One machine scheduling problem with fuzzy duedates

European Journal of Operational Research 79 (1994) 1-12 1 North-Holland

Theory and Methodology

One machine scheduling problem with fuzzy duedates

S a n g s u H a n

Graduate School of Science and Technology, Kobe University, Rokkodai, Kobe 657, Japan

Hiroaki Ishii Department of Mathematical Sciences, Faculty of Engineering, Osaka University, Suita, Osaka 565, Japan

Susumu Fujii Graduate School of Science and Technology, Kobe University, Rokkodai, Kobe 657, Japan

Received August 1988; revised December 1992

Abstract: This paper considers a generalized one machine maximum lateness problem with fuzzy duedate and controllable machine speed. There are one machine and n jobs J1, J2, .- . , Jn to be processed on the machine. A membership function is associated with each job Ji, which describes the degree of satisfaction with respect to completion time of Ji" Further machine speed can be changeable jobwise. The objective is to find an optimal schedule and optimal jobwise machine speeds and to minimize the total sum of costs associated with jobwise machine speeds and dissatisfaction with respect to completion times of jobs. We propose a polynomial time algorithm for solving the above problem under some assumptions.

Keywords: Fuzzy duedate; Variable machine speed; Scheduling; Combinatorial analysis

I. Introduction

Up to now, many scheduling problems with duedates have been considered. For example, Jackson [9] solved a one machine maximum lateness minimization problem by using the earliest duedates first rule in O(n log n), computational time. But introduction of different ready times makes the problem NP-hard in the strong sense [4]. If the preemptions are allowed, however, the parallel multi-machine problem can be solved efficiently by utilizing a maximum flow algorithm on the special network [5]. The open shop problem with maximum lateness criterion also can be solved by a LP formulation in a preemptive case by [2]. Especially, a two machine open shop problem with maximum lateness criterion can be solved in

Correspondence to: Dr. H. Ishii, Department of Mathematical Sciences, Faculty of Engineering, Osaka University, Suita, Osaka 565, Japan.

0377-2217/94/$07.00 © 1994 - Elsevier Science B.V. All rights reserved SSDI 0 3 7 7 - 2 2 1 7 ( 9 2 ) 0 0 4 4 0 - N

2 S. Han et al. / Fuzzy duedate scheduling

O(n log n) [10] in a preemptive case. For details, see recent survey papers [1,3]. However, in some situations, duedates are not rigid and some violations may be accepted. As an example, consider the manufacturing company with foreign and domestic demands. Goods for export must be produced strictly by the departure date of the ship for their destination since, if late, the company must wait for a month or more until the next suitable ship. But for goods of domestic use, the express delivery can recover a little of the delay, but its extra cost reduces the level of satisfaction of the company.

Considering these cases, this paper investigates a one machine scheduling problem with fuzzy duedates and jobwise changeable machine speeds. As for changeable machine speeds, we already considered a two machine open shop case [6], a uniform machine case [7], a mixed shop case [8], etc. Recently Nowicki and Zdrzalka considered a related problem [12]. Fuzzy duedates and changeable machine speeds reflect the following actual problems which often occur in the printing office. The lack of concreteness in regard to the duedates is caused by the variable time required in the proofreading on the side of the customer. If printer's assistant helps with the proofreading, the necessary time is halved.

Under the above assumptions, the objective is to find an optimal schedule and jobwise machine speeds which minimize the total sum of costs associated with dissatisfaction of all job completion times and jobwise machine speeds.

Section 2 introduces fuzzy duedates with membership functions describing the degree of satisfaction with respect to job completion times and formulates the problem P. P is first transformed into equivalent problem P. Then P is divided into subproblems P1, P2, . . . , ffq. Section 3 proposes an algorithm solving each subproblem F t and gives a polynomial solution procedure for/~. Section 4 gives an example illustrating the flow of the solution procedure for P. Finally, Section 5 summarizes this paper and discusses further research problems.

m i ( C i) =

2. Problem formulation

This paper makes the following assumptions. 1) There are one machine and n jobs J1, J2 . . . . , Jn which are to be processed on the machine. 2) Machine speed is changeable for each job. 3) A degree of satisfaction with respect to the completion time is associated with each job, which is

denoted with a certain membership function of a fuzzy set defined on R + (nonnegative part of the real numbers). This reflects the actual situation that earlier completion will give us more satisfaction.

4) The objective function consists of costs related to minimal satisfaction (maximal dissatisfaction) with respect to completion times of jobs and jobwise speeds. That is, if we speed up the processing of jobs, completions of jobs become earlier and the degree of satisfaction is improved. But it makes the running costs and burden of the machine larger, and furthermore, the quality of products may become worse. So we must balance them. We seek the optimal schedule and optimal jobwise speeds of the machine minimizing the abovemen-

tioned objective function. First we list up notations used in this paper.

Speed when machine processes job Ji" Processing amount of job Jj (processing time at unit speed of the machine). The completion time of job Ji" The membership function denoting degree of satisfaction with respect to Ci, which is defined as follows:

1,

m i (C j )= 1 - ( C i - d i ) / e i ,

0,

where e i, d i are nonnegative constants. Further we use the notation sj -- l / s ; for sake of convenience.

S. Han et al. / Fuzzy duedate scheduling 3

Under the above setting, we consider the following problem P:

(e) Minimize - a 0 m!n mj(Cj ) + ~ ars; J j ~ l

subject to s ; > 0 , j = 1 , 2 . . . . . n,

where a r, j = 0 , 1 . . . . , n , are positive constants. Now let min{mr(Cr) [ j = 1, 2 . . . . , n} = t. Then the inequalities

Cj<d~+(1- t ) e j , j = l , 2 . . . . . n, (1)

must hold. Note that the right-hand side of each inequality corresponds to the ordinary duedate of job Jr" Let ~-(k) denote the k-th job index of a schedule ~-. Then for 7r,

C=(k, = E P=,,,S~r(,,, k = 1, 2 , . . . , n . (2) j=l

After variable transformation, pi=~rsr, j= 1, 2 , . . . , n , we set ar=ajp.i, j= 1, 2 . . . . . n, and ~= 1 - t . Then P is equivalent to the following problem fi:

( P ) Minimize ao-t + ~ (dJpj) r= l

k

subject to ~ pro(j) ~ d~r(k ) + tear(k), j = l

p j > O , j= l, 2, . . . ,n, O < t < l ,

k = 1, 2 , . . . , n ,

~ / / ,

where / - / i s the set of all permutation schedules. Note that pj. is the actual processing time of Jr" When is fixed, fi seeks an optimal schedule and optimal processing time of each job J~. minimizing E~= l(~r/pr) under duedates d r + ~er, j = 1, 2 . . . . . n. In this case, an optimal schedule is obtained by processing jobs in a nondecreasing order of d r + te r, j = 1, 2 . . . . . n [13].

Defining

-tjk=(dr--dk)/(ek--ej) f o r j ~ k

and sorting different trk such that 0 < t~k < 1, let

0 = ~ 0 < ~ 1 < . . . < ~ q = l .

By dividing the interval [0, 1] into subintervals T/---[ti-1, ti], i = 1, 2 , . . . ,q, we introduce the following subproblems Pt, 1 = 1, 2 . . . . . q, since a certain schedule ~'t is always optimal on T t, l = 1, 2 . . . . . q:

n

(Pl) Minimize ao-t+ E(~r /p j ) j=l

k

subject to ~., P~rt(j) < dTrt(k ) q- -te~rt(k), k = 1, 2 . . . . . n, j=l

t ~ T t , p r > 0 , j = l , 2 . . . . . n.

Thus the best solution among optimal solutions of P~, Pe . . . . , fiq is an optimal solution of fi and the corresponding schedule is an optimal schedule.


3. Solution procedure of

First we propose a solution procedure of subproblem /~t. For convenience, we suppress subscript l in order that any confusion does not occur. Further let its optimal schedule be ~r t ( j )= j for easiness of explanation by changing the job number if necessary.

Now define functions fO), h(~) as follows:

f (~) = min E (~j/pj)l E P j < d k + t e k , P , > 0 , k = l , 2 . . . . ,n , j = l j = l

h ( t ) = a j + f ( t ) .

Theorem 1. f O ) / s nonincreasing and a convex funct ion of-t ~ T t.

Proof. Nonincreasingness of f(~) is clear and so we show its convexity only. For ~,,, ~a ~ Tt, let

n n

r ( L ) = E = E j = l j = l

respectively, and for ~ = At~ + At~ (0 < A < 1, A = 1 - A), let

n

= E j=l

Then since

k k k

E (,~p; + ~p:) =,~ Ep7 +X Epfl <_,~(d, +~,ek)+X(dk +~t3ek)=d k +-t;te k, j r 1 j = l j = l

P7 = APT + APfl is feasible for t = ?~. Thus

n n n n

f(~a)= E (~j/p~) <- E (~j/p/) <-,~ E (~j/pT) + X E (~j/p~)=,~f(~,~) + Xf(~), (3) j = l j = l j = l j = l

where the second inequality holds from the convexity of the function 1Ix . (3) shows convexity of f ( t ) . []

Theorem 1 implies h 0 ) is convex. Now we propose a calculation method of f( tc) for each tc ~ Tt. First we introduce the following Lagrange function L(.) .

L ( u 1 . . . . . Un, P l . . . . . pn ) = ( ~ j / p j ) - t - Uk E p j - d k - L e k , ( 4 ) j = l k = l j = l

where u 1, u 2 . . . . . . u n are Lagrange multipliers and restricted to be nonnegative. Let us define

L ( u 1, u 2 . . . . . un) = min{L(u 1, u 2 . . . . . un, Pl , P2 , . . . , P~)I Pl , P2 , - . . , P~ > 0}.

Then

f(~¢) = max{L(u, , u2 . . . . . un) lu , , u 2 . . . . ,u~ > 0}


by the theory of convex programming. Since it holds that

L(Ul, . . . ,Un, Pl . . . . ,Pn) =

>

(aj/pj) + Uk ~., Pe - dl, - -tcek j = l k = l j = l

(aJpj) +pj u k - (dk +-t¢ek)u k j=l k-1

~ 2 ~ ~ u e - ~ (dk+-tcek)u k j = l k=j k = l

by the relation between arithmetic and geometric means,

L(Ul . . . . 'un) = 2j=1 ~ ~J '~/ k=j ~-" uk -- k=l~" Uk( dk +'reek)

and (6) is given by

pj = aj U k j = 1, 2,. . . , n.

Hence let

Yk= U j, k= l, 2 , . . . ,n . J

Then

_ _ 2 uj=yff-y2+l, j = l, 2 , . . . , n - 1 , u , - y , ,

and

y l>y2> . . . >_yn>O,

by the nonnegativity of u j, j = 1, 2 , . . . , n. Substituting (7) into (6), it follows that

n--1

L ( U l , ' " , U n ) = 2 ~ a~jy~ - E (YZ-YZ+l) ( - tce j + d j ) j = l j = l

n

= - - ( y 2 ( t c g 1 + d l ) - 2 ~ - l y , ) - y ' [ y 2 l - t c ( e , - e , _ l ) + ( d j - d , _ , ) } - 2 ~ T y , ] j = 2

n

= _ ~ {~ tc (e j -e , -1 )+dj -d j_ l }[y , - ~ / { t c ( e , - e , - 1 ) + d j - d j _ , } ] 2 j=2

-(-tee 1 + d l ) { Y l - ~ l / ( - t c e I +dl)}2+al/(-tc e, +dl)

+ ~ a J{ tc (e , - ej_,) + d r - dj_ 1} j = 2

(5)

(6)

(7)

(8)

y2 . . . . . y ° ) . (9)


Now we define at/,/3; and y;, j = 1, 2 . . . . . n, as follows:

yj=~r~7~, j = l , 2 . . . . . n,

/ 31= t~e l+dz , / 3 ; = t ~ ( e ; - e / _ l ) + d ; - d ; _ 1, j = 2 , 3 . . . . ,n ,

a; = yJ /3; , j = 1, 2 . . . . . n.

Using these notations,

n g( Y,, Y2, . . . , Y.) = - /3;( y / - a;)2 + y . r f / /3;"

j = l j = l

Theorem 2. (i) f f a; _> atj+ 1, 1, 2,. . , n 1, then f ( t~) " 2 • - = ~/= 1Y) / /3 /by setting yj = atj, j = 1, 2 , . . . , n. (ii) i f for a certain i, ati < ati+l, then a solution giving f(-t~) satisfies Yi =Yi+r

Proof, The former part, (i), is clear. For the latter part, (ii), it is sufficient to show ~ with Yi > Yi+I never maximizes g(Y l , Y2 . . . . , yn).

First note that y must satisfy yj >Yi+l, J = 1, 2 , . . . , n - 1, from (8). So we cannot set yi = a~, Y~+I = ai+ 1. We show (ii) by checking the following cases.

Case 1: Yi+I -> a~. When Yi ~ ai, g( . ) increases, i.e., setting Yi = Yi+l produces a better solution than

Case 2: Y'i > ai >Yi+I- As Yi J, ai and Yi+l $ ai+l, g ( ' ) increases,i.e., by decreasing Yi and increasing Y~+I, the value of g( . ) is improved. But due to the constraint Yi-> Y~+I, the above change produces the solution yi = Y~+I, and it is better than Y.

Case 3: ati > Yi" Case 3 implies

ati+ 1 > ati ~-~ Yi > Yi+ 1" (10)

Again, as y~ T a~ and Y~+I ? a/+l, g ( ' ) increases. By (10) and the fact that Y~+I cannot increase above yi, Y,+I = Y/is necessarily reached by the process. []

When case (ii) of Theorem 2 occurs, i.e., there exists i such that a i < ati+l, we can update yj, a; , /3/ and y; as follows:

y ; = y j , f l ;= f l ; , y ; = y j ( a ; = a ; ) , j = l , 2 . . . . , i - 1 , ( l l a )

t t ' ( •i "k- ]ti+l ) (11b) Yi =Yi , /3i =/3i-[-/3i+1, Ti ~ t i " b T i + l ate= /3i..[_/3i+1 ,

Y;=Y/+I, f l ;= f l ;+ l , Yj=Y/+z ( a t ; = a i + l ) , j = i + l . . . . . n - 1 . ( l l c )

The above updating of (11) induces the updating of g(.), and the results are as follows:

n - 1 n - 1 g(Y;, " , Y ' - I ) = - E f l ; ( Y ; - a ' ) 2 .. ; + y , y /2 / f l ; . (12)

j = l j = l

For the revised g(.), we check the condition of Theorem 2 again. Note that g (y~ , . . . , y ' _ 1) is the very same type of g ( Y l , Y2 . . . . . Yn)" After each check, either the number of variables is decreased by one or condition (i) of Theorem 2 is reached. Therefore after at most n - 1 updatings of g(-), condition (i) of Theorem 2 occurs, and f( tc) and the corresponding (y/) are found• The type of g( . ) is not changed independent of the value to. Of course, aj is a function of t and so we describe aj with aj(t). According to the value of 7, the ordering of a j ( t ) is changed, i.e., which condition (i) or (ii) of Theorem 2 occurs is changed.


Here we denote the optimal value of t with t* and its corresponding optimal solution by (y7)" Now we are ready to give an algorithm for Pt, l = 1, 2 , . . . , q, which is based on the same principle as Megiddo's [11].

Algorithm for Pt Step 0. Set

Step 1. Step 2. Step 3.

Step 4.

a l ( t ) = ~(~-l/(~e 1 + d l ) ( = ' ~ , ( t ) / f l ( t ) ) ,

a t ( - t ) = ~ / { - t ( e t - e t _ l ) + d t - d t _ , } (=yj(- t) / f t ( - t )) , j = 2 , 3 . . . . ,n ,

and s = 1. Find ~s satisfying as (? )= %+1(?). If ~s ~ Tl ' then go to Step 4. Otherwise, go to Step 3. Calculate f O s) and corresponding y ' . Next calculate its subdifferential 0f(~s). If 0fO s) = - a 0 , then set t* = ~s and calculate ( p * ) from (y:) , and terminate. If min{p I p ~ af(ts)} > - a 0, then set T t = T t A [tl_l, ~s] and go to Step 4. If max{p Ip ~ af0s)} < - a 0 , then set T t = T t N [~s, tt] and go to Step 4. If a s > a~+ 1, then go to step 5. Otherwise, set

Yt=Yt, f i t= f i t , Yt=Yt ( a t = a t ) , J = l , 2 . . . . . s - l ,

l Y , = Y s + l , f s = f l s + f s + l , ")/s = ')/s + "Y s + l [ a s

Yj = Yj+I, f : = f i+1, Yj = Y:+l (a t=ai+l) ,

and n = n - 1. Go to Step 6.

Ys + Ys+ l )

~ s q - f s + l '

j = s + l . . . . , n - l ,

Step 5. Set s = s + 1. If s = n, go to Step 7. Otherwise, return to Step 1. Step 6. If s = n, go to Step 7. If s = 1, return to Step 1. Otherwise set s = s - 1 and return to Step 1. Step 7. Based on all informations about Yi, ay, j = 1, 2 , . . . , n, seek for ~0 satisfying ( d f ( D / d D = - a 0. If

0 * * there exists such ~0, then set t* = ~ , calculate ( p j ) from ~ and terminate. Otherwise, set

tt-1 if ( d f ( ~ ) / d ~ ) > - a 0 on the current Tt,

tl if ( d f ( t ) / d t ) < - a o on the current Tt,

then calculate ( p * ) from ?* and terminate.

Solution procedure for (I) Solve all PI according to the above algorithm.

(II) Choose the best solution among optimal solutions of fit's and calculate an optimal schedule and optimal jobwise machine speeds from it.

Theorem 3. If the nonlinear equation

Fi/( D i + Eix) 2 =A i=1

is solved in O(n z) computational time, our procedure solves fi in O( n 4) computational time.

Proof. Validity is clear from the above discussions. For the complexity, first we show the complexity of our algorithm for Pt. the computational time of each step is as follows: (a) O(n) to calculate f(~s) in Step 3. (b) It takes O(n) from Step 1 to Step 6. (c) From (a) and (b), O(n 2) till Step 7. (d) By assumption, it takes O(n 2) in Step 7.


In total, O(n 2) for each F t. Since the number of Pt is at most O(nZ), O(n 4) computational time is enough to solve fi by our procedure. []

4. Numerical example

Consider the job set with the following data:

J /3j d j- e r a r

J1 4 4 4 2 J2 5 1 6 3 J3 2 4 5 5 J4 3 5 1 1 J5 6 3 2 4

Further membership functions follows:

(11,_ C~_<4, ml(Cl) = l ( c 1 -- 4), 8 _> C 1 _> 4,

~0, C1>-8 ,

(11' C 2 < 1 ' m 2 ( C 2 ) = - 1 ( C 2 - 1), 7 _> C 2 >__ 1,

~0, C2>_7,

m3(C3) = -- ~(C 3 - 4), 9 >__ C 3 _~ 4,

~0, C 3 >_9,

/7/4(C4) = l ( c 4 - 5), 6 > C 4 >_ 5,

~0, C4>__ 6,

l 11 C 5 <3 , rns(Cs) - 3) 5 > C s >_ 3, = -l(c , _

I0 , C5 - 5,

and a o = 10.

describing satisfaction with respect to completion time C r are as

/Sj represents the basic processing time in hours, at the base operating speed of the machine. For example, job J~ will take 4 hours if the machine is operated at the base speed, dj is the formal duedate in hours, and if the completion time C r is not greater than d r, the customer of Jr is satisfied completely, i.e., the degree of dissatisfaction is 0. But if late, the degree of satisfaction decreases, i.e., the degree of dissatisfaction increases. But, the increase rate of dissatisfaction when the completion time of job Jr is late one more hoar, is e71 and if its completion time is very late and over d r + er, the degree of satisfaction is 0, i.e., the degree of dissatisfaction is 1. For example, C 1 becomes one hour later, the degree of satisfaction decreases at the rate 1/4. Each decreasing rate reflects the customer and factory sides affairs. So the factory side will balance the satisfaction level among customers, i.e., maximize the minimum satisfaction with respect to job completion times, a; is the cost to speed up the processing of


job Jj, and it represents an extra cost of electrical power a n d / o r the cost for hiring some part-timers, a o evaluates the gain at unit increase of minimal satisfaction, which evaluates the value when delay from the appointed time to the customer is little. (In actuality, this value may be fuzzy.)

After variable transformation PI =pls~ = 4Sl, P2 =p2s2 = 5s2, P3 =p3s3 = 2s3, P4 =P4S4 = 3s4, P5 = ,~5s5 = 6s 5 and t = 1 - t, we set al =,°la~ = 8, ~2 = 15, a3 = 10, ~4 = 3, a5 = 24.

( P ) Minimize t + ( 8 / p 0 + (15/p2) + (10/p3) + (3 /p4) + (24/p5)

subject to P~.~I) < d~.~) + te=~l),

P~-(1) +P~r(2) --< d~(2) q- te.n-(2),

P~r(1) + PTr(2) + P~r(3) --< dz,(3) + teTr(3),

P~r(1) +P~r(2) +P~'(3) +P~-(4) < d~r(4) + te~-(4),

P~r(1) + PTr(2) + PTr(3) + P¢r(4) + P~'(5) ~ drr(5) + left(5),

p l , p z , p 3 , P a , P s > O , 0 _ < t ~ l , 7 r ~ / / .

Then the critical values of t such that the order of dj + ~jej, j = 1, 2, 3, 4, 5, among the interval [0, 1] are t14 ½, t24 4, t25 1, t34 a. So P is divided into P1, P2, P3, P4, P5 corresponding to T 1 = [0, 1], T2 = [1, 1], T3 = [1, 1], T4 = [ 1 5] and T 5 = [4_, 1]. That is the best solution among optimal solutions of P~, P2, P3, P4, Ps, is an optimal solution of P.

We show the solution process of fix only. Those of the other subproblems are similar. Since

d2+-te2<ds+-tes<dl+tel<d3+-te3<d4+-tea for any ~ ~ T1,

the optimal processing order of jobs is J2 ~ J 5 -~J1 - ' J 3 - ' J4 , that is, ~'1(1) = 2, rrl(2) = 5, 7rl(3) = 1, • "1(4) = 3, ~r1(5)= 4. That is,

(P l ) Minimize } + ( 8 / P l ) + (15 /P2 ) + (10 /P3) + (3 /P4) + (24 /p5)

subject to P 2 < l + 6 t ( = d 2 + t e 2 ) ,

p 2 + P s < 3 + 2 t ( = d s + t e 5 ) ,

p2+ps+Pl<4+4-t ( = dl + tel),

p2+Ps+Pl+P3<4+5-t ( = d3 + te3),

p2+P5+Pl+P3+P4<5+-t ( = d4 + tea),

Pl,P2,P3,P4,P5 > 0 , t ~ [0, 1].

For sake of convenience, we reindex job number as follows: J1 ~ J3, J2 ~ J1, "]3 ~ J4, J4 ~ .]5, J5 ~ J2. Step 0: Set

O~l(t) 6 ~ + 1 ' a2 ( t ) - 4 ~ + 2 ' a3(~) 2~+1'

s = 1 (n = 5) and TI = [0, 0.250].


Step 1: F rom a l ( ~ ) = o~2(i),

2 ac id - i 1 = = 0.063.

4V~- + 6 2 ~

Since ~1 ~ T1 ' go to Step 3. Step 3: In this case,

f ( i l ) = P + 5 '

d ( 17~-+ 2 v ~ - + ~ / 8 + 17 ] -0+v~) 2

d t 1 { f ( i l ) } = ( i 1 --[- 5) 2 < - a 0 ( = - 1 0 ) ,

that is, subgradient is less than - a 0. Thus, we update T~ ,,-- T 1 n [i 1, t l ] [0.063, 0.250]). Go to Step 4. Step 4: Since a l (D < a2(~) for any ~ ~ T1, update

]/l"b~2 1 ~ " [ - 2 ~ 1(i) - - =

/~1 q- ~2 27 + 3 '

a z ( t ) ~ a3(t) , n = 4. Go to Step 6. Step 6: Since s = 1, re turn to Step 1. Step 1: Solve a l (D = az(~) and calculate its solution

3N-li - 1 = = - 0.024.

2( 1¢~- + 2 ~ - vr8)

Step 2: Since i 1 g~ T1 ' go to Step 4. Step 4: Since al(~) > a2(}) for all } ~ T1, go to Step 5. Step 5: Set s = s + 1 ( = 2). Since s < n, re turn to Step 1. Step 1: F rom a2(i)= O~3(t),

2 -- 0.905. - 2 ¢ Y 6

Step 2: Since ~2 ~i~ T1, go to Step 4. Step 4: Since a2( i ) < a3(i) for any t ~ T 1, update

• 2+ 3 ¢g + 1¢i-6 2(i)

f12 + f13 3t + 1

ot3(t) *- a4(t) , n = 3. Go to step 6. Step 6: Since s < n, set s -- s - 1 ( = 1) and re turn to Step 1. Step 1: F rom a l ( } ) = a2( i ) , }1= 0.642. Step 2: Since i I ~ 7"1, go to Step 4. Step 4: Since al(~) < a2(t) for any ~ ~ 7"1, upda te

1(i) ~1 + f12 5i + 4

([0, 0.250] n [0.063, 0.250] =

a2( t ) ~ a30 ) , n = 2. Go to Step 6.


J5 J~ J2 31 J3

2.376 3.216 5.094 6.465 7.998

F igu re 1. An op t ima l schedu le of the example in Sec t ion 4

Step 6: Since s = 1, re tu rn to Step 1. Step 1: F r o m oq(D = a 2 ( t ) , ~1 = 0.116. Since ~1 ~ T1 ' go to Step 3. Step 3: In this case,

f ( ~ l ) = t i "[- 5 ,

d ( l y e - + 2 ¢ ~ + ~ / ~ - + l ~ f ~ + ~ - ) 2

d t 1 { f ( ~ l ) J = _ ( P + 5 ) 2

upda te T 1 ~ [0.116, 0.250]. G o to Step 4. Step 4: Since a l ( t ) < aE(-t), upda te

4 1 ( ~ ) = ~ + 5 '

< - a 0 ( = - 1 0 ) ,

n = 1. G o to Step 6. Step 6: s = n and go to Step 7. Step 7: Seek }( ~ T 1) satisfying

d (~/~- + 2vC~- + ,f8 + lV'~ + vr3-) 2 ~ - { f ( t ) } = -- (~ + 5)2 = - a o ( = - 1 0 ) .

So we set J* = i = 0.216. T h e n obtain f O * ) = 52.161, y~' = 3.162, h O * ) = 54.322. Fur the r

p~' = 1.225, s~ = 0.245, p~' = 1.549, s~' = 0.258, P*3 = 0.894, s~ = 0.224, P4* = 1.000, s~ = 0.500, P5* = 0.548, ss* = 0.183.

Similarly, we obta in h ( t * ) = 51.347 f rom if__2, h ( t * ) = 51.347 f rom P_-3, h( t *) = 44.609 f rom P4, h(} *) = 42.009 f rom /'5.

F r o m the above results, an opt imal solution of P5 becomes an opt imal solution of ft. Correspondingly , an opt imal scheduling o rder is J5 ~ J4 ~ J2 ~ J , ~ J3 and an opt imal mach ine speed for each job is

r* ) - 1 , ¢ , - 1 = s, = ( s ~ ' = ( 0 . 3 9 6 ) - = 2 . 5 2 5 , s 2 = (s~ ' ) = ( 0 . 2 8 0 ) - 1 3.571,

'* ) - ' = 2.660, s 4 = ( S ~ ) = ( 0 . 3 4 3 ) - ' = 2.915, s 3 = (s~' = ( 0 . 3 7 6 ) - ' , , -1

r* - 1 1 s 5 = ( s~ ' ) = ( 0 . 7 6 7 ) - = 1.304.

Finally, the opt imal sat isfact ion t * = 0.8 and details of an opt imal schedule are shown in Figure 1.

5. Discussion

This p a p e r cons idered the one machine scheduling p rob l em with fuzzy duedates . T h o u g h a l inear m e m b e r s h i p funct ion is not enough to descr ibe an actual s i tuat ion of flexible duedates , we think it is a


first attempt to generalize an ordinary scheduling problem into a more flexible model by fuzzy duedates. In this sense, other factors such as processing times may be also fuzzified. But our solution procedure does not utilize this structure of fuzziness enough. So refinement of our model may be the first effort to be done.

Whether investigations of a so-called 'fuzzy scheduling model' are successful or not depends greatly on the degree of difficulty of the original model. In summary, we think this generalized area of scheduling theory may be fruitful and interesting, and many issues remain without any study. Especially our model is simple and it may not reflect actual manufacturing situations and so more general models of scheduling problems are to be fuzzified and solved.

References

[1] Blazewicz, J., "Selected topics in scheduling theory", Annals of Discrete Mathematics 31 (1987) 1-60. [2] Cho, Y., and Sahni, S., "Preemptive scheduling of independent jobs with release and duedate on open, flow and job shops",

Operations Research 29 (1981) 511-522. [3] Dempster, M.A.H., Lenstra, J.K. and Rinnooy Kan, A.H.G., Deterministic and Stochastic Scheduling, Reidel, Dordrecht, 1982. [4] Garey, M.R., Johnson, D.S., Simons, B.B., and Tarjan, R.E., "Scheduling unit tasks with arbitrary release times and

deadlines", SlAM Journal on Computing 10 (1981) 256-269. [5] Horn, W.A., "Some simple scheduling algorithms", Naval Research Logistics Quarterly 21 (1974) 177-185. [6] Ishii, H., and Nishida, T., "Two machine open shop scheduling problem with controllable machine speeds", Journal of the

Operations Research Society of Japan 29 (1986) 123-131. [7] Ishii, H., Martel, C., Masuda, T., and Nishida, T., "A generalized uniform processor system", Operations Research 33 (1985)

346-362. [8] Ishii, H., Masuda, T., and Nishida, T., Two machine mixed shop scheduling problem with controllable machine speeds",

Discrete Applied Mathematics 17 (1987) 29-38. [9] Jackson, J.R., "Scheduling a production line to minimize maximum tardiness", Research Report 43, Management Research

Project, University of California, Los Angeles CA, 1955. [10] Lawler, E.L., Lenstra, J.K. and Rinnooy Kan, A.H.G., "Minimizing maximum lateness in two machine open shop",

Mathematics of Operations Research 6 (1981) 153-158. [11] Megiddo, N., "Combinatorial optimization with rational objective functions", Mathematics of Operations Research 4 (1979)

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Journal of Operational Research 34 (1988) 208-220. [13] Smith, W.E., "Various optimizer for single stage production", Naval Research Logistics Quarterly 3 (1956) 59-66.

Documents

One machine scheduling problem with fuzzy duedates