Optimal sequencing of mixed models with sequence-dependent setups and utility workers on an assembly line

sequen e-dependent setups and utility workers

on an assembly line

May 22, 2009

Abstra t

This paperpresentsan integer programmingformulation for the

sequen -ingproblem in mixed-model assembly lines where the number of temporarily

hired utility workersand the numberof sequen e-dependent setups areto be

optimizedsimultaneouslythrougha ost fun tion. Theresultantmodeloers

an operational way to implement the utility work needed to avoid line

stop-pages, unlikeprevious papersaddressingthe goalof smoothingthe workload.

Thepresentresear hhasanimmediateappli ationtotheautomotiveindustry,

namelyto the ar-sequen ingproblem. Simulation resultsshow that the

pro-posed formulationleadsto theoptimumin areasonabletime forinstan esup

to15itemsandtosatisfa toryfeasiblesolutionsforsomeofthelargerproblems

we onsideredwithinamoderatetimelimit.

Keywords: Mixed-model assembly lines; Sequen ing; Spa ing onstraints;

Sequen e-dependentsetups;Utilityworkers;E onomi evaluation.

1 Introdu tion

Past resear h on the sequen ing problem in a mixed-model assembly line has

pri-marilyfo usedon two goals. Prior to the 1980's, themain goalwas (1): to smooth

the workload at ea h workstation through the line. This goal seeks to redu e line

stoppagesor ine ien ies like work ongestion or utility work (see Mitsumori,1969

or Xiaobo andOhno, 1997 for optimal models and for heuristi s, seeThomopoulos,

1967;Ma askill,1973;Sumi hrastetal.,1992;Smithetal.,1996andGottliebetal.,

2003).

TheemergingJIT on eptbythemid1980'shasraisedase ondgoalthat onsists

inkeepinga onstantrate ofpart usageto avoidlarge inventories. Goal (2)maybe

a hieved by: (2a)syn hronizing theprodu tion of ea hmodel withits demand and

(2b)levellingtheusageofea hpartatea hlevelofthemanufa turingpro ess. Ea h

goal(2a)and(2b)isusuallyexpressedasanobje tiveofminimizationofthevariation

(squareddeviations)ofthea tualprodu tion/usagefromthedesiredamount. Goals

(2a)and (2b)arestri tlyequivalent whenallprodu tsrequirethesamenumber and

mix of parts (see Miltenburg, 1989). Under this assumption, Goal (2)is optimally

Goal(2a)isoptimallysolvedbyKubiakandSethi,1991andbyBautistaetal.,2000;

heuristi methods tohandleGoal (2b)maybe foundinBautistaetal.,1996;Leuet

al.,1996 andMonden, 1998.

Goals(1)and(2)aresimultaneouslyaddressedinoptimalformulationsby

Kork-mazelandMeral,2001andbyDrexlandKimms, 2001 whoalsoprovide aheuristi

solutionapproa htotheproblembyusing olumn-generationte hniquestosolvethe

orrespondingLP-relaxation. Goals(1)and(2)areheuristi ally onsideredthrough

bi- riteriaapproa hesbyChoiandShin,1997;Tamuraetal.,1999 andbyKorkmazel

andMeral,2001.

Theadditionalobje tiveofminimizingtheset-up ost,namelyGoal(3),israrely

examined, possibly be ause it oni ts with both previous goals: a sequen e that

smooths the workloadneither ensuresa onstant rate ofpart usagenor a minimum

number of model swit hes (see Burns and Daganzo, 1997 for a dis ussion). A

heuristi approa hhandling both Goals(2)and(3)isdevelopedbyMansouri,2005.

Multi-obje tivemodelsin ludingthethreegoalsarequites ar e(Hyunetal.,1998).

The present paper onsiders Goals (1)and (3) inan optimal formulation ofthe

problemwheretheobje tivefun tionisa ostfun tionasso iatedwithhiring

tempo-rary/utilityworkerstoavoidlinestoppagesandwithmodelswit hesinthesequen e.

Theresear hpresentedherediersfromearlierstudiesinthesequen ingareain

sev-eral ways. First, as an e onomi ost fun tion is asso iated with the goals, the

hallenging task of assigning weights to the dierent obje tiveswhose measuring

unitsarenotidenti alisavoided(seeTamura etal.,1999orMurataetal.,1996for

adis ussiononmethodstosetweightstomulti-obje tive fun tions). Se ond,papers

thataddress the goal of smoothing the workload oftenexpress it asan obje tive of

minimizationofthetotalutilitywork(seeforinstan eHyunetal.,1998). Thislatter

obje tive onsistsinminimizingthenumberofhoursthatadditional workersshould

handletoavoidlinestoppagessonoindi ationonwhenandhowmanyworkersmust

temporarily be hired is provided. By ontrast, our formulation leads to assigning

utility workers to stations so as to pro ess some items in the sequen e. Spa ing

onstraintsareutilized hereto a ount for workstations apa itylimitations, likein

fewpapers (Smithet al.,1996; Choi and Shin, 1997; Drexl andKimms, 2001;

Got-tliebetal., 2003)although they aremore intuitive and operational than pro essing

time-basedrules.

Our work has a dire t appli ation to the ar sequen ing problem that onsists

in nding a sequen e of vehi les that minimizes the set-up osts while satisfying

apa ity ontraints along theassembly line. The set-up ostsof swit hingfrom one

model to another involve the ost of tuning up the ma hines and possibly the

ostof leaningspraygunsifthe olouris hanged. Assolvent pur hase onstitutesa

majorexpenseforautomotive bodyshops,thereisastrongin entive togroup

same- oloured vehi les to minimize the number of purges. In our experimental study,

we onsider this type of appli ation to generate instan es on whi h we apply the

proposedformulationthatissolvedusingthe ommer ialsoftwareMIPsolver

ILOG-Cplex9.0. Consideringthe optionof hiring utilityworkers insu h a ontext reates

The next se tion provides a general des ription of the sequen ing problem as

well as the usual onstraints relative to the demand satisfa tion and the apa ity

limitations. Se tion 3 presents the integer programming formulation designed to

a hieve Goals(1) and (3). Se tion 4 summarizes the optimization problemand the

assumptionswemade. These tionalsoin ludesausefultableofnotationsthereader

mayreferto. InSe tion 5wedes ribetheexperimentalframeworkandwe omment

thesimulation results. We on lude inSe tion 6.

2 Problem statement

Thisse tion isdedi ated to a general des ription of the sequen ing problem.

Con-straints relative to demand satisfa tion and apa ity limitations arethen provided.

Thenotationswewillusethroughout aresummarizedinTables8and9(seeSe tion

4).

2.1 General des ription

Inthe mixed-modelassembly line,some models of items have to be assembled over

ahorizon

T

(usuallya day) for whi ha total demand of

N

items must besatised. The y le time

θ

isdened as

θ = ⌊T /N ⌋

whi h means that items arelaun hed in xed

θ

time intervals to the assembly line.

The line is partitioned into

K

stations of three types. Subset

K

C

groups the stationswith axed apa ity andsubset

K

F

is omposedof exible stationswhose apa ity an be extendedwithutility workers temporarily hired. Stationsthat

nei-ther belongs to

K

C

nor

K

F

are those for whi h pro essing times never ex eed the y letime. Operationson su h stations involve doingexa tly thesame thing to all

items(installing safetybelts on allvehi lesis anexample).

A model or variant results from a ombination of options. Ea h option may

be fullypro essedonasinglestationor mayundergoseveraloperationsondierent

workstationstobe ompleted. Inthefollowing,wewill onsiderea hitemseparately

byletting

y

i,h

beabinaryvariablethattakesavalueof

1

ifitem

i

isinposition

h

in the sequen e,and

0

otherwise. Ourin entive isto allowany ombinationofoptions forthe ustomersothespa eof hoi es isnot limitedto apredeterminednumber of

variants.

Solving the sequen ing problem involvesnding an arrangement of the itemsin

thesequen e,saya

0 − 1

matrix

Y = {y

i,h

}

ofdimension

N × N

soastomeetseveral obje tiveslikeminimizingthenumberofutilityworkersandthesetups,levellingthe

utilization rate, subje t to a set of onstraints relative to the demand satisfa tion

and apa ity limitations.

2.2 Usual onstraints: demand satisfa tion and spa ing

require-ments

Werstprovidethe onstraintsrelativetothedemandsatisfa tion. Wethengivethe

2.2.1 Demand satisfa tion

The following onstraints ensure that the demand is satised over horizon

T

(ea h itemis sequen ed) and thatexa tly one produ t is assigned to ea h position inthe

sequen e(see for instan eDrexl andKimms, 2001)

N

X

h=1

y

i,h

= 1, i = 1, . . . , N,

(1)

N

X

i=1

y

i,h

= 1, h = 1, . . . , N.

2.2.2 General form of the spa ing onstraints

Spa ing onstraints are spe ied to avoidline stoppages. They di tate the spa ing

(i.e. thenumberofitemswithoutoption)betweentwo onse utiveitemswithoption.

Thespa ing onstraint

ν

o

: η

o

stipulates that `theremust be at most

ν

o

items with option

o

inanysequen e of

η

o

onse utive items'.

A spa ing onstraint for option

o

is required as soon as its pro essing time on one station

k

, denoted by

θ

o,k

is su h that

θ

o,k

> λ

k

· θ

, where

λ

k

is the number of permanent workers dedi ated to station

k.

When the previous inequality holds for several stations

k,

the spa ing onstraint for option

o

is di tated by the most apa itated station

k

o

pro essing this option

o,

with

k

o

∈ arg max

k∈K

C

∪K

F

{θ

o,k

},

and we usually have

ν

o

= λ

k

o

and

η

o

= ν

o

/d

o

,

where

d

o

is the average rate of demandfor option

o.

Consider for instan e one option

o = 1

and 3 stations

k = 1, 2, 3

. The y le time is

θ = 60

se . The number of permanent workers on ea h station is given by

λ

1

= 2; λ

2

= 4; λ

3

= 3

. We assume the demand rate for option 1 equal to 20%. Table1 providesfor ea h station

k

thepro essingtime

θ

1,k

andthevalueof

λ

k

· θ

.

Station

k

Pro ess. time

θ

1

,k

λ

k

· θ

Spa ing onstraint

ν

1

: η

1

1

127

120

2

200

240

3

190

180

3 : 15

Table1: Spa ing onstraintsforanoption

o

-illustration

Two stations

k = 1

and

k = 3

aresu h that

θ

1,k

> λ

k

· θ

, indeed for

k = 1

we have

127 > 120

andfor

k = 2,

we have

190 > 180

. Amongthesetwostations,station

k

1

= 3

isthestationforwhi h

θ

1,k

ismaximum. With

λ

3

= 3

wehave

ν

1

= 3

. Sin e

d

1

= 1/5

then

η

1

= 15

. For option 1, themost apa itated stationis station 3 and

thespa ing onstraintis

ν

1

: η

1

= 3 : 15

meaningthattheremustbeatmost3items withoption1 inany sequen eof 15items.

When a station

k

provides the tightest spa ing onstraint for more than one option,theseoptions aregrouped inasubsetandthetightestofthetightest spa ing

ommon assumption that ea h option is fully pro essed on a single station only

dedi atedto the installation of this option. Thus, inthefollowing,wewill make no

dieren ebetween option

o

and thestation

k

onwhi h theoptionistreated andwe willkeepthe notation

k

to designatetheoption pro essedon station

k.

Forinstan e,followingthepreviousexample,ifase ondoption

o = 2

isalsosu h thatstation3isthemost apa itatedstationwithaspa ing onstraint

ν

2

: η

2

= 3 : 4

, wethuswill onsideroption1or2asasingle`arti ial'option

o

pro essedonstation

k = 3

with a spa ing onstraint of

ν

o

: η

o

= 3 : 15

(sin e this is tighter than

ν

2

: η

2

= 3 : 4

).

Letting

δ

i,k

beaparameter thattakesavalueof

1

ifoption

k

isrequiredbyitem

i

and

0

otherwise, the spa ing onstraint is written as

h

X

l=h−η

k

+1

N

X

i=1

δ

i,k

· y

i,l

≤ ν

k

, h = 1, . . . , N ; k ∈ K

C

∪ K

F

.

(2)

Thespa ing onstraint(2)appliesforanyexiblestationaslongasnoutilityworker

operateson this station. This onstraint an alsoiteratively be dened as

m

k,h

= m

k,h−1

+

P

N

i=1

δ

i,k

· y

i,h

−

P

N

i=1

δ

i,k

· y

i,h−η

k

≤ ν

k

,

h = 1, . . . , N ; k ∈ K

C

∪ K

F

,

(3)

where

m

k,h

isthetotal numberofitemswithoption re orded intheset ofpositions (or window)

{h − η

k

+ 1, . . . , h} .

For thesake of larity, we will use expression (3) whenwe adaptthespa ing onstraint to the ase of exiblestations hosting utility

workers (see paragraph3.1.1, p. 8).

As an illustration, onsider

N = 6

items to be produ ed during horizon

T

. Four models are produ ed:

A, B, C, D

and models

B

and

D

require option

k = 1

. The spa ing onstraint is

ν

1

: η

1

= 1 : 2

. Table 2 provides a sequen e of models, where

h = −1, 0

are negative indi es hosen for theprevioushorizon

T − 1

. Indi es

h = 1, ..., 6

represents items'position during horizon

T.

For ea h position

h,

the value of

P

N

i=1

δ

i,k

· y

i,h

is simply equal to 1 if item in position

h

requires option

k

(with

k = 1

inour example)and zerootherwise. Constraint(2) onsistsin ounting the number of options

k

in any window of

η

k

items and he king if this number

is inferior to

ν

k

.

In onstraint (3), the number of items withoption inthe window

{h − η

k

+ 1, . . . , h}

equalsthe numberof items withoption intheprevious window to whi h we add the last element of the new window and from whi h we subtra t

therstelementoftheformer window, aswindows roll. Considering

h = 1

,we have item

D

inthe window

{0, 1}

thus

m

1,1

= 1

. For

h = 2,

we onsider window

{1, 2}

, we add

P

N

i=1

δ

i,k

· y

i,2

= 1

(last element of the new window) to

m

1,1

and sustra t

P

N

i=1

δ

i,k

· y

i,0

= 0

(rstelement oftheformer window)to obtain

m

1,2

= 2.

Notethe

Horizon

T − 1

T

h

...

−1

0

1

2

3

4

5

6

Item

D

C

D

B

A

B

D

C

P

N

i=1

δ

i,k

· y

i,h

1

0

1

1

0

1

1

0

Table2: Che kingthespa ing onstraints

2.2.3 Derivingthespa ing onstraintforstationswithasinglededi ated

worker

For the stations designed to host a single permanent worker

(λ

k

= 1),

spa ing on-straints anbederived from apa itylimitationsdened intermsofoperation time.

We note

θ

i,k

the pro essing timeof item

i

on station

k.

The operation time of the

h

th

item on station

k

is

P

N

i=1

θ

i,k

· y

i,h

.

We an reasonably assume that there are only two possible values for thepro essing time on station

k

:

θ

max

k

and

θ

min

k

whi h respe tively designates the operation time when option

k

is installed on any item

i

and when it is not, with

θ

min

k

< θ < θ

max

k

. Th e pro essing time an therefore be written as

P

N

i=1

θ

i,k

· y

i,h

= θ

max

k

P

N

i=1

δ

i,k

· y

i,h

+ θ

k

min

P

N

i=1

(1 − δ

i,k

) · y

i,h

,re alling

δ

i,k

= 1

ifitem

i

hasoption

k

and zero otherwise. Utilizing onstraint (1), we have

P

N

i=1

θ

i,k

· y

i,h

= (θ

max

_k

− θ

_k

min

)

P

N

i=1

δ

i,k

· y

i,h

+ θ

_k

min

. Thus, ifitem

i

has option

k,

itspro essing time onstation

k

equals

θ

max

k

;itequals

θ

min

k

otherwise. Let

R

max

k

be thedieren e between the total timededi ated to operations per-formed on station

k

and

θ.

This represents the extra time that an be spent on stationk, ompared to the average

θ.

To pro ess oneitemwithoption onstation

k,

this extratime

R

max

k

mustbe su hthat

R

max

k

≥ θ

k

max

− θ.

The ex ess workingtime omparedto the y letime,

R

k,h

,

after ompletion onstation

k

ofoperationsonthe

h

th

itemis given by

R

k,h

= max

(

R

k,h−1

+ (θ

k

max

− θ

min

k

)

N

X

i=1

δ

i,k

· y

i,h

+ θ

min

k

− θ, 0

)

.

(4)

The apa ityof station

k

isviolated bytheiteminposition

h

assoonas

R

k,h

>

R

max

_k

and

R

k,h−1

≤ R

max

k

, ∀ k ∈ K

C

∪ K

F

| λ

k

= 1

and

h = 1, . . . , N.

To illustrate, onsidera station

k

hosting asinglededi ated workerwith

θ

max

k

=

80

se .,

θ

min

k

= 50

se .,

θ = 60

se . and

R

max

k

, theextra timethat an be spent on station

k

is

R

max

k

= 40

se . Asoption

k

onsumesanadditionaltimeof

θ

max

k

−θ = 20

se . then 2 items withoption

k

may be produ ed onse utively. To at h up these

40

extra se onds, it is then ne esary to produ e 4 items without option, sin e the underuseof su h itemson station

k

is

θ − θ

min

k

= 10

se .

More formally, let us onsider the situation in whi h option

k

is required by the rst itemin the sequen e

(

P

N

i=1

δ

i,k

· y

i,1

= 1)

and

R

k,0

= 0

. From (4) we have

R

k,1

= θ

_k

max

− θ.

Ase onditemwithoptionmayfollowif

R

k,2

= 2(θ

max

k

− θ) ≤ R

max

k

. Finally, the maximum number

ν

k

of onse utive items with option

k

is su h that

ν

k

=

$

R

max

_k

θ

max

k

− θ

%

.

(5)

After

ν

k

onse utive items withoption, we have

R

k,ν

k

= R

max

k

and a number

µ

k

of itemswithoutoption mustfollowbeforesequen ingon eagain

ν

k

onse utive items requiringtheoption. Thisnumber

µ

k

issu hthat

(θ − θ

min

k

)µ

k

≥ (θ

k

max

− θ)ν

k

whi h nallyleads to

µ

k

=

&

ν

k

·

θ

max

_k

− θ

θ − θ

min

k

'

.

(6)

The spa ing onstraint (2) or (3) still applies with

ν

k

as determined by (5) and

η

k

= ν

k

+ µ

k

with

µ

k

dened by(6)forall stations

k

su hthat

λ

k

= 1

. Itshouldbe notedthatthisspa ing onstraintisane essaryandsu ient onditionnottoex eed

the apa ityonlywhen

ν

k

= 1

orequivalently when

R

max

k

< 2·(θ

max

k

− θ).

Otherwise, thespa ing onstraintisonlyasu ient ondition: awindow

{h − η

k

+ 1, . . . h}

may ontainmorethan

ν

k

itemswiththeoptionwithoutviolatingthe apa ity onstraint ofthe station,

R

k,l

≤ R

max

k

for all

l ∈ {h − η

k

+ 1, . . . h}

,with

R

k,l

dened by (4).

Letus onsideragaintheexampleofastation

k

hostingasinglededi atedworker with

θ

max

k

= 80

se .,

θ

min

k

= 50

se .,

θ = 60

se . and

R

max

k

= 40

se . From (5) and (6), we get

ν

k

= 2

and

µ

k

= 4

leading to

η

k

= 6.

The spa ing onstraint states ina stri t sense that after 2 onse utive items withoption, 4 items without option

mustbesequen edsoasto geta zeroex eeding pro essingtimewhi h allows again

for the sequen ing of 2 onse utive items with option and so on. The su ient

onditionnottoex eedthe apa itylimitationistosequen eatmost2vehi leswith

optionout of6 vehi les. Table3 displays a sequen eof items for whi h the spa ing

onstraint an be he ked for

h = 6

and

h = 7

. For

h = 6,

onstraint (2) gives

P

6

l=1

P

N

i=1

δ

i,k

· y

i,l

= 2 ≤ 2.

For

h = 7,

we get

P

7

l=2

P

N

i=1

δ

i,k

· y

i,l

= 2 ≤ 2.

Note the values of

R

k,l

never ex eed

R

max

k

= 40,

meaning that the apa ity limitation is not violated. However, pla ing an itemwithoption inposition

7

wouldhave led to

R

k,7

= 40 ≤ R

max

_k

,

witha numberof

3

items withoption in thewindow

{2, . . . , 7}.

Thisillustratesthe fa t thatthe spa ing onstraint isa su ient ondition to meet

the apa ity onstraint but it isnot ane essary ondition when

ν

k

> 1.

Position

l

1

2

3

4

5

6

7

P

N

i=1

δ

i,k

· y

i,l

0

1

0

0

0

1

0

R

k,l

10

30

20

10

0

20

10

Table3: Spa ing onstraint with no utility worker

3 Optimal formulation of the sequen ing problem

Theobje tivefun tion onsideredhereisa ostfun tioninvolvingtwoelements: the

without entailing line stoppages, somore grouping possibilities areavailable, hen e

redu ing thenumberof setups.

Werstprovidethe analyti aldes riptionoftheproblemofoptimizingthe

num-berofutilityworkerstobetemporarilyhired. Wethenturntotheformaldes ription

oftheproblemrelativeto theminimization ofthenumberofsetupsinthesequen e.

3.1 Obje tive and onstraints relative to utility workers

Paststudieshave onsideredthe obje tiveofminimizingtotalutilityworkexpressed

asaglobalworkingtimeandminimizedassu h,withoutevaluatingitse onomi

im-pa tasnowage ostsareintrodu ed (seeforinstan eHyunetal. 1998). Wehandle

theobje tive dierently: our in entive is to minimize thenumberof utilityworkers

required to implement sequen ing de isions taken for horizon

T

. Theresultant for-mulationgivesthe optimal assignment of utilityworkers on stationsand items. We

rst provide an adaptation of onstraints (3) to the ase of exible stations (from

the set

K

F

)

. We thenderive theoptimalnumberof utilityworkers.

3.1.1 Spa ing onstraints forstations with utility workers

Weassumethatea hexiblestationhosts asinglededi atedworker

(λ

k

= 1)

sothe spa ing onstraint

ν

k

: η

k

is dened by using Eq. (5) to determine

ν

k

and Eq. (6) togetthe valueof

η

k

= ν

k

+ µ

k

. Ea hexible station

k ∈ K

F

is on eivedto hosta singleutilityworker by y le time.

Let

w

k,h

beabinaryvariablethattakesavalueof

1

ifautilityworkerisrequired onstation

k

forthe iteminposition

h

and

0

otherwise. Weassumethat workers do not ollaborate on the same task: the item in position

h

is fully pro essed by the utilityworkerwhiletheregularworkernishestheoperationsonthe

(h − 1)

th

item.

A utility worker is required on station

k

to pro ess the item in position

h

in orderto avoida violationof apa itywhi ho urswhen

R

k,h−1

+ θ

max

k

− θ > R

max

k

. Theitemin position

h

isne essarilyan itemwithoption (see Eq. (4)). We assume the regular worker at hes up on thetotal ba klog

R

k,h−1

(while theutility worker fullyhandlestheiteminposition

h

). Thisassumptionmakespossibleanadaptation of the spa ing onstraints otherwise a reasoning in terms of ex ess working time

mustbeadopted andspa ing onstraintsmust be abandoned. Formally,we assume

R

k,h−1

≤ θ

(the ex ess working time an not be superior to the `average' timefor produ inganitem). Asastraightforwardupperboundfor

R

k,h−1

equals

(θ

max

k

−θ)ν

k

,

wenallyassume

(θ

max

k

− θ)ν

k

≤ θ.

Thisimpliesthatthepresen eofautilityworker on station

k

for item inposition

h

leads to a redu tion of the ex ess working time by an amount of

(θ

max

k

− θ)(ν

k

+ 1)

where

(θ

max

k

− θ)ν

k

is handled by the regular worker and

θ

max

k

− θ

isabsorbed bytheutilityworker himself. Thisisequivalentto ` an elling'

ν

k

+ 1

itemswithoptioninthe window

{h − η

k

+ 1, . . . , h}

sin ethenext iteminposition

h + 1

maybe hosenasiftherewasnoitemwithoptioninpositions

h − η

k

+ 1

to

h.

To illustrate thereasoning, onsider a station

k

with

θ

max

k

= 80

se .,

θ

min

k

= 40

se .,

θ = 60

se . and

R

max

aught up. The regularworker handlesthis rstitemand needsto work 20se onds

morethanthe y letimeto omplete theitem. Ifase onditemwithoptionistobe

pro essed, the umulated ex ess working time would be

40

se onds, whi h is twi e the extratime thatis allowed. A utility worker istherefore ne esary to handle this

se onditem, sowe have

w

k,2

= 1

. Theutilityworkerfullypro esses this item, thus absorbinganextratimeof20se . whiletheregularworker annishhisworkonthe

rst item, therefore absorbing also an extratime of 20 se onds. A third itemwith

option anthereforebepro essed, asthepresen eoftheutilityworkernallyallows

forabsorbingthe equivalent ofthe extratime neededtopro ess 2itemswithoption

(oneextratimeisdonebytheutilityworkerhimselfandtheotheronebytheregular

worker). Thepresen eoftheutilityworkerfor

h = 2

(se onditem)isthusequivalent to an elling

ν

k

+ 1 = 2

itemswithoptioninthewindow

{h − η

k

+ 1, . . . , h} = {1, 2}.

Thus,toadapt onstraint(3),wemustde rease

m

k,h

by

(ν

k

+ 1) w

k,h

andrepla e

P

N

i=1

δ

i,k

· y

i,h−η

k

with a proper binary variable, namely

q

k,h−η

k

,

that takes into a ount the fa t that a possible utility worker has led to the ` an ellation' of the

option,ifany,on the

(h − η

k

)

th

item. Letting

q

k,h−η

k

be su ha binaryvariable,the spa ing onstraints forexible stations an nowbe written as

m

k,h

= m

k,h−1

+

P

N

i=1

δ

i,k

· y

i,h

− q

k,h−η

k

− (ν

k

+ 1) w

k,h

≤ ν

k

,

h = 1, . . . , N ; k ∈ K

F

,

(7)

where

q

k,h−η

k

takesa value of

1

ifitem inposition

h − η

k

has theoption and none ofthe items inpositions

{h − η

k

, . . . , h − 1}

was handledbya utility worker, and

0

otherwise. Formally,thevariable

q

k,h−η

k

orrespondsto thefollowing denition

q

k,h−η

k

=

N

X

i=1

δ

i,k

· y

i,h−η

k

h−1

Y

l=h−η

k

(1 − w

k,l

).

(8)

Thisexpression islinearizedusing thethreeinequalities

q

k,h−η

k

≤

P

N

i=1

δ

i,k

· y

i,h−η

k

,

q

k,h−η

k

≥

P

N

i=1

δ

i,k

· y

i,h−η

k

−

P

h−1

l=h−η

k

w

k,l

,

2q

k,h−η

k

≤ 1 +

P

N

i=1

δ

i,k

· y

i,h−η

k

−

P

h−1

l=h−η

k

w

k,l

.

(9)

To illustrate, onsider the example in Table 4 for a station

k ∈ K

F

and

ν

k

=

µ

k

= 1

sothespa ing onstraint is

ν

k

: η

k

= 1 : 2.

Table4 providesthe value ofthe variables in luded inEq. (7)for asequen e of

3

itemsea h requiringtheoption.

For

h = 1

onstraint (3)gives

m

k,1

=

P

N

i=1

δ

i,k

· y

i,1

− 2w

k,1

,

sin e

m

k,0

and

q

k,−3

have negative indi es, they are ignored. Assuming there is no initial ondition, we

setto zeroallvariables withnegative indi esinEq. (7). However, initial onditions

(i.e. past de isions) will be taken into a ount in theexperiment sothis point will

be dis ussed further in Se tion 5. We have

m

k,1

= 1 ≤ ν

k

= 1

with

w

k,1

= 0.

Notethat

w

k,1

ould have been equal to 1without violatingthespa ing onstraint. However,the obje tiveistominimizethe ostasso iatedwithtemporary workersso

Position

h

1

2

3

P

N

i=1

δ

i,k

· y

i,h

1

1

1

m

k,h

1

0

1

w

k,h

0

1

0

q

k,h−η

k

/

/

0

Table 4: Spa ing onstraintforaexiblestation

k ∈ K

F

hostingautilityworker

variables

{w

k,h

}

will always take their minimum value while satisfying the spa ing onstraints. For

h = 2,

we have

m

k,2

= m

k,1

+

P

N

i=1

δ

i,k

· y

i,2

− 2w

k,2

= 0

with

m

k,1

= 1

and

w

k,1

= 1.

A utility worker is required to pro ess the se ond item as the spa ing onstraint is violated sin e items in position 1 and 2 both require

the option. For

h = 3,

we get

m

k,3

= m

k,2

+

P

N

i=1

δ

i,k

· y

i,3

− q

k,1

− 2w

k,3

with

q

k,1

=

P

N

i=1

δ

i,k

· y

i,1

Q

2

l=1

(1 − w

k,l

)

as dened by Eq. (8). We have

q

k,1

= 0

sin e therstitemhastheoptionbutautilityworkerwaspresentfor these onditemand

this leads to ` an elling' the option on item 1 and 2. We have

m

k,3

= 1 ≤ 1

with

w

k,3

= 0.

3.1.2 Optimal number of utility workers

In this paragraph we rst provide a formulation of the obje tive fun tion relative

to the ost of hiring temporary workers for horizon

T

whi h is usually a day of produ tion. Thisobje tivefun tionin ludesvariablesrepresentingthetotalnumber

ofutilityworkershiredperhalf-days. Wethenderivesuitable onstraintstolinkthese

variables to the binary variables

{w

k,h

}

ree ting the presen e (or the absen e) of utility workers onsome station anditem.

Let

W

1

(T )

bethenumberofa tivetemporaryworkersinthersthalfofhorizon

T.

Analogously,

W

2

(T )

designates thenumber ofutilityworkersrequired to pro ess items in the se ond half of horizon

T.

We nally dene

W

3

(T )

as the number of utility workers thatareneededto implement sequen ing de isions made for horizon

T

butphysi allypresentonlyinthenexthorizon

T + 1

. Theobje tiveistominimize

ω · [W

1

(T ) − W

3

(T − 1) + W

2

(T ) + W

3

(T )] ,

(10)

where

ω

is the ost of hiring a utility worker for one half-day, and the dieren e

W

1

(T ) − W

3

(T − 1)

represents the number of utility workers resulting from hiring de isions made for

T

and a tually present during the rst half of horizon

T.

The e onomi fun tion (10) therefore refers to hiring de isions aimed at implementing

the sequen ing of items for horizon

T

only. We shall now spe ifythe relationships between variables

{w

k,h

}

and variables

W

1

(T ), W

2

(T ), W

3

(T ).

Anytemporary workerrequiredonstation

k

hasa workingtimeof

θ

max

k

on ea h itemrequiringtheoption. Thisimpliesthatheisassignedtostation

k

foranumber of y letimesequalsto

β

k

=

θ

max

k

/θ

 .

Thusautilityworkerthathandlesonstation

k

the

h

th

iteminthe sequen ebe omesanewavailable forperformingoperationson

the

(h + β

k

)

th

item, ifneed be, eitheron the same station or on any other station,

β

k

≥ 2

sin e

θ

max

k

> θ.

From the assumption

(θ

max

k

− θ)ν

k

≤ θ

we made in the previous paragraph, we get

θ

max

k

/θ ≤ 1 + 1/ν

k

with

1 + 1/ν

k

taking a maximum value of

2

when

ν

k

= 1

. We therefore have

β

k

= 2, ∀k.

When

ν

k

> 1,

it should be notedthat the utility worker spends lessthan

2

y le times on station

k

sothe remaining time an be onsidered as a free time for moving from one station to

another.

Re allthehorizon isdividedinto

N

y letimes. During the

h

th

y letime,item

inposition

h − k + 1

ispro essedonstation

k.

Forinstan e,withnoinitial ondition,

N = 6

and

K

F

= {1, 2},

in the rst y le time, the rst item in the sequen e is pro essed on station

1

and no item is pro essed on station

2.

In the se ond y le time, the se ond item is pro essed on station 1 while item 1 rosses station 2 et .

Asautility workerstays

β

k

= 2

y letimesonanystationstation

k

ea h timehe is required, the sum

w

k,h−k

+ w

k,h−k+1

givesthe number of utility workers operating on station

k

in the

h

th

y le time. The sum

P

k∈K

F

(w

k,h−k

+ w

k,h−k+1

)

therefore representsthe total numberof a tiveutilityworkerson all stationsinthe

h

th

y le.

Itfollows that thenumberof utilityworkers neededinthe rsthalf ofhorizon

T

is

W

1

(T ) = max

h=1,...,[N/2]

n

P

k∈K

F

(w

k,h−k

+ w

k,h−k+1

)

o

. A similar reasoning holds

for

W

2

(T )

and

W

3

(T )

. Thelinearversionof the onstraintsasso iated with

W

1

(T ),

W

2

(T ), W

3

(T )

is

P

k∈K

F

(w

k,h−k

+ w

k,h−k+1

) ≤ W

1

(T ), h = 1, . . . ,



_N

2

 ,

P

k∈K

F

(w

k,h−k

+ w

k,h−k+1

) ≤ W

2

(T ), h =



_N

2

 + 1, . . . , N,

P

k∈K

F

P

h−k+1

l=h−k

l≤N

w

k,l

≤ W

3

(T ), h = N + 1, . . . , N + K − 1.

(11)

Let us note that at the end of horizon

T,

part of the sequen e that has been de idedin

T

isonly pro essed and ompleted in

T + 1.

Thethird inequalityin(11) ree ts the hiring de isions referred to the sequen ing of these items although the

orrespondingutilityworkerswill onlybe present atthebeginningofhorizon

T + 1.

3.1.3 Illustration

Let us onsider a simple example with

K = 3

stations, where stations

k = 1

and

k = 3

areexibleand

k = 2

isneitheraxed- apa itystationnoraexibleone. We assume that

N = 6

items are sequen ed during ea h horizon

T − 1, T

and

T + 1.

Three horizons are examined to larify the interdependen e between past, present

and future de isions. With two options (

k = 1

,

k = 3)

, 4 models of items are available: item

A

has no option, item

B

requires option

k = 1

only; option

k = 3

is ex lusively installed on item

C

and item

D

in ludes both options. We assume

ν

1

= ν

3

= 1

and

µ

1

= 1, µ

3

= 2.

Table 5 illustrates the satisfa tion of onstraints (7) for stations

k = 1

and

k = 3,

for ea hposition

h

where

h = −5, . . . , 0

are thepositions of items for whi h sequen ing de isions are made during horizon

T − 1,

de isions for

h = 1, . . . , 6

are made for horizon

T,

et . We indeed adopt the onvention that negative positions

h ≤ 0

refer to past sequen ing de isions whereas positions

h > N

are relative to futuresequen ing. Table 5also displaysthevalues of

{w

k,h

} .

For instan e, we have

w

1,2

= 1

meaning that a temporary worker must pro ess on station

1

the item in these ondpositioninthe sequen ewhi h isde ided duringhorizon

T.

Thisitema model

B

in ludesoption

k = 1

justlikethepreviousone (amodel

D

)whi hwould haveledtoaviolationofthespa ing onstraint

ν

1

: η

1

(i. e.

1 : 2)

ifnoworker ould have temporarily been hired.

P

N

i=1

δ

i,k

y

i,h

m

k,h

≤ν

k

= 1

w

k,h

q

k,h−η

k

Hor.

h

Model

k = 1

k = 3

k = 1

k = 3

k = 1

k = 3

k = 1

k = 3

−5

D

1

1

/

/

0

0

/

/

−4

A

0

0

1

/

0

0

/

/

T −1

−3

B

1

0

1

1

0

0

1

/

−2

D

1

1

0

1

1

0

0

1

−1

C

0

1

0

0

0

1

0

0

0

C

0

1

0

1

0

0

0

0

1

D

1

1

1

0

0

1

0

0

2

B

1

0

0

0

1

0

0

0

T

3

A

0

0

0

0

0

0

0

0

4

B

1

0

1

0

0

0

0

0

5

D

1

1

0

1

1

0

0

0

6

C

0

1

0

0

0

1

0

0

7

D

1

1

1

1

0

0

0

0

8

B

1

0

0

1

1

0

0

0

T +1

9

A

0

0

0

1

0

0

0

0

10

D

1

1

1

1

0

0

0

1

11

B

1

0

0

1

1

0

0

0

12

A

0

0

0

1

0

0

0

0

Table 5: Spa ing onstraintsfortwof lexiblestations

Table 6showsfor ea h y letimethe items positions thatarepro essedon ea h

station. For instan e, itemin position

h = −5

ispro essed on station

k = 1

inthe rst y le time of horizon

T − 1.

This item leads over station

k = 2

in the se ond y letime. It undergoesoperations on station

k = 3

in the third y letime and is nally ompletedinthefourth y letime,asindi atedinthepenultimate olumnof

Table6. Re allthata utilityworkerrequired onstation

k

toperform operations on the

h

th

itemisbusyfor

2

y letimes,

∀k = 1, 3.

Thea tivityofevery utilityworker is represented by a re tangle in Table 6. As an example,

w

1,−2

= 1

is symbolized byare tanglearoundpositions

−2

and

−1

indi atingthatthe orrespondingutility worker is busy on station

1

inthe fourthand fth y letime of horizon

T − 1.

We alsoreportinea hre tanglethehorizoninwhi hthehiringde isionhasbeen taken.

Withsu ha table,we anvisualize thea tivity ofutilityworkers inthetimesothe

number of ne essary workers an easily be derived. The last olumn provides the

numberof a tive workers per y le time. Considering horizon

T,

one utility worker is a tive on workstation

k = 3

in the rst y le time and he is engaged for

2

y le times. Thismeans that this worker an not beallo ated to station

1

inthe se ond y le time to perform operations on the item in position

2,

thus

2

utility workers mustbe a tive inthe se ond y letime.

Horizon Cy le

k = 1

k = 2

k = 3

item(pos.) u. workers

1

−5

0

2

−4

−5

0

T − 1

3

−3

−4

−5

0

4

−2 (T − 1)

−3

−4

−5

1

5

−1

−2

−3

−4

1

6

0

−1

−2

−3

0

1

1

0

−1 (T − 1)

−2

1

2

2 (T )

1

0

−1

2

T

3

3

2

1 (T )

0

2

4

4

3

2

1

1

5

5 (T )

4

3

2

1

6

6

5

4

3

1

1

7

6

5

4

0

2

8 (T + 1)

7

6 (T )

5

2

T + 1

3

9

8

7

6

2

4

10

9

8

7

0

5

11 (T + 1)

10

9

8

1

6

12

11

10

9

1

12

11

10

12

11

12

Table6: Itemspositionsonea hstationanda tivityofutilityworkers

Applying onstraints (11) to our example yields

w

1,0

+ w

1,1

+ w

3,−2

+ w

3,−1

≤

W

1

(T )

for

h = 1,

wherethevalueofthesum

w

1,0

+w

1,1

+w

3,−2

+w

3,−1

anbereadin thelast olumn ofTable6intherowrelative totherst y letimeofhorizon

T.

We get

W

1

(T ) ≥ 1, W

1

(T ) ≥ 2, W

1

(T ) ≥ 2

for

h = 1, 2, 3

respe tively. Thisleads to an optimal value

W

∗

1

(T ) = 2,

assuming thesequen e ofitems given for

T

inTable5 is anoptimalone given the ostparameters. As

W

1

(T )

isto be minimized,ittakesits minimumpossiblevalue. Similarlyweobtain

W

∗

2

(T ) = 1.

Thethirdinequalityin(11) appliesfor

h = 7, 8.

Wehave

w

1,6

+ w

3,4

+ w

3,5

≤ W

3

(T )

for

h = 7

and

w

3,5

+ w

3,6

≤

W

3

(T )

for

h = 8.

Notethat

W

3

(T )

isonlylinked tovariables

{w

k,h

}

asso iatedwith itemspositionsthatdonotex eed

N = 6.

Wehave

W

∗

3

(T ) = 1.

Theoptimalnumber of utility workers whose e onomi onsequen e is attributable to the sequen ing

de isions made for horizon

T

equals

W

∗

1

(T ) − W

3

∗

(T − 1) + W

2

∗

(T ) + W

3

∗

(T ) = 3.

In the rst half of

T

,

2

operators are required but one of them has been de ided in

T − 1

so its wage is in luded in thee onomi fun tion related to de isions made for horizon

T − 1.

A single operator is required inthese ond half-dayand a utility worker thatwill be present only inthenext dayis a tually paidinhorizon

T.

Many operationsne essitate sequen e-dependent setups. For instan e, inthe

auto-motive industry, vehi les of a same olour are grouped in an attempt to minimize

the ost of leaningsprayguns. We will fo us here on su h setup osts and we will

onsider a single paint station. The formulation below may easily be generalized

to anynumber ofstations and anytype of setup osts(see Giard, 2003). A typi al

setup ostinvolvesthe ostasso iatedwithma hinetuneupstoswit hfromamodel

toanother, e.g. atwo-door vehi letoa four-door one. Vehi lesof asame olourare

groupedinbat hesbutthe bat hsizehasanupperboundfortworeasons. First,

af-tersome vehi les,spraygunsgetpaint-en rusted andneed apurge. Se ond, a olour

hange is required to avoid eye tiredness responsible for a less ee tive paint aw

dete tion. Welet

L

bethe maximum number of ars inanyidenti al- olour group. The binary variable

u

h

takes a value of

1

if the olour of the ar in position

h

is dierent from the olour of the previous one inthesequen e (position

h − 1)

and

0

otherwise. Thesumofvariables

u

h

overthe wholesequen eexpressesthenumberof purges. Letting

γ

betheunit ostof leaningpaint fromsprayguns (solvent,labour andstation downtime), the obje tive isto minimize

γ ·

N

X

h=1

u

h

.

(12)

The onstraint on the maximum group size is equivalent to enfor e at least one

olour swit h between positions

h − L + 1

and

h, ∀h

. Thisiswritten as

h

X

l=h−L+1

u

h

≥ 1, h = 1, . . . , N,

(13)

whi hindi atesthenumberof olour hangesanysubsequen eof

L

vehi lesmust be at leastequal to one.

We must impose additional onstraints on variables

u

h

so they take a value of

1

ea h time there is a olour swit h in the sequen e, and zero otherwise. The olour index of vehi le

i

is denoted by

ρ

i

so the olour index whi h is applied to the vehi le in position

h

is

P

N

i=1

ρ

i

· y

i,h

. Variable

u

h

will take a value of 1 if a olour hange o urs between position

h − 1

and position

h,

whi h is translated by anonzerodieren e of olour indi esbetweenposition

h

andposition

h − 1

,thatis

P

N

i=1

ρ

i

· y

i,h

−

P

N

i=1

ρ

i

· y

i,h−1

6= 0.

This onditionisfullledwhenthetwo following inequalitieshold:

P

N

i=1

ρ

i

· y

i,h

−

P

N

i=1

ρ

i

· y

i,h−1

≤ P · u

h

, h = 1, . . . , N,

P

N

i=1

ρ

i

· y

i,h

−

P

N

i=1

ρ

i

· y

i,h−1

≥ −P · u

h

, h = 1, . . . , N.

(14)

To illustrate, onsider

N = 6

vehi les and

P = 2

olours (an index value of

1

is hosen for the blue olour and a value of 2 orresponds to the red hue). The

maximum number of ars with same olours in a sequen e is

L = 3

. In table 7, we give the values of relevant variables for a given sequen e of vehi les for whi h

oloursareindi ated. Forea hposition

h,

thevalueof

P

N

i=1

ρ

i

· y

i,h

simplygivesthe olourindexofthevehi leinposition

h

(thisisgiveninthe

4

th

rowofTable7). The

dieren e

P

N

i=1

ρ

i

· y

i,h

−

P

N

i=1

ρ

i

· y

i,h−1

betweenthe olourindi esinposition

h

and

h − 1

isnon zerowhenthe olourhas hanged between the urrent positionandthe previousone. Thisdieren e isgiven for ea hposition

h

inthe

5

th

rowof thetable

( olour dieren e). During horizon

T,

this happens in positions

h = 1

and

h = 3

(and alsofor

h = 0

whi h is the last position inhorizon

T − 1)

. Appropriate values for

u

h

arethenobtainedutilizing onstraint(14). Forinstan e, inposition

h = 3,

we have

P

6

i=1

ρ

i

· y

i,3

−

P

6

i=1

ρ

i

· y

i,2

= −1

andwe want

u

3

su hthat

−2u

3

≤ −1 ≤ 2u

3

whi himplies

u

3

= 1.

Thus,a olourdieren ene esarilyimpliestheimplementation ofa olour hange. ThelastrowofTable7 he ks onstraint(13). This onstraintis

violated for

h = 6

asno olour hange hasbeendone for thelast 3vehi les sothere is more than 3 onse utive vehi les with same olour. Letting

γ = 10

euros be the unit ostof apurge, thetotal ostof leaninggunsduringhorizon

T

amountsto

20

eurosas

P

6

h=1

u

h

= 2.

Horizon

T − 1

T

Position

h

...

−2

−1

0

1

2

3

4

5

6

Colourname red red blue red red blue blue blue blue

Colourindex

2

2

1

2

2

1

1

1

1

Colourdieren e

0

−1

1

0

−1

0

0

0

Colour hange

(u

h

)

0

1

1

0

1

0

0

0

P

h

l=h−2

u

h

2

2

2

1

1

0

Table7: Constraintson olour hanges-illustration

4 Summary: notations, assumptions and the

optimiza-tion problem

Tables8and9providethenotationsweusedsofar. Theoptimizationproblemnally

onsistsinminimizingthe sumoftheobje tivefun tionsin(10)and (12)subje tto

onstraints (1), (3) for all xed- apa ity stations in the set

K

C

, (7) for all exible stationsin

K

F

,(8), (9),(11), (13), (14). We madethefollowing assumptions.

Assumption 1. There are only two possible values for the pro essing time on

station

k, θ

max

k

and

θ

min

k

where

θ

max

k

= max

i=1...,N

{θ

i,k

}

(with at least one item

i

requiringoption

k

)and

θ

min

k

= max

i=1...,N

 θ

i,k

| θ

i,k

< θ .

Thus, installing option

k

takes always the same pro essing time whatever the itemtype. Allitemswithoutoption

k

areassumedtohavethesame pro essingtime

θ

_k

min

onstation

k

althoughsomeofthemmayrequireazerotimewhi ha tuallyleads toagreaterde reaseintheex essworkingtimethantheonewe onsidered. Without

this assumption howeverthe logi ofspa ing requirements onexible stations must

be abandonedinfavor ofa reasoninginterms ofpro essingtimes.

Assumption 2. For any option

o

pro essed on several stations

K

o

with

K

o

=

 k| θ

o,k

> θ

T

Sequen inghorizon,usuallyaday

N

Numberofitemstobesequen edover

T

θ = ⌊T /N ⌋

Cy letime

K

Setofallstationsintheline

K

C

Subsetof xed- apa itystations

K

F

Subsetof exiblestations

O ≡ K

C

∪ K

F

Setofoptions

λ

k

Numberofdedi atedoperatorsonstation

k

ν

k

: η

k

Spa ing onstraint: `atmost

ν

k

itemswithoption

k

outof

η

k

onse utiveitems'

δ

i,k

Booleantakingavalueof

1

ifitem

i

requiresoption

k

and

0

otherwise

θ

i,k

Pro essingtimeofitem

i

onstation

k

θ

max

k

Pro essingtimeonstation

k

ofanyitemrequiringtheoption

k

θ

min

k

Pro essingtimeonstation

k

ofanyitemwithoutoption

k

R

max

k

Maximumex essworkingtimeallowedonstation

k

µ

k

Numberofitemswithoutoptiontobesequen edafter

ν

k

onse utive

itemswithoption

k

toavoidex eeding apa ity,when

λ

k

= 1

β

k

Numberof y letimesautilityworkerspends onstation

k

to pro ess

oneitemwithoption

ω

Costofhiringautilityworkerperhalf-day

γ

Set-up ost

L

Maximumnumberof onse utiveitemswithsame olour

P

Numberof olours

ρ

i

Colourindex hosenforitem

i, 1 ≤ ρ

i

≤ P

Table 8: Notationsforparameters

Itshouldbenotedthatwhenautilityworkerisallowedtoworkonsu hstation

k

, thisrelaxesthespa ing onstraintsoallstationsin

K

o

mustbe onsideredseparately. A simplifying assumption onsists in onsidering that any option is fullypro essed

bya singlespe ialized workstation.

Assumption3. Ea h exible stationhasasinglededi ated workerand mayhost

asingleutilityworkerper y letime. Assumption4. The maximumex ess working

time,

(θ

max

k

− θ)ν

k

,

without violating the apa ity of any exible station

k

is su h that

(θ

max

k

− θ)ν

k

≤ θ

so the regular worker fully absorbs the ex ess working time whilethe utilityworkersolely pro essesan itemwithoption.

y

i,h

∈ {0, 1}

Binarythat takesavalueof

1

ifitem

i

ispla edinposition

h

m

k,h

∈ N

Numberofitemswithoption

k

in thewindow

{h − η

k

+ 1, . . . , h}

w

k,h

∈ {0, 1}

Binarythat takesavalueof

1

ifautilityworkerisrequiredonstation

k

topro essthe

h

th

item

q

k,h

∈ {0, 1}

Binarythat takesavalueof

1

ifiteminposition

h

hastheoptionand

noneoftheitemsin positions

{h − η

k

, . . . , h − 1}

ishandledbya utilityworker

W

1

(T ) ∈ N

Numberofa tivetemporaryworkersinthersthalf ofhorizon

T

W

2

(T ) ∈ N

Numberofutilityworkersrequiredtopro essitemsinthese ondhalf

ofhorizon

T

W

3

(T ) ∈ N

Numberofutilityworkersneededtoimplementsequen ingde isions

madefor

T

but physi allypresentonlyinthenexthorizon

T + 1

u

h

∈ {0, 1}

Binarythat takesavalueof

1

ifaset-upisin urredto pro essthe

h

th

item(e.g. a olourswith)

Table9: Notations

5 Computational study

We rst dis uss the initial onditions. We then des ribe the experimental design.

We nallyturnto the omment of theresults.

5.1 Initial onditions

In pra ti e, sequen ing de isions are made on a rolling horizon basis: the sequen e

implementedwithinhorizon

T − 1

providesinitial onditionstothesequen ing prob-lem for horizon

T.

In the experiment, we have generated a random sequen ing of itemsfor horizon

T − 1

andwehave optimized thesequen ing forhorizon

T.

Inthe random sequen ing, we have in orporated some rules in order to obtain a feasible

sequen e that satises the spa ing onstraints for xed- apa ity stations only. We

then have omputed the values of all variables orresponding to this feasible

ran-domsequen e,someofthembeingin ludedasinitialparametersintheoptimization

program relative to horizon

T.

5.2 Experimental design

Thenumber

N

ofvehi lestobesequen edduringthehorizon tookthevaluesinthe set

{10, 15, 20, 25, 30, 35}.

For

N = 10

and

15,

thenumber

P

of olourswas setto

3

withamaximumof

L = 3

onse utive arswithsame olour. For

N = 20

and

25,

we hose

P = 4

and

L = 5.

Finally,for

N = 30

and

35,

we set

P = 5

and

L = 6.

Note thatfor ea hvalueof

N,

parameters

P

and

L

aresu h thatareasonable numberof subset of onse utive identi al- olour ofvehi les maybe obtained.

K

F

of exible stations took the values of

2

and

4,

i.e.

(K

C

, K

F

) = {(1, 2), (2, 4)}

leading to a total number of either

3

or

6

options. We set

ν

k

= 1

for all

k,

sothe spa ing onstraint for exiblestations is a ne essaryand su ient onditionnot to

ex eed the apa ity. We let

η

k

U {2, 3, 4, 5}

for all

k

so we get reasonable values forthe spa ing onstraints withrespe tto thenumberof itemsto besequen ed.

Letting

N

k

bethenumberofvehi leswithoption

k,

weset

N

k

= 0.75N ν

k

/η

k

for all

k ∈ K

C

and

N

k

= 2N ν

k

/η

k

for all

k ∈ K

F

.

This makes easier the sear h for a feasiblesequen e satisfying thespa ing onstraints for xed- apa ity stations while

makingne essarythe appointment ofsome utility workers.

Ea h item is uniformly pi ked at random to be endowed with option

k

until the number of itemswith option

k

rea hesthe valueof

N

k

.

The olours of ars are identi allyindependentlyuniformlydistributedoverthesetofpossible olours. Thus

thereis on average a proportion

1/P

of ars ofea h olour. In a ordan e withthe fren h wage levels we set

ω = 70

euros. In thefren h automotive industry, the ost of leaning the sprayguns approximates

10

eurossoweset

γ = 10

throughout.

For ea h problem dened in terms of values for

N

and for the pair

(K

C

, K

F

),

we made 3 repli ations and we obtained

30

instan es sin e we only onsidered

(K

C

, K

F

) = (1, 2)

for

N ≥ 30.

Intheoptimizationprograms,thespa ing onstraints forexible stations werenally written as

h

X

l=1

N

X

i=1

δ

i,k

· y

i,l

− (1 + ν

k

) · w

k,l

!

−

h−η

k

X

l=1

q

k,l

, h = 1, . . . , N ; k ∈ K

F

,

(15)

instead of using (7) whi h would lead to a higher number of variables with the

introdu tionofthe `intermediary'variables

{m

k,h

}.

Analogously,weusedthespa ing onstraints for apa itated stations asgiven by(2).

5.3 Results

Table 10 reports the results for the small instan es (up to 15 vehi les) for whi h

we get an optimal solution ina reasonable timethrough theuse of the ommer ial

software MIPsolverILOG-Cplex 9.0. The rstthree olumnsgive thevalues of

N

,

K

C

and

K

F

whi h dene a type of instan e. The next two olumns provide the number of variables and of onstraints. For ea h type of problem and ea h of the

threerepli ations,wereportthequalityoftherstfeasiblesolutionmeasuredasthe

relative deviation (gap)to the optimalvalue. The penultimate olumn displaysthe

exe ution time to rea h the rst feasible solution whereas the last olumn exhibits

the time to obtain the optimum. Computing times are expressed in CPU se onds

andrefer to aBi-Xeon 3.4GHz with4Goof mainmemory.

Ex ept for one of the largest problems, it takes less than 2 se . to get a rst

integer (feasible)solution. Thequalityof these rstfeasible solutions is rather

het-erogeneous, with a deviation to the optimal value ranging from about 15% to 75%

withno obvious relationship between the deviation and theproblem size. The

exe- utiontimeto rea htheoptimumseemsto stronglydependonthestru tureofea h

N

K

C

K

F

Var. Const. Gap(%) Time Time, Opt.

10

1

2

151

166

34.38 0.01 0.67 32.26 0.01

0.27

72.73 0.03

0.65

10 2 4 189 273 19.15 0.98 14.33 17.02 1.99 79.99 35.90 1.26 240.65 15 1 2 301 251 54.55 0.04 3.83 75.00 1.41 648.98 57.58 0.03 0.23 15 2 4 359 413 18.03 0.24 66.17 46.00 15.87 1026.38 14.52 0.12 47.19

Table 10: Feasible andoptimal solution tosmall instan es

the largest time is asso iated with the biggest problem and the smallest instan es

arealsothe fastest problemsto be solvedto optimality.

Table 11 displays the results for larger instan es (up to 35 vehi les) for whi h

Cplex wasnot able to nd theoptimal solution within the timelimit of 7200 CPU

se onds. Like in the previous table, the rst ve olumns des ribe the instan es.

The next two olumns report the quality of the rst feasible solution and thetime

required byCplex to rea h it. Thequalityis measuredbythe relative gap between

the rstfeasiblesolution andthebestbound. The penultimate olumn displaysthe

gapbetweenthebestintegersolutiondeliveredbyCplexafter7200se . andthebest

bound. Thelast olumn providestheimprovement oftheobje tive valueafter 7200

se . ompared to the rst feasible solution(for instan e, thebestobje tive valueis

23.40% lowerthan therst feasiblesolution to therstproblem).

For instan es up to 20 vehi les Cplex delivers solutions of good quality within

the 7200-se ond timelimit. For 13 ases out of 18, Cplex is able to provide a rst

feasible solution in less than 3 min. Even if the solution quality after 7200 se .

rapidly degrades aslarger instan es are onsidered, the solution to 5 instan es over

18exhibitsagapbelow5%. Inall ases,therstfeasiblesolutionissigni antly

im-provedduringsubsequent iterationswithade reaseintheobje tivevalueof47.75%

onaverage.

6 Con lusion

Inthispaperwe have developedan optimalformulationfor thesequen ing problem

where simultaneous minimization of the number of utility workers and setups is

desired. The resultant optimization model is more realisti than those of earlier

papersdealingwithsu hgoalssin eitprovidesanoperationalwaytoimplementthe

utility work needed to avoid line stoppages. Although theproposed approa h here

an hardly deal with real-size problems, it may lay down the basis for developing

N

K

C

K

F

Var. Const. Gap(%) Time Gap(%) Imp. (%) 20 1 2 501 336 26.09 0.06 2.17 23.40 40.63 0.20 3.13 36.36 65.63 0.15 3.13 60.61 20 2 4 579 553 60.38 468.59 5.00 34.92 49.06 4.06 14.46 29.51 50.94 14.60 16.67 14.29 25 1 2 751 420 275.00 11.09 150.00 50.00 291.67 93.34 95.61 95.83 66.67 0.63 3.03 61.76 25 2 4 849 693 139.39 1703.85 77.67 33.90 80.00 424.21 70.00 5.88 78.79 23.45 66.67 7.27 30 1 2 1051 505 308.33 12.64 158.33 58.06 308.33 16.37 108.33 96.00 308.33 163.80 108.33 96.00 35 1 2 1401 590 225.00 735.13 158.33 25.81 140.74 16.55 39.06 58.54 361.54 422.38 169.23 71.43

Table 11: Feasibleand bestinteger solutionto larger instan es

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