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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
…≤
(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.
•To
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
):
Min
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
Res
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)
Fill
ing
star
tsP
acki
ng s
tart
sF
illin
g en
dsP
acki
ng e
nds
Bat
ch s
ize
Man
ufac
-tu
ring
rate
Pac
king
ra
te
Nee
d to
enf
orce
ca
paci
ty c
onst
rain
t he
re o
nly
()
0
,,
,,
,cu
mu
lativ
e
al
l,
1
al
l,
),1,1(,
,cu
mu
lativ
e
all
,
al
l,
s.t.
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
•A
t mos
t one
pro
duct
man
ufac
ture
d on
eac
h da
y.
•D
eman
ds f
or e
ach
prod
uct o
n ea
ch d
ay.
•M
inim
ize
setu
p +
hol
ding
cos
t.
0 ,
}1,0{ ,
,
al
l ,1
, al
l ,
, al
l
,
, al
l
,
, al
l
,1,
all
,1
, al
l
,
, al
l
,
, al
l
,
s.t.
min
1,
1,
1,
1,
1,,
≥∈
=≤≤≥
−+
≥−
≤≤−
≥+
=+
+
∑∑∑
−−
−
−
−
≠
itit
ijtit
itiit
itit
jtijt
ti
ijt
jtt
iijt
ti
it
itit
ti
itit
itit
itt
iit
ij
ijtij
itit s
xz
y
ty
ti
Cy
xt
iy
ti
yt
iy
yt
iy
zt
iy
zt
iy
yz
ti
sd
xs
qs
h
δ
δδδ
δ
Inte
ger
pro
gra
mm
ing
mo
del
(Wo
lsey)
Man
y va
riabl
es
()
()
ti
xi
y
ti
sC
x
ti
sd
xs
sh
q
itt
itit
itit
itt
iti
iti
yy
tt
, al
l ,
0
, al
l ,0
,0
, al
l ,
s.t.
min
1,
1
=→
≠≥
≤≤
+=
+
+
−
∑∑
−
Min
imiz
e ho
ldin
g an
d se
tup
cost
s
Inve
ntor
y ba
lanc
e
Hyb
rid
mo
del
Pro
duct
ion
capa
city
Exa
mp
le o
f F
aste
r S
olu
tion
:P
rod
uct
co
nfig
ura
tion
•F
ind
optim
al s
elec
tion
of c
ompo
nent
s to
mak
e up
a p
rodu
ct,
subj
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.
•M
inim
ize
cost
usi
ng lo
gic-
base
d B
ende
rs d
ecom
posi
tion.
()
ie
ix
Di
xt
jS
Dt
jR
t
C
jij
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
ible
sch
edul
e fo
r m
achi
ne
. T
hen
jobs
ca
nnot
all
be a
ssig
ned
to m
achi
ne
.
Sup
pose
in fa
ct th
at s
ome
subs
et
of th
ese
jobs
ca
nnot
be
assi
gned
to
mac
hine
.
The
n ad
d th
e co
nstr
aint
}|
{i
xj
j=
)(
so
me
for
x
Jj
ix
ij
∈≠
)(x
J i
i i
i
Kk
ix
Jj
ix
jS
Dt
jR
t
C
ki
j
jj
xj
jjj
jx
j
j
,,1
, al
l ),
(
som
efo
r
al
l,
al
l,
s.t.
min
…=
∈≠
≤+≥
∑
Thi
s yi
elds
the
mas
ter
prob
lem
,
Thi
s pr
oble
m c
an b
e w
ritte
n as
a m
ixed
0-1
pro
blem
:
}1,0{
al
l
},{
min
}{
max
,,1
, al
l ,1
)1(
al
l ,1
al
l,
al
l,
s.t.
min
∈
−≤
=≥
−≥
≤+≥
∑∑∑
∑
∑
= ij
jj
jj
ijj
ij
ix
jij
iij
ji
ijij
j
jjij
ijij
y
iR
Sy
D
Kk
iy
jy
jS
yD
t
jR
t
yC
k j
…
Valid
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
131
4
145
3
152
3
163
3
172
6
Item
Dur
a-
tion
Labo
r
182
7
191
4
201
4
211
4
222
4
234
7
245
8
252
8
261
3
271
3
282
6
291
8
303
3
312
3
321
3
332
3
342
3
Pro
blem
dat
a
1 →
2,4
2 →
33
→5,
74
→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.
,3,3
8),3,,4,4(
),2,,4,3(
),,
(cu
mul
ativ
e
etc.
,43
subj
ect t
o
min
14
12
34
1
21
+≥
+≥
+≥
+≥
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
ura
tion;
Act
ivity
a[t in
Ta
sks]
(du
ratio
n[t])
;D
iscr
ete
Re
sou
rce
re
s(8
);A
ctiv
ity m
ake
spa
n(0
);m
inim
ize
ma
kesp
an
.end
sub
ject
to
fora
ll(t in
Task
s)a
[t] p
rece
de
s m
ake
spa
n;
fora
ll(p
in s
etO
fPre
ced
en
ces)
a[p
.be
fore
] p
rece
de
s a
[p.a
fte
r];
fora
ll(t in
Task
s)a
[t] re
qu
ire
s(d
em
an
d[t])
re
s;};
III. A
lgor
ithm
ic Id
eas
Do
mai
n R
edu
ctio
n f
or
Ele
men
t
)),
,,
(,(
elem
ent
1z
xx
yn
…
can
be p
roce
ssed
with
a d
iscr
ete
dom
ain
redu
ctio
n al
gorit
hm th
at m
aint
ains
hvp
erar
c co
nsis
tenc
y.
othe
rwis
e}{
if
}|
{
{j
j
j
y
j
x
yz
x
xx
yy
Dj
xz
z
D
jD
DD
DD
jD
D
DD
D
=←
∅≠
∩∩
←
∩←
∈∪
Dom
ain
of z
Exa
mpl
e...
)),
,,
,(,
(el
emen
t4
32
1z
xx
xx
y
The
initi
al d
omai
ns a
re:
}70,
50,40{
}90,
80,50,
40{
}20,
10{
}50,
10{
}4,3,1{
}90,
80,60,
30,20{
4321
======
xxxxyz
DDDDDD
The
red
uced
dom
ains
are
:
}70,
50,40{
}90,
80{
}20,
10{
}50,
10{}3{
}90,
80{
4321
======
xxxxyz
DDDDDD
Co
ntin
uo
us
rela
xatio
n o
f ele
men
t
is tr
ivia
l.
The
con
vex
hull
rela
xatio
n is
{}
{} i
ii
ic
zc
max
min
≤≤
11
11
+−
≥
+
−
+−
≥
−
∑∑
∑∑
∈∈
∈∈
kz
mmx
kz
mmx y
y
yy
Di
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
…
If0
≤x i
≤m
for
alli
, the
n th
e co
nve
x h
ullr
elax
atio
n of
is
∑∑
∈∈
≤≤
−−
yy
Dj
jD
jj
xz
mk
x)1
(
plus
bou
nds,
whe
rek =
|Dy|
.
Exa
mpl
e...
50
50
)),
,(,
(el
emen
t
21
21
≤≤
≤≤
xxz
xx
y
The
con
vex
hull
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
Exa
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
e1
nyy…
can
be p
roce
ssed
with
an
algo
rithm
bas
ed o
n m
axim
um
card
inal
ity b
ipar
tite
mat
chin
g an
d a
theo
rem
of B
erge
.
Do
mai
n R
edu
ctio
n fo
r All-
diff
eren
t
Con
side
r th
e do
mai
ns
}6,5,4,3,2,1{
}5,1{
}5,3,2,1{
}5,3,2{}1{
54321
∈∈∈∈∈
yyyyy
y 1 y 2 y 3 y 4 y 5
1 2 3 4 5 6
Indi
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
.
Mar
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
nalit
y bi
part
ite m
atch
ing.
y 1 y 2 y 3 y 4 y 5
1 2 3 4 5 6
Mar
k ed
ges
in a
ltern
atin
g pa
ths
that
sta
rt a
t an
unco
vere
d ve
rtex
.
Mar
k ed
ges
in a
ltern
atin
g cy
cles
.
Rem
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
