wn

to e

ngin

eerin

g an

d op

erat

ions

res

earc

h co

mm

uniti

es.

•S

olve

rs th

at fu

lly in

tegr

ate

MIP

and

CP

will

be

deve

lope

d (3

-5 y

ears

).

•H

euris

tic m

etho

ds w

ill b

e in

tegr

ated

with

CP

.

•M

IP/C

P/h

euris

tics

will

be

rega

rded

as

a si

ngle

te

chno

logy

.

II. B

asic

Con

cept

s an

d P

robl

em

For

mul

atio

n

A S

imp

le E

xam

ple

}4,

,1{

},

,{

diff

eren

t-

all

30

25

3su

bje

ct t

o

48

5m

in

32

1

32

1

32

1

…∈

≥+

++

+

jx

xx

xxx

x

xx

x

We

will

illu

stra

te h

ow s

earc

h, in

fere

nce

and

rela

xatio

n m

ay b

e co

mbi

ned

to s

olve

this

pro

blem

by:

•co

nstr

aint

pro

gram

min

g

•in

tege

r pr

ogra

mm

ing

•a

hybr

id a

ppro

ach

1. S

olv

e as

an

inte

ger

pro

gra

mm

ing

pro

ble

m

Sea

rch:

Bra

nch

on v

aria

bles

with

frac

tiona

l val

ues

in

solu

tion

of c

ontin

uous

rel

axat

ion.

Infe

ren

ce:

Gen

erat

e cu

ttin

g pl

anes

(co

verin

g in

equa

litie

s).

Rela

xatio

n:C

ontin

uous

(LP

) re

laxa

tion.

Rew

rite

prob

lem

usi

ng in

tege

r pr

ogra

mm

ing

mod

el:

Let y

ijbe

1 if

x i=

j, 0

oth

erw

ise.

ji

y

jy

iy

ijy

x

xx

x

xx

x ij

iij

jijj

iji

, a

ll},1,0{

4,,1

,1

3,2,1,1

3,2,1,

17

42

4su

bje

ct t

o

53

4m

in

3 14

1

5

1

32

1

32

1 ∈

=≤

==

==

≥+

++

+

∑∑

==

=

Co

ntin

uou

s re

laxa

tion

ji

yx

xx

xx

xx

xx

jy

iy

ijy

x

xx

x

xx

x

ij

iij

jijj

iji

, al

l1

,0

8

445

4,,1

,1

3,2,1,1

3,2,1,

17

42

4su

bje

ct to

53

4m

in

32

1

32

31

213 14

1

4

1

32

1

32

1

≤≤

≥+

+≥

+≥

+≥

+

=≤

==

==

≥+

++

+

∑∑

==

=

Rel

ax in

tegr

ality

Cov

erin

g in

equa

litie

s

Bra

nch

an

d b

ou

nd

(B

ran

ch a

nd

rel

ax)

The

incu

mb

en

t so

lutio

nis th

e be

st fe

asib

le s

olut

ion

foun

d so

far.

At e

ach

node

of t

he b

ranc

hing

tre

e:

•If

The

re is

no

need

to

bran

ch f

urth

er.

•N

o fe

asib

le s

olut

ion

in th

at s

ubtr

ee c

an b

e be

tter

th

an th

e in

cum

bent

sol

utio

n.

•U

se S

OS

-1 b

ranc

hing

.

Opt

imal

va

lue

of

rela

xatio

n ≥

Valu

e of

in

cum

bent

so

lutio

n

5.4

90

00

1

2/12/1

00

2/12/1

00

=

=

z

y

y 12 =

1y 1

3 =

1y 1

4 =

1

2.5

00

01

0

8.00

02.0

01

00

=

=

z

y

4.50

00

01

09.0

01.0

10

00

=

=

z

y

y 11 =

1

Infe

as.

Infe

as.

Infe

as.

z =

54

z=

51

Infe

as.

z =

52

50

02/1

02/1

10

00

00

10

=

=

z

y

Infe

as.

500

02/1

2/1

01

00

10

00

=y

8.5

00

10

0

15

/1

30

01

5/

2

00

10

=

=

z

y

Infe

as.

Infe

as.

Infe

as. Infe

as.

2. S

olv

e as

a c

on

stra

int p

rog

ram

min

g p

rob

lem

}4,

,1{}

,,

{di

ffere

nt-

all

302

53

48

5

32

1

32

1

32

1

…∈

≥+

+≤

++

jx

xx

xxx

xz

xx

x

Sta

rt w

ith z

= ∞

.W

ill d

ecre

ase

as fe

asib

le

solu

tions

are

fou

nd.

Sea

rch:

Dom

ain

split

ting

Infe

ren

ce:

Dom

ain

redu

ctio

n R

ela

xatio

n:C

onst

rain

t sto

re (

set o

f cu

rren

t var

iabl

e do

mai

ns)

Co

nst

rain

t sto

re ca

n be

vie

wed

as

cons

istin

g of

in-d

omai

n co

nstr

aint

s xj∈

Dj,

whi

ch f

orm

a r

elax

atio

n of

the

prob

lem

.

Glo

bal c

onst

rain

t

Dom

ain

redu

ctio

n fo

r in

equa

litie

s

•B

ound

s pr

opag

atio

n on

302

53

48

5

32

1

32

1

≥+

+≤

++

xx

xz

xx

x

impl

ies

For

exa

mpl

e,30

25

33

21

≥+

+x

xx

25

812

305

23

303

12

=−

−≥

−−

≥x

xx

So

the

dom

ain

of x 2

is r

educ

ed t

o {2

,3,4

}.

Dom

ain

redu

ctio

n fo

r al

l-diff

eren

t (e.g

., R

ég

in)

•M

aint

ain

hype

rarc

con

sist

ency

on

},

,{

diffe

rent

-al

l3

21

xx

x

Sup

pose

for

exam

ple:

Dom

ain

of x 1

Dom

ain

of x 2

Dom

ain

of x 3

12

12

1234

The

n on

e ca

n re

duce

th

e do

mai

ns:

12

12

34

•In

gen

eral

, so

lve

a m

axim

um c

ardi

nalit

y m

atch

ing

prob

lem

and

app

ly a

theo

rem

of

Ber

ge

z =

∞1234

234

1234

34

23

234

234

12

infe

asib

lex

= (

3,4

,1)

valu

e =

51

z =

∞z

= 5

2

Domain of x1

Domain of x2

Domain of x3

D2=

{2,3

}D

2={4

}

Dom

ain

of x 2

D2=

{2}

D2=

{3}

D1=

{3}

D1=

{2}

infe

asib

lex

= (

4,3

,2)

valu

e =

52

3. S

olv

e u

sin

g a

hyb

rid

ap

pro

ach

Sea

rch

:

•B

ranc

h on

frac

tiona

l var

iabl

es in

sol

utio

n of

re

laxa

tion.

•D

rop

co

nstr

aint

s w

ith y ij’s

. T

his

mak

es r

elax

atio

n to

o

larg

e w

itho

ut m

uch

imp

rove

men

t in

qua

lity.

•If

vari

able

s ar

e al

l int

egra

l, b

ranc

h b

y sp

littin

g d

om

ain.

•U

se b

ranc

h an

d bo

und.

Infe

ren

ce: •

Use

bou

nds

prop

agat

ion

for

all i

nequ

aliti

es.

•M

aint

ain

hype

rarc

con

sist

ency

for

all-

diffe

rent

co

nstr

aint

s.

Rela

xatio

n:

•P

ut k

naps

ack

cons

trai

nt in

LP

.

•P

ut c

over

ing

ineq

ualit

ies

base

d on

kn

apsa

ck/a

ll-di

ffere

nt in

to L

P.

}4,

,1{

8

546

},

,{

diff

eren

t-

all

30

25

3s.

t.

48

5m

in

32

1

32

31

21

32

1

32

1

32

1

…∈

≥+

+≥

+≥

+≥

+

≥+

+≤

++

jx

xx

x

xx

xx

xx

xx

xxx

x

zx

xx

Mo

del

for

hyb

rid

ap

pro

ach

Cov

erin

g in

equa

litie

s ge

nera

ted

with

all-

diff

Gen

erat

e an

d

pro

pag

ate

cove

ring

in

equa

litie

s at

ea

ch n

od

e o

f se

arch

tree

z =

∞ 2 3 4

3 4

2 3 4

4

3

2 4

x=

(3

,4,1

)va

lue

= 5

1

z=

∞ne

w c

ove

rs:

x 1+

x2≥ 7

x 1+

x3≥ 6

x 2+

x3≥ 5

z =

52

infe

asib

lex

= (

4,3

,2)

valu

e =

52

x=

(3

, 3,

3)

z =

51

x 2 =

3x 2

=4

2 3

4 4

1 2 3

x 1=

2x 1

=3

x=

(2

,4,2

)va

lue

= 5

0

Opt

imiz

atio

n a

nd C

onst

rain

t P

rogr

am

min

g C

ompa

red

•C

onst

rain

t typ

es.

•O

ptim

izat

ion

may

be

supe

rior

whe

n co

nstr

aint

s co

ntai

n m

any

varia

bles

(an

d ha

ve g

ood

rela

xatio

ns).

•C

onst

rain

t pro

gram

min

g m

ay b

e su

perio

r w

hen

cons

trai

nts

cont

ain

few

var

iabl

es (

and

prop

agat

e w

ell).

Op

timiz

atio

n a

nd

CP

Co

mp

ared

•E

xplo

iting

str

uctu

re

•O

ptim

izat

ion

relie

s on

dee

p an

alys

is o

f the

m

athe

mat

ical

str

uctu

re o

f sp

ecifi

c cl

asse

s of

pro

blem

s,

part

icul

arly

pol

yhed

ral a

naly

sis,

whi

ch y

ield

s st

rong

cu

ttin

g pl

anes

.

•C

onst

rain

t pro

gram

min

g id

entif

ies

subs

ets

of

prob

lem

con

stra

ints

that

hav

e sp

ecia

l str

uctu

re (

e.g.

, all-different, cumulative

) an

d ap

ply

tailo

r-m

ade

dom

ain-

redu

ctio

n al

gorit

hms.

Op

timiz

atio

n a

nd

CP

Co

mp

ared

•R

elax

atio

n an

d in

fere

nce.

•O

ptim

izat

ion

crea

tes

stro

ng r

elax

atio

ns w

ith c

uttin

g pl

anes

, Lag

rang

ean

rel

axat

ion,

etc

. T

hese

pro

vide

bou

nds

on th

e op

timal

val

ue.

•C

onst

rain

t pro

gram

min

g ex

ploi

ts th

e po

wer

of i

nfer

ence

, es

peci

ally

in d

omai

n re

duct

ion

algo

rithm

s. T

his

redu

ces

the

sear

ch s

pace

.

Op

timiz

atio

n a

nd

CP

Co

mp

ared

•M

odel

ing

styl

e

•O

ptim

izat

ion

uses

dec

lara

tive

mod

els

that

can

be

solv

ed w

ith a

var

iety

of a

lgor

ithm

s. B

ut t

he

lang

uage

is h

ighl

y re

stric

ted

(e.g

, in

equa

lity

cons

trai

nts)

.

•C

onst

rain

t pro

gram

min

g m

odel

s ar

e fo

rmul

ated

in

a q

uasi

-pro

cedu

ral m

anne

r th

at g

ives

the

use

r m

ore

oppo

rtun

ity t

o di

rect

the

solu

tion

algo

rithm

. B

ut th

e m

odel

is m

ore

clos

ely

tied

to th

e so

lutio

n m

etho

d.

Con

sist

enc

y

•A c

onst

rain

t set

is co

nsis

tent

if ev

ery

part

ial

assi

gnm

ent t

hat v

iola

tes

no c

onst

rain

t is

feas

ible

(i.

e., c

an b

e ex

tend

ed to

a fe

asib

le s

olut

ion)

.

•C

onsi

sten

cy is

not

the

sam

e as

feas

ibili

ty.

•C

onsi

sten

cy m

eans

that

all

infe

asib

le p

artia

l as

sign

men

ts a

re e

xplic

itly

rule

d ou

t by

a co

nstr

aint

.

•F

ully

con

sist

ent c

onst

rain

t set

s ca

n be

sol

ved

with

out b

ackt

rack

ing.

Gen

eral

Co

nsi

sten

cy

Con

side

r th

e co

nstr

aint

set

{0,1

}

01

10

01

10

01

∈≥

−≥

+

jx

xx

xx

It is

not

con

sist

ent,

beca

use

x 1=

0 v

iola

tes

no

cons

trai

nt a

nd y

et is

infe

asib

le (

no s

olut

ion

has

x 1=

0).

Add

ing

the

cons

trai

nt x 1=

0m

akes

the

set

con

sist

ent.

{0,1

}

sco

nst

rain

tot

her

01

10

01

10

01

≥−

≥+

jx

xx

xx

01

=x

11

=x

subt

ree

with

299no

des

but n

o fe

asib

le s

olut

ion

By

addi

ng t

he c

onst

rain

t x 1

= 0

, the

left

subt

ree

is

elim

inat

ed

Hyp

erar

c C

on

sist

ency

•A c

onst

rain

t set

is hy

pera

rc c

onsi

sten

tif e

very

va

lue

in e

very

var

iabl

e do

mai

n is

par

t of

som

e fe

asib

le

solu

tion.

•T

hat i

s, th

e do

mai

ns a

re r

educ

ed a

s m

uch

as

poss

ible

.

•If

all c

onst

rain

ts a

re “

bina

ry”

(con

tain

2

varia

bles

), h

yper

arc

cons

iste

nt =

arc

con

sist

ent.

•D

omai

n re

duct

ion

is C

P’s

big

gest

eng

ine.

Gra

ph c

olor

ing

prob

lem

s th

at c

an b

e so

lved

by

arc

cons

iste

ncy

mai

nten

ance

alo

ne.

A M

ode

ling

Exa

mpl

e

Tra

velin

g sa

lesm

an p

robl

em:

Let c

ij=

dis

tanc

e fr

om c

ity ito

city

j.

Fin

d th

e sh

orte

st r

oute

that

vis

its e

ach

of

n c

ities

ex

actly

onc

e.

Inte

ger

Pro

gra

mm

ing

Mo

del

}1,0{

},

,1{,

disj

oint

all

,1

al

l,1

al

l,1

subj

ect t

o

min

imiz

e

⊂≥

==

∑∑

∑∑∑ ∈∈

ijVi

Wj

ij

jij

iij

ijij

ij

x

nW

Vx

ix

jx

xc

Let x

ij=

1 if

city

iim

med

iate

ly p

rece

des

city

j,

0 ot

herw

ise

Sub

tour

elim

inat

ion

cons

trai

nts

Co

nst

rain

t Pro

gram

min

g M

od

el

Let y

k=

the

kth

city

vis

ited.

The

mod

el w

ould

be

writ

ten

in a

spe

cific

con

stra

int

prog

ram

min

g la

ngua

ge b

ut w

ould

ess

entia

lly s

ay:

},

,1{)

,,

(di

ffere

nt-

all

subj

ect t

o

min

imiz

e

1

1

ny

yy

c

k

n

ky

yk

k

∑+

Varia

ble

indi

ces

“Glo

bal”

cons

trai

nt

Ass

ignm

ent

prob

lem

with

two

linke

d fo

rmul

atio

ns

jj

x

yx

jy

ji

al

l ,

s'on

s

cons

trai

nt

s'on

s

cons

trai

nts.

t.

obje

ctiv

e

som

em

in

=

x i=

em

ploy

ee a

ssig

ned

time

sloti

y j=

tim

e sl

ot a

ssig

ned

empl

oyee

j

Mul

tiple

For

mul

atio

ns

Mu

ltipl

e F

orm

ula

tion

s

•T

he li

nkag

e of

two

mod

els

impr

oves

con

stra

int

prop

agat

ion.

•H

ere

a va

riabl

e (r

athe

r th

an a

con

stan

t) h

as a

var

iabl

e in

dex.

Ele

me

nt c

onst

rain

t

The

con

stra

int

5≤

ycca

n be

impl

emen

ted:

)),

,,

(,(

elem

ent

51

zc

cy

zn

…≤

Ass

ign

zth

e yt

h va

lue

in th

e lis

t

5≤

yxca

n be

impl

emen

ted:

The

con

stra

int

)),

,,

(,(

elem

ent

51

zx

xy

zn

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(thi

s is

a s

light

ly d

iffer

ent

cons

trai

nt)

Add

the

cons

trai

nt

z=

xy

Cum

ula

tive

con

stra

int

()

Lr

rd

dt

tn

nn

),,

,(

),,

,(

),,

,(

cum

ulat

ive

11

1…

……

•U

sed

for

reso

urce

-con

stra

ined

sch

edul

ing.

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tal r

esou

rces

con

sum

ed b

y jo

bs a

t any

one

tim

e m

ust n

ot e

xcee

d L.

Job

sta

rt ti

mes

Job

du

ratio

ns

Job

res

ou

rce

req

uir

emen

ts

Min

imiz

e m

akes

pan

(no

dead

lines

, all

rele

ase

times

= 0

):

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mak

espa

n =

8

L

1 23

4 5

time

reso

urce

s

Cu

mu

lativ

eC

on

stra

int

()

237

),2,2,3,3,3(),5,5,3,3,3(

),,

,(

cum

ula

tive

s.t.

min

51

51

+≥

+≥

tz

tz

tt

z ⋮

Job

star

t tim

esD

urat

ions

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ourc

es u

sed

L

Pro

duct

ion

Sch

edu

ling

Cap

acity

C1

Cap

acity

C2

Cap

acity

C3

Man

ufac

turin

gU

nit

Sto

rage

Tank

sP

acki

ngU

nits

Fill

ing

of S

tora

geTa

nk

Leve

l

tu

t +

(b

/r)

u +

(b

/s)

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ing

star

tsP

acki

ng s

tart

sF

illin

g en

dsP

acki

ng e

nds

Bat

ch s

ize

Man

ufac

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ring

rate

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king

ra

te

Nee

d to

enf

orce

ca

paci

ty c

onst

rain

t he

re o

nly

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0

,,

,,

,cu

mu

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e

al

l,

1

al

l,

),1,1(,

,cu

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e

all

,

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l,

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min

11

≥≥

+

−+

=≥

+≥

jj

nn

ii

iii

i

iii

ii

jj

jjj t

u

pe

sbsb

u

iC

us

rsb

it

sbu

v

mv

tj

Rt

jsb

uTT

……

Mak

esp

an

Job

rel

ease

tim

e

mst

ora

ge ta

nks

Job

du

ratio

n

Tan

k ca

pac

ity

pp

acki

ng u

nits

Exa

mp

le o

f M

od

el S

imp

lific

atio

n:

Lot s

izin

g &

sch

edu

ling

Day

:1

2

3

4

5

6

7

8

AB

A

Pro

duct

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t mos

t one

pro

duct

man

ufac

ture

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eac

h da

y.

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eman

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or e

ach

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ay.

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,

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l ,1

, al

l ,

, al

l

,

, al

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, al

l

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,

, al

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,

, al

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,

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1,

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ti

xi

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ti

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xs

sh

q

itt

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itit

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iti

iti

yy

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, al

l ,

0

, al

l ,0

,0

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ldin

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tup

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ntor

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f F

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r S

olu

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co

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•F

ind

optim

al s

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of c

ompo

nent

s to

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e up

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rodu

ct,

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ect t

o co

nfig

urat

ion

cons

trai

nts.

•U

se c

ontin

uous

rel

axat

ion

of e

lem

ent c

onst

rain

ts a

nd r

educ

ed

cost

pro

paga

tion.

Com

puta

tiona

l Res

ults

(O

tto

sso

n &

Th

ors

tein

sso

n)

0.010.1110100

1000

8x10

16x2

020

x24

20x3

0

Pro

ble

m

Seconds

CP

LEX

CLP

Hyb

rid

Exa

mp

le o

f F

aste

r S

olu

tion

:M

ach

ine

sch

edu

ling

•S

ched

ule

jobs

on

mac

hine

s th

at r

un a

t diff

eren

t sp

eeds

and

in

cur

diffe

rent

cos

ts.

•E

ach

job

has

a re

leas

e tim

e an

d de

adlin

e.

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inim

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cost

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gic-

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d B

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rs d

ecom

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tion.

()

ie

ix

Di

xt

jS

Dt

jR

t

C

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jjj

jx

j

jjj

jx

j

j

a

ll,

1,),

|(

),|

(cu

mu

lativ

e

a

ll,

a

ll,

s.t.

min

==

≤+≥

A m

odel

for

the

prob

lem

:

Rel

ease

dat

e fo

r jo

b j

Cos

t of

assi

gnin

g m

achi

ne

x jto

job

j

Mac

hine

ass

igne

d to

job j

Sta

rt ti

mes

of j

obs

assi

gned

to

mac

hine

i

Sta

rt ti

me

for

job j

Job

dura

tion

Dea

dlin

e

For

a g

iven

set

of a

ssig

nmen

ts

t

he s

ubpr

oble

m is

the

set o

f 1-

mac

hine

pro

blem

s,x

()

ie

ix

Di

xt

jij

jj

al

l,

1,),

|(

),|

(cu

mu

lativ

es.

t.

0m

in

==

Fea

sibi

lity

of e

ach

prob

lem

is c

heck

ed b

y co

nstr

aint

pr

ogra

mm

ing.

O

ne o

r m

ore

infe

asib

le p

robl

ems

resu

lts in

an

opt

imal

val

ue ∞.

Oth

erw

ise

the

valu

e is

zer

o.

Sup

pose

ther

e is

no

feas

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sch

edul

e fo

r m

achi

ne

. T

hen

jobs

ca

nnot

all

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ssig

ned

to m

achi

ne

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Sup

pose

in fa

ct th

at s

ome

subs

et

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ese

jobs

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nnot

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assi

gned

to

mac

hine

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The

n ad

d th

e co

nstr

aint

}|

{i

xj

j=

)(

so

me

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x

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ix

ij

∈≠

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ki

j

jj

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j

j

,,1

, al

l ),

(

som

efo

r

al

l,

al

l,

s.t.

min

…=

∈≠

≤+≥

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s yi

elds

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mas

ter

prob

lem

,

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s pr

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an b

e w

ritte

n as

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ixed

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}1,0{

al

l

},{

min

}{

max

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al

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= ij

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ix

jij

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j

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iy

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t

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k j

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co

nstr

aint

ad

ded

to

impr

ove

perf

orm

ance

Com

puta

tiona

l Res

ults

(J

ain

& G

ross

ma

nn

)

0.010.1110100

1000

1000

0

1000

00

12

34

5

Pro

ble

m s

ize

SecondsM

ILP

CP

OP

L

Hyb

rid

Pro

blem

siz

es

(jobs

, mac

hine

s)1

-(3

,2)

2 -

(7,3

)3

-(1

2,3)

4 -

(15,

5)5

-(2

0,5)

Eac

h da

ta p

oint

re

pres

ents

an

aver

age

of 2

inst

ance

s

MIL

P a

nd C

P ra

n ou

t of

mem

ory

on 1

of t

he

larg

est i

nsta

nces

Enh

ance

men

t Usi

ng “

Bra

nch

and

Che

ck”

(Th

ors

tein

sso

n)

Com

puta

tion

times

in s

econ

ds.

Pro

blem

s ha

ve 3

0 jo

bs, 7

mac

hine

s.

020406080100

120

140

12

34

5

Pro

ble

m

Seconds

Hyb

rid

Bra

nch

& c

heck

CP

Mo

del

ing

Exa

mp

le

•W

ill u

se IL

OG

’s O

PL

Stu

dio

mod

elin

g la

ngua

ge.

•S

hip

load

ing

prob

lem

•Lo

ad 3

4 ite

ms

on th

e sh

ip in

min

imum

tim

e (m

in

mak

espa

n)

•E

ach

item

req

uire

s a

cert

ain

time

and

cert

ain

num

ber

of

wor

kers

.

•To

tal o

f 8 w

orke

rs a

vaila

ble.

•E

xam

ple

is fr

om O

PL

Stu

dio

3.5

Lang

uage

Man

ual,

pp 3

08-3

10.

Item

Dur

a-

tion

Labo

r

13

4

24

4

34

3

46

4

55

5

62

5

73

4

84

3

93

4

102

8

113

4

122

5

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4

145

3

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3

163

3

172

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Item

Dur

a-

tion

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r

182

7

191

4

201

4

211

4

222

4

234

7

245

8

252

8

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3

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3

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3

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Pro

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dat

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1 →

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2 →

33

→5,

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→5

5 →

66

→8

7 →

88

→9

9 →

109

→14

10 →

1110

→12

11 →

1312

→13

13 →

15,1

614

→15

15 →

1816

→17

17 →

1818

→19

18 →

20,2

119

→23

20 →

2321

→22

22 →

2323

→24

24 →

2525

→26

,30,

31,3

226

→27

27 →

2828

→29

30 →

2831

→28

32 →

3333

→34

Pre

cede

nce

cons

trai

nts

Thi

s is

act

ually

a p

robl

em th

at u

ses

the

cum

ulat

ive

cons

trai

nt.

()

etc.

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8),3,,4,4(

),2,,4,3(

),,

(cu

mul

ativ

e

etc.

,43

subj

ect t

o

min

14

12

34

1

21

+≥

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tt

tt

ttt

z,

tzz

……

int ca

pa

city

= 8

;in

t n

bT

ask

s =

34

;ra

ng

e T

ask

s 1

..n

bT

ask

s;in

t d

ura

tion

[Ta

sks]

= [3

,4,4

,6,…

,2];

int to

talD

ura

tion =

su

m(t

in T

ask

s) d

ura

tion

[t];

int d

em

an

d[T

ask

s] =

[4

,4,3

,4,…

,3];

stru

ct P

rece

den

ces

{in

t b

efo

re;

int a

fte

r;

} {Pre

ced

en

ces}

se

tOfP

rece

de

nce

s =

{<

1,2

>, <

1,4

>, …

, <

33

,34

> }

;

sch

ed

ule

Ho

rizo

n =

to

talD

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ivity

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sks]

(du

ratio

n[t])

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iscr

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sou

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s(8

);A

ctiv

ity m

ake

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n(0

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kesp

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fora

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em

an

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lgor

ithm

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eas

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mai

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or

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elem

ent

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xx

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can

be p

roce

ssed

with

a d

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dom

ain

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gorit

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at m

aint

ains

hvp

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rwis

e}{

if

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{j

j

j

y

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x

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ain

of z

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1z

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xx

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initi

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ains

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50,40{

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80{

}20,

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10{}3{

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80{

4321

======

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Co

ntin

uo

us

rela

xatio

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f ele

men

t

is tr

ivia

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con

vex

hull

rela

xatio

n is

{}

{} i

ii

ic

zc

max

min

≤≤

11

11

+−

+

+−

∑∑

∑∑

∈∈

∈∈

kz

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kz

mmx y

y

yy

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Di

iii

Di

Di

iii

)),

,,

(,(

elem

ent

1z

cc

yn

)),

,,

(,(

elem

ent

1z

xx

yn

…ha

s th

e fo

llow

ing

rela

xatio

n

prov

ided

0 ≤

x i≤

mi (

and

whe

rek

= |D

y|).

)),

,,

(,(

elem

ent

1z

xx

yn

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≤x i

≤m

for

alli

, the

n th

e co

nve

x h

ullr

elax

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is

∑∑

∈∈

≤≤

−−

yy

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jD

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mk

x)1

(

plus

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whe

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|Dy|

.

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mpl

e...

50

50

)),

,(,

(el

emen

t

21

21

≤≤

≤≤

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xx

y

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con

vex

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rela

xatio

n is

:

50

50

5

21

21

21

≤≤

≤≤

+≤

≤−

+ xxx

xz

xx

Ifth

e ab

ove

rem

ains

val

id a

nd w

e ha

ve

40

1≤

≤x

20

45

92

04

52

12

1+

+≤

≤−

+x

xz

xx

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mp

le:

x y, w

here

Dy=

{1,

2,3}

and

50

40

30

321

≤≤

≤≤

≤≤

xxx

Rep

lace

x y w

ith z

and

elem

ent(y

,(x1,

x 2,x

3),z

)

Rel

axat

ion:

47

12

04

71

24

71

54

72

04

71

20

47

12

47

15

47

20

32

13

21

32

13

21

10

++

+≤

≤−

++

++

≤≤

−+

+x

xx

zx

xx

xx

xz

xx

x

Do

mai

n R

edu

ctio

n f

or A

ll-d

iffer

ent

),

(nt

alld

iffer

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nyy…

can

be p

roce

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with

an

algo

rithm

bas

ed o

n m

axim

um

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inal

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ipar

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d a

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rem

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erge

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mai

n R

edu

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r All-

diff

eren

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Con

side

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mai

ns

}6,5,4,3,2,1{

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54321

∈∈∈∈∈

yyyyy

y 1 y 2 y 3 y 4 y 5

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cate

dom

ains

with

edg

es

Fin

d m

axim

um c

ardi

nalit

y bi

part

ite m

atch

ing.

Mar

k ed

ges

in a

ltern

atin

g pa

ths

that

sta

rt a

t an

unco

vere

d ve

rtex

.

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k ed

ges

in a

ltern

atin

g cy

cles

.

Rem

ove

unm

arke

d ed

ges

not i

n m

atch

ing.

Indi

cate

dom

ains

with

edg

es

Fin

d m

axim

um c

ardi

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y bi

part

ite m

atch

ing.

y 1 y 2 y 3 y 4 y 5

1 2 3 4 5 6

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k ed

ges

in a

ltern

atin

g pa

ths

that

sta

rt a

t an

unco

vere

d ve

rtex

.

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k ed

ges

in a

ltern

atin

g cy

cles

.

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ove

unm

arke

d ed

ges

not i

n m

atch

ing.

Do

mai

n R

edu

ctio

n fo

r All-

diff

eren

t

Dom

ains

hav

e be

en r

educ

ed:

}6,5,4,3,2,1{

}5,1{

}5,3,2,1{

}5,3,2{}1{

54321

∈∈∈∈∈

yyyyy

}6,4{

}5{

}3,2{

}3,2{

}1{

54321

∈∈∈∈∈

yyyyy

Do

mai

n R

edu

ctio

n f

or

Cu

mu

lativ

e

Use

“ed

ge f

indi

ng”

tech

niqu

es.

Rel

axat

ion

of

cum

ula

tive

Whe

re t =

(t 1

,…,t n

)ar

e jo

b st

art t

imes

d=

(d 1

,…,d

n)ar

e jo

b du

ratio

nsr

= (

r 1,…

,rn)

are

reso

urce

con

sum

ptio

n ra

tes

Lis

max

imum

tota

l res

ourc

e co

nsum

ptio

n ra

tea

= (

a 1,…

,an)

are

earli

est s

tart

tim

es

()

Lr

dt

,,

,cu

mul

ativ

e

One

can

con

stru

ct a

rel

axat

ion

cons

istin

g of

the

follo

win

g va

lid c

uts.

If so

me

subs

et o

f job

s {j1,

…,j k

} ar

e id

entic

al (

sam

e re

leas

e tim

e a 0

, dur

atio

n d0,

and

res

ourc

e co

nsum

ptio

n ra

te

r 0),

then

[] 0

210

)1(

2)1

(1

dQ

Pk

Pa

Pt

tkj

j+

−+

+≥

++⋯

is a

val

id c

ut a

nd is

face

t-de

finin

g if

ther

e ar

e no

dea

dlin

es,

whe

re1

,0

=

=Qk

PrL

Q

The

follo

win

g cu

t is

valid

for

any

subs

et o

f job

s {j

1,…

,j k}

i

k i

ij

jd

Lri

kt

tk

∑ =

+−

≥+

+1

2121)

(1⋯

Whe

re th

e jo

bs a

re o

rder

ed b

y no

ndec

reas

ing

r jd j

.

Ana

logo

us c

uts

can

be b

ased

on

dead

lines

.

Exa

mp

le:

Con

side

r pr

oble

m w

ith fo

llow

ing

min

imum

mak

espa

n so

lutio

n (a

ll re

leas

e tim

es =

0):

Min

mak

espa

n =

8

L

1 23

4 5

time

reso

urce

s

0

6

23

3

5,5

,3,3

,3s.

t.

min

765

43

21

745

43

2

145

43

21

32

1

54

32

1

≥≥

++

++

≥+

++

≥+

++

≥+

++

++

++

≥ jt

tt

tt

t

tt

tt

tt

tt

tt

t

tt

tt

tzz

Rel

axat

ion:

Res

ultin

g bo

und:

17

.5m

akes

pan

≥=

z

Fac

et d

efin

ing

Rel

axin

g D

isju

nct

ion

s o

f Lin

ear

Sys

tem

s

()

kk

kb

xA

≤∨

(Ele

men

tis a

spe

cial

cas

e.)

Con

vex

hull

rela

xatio

n.

0

1

al

l,

=

=≤

kkkk

kk

kk

y

y

xx

ky

bx

Ak

Add

ition

al v

aria

bles

nee

ded.

Can

be

exte

nded

to n

onlin

ear

syst

ems.

“Big

M”

rela

xatio

n

0

1

al

l),

1(

=−

−≤

∑ kkk

kk

k

y

y

ky

Mb

xA

k

Whe

re (

taki

ng t

he m

ax in

eac

h ro

w):

k ik i

k ik i

xk

k ib

kk

bA

xA

M−

≤=

}'

al

l,

|{

max

max

''

Thi

s si

mpl

ifies

for

a di

sjun

ctio

n of

ineq

ualit

ies

whe

re 0

≤x j

≤m

j

()

kk

K kb

xa

≤=∨ 1

∑∑

==

−+

K k

kkK k

kkK

Mbx

Ma

11

1w

here

{}

jj

k jk

ma

M∑

=,0

ma

x

Exa

mp

le:

≤=∨

≤=∨

=1

080

ma

chin

e

larg

e

550

ma

chin

e

sma

ll

0

ma

chin

e

no

xz

xzx

Out

put o

f mac

hine

Fix

ed c

ost o

f mac

hine

Con

vex

hull

rela

xatio

n:

0,

110

5

80

50 3

2

32

32

32 ≥≤

++

≤+

yy

yy

yy

x

yy

z

x

z

Big

-M r

elax

atio

n:

0,

1

8050

55

51

0

10

10 3

2

32

32

3

2

32 ≥≤

+≥≥+

≤−

≤+

yy

yy

yz

yz

yx

yx

yy

x

x

z

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