In*ernalional.lournal of Production Economics, 24 ( 1991 ) 77 9u 77 Etsevicr

Feeding parts to an assembly line by several production units

Dim :,t:'i '.~;3er.k,, Dinzburg a n a Anato iy G o l d i ' d d

~epartm.,nt q~ tndustrial Engineering and Manacza~h.. at, Ben-Gurion Univer.~'~tl' OJ the Net,,ev, Beer-';!wra, L~rat,I

At, s tract

Tim problem of production control wi,.h disturbances fur processing assembly-line parts is introduced. A flexible man- ~facturing system is considered inc'. ,aing an assembly-line ¢,ith several .~upplementary units, each unit producing different parts for the line. Each unit has scv ,ral possible speeds wlfich are subject t,a dislurlran~.es. The o,tputs brahe unit:; are not

• transferable, t.t.. a unit cannot pet,duce parts for other units. For each unit's output there is a pregiven deterministic rate of demand of the assembly-line If at the routine time point there i~ a shortage of any types of parts the whole FMS is in shortage. Gi',en the deterministic rates of demand for each ur-it and the confidence probability for the FMS not to be in shortage, the decision-maker determines at the routine control point botq the speeds 1o be introduced at that point for a!! unit':, and ;.be next con;.rol point. The objective functions are: to minimize the operating costs duriug the planning horizon and to minimize ~he number of con,rol points. ,a stochastic optimization formulation is I. resented followed by a heuristic solution•

1. ~ntroduetion

The p r ob l e m associa ted .vith s~,nchronizing several p roduc t i on 0rocesses has been d iscussed exten- sively in the l i terature. Var ious research has been unde r i aken an the area o f " j u s t - i n - t i m e s y s t e m s " [ 1 ], s torage and retr ieval sys tems [2,3 ], flexible m a n u f a c t u r i n g ~ells [ 4 - 6 ], var ious conveyor and assembly sys tems [ 7 - 1 0 ] , etc. T h e p r o b l e m b e c o m e s hil~l~ diff icul t v, hen the processes to be synch ron i zed are unde r ra ,adom d is tu rbances . Th i s of ten happep.s in no t fully a a t o m a t e d ( s e m i a u t o m a t e d ) p r o d u c t i o n sys tems where tbe ou tpu t cannc~ bc c ,>~t inuousb m e a s u r e d ,m line. Such semia-':~om?.ted, e "n ma- chh ,e sys tems cover a broad spec t rum o f m a , ~ d a ~ m r i n g sys tems (bu i ld ing proiects , min ing , metal - lurgy, research and d e v e l o p m e n t projects , text ~.. indus t ry , etc. :. Tttc7 usually have several poss ib le speeds which are subjec t to rar~dom dismrb~:~, ' , : . For lbv~e syst: ,as 'J~c ot~tpu~ can be measured c, n ly

in prese t con t ro l i:."ints as it is impos~ibl,~ or too cost ly to m e a s m e it cont inuous ly . Go lenko° G inz bu r g et at. [ 7 ] have e x a m i n e d the case o f a p roduc t i on sys tem which inc ludes a deter~

m i n i s d c assembly- l ine and a supp]eme,:;~aty p rod : t : ' don unit under r a n d o m di:t::rk~.ncca to process par ts tbr the hne . The uni t has several poss ib le ';:-sleds wi th given dizir, bu t t cn ~vrictions. k is a s sumed that a!! speeds are vaad,~m vo,iable's, no,, ,- :>. % % _. . . . . . ,~,~ t~_...,~.i ,~e cc:~!,.'o~ poin! nnd t -n l a in cons tan t unti l the tmxt con tm! pei~t . The :,ystem opera tes u n d e r per iodical con t ro l and dec i s io r -maKing is car- r ied out u n d e r a chance cons t r a in t that ensures that at any t ime poin t there will not be a shor tage o f pa=-ts for the l ine at a given conf idence level. G iven the rout ine eor, trol point , the actuai a ccumula t ed p roduc t i on obse r ve d at that pot:at, the de t e rmin i s t i c rate o f d e m a n d o f the assemEly-.line and the con- f idence probabi l i ty not to be in on- l ine shor tage o f the parts , the d - c i s i o n - m a k e r de t e rmines bo th the speed to be in l :oduc~d and ~.he next contro! peip*

This pap,q is a .qwIher d e v e ! o p m e n t o f s imi lar p rob lems . A flexible manufac tu r ing sys~.em i~, cons id- e red iac lud ing an assembly- l ine wi th several s u p p l e m e n t a w units , each u r i t p roduc ing .~ittdrent par ts *."or ~..t!e !in.e. E~r'h H nit ha~ qeveral p a ~ i b i e speed~ which are ~uhjecl to di¢im h.',-aces. The ou tpu t s e i ' t h e

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units are n(~: t.,a::sfcrable, i.e.. a vnit cannot produce parts for other units. For each unit 's output there is a p"egiven dctccministic rate of demand O" the assembly-line. If at the routine t ime point there is a shortage of any types of ',x;r t~- '-V,e whole FMS i~ in shortage. Given the deterministic rates of demand for each unit and the eo~afidence prob~h;r;~y io~ ihc FMS ,J3t to bc in sho~'tage, the decision-maker determines at the routine control point both ihc speeds to be introduced at that p tint for all units and the ne~t contrO i~,i~,t. The c, biectiv,~ fi~nczlo.".s .,re: to minimize the opz~atir, 5 cc, st~ du~ing the plan~iing horizon and to minimize the numbe~ of control p.~ints. A stochastic optimization formulat~zn iz pre- sented ~bllowed by a heuristic solution,.

2. The problem

The FMS considered here includes an assembly line and several ...,,,-,,,,.,,.,,,,~,,o..." ~ . . . . . . . . . . . . . . . . i,:~.-:;-.:-~.,'~- ~mits processing different parts for the line. Each unit has several positive speeds which are sub~eet to randt, w. disturbances and may affect the production process. Those disturbances may occur several times between two adiacent control points (random breakdowns of equipment, strikes or illness among; persov.neL non-delivery of raw materials in time, etc.) In order to simplify the problem we assume tbe ex~'eme ::use, ~.,amely, that disturbances occur only once w.VA!e adjusting the speed at the control poinT_

w.stnbut.oa~, o, Later oa the sv, eed ren~ains co~lstar~, until the next control point. Given the probability "" " ~ " the production speeds ior all units, the operating costs of using each speed per tin~: unit, the planning horizoil, the confidence probability not to be in on-line shortage of the paris and ii~e deterministic rates of demand for each type of parts to supply the as,:embly line, we have to d e t e n , , , c the u, . ,~,~, ,-u,a, , , ,g policy. Namely, the management must determine control points which are common for all u~_fits. At every control point the decision-n-~:ker observe~ toe amou,ns pfouucetl oy itl~ utlitn and dete~ mine3 both the next control point and the proper speed for each unit.

Each unii should avoid su~!us speeds since the parts may pile up and cause a log-jam of components i,~ the lirae. All speeds for all units are assumed to be positive, i.e. all parts are processed with positive ,~ates omy. It is also assumed that for all units the manufacturing process cannot be stopped until the c~v.~ of the planning horizon, i.e. one of the possible speeds has to be introduced at any stated moment until the final date. For all units the control t ime and the speed reset t ime are ,~egligible and the setup costs are fixed independent of the soeed and oDhe unit.

The objective functions introduced are to minimize the operating costs of all units from the beginning ,.,~_, . . . . . . . ,,~ ~r,, Ena'., ,.'"ate a~:d to m-:nimizc ~he nurabe~- of control noints.. It is a~sumed that i.r.crcasin£ the speed results in increasing the operating costs. Moreover, the ratio of the operating costs for two differ- ent vosslbk -'pecds e;iceeds the ratio of the corresponding average speed values. This prevents using unnec~ :,z.~ry large speeds.

Tc proceed t 'u~hen we ~:troduce the following notations: T p = the end of the planning horizo,~. L = the number ofproducfion units. l/~(t) = ~:c:ual am~,unt of pa~s produced by/ th unit up to moment t, !=- 1,2 .... L.

z': --. ~he ttil rac:5~, x~ sv,~.:_; of ~ l i t ' ~ ," "" ~';* ! .expn~ ~-~ ,~ .q~.n ..... v pe r t ime unit, j = 1 9 ,11 1=1~2 ..... Lo u,, = the average value of ~.,ced no. ~"~ = tl~e t deterministic) rate of the assembly line demand for par;.s produced by unit number 1,

expresse( ~- m quantit., per t ime unit.

v = the ~ ~ambe, of control points (not including TO). i = the 'index ,',f the speed i~.:~oduced at the ith control putat t, foi unit / , 1 ,< k~, ~< m~; thus, kt, is

an ordiw...., hi, tuber :.~nditioned on oqr decisions. -! = th ,r,i,~i~,~a~ : i~e span between two conse,'utive cont~ot points t, and t,. ~ (excel:~t for the

c a s e l , , , - )

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3 = the value of the closeness to the : ~e date. p = the ceafidence probability value not to be in on-line shortage of the parts. c o = the co,;t of using speed v 0 per time unit. S = op~'~adng costs.

Vahles Wt° d and p ";re constant and given erAernally, and vaiu ,. e:0 are sorted in ascending order separately for each 1.

Decision variables, thus, are ti and kn, 1 -<. 1 ~< L, 1 <~ i <.< N. Note that choosing :.ndex kn at point ti results in introducing random speed v,,: j=.tp~.

The problem is to minimize the operating costs L N

F = m i n S = mi,~ ~ ~_~ C l k t , ( i t + l _ i i ) (1~,, , , * , :l : . ; f ( l l , .~L'.. t . " ] ' : 1

and to minimize the number of control points

min N ( l b ) {t,. ~,,',

Subject to

Pr [V-Ct( t ) )Wd] ) p VO<<.t4T P (2) I

and eontro! point re,itricdons"

tt .--0 (3 )

tu+l = T e (4)

t~<t~+t<T e, i=1 ..... N (5)

t , ~ . t - t i ~ d , i - 1 ...... N - I (6)

qhis is z stocha,fic optiraizatioi, problem with a chance coi'~traint (2) and a random number of constraints and variables. When the number of control points increases, the confi:e ~¢e probability may increzz.e, b,Jt cootrz] becomes cenL;aao~s. To prevent cont inuous , ontrol, which is not typical for sem- iautomated sys*ems, the minimal time spa*a between two consecutive control points, d, was incorpo- rated into the model.

The stochastic nature ~,fthe problem makes it too dS~cul',. ~..~ solve, .~o that a heuristic ape:roach ~¢.ems adequate. The heuristic alg,J,, :t!~n, is outlined in th.~ :.ext section.

3. A heuri~d~ nigor:,th_~

The algorithm proposed considers the system s~ate at the control l~oiut a~ beior~ging to o_,.~ ,;'. two ~mtu~!!y di~j:.'_:'..i _~_rbr ,~*¢" ~hnrt~o,~ -r_~urp!0_S. !t ~ the routine com~oi ~oim ~erc ~s a sho~,a~e of any types of parts the wh~le system is considered ~o be in shortage. ,~n this situation the .-lg3ri~hm looks for the ~ombi:~tion of units ' speecis allowing to make up for shorLge as cas, as possible, bm wltno~:t ;',tro- :uciag unneccssary h;~h ~eeds for ~utpl~l~ units T:,~ ~ex: cc::tro ~, point ;,: to be determined by the

F,,'egiven probability not to be in on-line skortage. Ti:~s orobai;ility is a z ' : 2":~.~ ~f ~r.'~abi!ities of in- <!i',id~al production units ne: t~ % ia shortage, since th,:? units are i.qdepende,t. The s~:eed-vector con- s~;:-r~d may ;nciude a a'ela~ivt. ,,.,, ~t;cc,~ cL~.~en for st:tpba.~ unit and a high ~peed f:,~ n rod~:,",: i um~ which is in ~qe, aage l 'he probabi!ity ok'the fi~st not to t:e ir~ sEo'tage will then be a decr~a~i:,~f~ (aon- i~:cr~;,sh~g ) f,.~ncfion of :it:: ¢, while t~e respecfiw." probaoility of th~ second will be an incre~:,~V- (non- decrea~i,',~ ~ fi'-,el~on of tim. e. Henc:, the p,ob'~bility ofa ~:yste:n as a whnle not to be in on-line ',~ :rtage turn~ 3ut, generally, to be ,t ,,,~z:on.,~notonic fuactioe...f time lbr ti'_ ~ speed-vector considered.

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In case of SOl-plus the decision-maker tries to, introd,~ce lower speeds in order to minimize the ovezall cast of production to meet criterion ( l a ) but without .~z.L__ng,'~i tbe moment when relation (2) fails ~o be satisfied• Again, different speed vectors arc considered. For each particular vector the time span between the current and the next control poir~t is maximized subject to (2). Next, the speed vectors are compared and the one resulting in the lo,vcst sum of operating costs of all production units per unit time is chosen.

The steps of the algorithm are as follows: .~tep i. At time t = 0 the input data are given: (a) the end of the pianning horizon T e, (b) the speeds

vo, j-- l,..•,ml, :i~= 1 ..... L, and (c) the constant values W~, p, dand clj Set tt =0 , i.e. the first control point is determined at moment t=0 . Note that V~(0) may differ from

zero. Step 2. Each production u~_it may be either controded or an . . . . . . . . led . If at some control point

t the relation V.Y,(t)~ WzT e holds, the production unit number l is declared to be uncontrolled. For cach uncontrolled unit speed vt~ (i.e. speed with lowest average value #,.j ~ ~s introduced. Denote the number of controlled units by Lc. In what follows, only controlled units will be considered.

If i = 1 then set S = 0. Otherwise set i = i + 1; set S = S + v/-=~ ctk~.,_, ( t , - 6-~ ). Observe P~(t:)for each l. If V { ( t ) ) W i T e, introduce speed vtj for unit l. Count the number of con-

trolled ui1?ts. I f L c = 0 then reset t~+~= TO; go to Step 13. ,Step 3 If t,> T e - & go to Step 12. Step 4. If Vf(6) >_-Wtt~ for each l, go to Step 8. Step 5. Denote the speed-vector (i.e. a particular combination of units' speeds) by 3". Solve the two-

r . . stage optimization problem as mllow~. Stage 1. For each Jseparateiy determine the next control point ,s ,,+ ~to minimize the objective function:

F~---min [tT,'+,-t~] (7) t > t i

Subject to) L

1-~ Pr{ i/-~(t,) + ~',;,,, (6+, -t~) >i WI6+, } >~p (8) / = 1

6+~- t ,>d (9)

t,+~ <~T e ( i 0 )

The solution of problem ( 7 ) - ( 10 ) is described in the Af~pendix. Stage 2. Choose the optmaal spced-vector u to minimize:

Fi =ra in [FSo] J

Step 6. If there is no speed-vector for which the solution of problem ( 7 ) - (10) can be found, i•e. if .,%r all speed-vectors J relation

L

Pr{ V;(1,) + " -,k,, ,"°~+, -t,)>_,~'v)t~+,}<p (11) l =

ho!c~s, consider a vector of speeds with maximal averages. Next, look for units, for which introducing speed with maximal average valve, i.e• kt~= m,, would result in probability equal tc one not to be in shortage on the due date,

Pr{ vfl( li) 4r VI.~.,,( TO-6 ) >1 W~T e) = 1 ( 12 )

If such units are identif ied t~en find f~r each one of them the lowest kt,-value for which (12) bolds. Introduce the resulting s,,eed-vector and set t~+~ = To; go to Step 13.

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Iftlzc solution of problem ( 7 ) - ( 1 0 ) is obtained for r> 0 different speed-vectors (J~, J2 ..... Jq ..... J , ) , resulting in the same t,+ ~, choose speed-vector J~ which minimizes the objective function

L

F(, :) = m i n ~ C/k,, q 1 = 1

Step 7. Introduce speed-vector J., set t~+ ~ and go to Step 2. Step 8. Solve the optimization problem for each speed-vector J separately as follows: determine the

next point t[+ t in order to maximize the objective function

F J = m a x [t[+, -t~] (13) ;2- h

subject to (8) , (9) and (10). The bisection method [ 11 ] may be used to obtain the ootimal solution. Note that maximization of tV..c t ime span between adjacent control points li ann t,+~ reflects the

second objective of minimizing the number of control points N. Step 9. "~ g no vector J is found f ~ which problem ( 13 ) has a soiution, proceed with Step ! 0. Otherwise

go to Step I 1. Step 1O. Since all the units are in surplus, it fo'&;ws that the probability of the system as a whole not

be in shortage is a non-increasing function of time. Therefore, problem ( 13 ) has no solution subject to constraint (9) , i.e. relation max i7 ~ < y holds. 7be *ituation may exist if p is close to one and d is large enough.

Set ti+ 1 = ts+d. Consider a vector of speeds with rr:aximal averages. Determine units for which intro- ducing . . . . a ,,.:,h , -~ . , :~ ," . . . . . . . . . . . ~,,,~ ; o " . . . . . i.~p~;,~ o~ , : ;,,,y ,,,,, ,,~ be ;n sho~age ~* : , . ,,

Pr{ V(,(t,) +Vtk,,d>~ Wit,+, } = ! (14)

If such units are identified, then find for each one of them the lowest k,, value for which (14) holds. introduce the resulting speed-vector and go to Step 2.

Step 11. If problem (13) can be solved for r > 0 different speed-vectors (Jl, J., . . . . . Jq ... . . Jr), choose speed~vector J. which minimizes the objective function

L

F2 (:) =ra in ~ clk, q / = I

Introduce speed-vector J_-, set t,+ ~ = t ~ and go to Step 2. Step 12. Set t~+ t = T e. Consider all speed-veer..rs for which chance constraint (8) is satisfied. If r > 0

such speed-vectors are identifi,zd, choose vector d_ wbich minimizes the objective function,

L

F~ :) = m i n ~ Clk,,

Introduce speed-vector .-_. and proceed with Step 13. If no speed-vector satisfying (8) can be determined, consider a vector of speeds with maximal in-

dices. Determine units, which ce~ainly will not be in shortage at t,+t. If such units are identified, then find for each one of them the lowest ~, value for which (14) holds. Introduce the resulting speed-vector and proceed with Step 13.

Step 13. Set S = S + Y ~= ~ ctk,, ( T P - ti). End of the algorithm.

A simplified schematic description of the algorithm is presented on Fig. 1.

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St a rt with = I. tl=0,5=O i

At current control r {,~ point t i observe V/(~), ~ 1 E / v C L , for all units I I which had not reached the r tergel, s at tj_ I

If for some units ! IEL c (~ Vl~ ~) >-WT P holds, exclode I these units from L c and introduce for them minimal I speeds Vll unti l T P I

no yes 1 Solve (see Appendix) optiinizatioi~ x Using bisectioh t,~athod solve (,6~ IProblem (7-10) and determine next I optimization problen~ (8-I0, 13) I

and determine next c. ~ A!-Oi point j control point ~÷! and speeds Vl] " i J = kl i __1 t i+ ! and speeds Vlj , ! = k l i ]

i I

.2- i . . . .

I I

2 [ # ='+ ! ~'i end of the algorithm

Fig. 1. Heuristic control algor,tht~

i Calculate S=S +l= ~I c,>,~h ( TP- I' ) I

4. Numerical example

Let us consider the following example: two production units, each one with two possible speed~ ,mi- formly distributed

v i i~U(3 ,5 ) , # l l = 4 , e l l = l ; v12~U(4,8) , #12=6, c:2=1.6 vz,~O(2,6), a2i=4, c2i=!; v22~U(3,7), p 2 2 = 5 , C22=1.4

The input data (given externally) are as follows:

Wl=4; 1,I/2=3.8; Te=13; d=2; V{(0)=0; VI2(0)=4; p=0.9

The solution according to the algmithm outlined above is as follows:

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Stage 1

Since V~(0) = 0 and vf2 (0) = 4 < Wz× Te= 3.8X 13=49.4, both units are controlled. Set i= 1, S=0. The system state is su~lus, therefore Step 8 of the algorithm is applied. Solve problem (13) subject te (8), (9) and (10).

Since V { (0) = 0 it follows (see Appendix) that the probability of the first refit not to be in shortage at time t ( t> t~ ) becomes constant, P(t) = 1 -Fo(W). For speed v~ ~, P(t) = 1 -0 .5 (4-- 3) =0.5. For speed vt z, W= v' = 4 and P( t ) = i. Therelbre speed v~ 2 is the only choice for the first unit so that relation (8) may be satisfied.

Applying (A8) and (AI 1 ) we obtain for .~he ~econd unit:

zt [ ~. \ "2,: r, =9+3 . -~_2=2-22 , P(t)=l-O.25~3.8--~t- 2 ) = 0 . 9 : : , t /= 2.86

3 . 8 _ 3 = 5 , P ( t ) - ! - 0 . 2 5 3 . 8 - - 3 = 0 . 9 ~ t 2 = 1 0

Thus, the solution of problew ( 13 ) is found for 2 speed vectors, Jj = (2,1 ) and J2 = (2,2). Therefore Step 11 is applied,

F2 = m i n [ (c~2 +C2l ), (c~2 +c22) ] =c~ +cz~ =2.6

Therefore, vector (2,1) is chosen, and t2=2.86. Sampling two random numbers, 0.8 and 0.3, results in th : following actual production:

V~(2.86) = 0 + [4+ ( 8 - 4 ) ×0.8] X2.86=20.59

vf(2 .86) = 4 + [2+ ( 6 - 2 ) X0.3] X2.86= 13.15

Stage 2

S~t i=2, S = 0 + (c~2+ c2~ ) ( t2-h)=7.44. At control point t2=2.86 both units are cGm;o~ed, since vf~ (2.86) =20.59< W~ TP=52 and VYz (2.86) = i3.15 < W2 Tv=49.4. The system state is surplus, since for the first unit A V= 20 .59- 4 × 2.86 = 9.15 > 0 and for tke second unit A V= 13. i 5 - 3.8 × 2.°6 = 2.28 > 0. Step 8 is _~ppiied. The results tbr the first unit are as follows:

915 v , . ~ : , q = 2 . 8 6 + ~ 3 = t 2 . 0 1 ~ P ( t ) = l for2.86<t~<12.01

vl2:P(t)=l since W = v ' = 4

For :he second ~,~it we obtain:

V21 : ~" I = 2 . 8 6 " + ~ = 4 . 1 3 , P( t ) = 1 - 0 . 2 5 ( 3 . 8 - 2 2 t 8 - 2 )

Fc~ +2 taken alone the requirement P(t) =0.9 i,.aplies At= 1.63 < d=2.

vz2: Zl =2 .86+3 . =5 .7! , P( t ) = 1 -0 .25 3.8-~ - 3

For v2z taken alone the requirement P( t ) = 0.9 implies At = 5.7. We see that P( t ) = 1 for v~,. and v:,_ on the time intervM (2.86 + 2 := 4.86, 2.86 + 5.7 = 8.56). Therefore

problem (13) can be solved for 2 speed vectors, J~ = (1,2) and J2= (2,2). Step 11 is applied,

F2 = m i n [ (etl "~-C22), (el2 "~'C22) ] =Cll +e22 =2.4

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Therefore, vector (1,2) is chosen, and t3=8.56. Sampling two taadom numbers, 0.4I and 0.12, re- suits in the following actual production:

V.C(8.56) =20 .59+ [3+ ( 5 - 3 ) ×0.41 ] ×5 .7=42 .36

V~(8.56) = 13.15+ [3+ ( 7 - 3 ) ×0.12] × 5.7=22.99

Stage 3

Set i---3, S = 7 . 4 4 + (Ctt+C22) ( t3-- tz)=21.12. At control point t3=8.56 both units are controlled, since 42.36 < W~ T P= 52 and 32.99 < IV,_ T "~= 49.4. The system state is surplus, sin,:e for l!,.e first unit A V = 4 2 . 3 6 - 4 × 8.56=8.12 and for the second unit AV= 32 .9q-3 .8 × 8.56=0.46. Therefore, Step 8 is applied.

Using (A8) we obtain for the first unit:

6 812 V~l: z~ =8.5 + ~ = 16.68

v~2:P(g)=l s i n c e W = v ' = 4

Thus, on the interval (t3, 7 r ~ me probability P ( t ) = 1 Ibr both v~ ~ and v~ 2- The results for the second unit are as shown next:

v2 t : r~=8 .56+ . _ =8.82, P ( t ) = l - 0 . 2 5 3 . 8 - - 2

For v2~ taken alone the requirement P(t ) =0.9 implies At=0.39 <d=2 .

, 0 46 n 1 ~ 1-0.~5(" ~ _ 0At46_3") /')22: TI = 8 - 5 6 n r - ~ - - 3 3 " ~ ' 7 " "-e P ( t t = "a

For v22 taken alone the requirement P ( t ) = 0.9 implies At = I. 15 < d = 2. It comes out that for the second unit taken alone P(.-'3+d) < p for both v21 and v;:,. Hence, problem

(13) has no solution subject to constraint (9). Therefore, Step i0 is applied. Set ;~.-: ts+d--10.56. Vector of speeds with maximal averages is (2,2). Introducing speed with maximal average, v~2, for the first unit implies eertain'y (for this unit) not to be in shortage at t4, see (14). This certainty still holds for the speed with the lower average value, i.e. v~ ~. Consequently, the resulting speed vector is (1,2). Sampling two random numbers, 0.33 and 0.6, results in the following actual production:

V{(10.50) =42 .36+ [3+ ( 5 - 3 ) X0.33 ] ×2=49 .68 ,

Vr(10.56) =32 .99+ [3+ ( 7 - 3 ) ×0 .6] ×2=43 .79 .

Stage 4

Set i = 4, S = 21.12 + (c~ ~ + c22 ) ( t4 - t3 ) = 25.92. At control point t4 ~ 10.56 both units are still co n- trolled: 49.68<52 and 43.79<49.4. The system is in surplus: for the first unit A V = 4 9 . 6 8 - 4 × . . . . . . . . "~ ~J.~ = ,.',~, an~, fm the second u~' t A V= 43 .79-3 .8 X 10~56 = 3.66. Therefore, Step 8 is applied. The results for the first unit are as follows:

7 44 vi i : zt = 1 0 . 5 6 + ~ = 18.0

v t 2 : P ( t ) = l s i n c e W = v ' = 4

85

It follows that on the interval (% 7re) the probability P(t) = 1 for both Vl~ and ~':2. For the second unit we obtain:

v2,: ~, =10 .56+ . ~ = 12.59, P ( t ) = 1 - 0 . 2 5 ( 3 . 8 - ~ t 6 - 2 )

For vz~ taken alone the requiremen~ P( t ) ---0.9 ;.replies At= 2.61 aad t5 = 13.15 > To-= 13.9.

v22: r~ = 1 0 . 5 6 + ~ + 15.14, i.e. P ( t ) = 1 f o r t < T e

Therefore, for each one of 4 possible speed vectors, the solution of problem (13) resuks in t5 = T e= i 3. The vector optimizing F2 is (1,1), since cl~+c2~ = 2. Sampling two random numbers, 0.25 and 0.71, results in the following actual production at the planning horizon:

V~( 13 .0 )=49 .68+ [3+ ( 5 - 3 ) × 0 . 2 5 ] X 2.44= 58.22

Vr(. • ~.,, J - - , ~ . ' a ~ _ ~ a 7o~a_. ~ [ 2 + (6 - 9 ) , , ×0.71 ] ×2 .44=55.60

Finally, calculate:

S = 2 5 . 9 2 + (c~ +c2t ) (is - t 4 ) =33.8

5. Conclusions

The problem of controlling an FMS including an assembly-line and several supplementary produc- tion units is considered. A heuristic control algorithm is suggested. Further research can be done in two major directions. First, to develop the algorithm for the case when speeds are stochastic processes which vary at any point between two adjacent control points. " . . . . a o,. . . . . . . . to make intensive simulation to im- pro ee the algorithm.

6. Acknowledgements

This researcl~ is partially supported by the Paul Ivanier center f'~r robotics and production management. ,

Appendix

The appendb, contains a description of the algorithm for solving problem ( 7 ) - ( ! 0 ) . First, some additional notations are introduced and then two lemmas are developed.

Consider pfc(Liictiofi •ilit at ~.v~luu, pOiiat i i. For siniplicit) we omit both mu,cles" of the unit, l, ~" • ~tti~d

of the random si~¢ed, j . T~, = f o l ! ~ . i n g t e ~ s ".-~re defined: V/(t) = actual 'amount of parts produced by the unit up to moment t. v = the rm,dom speed of the unit. v' = the minimal rate of speed v. v" = the maximal rate of speed v, which is assumed to be finite• F~.(. ) = the cumulative distribution function ofv. g,: = the (detemfinist ic) rate of the assembly line demand f,~r parts produced by the unit. P ( t ) = the probability of the unit not to be in shortage at t ime t (t > t,).

In these notations P ( t ) is given by (see (8) above ):

P ( t ) =Pr{ V'-(t~) + v ( t ~t , ) ~- ~J.~} (A1)

Adding - Wt~ to both sides of the inequality in ( A I ) and rearranging the terms, we obtain for t> t,,

86

,vnere At = t - t~, A V, = v~'(t~) - Wt~. Let IA (to) denote the indicator function, defined by

t ; ~ ) _ ) ' 1 ifog~A ~ " - ~0 ifoJCA

Consider the unit in shortage at control point t~. Define:

To = t i + ~ i f v " > W

t~+ -I AV" ! i f v ' > W TI = v' - W

(A2)

(A3)

(A4)

Lemma 1. by

PO) =o

P(t)=(1-F~,{W+~})I(ro.~o>(t)

_ ~ I AV'~ • )"-'.'il I P(t)_(1-.,.{W-r AL t)¢r°'r')(t)+itr"°°)(t)

If the unit is in shortage at control point t,, then P(t) is a non-decreasing function o f t given

ii :I'~< W (A5)

i fv ' ~ W<v" (A6)

if W< v' (A7)

Pr,mf. For the unit ip shortage P(t~) = 0 and A I~ = - I A ~ i- For t > ti

due to (A2). The term W + I A V,I/At decreases wuh time, which implies that F,,{ W+ I A VA/At} is a non-h,~.:~,~u,r~ ,uu~ttun of time. Therefore, P(t) is a non-decreasing function of time.

If v" <. W,ben v" -<.< W+ [ AI~I/At for every t> ti, i.e, the argument ofF~( . ) is greater than or equal to the maximal rate of speed v and. hence .~( W+ 1A IT, l/At} = 1. It follows that P( t ) = 0, i.e. ( A5 ) is shown.

If v" > W then t~< To implies that

W+ IAVjl >~ W+ I A V ' I - v " At To - t~ -

The latter stems from the dei,.~i , lea ~,;" To (A3). Hence, P ( t ) = O for t ~< To, and, ~sing (A2), we obtain (A6).

If v' > W, then t> T~ leads to

I AI~ I [ A V~ ] 1 L [ / * . I _ w + v' t ' t , ~ * T i - - t .

t a r ~ ~ i

The last equality fo!lows from (A4). Therefore, the argument ofF~( . ) is smaller than the minimal rate ef speed v and, hence, P ( t) -- 1 for t > T~. Taking into account (A6), we obtain (A7).

Next, consider the unit in surplus at control point t,. When AV,=0 it follows from (A2) th=* the

87

probability of the unit not to be in shortage at t ime t ( t> t~) becomes constant, P(t) = I - ~ { W}, i.e.

P ( t ) = l i fW<v '

O < P ( t ) < 1 i f v '<W<v"

P(t)=O "¢ . i h l,, .~ H l

Finally, consider the case whe~ the actual amount of parts produced by the unit is greater than, and aot equal to, the amount to be required by the assembly line up to moment t~, i.e. A Vi> O. Define:

% =t , + ~ v ; - v ' i f v ' < W (AS)

+ AV,. i f ¢ ' < W (A9) to= t , ~V-v"

Lemma 2. of t given by

P ( t ) = l

P( t )=l t , , . , , i ( t )+(1-F,~{w-A~L})I ( ... . )(t)

P(t)=It,,.~,i(t)+(l F ~ AV~)~ - ( , ) t,

If the unit is in surplus at control point t., and AV,> 0, then P(t) is a non-increasing function

i fv ' >i W (AI0)

if v" >~ W>v' ( A l l )

if W> v" (A12)

Proof. For the unit in surplus P(tj) = 1. For t > ti:

, . , l

due ~o (A2). The term W - A VdAt increases vdth time, which implies that F~{ W - A VJAt} is a non- decreasing function of time. Consequently, P ( t ) is a non-increasing function of time.

If ¢ >t W, then :/>I W - AVg.~At for every t> ti, i.e. the argument ofFv(" ) is smaller than or equal to the minimal rate of speed v and, hence, P(t) = 1, so that (AI0 ) is obtained.

If v' < Wthen t < z, implies that

The last equality follows from the definition ofzi , (A8). Hence, P(t) = 1 for t ~ z~, and, using (A2), we obtain (A11 ).

i f v': < ~, then reiation , r z

IV-- > W-- zivi =v" T o - t -

holds for every t> Zo. Consequently,. the argument of F~(. ) is greater than the maximal rate of speed v and. hence, P ( t ) = 0 for t > "Co. Using (A 11 ) we ob!ain (A 12 ).

The algorithm for solution of problem ( 7 ) - (19) can now be described. Speed-vector J is considered. The nex*; control point t,a+, is to be found as the earliest t ime point (7) within the time interval (5-+ d, Te), for which (8) is satisfied, i.e. ~t which the system will not be in on-line shortage with the pregiven

88

probability. This probabi]ky is a product of probabilities of individual units not to be in shortage, since the units are independent. The system is in shortage at control point ti, which means that at least one unit is in shortage. Generally, there are units in shortage and others which are in surplus. The probabil- ity of the unit not to be in shortage at time t> t , P( t ) is a non-decreasing function of time if the unit is i n shc, rtage at ti (by Lemma 1 ), and it is a non-increasing function of t ime if the unit ,.'s in surplus at t~ (by Lemma 2). Hence, the probability of a system as a whole not to be in on-line shortage turns out, generally, to be a non-monotot, ic function of time°

To facilitate the treatment, we first examine the existe_n_ce and find va!oes of To.. TI (unit in shortage), z~, Zo (unit is definitely in surplus, AV,> 0), or calculate constant P(t) ( i f ~ V . = 0 ) .

!t follows from Lemma 1 that in the case of shortage for any unit there exist three alternative cases: (A) there are two values To, TI; (B) there is only one value To; (C) there are no values To, T~.

When the unit is definitely in surplus (A I/)> 0), k fc.llows from Lemma 2 that three alternative op- tions are possible: (A) there exist two values Zo, zt; (B) there exists only one value zt; (C) there are no values zo, zt.

Let us take all the values of To, T~. z~, Zo for all units falling within the interval, (t~+d, TO), if any, and arrange them in an ascending order. Denote the number of these values by n, n = O. ! ..... As a result, time-interval ( t i+ d, TP) is subdivided into n + 1 consecutive subintervals, with the boundaries: to = ti+ d,

P (o~,

I E - ~ ( s u r p l u s )

i - - " . . . . . . . i - - - ( ~ o r t a g o

0 9

I

unit3

' ~ ! '--~L ~ -- --I , , (su rp lus ) oo~- ? ;_ . . . . . . . . 2q_ . _ _

, o l ~ ¢ . . . . ~, . . . . . . . = P.(o !

o o L . . . . . ~' . . . . . . . . . ~ - - ~ u ~

' ° i ~ - - r 3 . . . . . ~- - - i , [ I i ', I ~ I

' r 09

t 0 t, t~ t ; "~ T P '

Fig. 2. A graphical i l lust rat ion o f the a lgor i thm fo r so[ving prob lem ( ? ) - ( ! 0 )

o 9

11 . . . . . t .... . . t~+ 1 = TO. Each of the t ime points, t,, t2 . . . . , t~, is a boundary-point of a region, within which the probabiii~y, P ( t ) , of one of the units is t ime dependent. Therefore, the number of t ime-depeedent factors in the probabil i ty of the system not to be in on-line shortage is a constant w:&in each subinterval (t-_ i, t_-) for z = 1, 2 . . . . , n + 1. Therefore, it is convenient to examine each subir~terval (t-_ ~, t:) sepa- rately with respect to condi t ion (8) .

The general idea of the algorithm is illustraaed in Fig. 2. Let P~ (t) denote the probabili ty of the system not to be in shortage at t ime t, i.e. the left side of (8) . Let P_ (t) denote the probabili ty of all those units, which are in shortage at t,, not to be in shortage at t ime t. Let L+ denote the number of production units in surplus at G if any. In the general ease L+ >0 , let P+ (t) denote the probabili ty of all those units, which are in surplus at G not to be in sh6rtage at t ime t. Be:rag a laroduct ',~f ~on-;mcreasing functions of time, P+ ( t) is in itself a non-increasing function of time. If P+ (t :_ 1 ) <P, then, obviously, problem ( 7 ) - (10) has no solution for t > t~_ l-

Consider_ now the routine subinterval, (t-_ i, t~), and assume, P+ (t :_ ~)>~p and ,°~,_._~' J' <p. Let's f ind all local maxima of P~(t) within the interval (t~_ 1, t_-). Denote the number of these maxima by M, , ~ = n , t , D e n o t e " - ~ . . . . : ^ - . . . . . . . . . me ,~,~t~t,,~ o fy ' s ma x i mu m by t~.. Arrange these maxima in an ascending order of their locations: t'~, t~ . . . . . t~t. Subdivide the interval (t_._ 1, t-) into M + 1 consecutive local t ime :'nter- vals, with the boundaries: t ~ = t~_ ~, t '~ . . . . . t ~ . . . . . t ~t, t i t + l = t:. Note, that P~ (t ~ ) ~ p for some y ( y = I, 2 .... , M + 1 ) is the necessary and the sufficient condi t ion for the solution of ( 7 ) - ( 1 0 ) to exist in (t :_ ~, t_.], provided it d~es not e~ist in [L +d , t~_ 1 ]. To show necessity, notice that P~(t'o ) <p and P~(t) >_.p for some t in ( t ._ :, t- ] imply ( i ) P~ ( t b+ i ( >/P. ,~r ( i i ) the function P~ (t) has a max imum in (t_-_ 1, t: ], or ( i i i ) both ( i ) and ( i i ) . On the other hand P~(t'y)~, in.plies that the solution, i.e. m i n i m u m t for w , , c a ~ , ~ .~-p, exists in the interval (t_,_ 1, G] and, Mace, "t exists in the inter~ a~. ~",- ,. !.~o (-)bviou~hz_.

t P ¢ ~' ) -'Z.,~ ~- "~ r~ z , , , < ~ huply ti~,~ &ere is no somuon m the interval (ty_ l t,, ]. Assume that funct ion P~(t) is a cont inuous one. Then the bisection method [ ! 1 ~ may be used to

obta in the opt imal solution. To our opin ion this assumption is realistic enough. On the basis of the above considerations, we can specify the algorithm for solving problem ( 7 ) - ( l 0)

as follows:

Step 1. Set flag indicat ing that the solution, t*, has not been found. Step 2. For each one of L units examine existence and find values of To, T~, zt, %. Step 3. Sort values of To, TE, z~, Zo0 falling within the interval (t~+d, TP), in an ascending order. Let

n such values are found, n = 0, i,2 .. . . . 2L. Denote sorted values by tl,t2 .. . . . G. Set to = t, + d, t~ ~ ~ = T e. Step 4. Calculate P+ (to). ~fP+ Uo) < o then go to Step 16. Step 5. Calculate P_ ~ , ~ Set 7~(t~)=_:~+ (to) P_ (to). If P~(to)<p, proceed with Step 6. Otherwise

set t* = to. set flag indic~ ,trig that the solution has been found, go to Step 16. Step 6. Set z = 1. Step 7. l f z > n + 1 then go to Step 16. Step 8. Consider interval (t_._ ~, t-). If v" <~ W in this interval for any uni t in shortage or i O~ t-, then

set z=z+ !, ge .*o Step 7. Otherwise proceed with Step 9. Step 9. F ind all local maxima of P~(t) in interval (t~_ ~, t ) . Let the number of*hose maxima be M,

_~!--0, :~ . . . Ar;are, c the maxima m an ascending order of their locations. Denote locatio~, oi y's maxi- m u m by t~,. Set t~ = G- ~- Set t~+ 1 = t~.

Step 19. Set y = 1. Step 11. I f y > M + 1 then set z=z+ 1, go to Step 7. Otherwise proceed with Step i2. Step 12. Calculate P+ (t~) and P_ (t?i.). Set P~(t' r ) =P+ (t~,) P_ (t'y). i f P~(t~, ) < p then go to Step 14. Step ! 3. F ind solutio~ t* by the bisection method on the interval (t~,_ ~, t'y ]. Set a flag indicating that

the solution has bee~ ~ u ~ d . Go to Step 16. Step 14. i f P+ (t~) < p then go to Step 16. Step 15o Set y=y+ i~ Go to Step 11.

90

Step 16. I f the so lu t ion has been f~und , t hen t* is a so lu t ion . O t h e r w i s e the re is no so lu t ion o n the

interval ( t , 4 d, Tel. E n d o f the a lgor i thm.

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