Single-machine earliness-tardiness scheduling about a common due date with tolerances

international journal of

production economics

ELSEVIER Int. J. Production Economics 42 (1995) 217-227

Single-machine earliness-tardiness scheduling about a common due date with tolerances

Michael X. Weng”, Jose A. Venturaby*

“Department qf Industrial and Management Systems Engineering, University of South Florida, Tampa, FL 33620-5350, USA

bDepartment of Industrial and Manufacturing Engineering, 207 Hammond Building, The Pennsyloania State University, Unit~ersi& Park,

PA 16802, USA

Received I October 1994; accepted 6 September 1995

Abstract

In this paper the problem of minimizing the mean absolute deviation (MAD) of job completion times about a given common due date with different sizes of tolerance in an n-job, single-machine scheduling environment is considered. We describe some optimality conditions and show that the problem is NP-complete. A heuristic algorithm is proposed to find an approximate solution to the problem. A pseudo-polynomial dynamic programming algorithm is also proposed to optimally solve the problem if the agreement condition defined in the paper holds. The dynamic programming algorithm can also be used to find a lower bound for the problem by adjusting the tolerances of some jobs. An example is provided to illustrate the proposed methods. A computational study on randomly generated test problems is conducted to investigate the performance of the proposed heuristic and the quality of lower bounds generated by the dynamic programming method. Computational results are reported.

Kevwords: Single-machine scheduling; Job completion; Dynamic programming algorithm

1. Introduction

There are many practical situations in which production schedules must be evaluated with respect to both earliness and tardiness costs. This scheduling objective is consistent with the recent increasing interest in Just-In-Time (JIT) production. Consequently, the literature in scheduling against due dates to penalize both earliness and tardiness has grown rapidly. For a comprehensive review of machine scheduling with costs due to

* Corresponding author

early or late completion ofjobs with respect to their due dates, refer to, for example, Raghavachari [l], and Baker and Scudder [Z]. A relevant literature review is given below.

Kanet [3] introduced the single-machine scheduling problem of minimizing the mean unweighted absolute deviation (MAD) of job completion times from a given common due date d, where all jobs are available for processing at time zero. Under the assumption that the given common due date is greater than or equal to the total processing time of all jobs, he proposed a simple polynomial time algorithm to generate an optimal schedule in which the (second) shortest job completes at d if the

0925.5273/95/$09.50 #$I 1995 Elsevier Science B.V. All rights reserved

SSDI 0925-5273(95)00200-6

218 M.X. Weng, J.A. Ventur-aflnt. J. Production Economics 42 (1995) 217-227

number ofjobs is odd (even). Sundararaghavan and Ahmed [4] noticed that Kanet’s algorithm cannot solve the MAD problem if d is restrictively small, since the optimal schedule generated by Kanet’s algorithm may require the first job to start before time zero which violates the zero-ready-time assumption. This notion gives rise to the restricteli and unrestricted versions of the MAD problem. Panwalker et al. [S] investigated the single- machine problem of minimizing the weighted sum of due date, earliness and tardiness penalties, where the due date is a decision variable. They proposed an O(n log n) algorithm that optimally solves the problem. The optimality conditions for the MAD problem were discussed by Raghavachari [6], Hall [7], and Bagchi et al. [S]. Bagchi et al. found the critical due date d* such that the MAD problem is restricted if d < d*. The unrestricted version of the MAD problem is equivalent to a special case of the problem studied by Panwalker et al. While the unrestricted version of the MAD problem can be solved in polynomial time, the restricted version was independently demonstrated to be NP-complete in the ordinary sense by Hall et al. [9] and Hoogeveen and van de Velde [lo].

An obvious extension of the MAD problem in single-machine scheduling is the mean weighted absolute deviation (MWAD) problem. This problem was shown to be NP-complete by Hall and Posner [ll]. A pseudo-polynomial algorithm for identifying an optimal schedule to the MWAD problem was proposed in [l 11. Bagchi et al. [S] discussed a special case of the MWAD problem, where all jobs have equal earliness and tardiness weights. They presented a polynomial algorithm for determining the critical common due date d*. If d 3 d*, the special case is unrestricted and can be solved by their efficient algorithm.

Cheng [12] discussed another extension of Kanet’s problem where no penalty will occur for a job if the deviation of its completion time from the given common due date does not exceed some predetermined constant which is called due date tolerance [2]. This extension is called the MAD problem with tolerance. For consistency, Kanet’s problem is referred to as the MAD problem (without tolerance). Cheng investigated the case in which there is a due date tolerance common for all jobs.

However, Cheng set the due date tolerance so small that for any schedule at most a single job can avoid the penalty. Under this assumption, he proved some dominance properties which are sufficient for determining an optimal schedule. Dickman et al. [13] discussed the availability of multiple optimal schedules for Cheng’s problem. Weng and Ventura [14] investigated the MAD problem with a due date tolerance of arbitrary size common for all jobs. They discussed some optimality properties and proposed a polynomial algorithm to solve the unrestricted version of the problem.

In this paper, we consider the MAD problem with tolerance which varies from job to job. That is, the due date tolerance for a job can differ from that for another job. We describe some optimality conditions and show that this problem is NP-complete.

2. Notation and problem description

The following notation is used throughout the paper:

; number of jobs. 11, 2,. , n), job index set

Ji job i

Pi processing time of job i d common due date for all jobs

ti due date tolerance for job i

ci completion time of job i

Ei max {O, (d - ti) - C,}, earliness of job i

Ti max (0, Ci - (d + ti)}, tardiness of job i E set of jobs that are completed on or before d T set of jobs that are started on or after d

In addition, assume that all jobs are available at time zero and preemption is not allowed. Without loss of generality, it is also assumed that d, pi and ti, for all i, are all integers. Then, the scheduling problem concerned in this paper can be defined as follows:

Problem P. Given n,jobs with processing times (pI, p2, , p,>, a common due date d and due date toler-

ances {tI, tz, . , t,},$nd a schedule which minimizes

the following sum of earliness and tardiness penalties:

Z = C(Ei + Ti). (1) isN

M.X. Weng, J.A. VenturalInt. J. Production Economics 42 (1995) 217-227 219

The objective function defined in Eq. (1) takes into account the fact that no penalty will incur for job Ji if it is completed within its delivery window [d - ti, d + ti], and when job Ji is completed outside its delivery window, its earliness (tardiness) cost is proportional to the early amount with respect to d - ti (the late amount with respect to A + ti). d - ti and d + ti are called, respectively, the early and late delivery time for job Ji. As pointed out by Baker and Scudder [2], this measure of job earliness and tardiness costs is more conventional and consistent than the one used by Cheng [12], where the penalty is proportional to the early or late amount with respect to the common due date d when jobs are completed outside their delivery window.

3. Preliminaries

It is straightforward to verify that there exists an optimal schedule in which there is no idle time between any two consecutive jobs. Thus, a schedule is completely determined by a start time and a se- quence of jobs. The next two theorems further characterize an optimal schedule.

Theorem 1. Assume that, for jobs Ji and Jj, pi < pj and ti d tj. Then, there exists an optimal schedule to Problem P such that

(a) job Jj precedes job Ji, if both jobs Ji and Jj are completed on or before d, and

(b) job Ji precedes job Jj, if both jobs Ji and Jj are started on or after d.

Proof. (a) Let S be an arbitrary schedule. Assume that in schedule S, both jobs Ji and Jj are completed on or before d but job Ji precedes job .Q. Let A denote the set of jobs that precede job Ji, B the set of jobs that succeed job Ji and precede job Jj, and C the set ofjobs that succeed job 4. This situation is depicted in Fig. l(a), where s is the start time of job Ji and P is the total processing time of jobs in set B. Note that the set B is empty and P = 0 ifjobs Ji and Jj are adjacent. Now construct a new schedule S’ by interchanging jobs Ji and Jj in schedule S, as depicted in Fig. l(b).

Since the completion time of each job in sets A and C is identical in schedules S and S’, the net change in total cost by interchanging jobs Ji and Jj is given by

AZ = z’ - z = E; - Ei + ES - Ej + &,

where AB denotes the change of cost caused by jobs in set B. We now show that AZ d 0.

Since Pi d Pj, Ck < CL, for all k E B. This implies As d 0. Then, the proof is completed if it can be

c-l I

d

I I

I I I I I 1 A lil B Ijl C

(a). Schedule S.

I I A j Bi C

I I

I

0 s

(b) . Schedule S'.

Fig. 1. Schedules S and S’.

220 M.X. Weng, J.A. Ven~urn,!lnt. J. Produclim Economics 42 (1995) 217-227

shown that El - Ei + Ej - Ei < 0. We consider Theorems 3 and 4 presented below provide rules four cases. that can be used to order adjacent jobs in sets E and

CUSS 1: d - tj d s + pj. T, respectively. For this case, job J, incurs no cost in both sched-

ules S and S’. In addition, Ei 3 Ei since Ci d C:. Therefore, Ei - Ei + ES - Ej < 0.

Case 2: d - tj > s + pj and d - ti 6 s + pi + P + pj. In this case, job Ji incurs cost only in sched-

ule S and job Jj incurs cost only in schedule S’. Therefore, Ei = d - ti - (S + pi), Ei = Ei = 0, and E[i = d - tj - (s + pj). This results in the following:

Theorem 3. Let Ji and Jj he adjacent ,johs in an arbitrary schedule S and c he the time at which both Ji und Jj are completed. Let k = arg max Ipi + ti,

pj + tj} and r = arg min (pi + ti, pi + tjl. If both jobs Ji and Jj start on or qfter d, then it is pwferahle

to huce job Jk come,fir.st except when

Pk + tk > PI + b, PI > Pk and

d > c + tk + pk - p,., EI - EL + Eii - Ej = (ti - tj) + (pi - pi) ~ 0.

CUSS 3: s + pj < d - tj < s + pi + P + pj and s + pi + P + pj < d - ti. In this case, job Ji incurs

cost in both schedules S and S’, and job Jj incurs cost only in schedule S’.

That is, E, = d - ti - (S + pi), Ei = d - t, ~

(S + pi + P + pj), Ej (s + pj). Therefore,

E; - Ei + El - Ej = d

where the inequality

= 0, and ES = d - tj -

_ tj - (S + 2pj + P) < 0,

follows since CE - tj .$ s + pi + P + pj and pi < pp

CLZX 4: d - tj > s + pi + P + pj. In this case, both jobs J, and Jj incur costs in both schedules S and S’. This implies that Ei = d - ti - (S + pi). EI = d - ti - (S + pi + P + pj), Ej = d - tj - (S + pi

+ p + Pjh and ES = d - tj - (s + pj). Hence, EI - Ei + ES - Ej = (pi - p,i) d 0.

The four cases examined above include all possible cases that can happen. Therefore, it has been shown that schedule S’ is at least as good as schedule S. Since S is an arbitrary schedule, the proof is complete.

(b) The proof for this part is similar to that for

(a) 0.

Theorem 2. [f pi d pj implies t; < tj,,fi)r ~11 Ji and Jj, then there exists an optimal schedule to Problem P in which jobs in set E are in LPT (longest processing time,first) order and ,johs in set T are in SPT (shortest processing time first) order.

Proof. It follows immediately from Theorem 1. 0

in vvhich ca,se ,joh J, should come jirst.

Proof. Assume that Ji precedes Jj in S. Consider the schedule S’ that is identical to S except that jobs Ji and Ji are interchanged. We seek conditions to determine how the two jobs should be ordered. Rather than comparing z for both schedules, it will suffice to compare the contributions to total cost that come from jobs Ji and Ji, since the total contributions of the other jobs are the same in both schedules. Note that s < d - pi - pi implies both jobs are completed no later than din S and S’. Thus let

Zij = E,(S) + E,(S) = max(0, d - ti - C + pi)

+ max{O, d - tj - C)

and

Zji = EAS’) + E,(S’) = max{O, d - t,j - c + pj)

+ max{O, d - ti - c).

Then job Ji should precede job Jj if Zij < Zji, and job Ji should precede job Ji, otherwise.

Assume that pi 3 pj. NOW we find the conditions under which zij 6 Zji. There are several cases to consider.

Case 1: ti >, tj, By Theorem 1, Zij < Zji. Note that in this case, p; + ti 3 pj + tj.

Cuse 2: ti < tj.

Case 2.1: pi + ti 3 pj + tj

Zij < Zji, since max (0, d - ti - c + pj)

< max {O, d - tj - c + pi> and max {O, d - tj - C]

< max{O, d - ti - c).

M.X. Weng, J.A. VenturalInt. J. Production Economics 42 (1995) 217-227 221

Case 2.21 pi + ti < pj + tj.

Case 2.2.1: d > tj + C.

Zij = (d - ti - C + Pj) + (d - tj - C)

= 2d - ti - tj - 2~ + Pj

and

preferable to have the job with the smaller tolerance

come first except when

s - d + max{pi, pj} > max{ti, tj},

in which case the shorter job should come first.

Zji = (d - tj - c + pi) + (d - ti - C)

= 2d - ti - tj - 2~ + Pi.

Therefore, zij < Zji, since pi > pj Furthermore, Zij = Zji if and only if pi = pj.

Proof. It follows from Baker [15, p. 311. 0

Theorem 5. There exists an optimal schedule to Problem P in which some job, say job k, is completed at d - tk Or d + tk.

Case 2.2.2: ti + c < d < tj + C.

Zij = d - ti - c + pj,

and

zji = maxf0, d - tj - c + pi> + (d - ti - c).

Proof. Let S* be an optimal schedule and let m and m’ be the number of early and tardy jobs in S*, respectively. Assume that no job in S* is completed at one end of its delivery window. If the schedule S* is shifted earlier by E (a positive small number), the net change in total cost is

Thus, Zij > Zji, unless d > tj + c + pj - pi. Note that d > tj + c + pj - pi is reduced to case 2.2.1 if

Pi = Pj.

AzL = (m - ml)&.

Case 2.2.3: d < ti + c.

Similarly, if the schedule S* is shifted later by E, the net change in total cost is

Zij = max{O, d - ti - c + pj} 2 Zji AzR = (m’ - m)E.

= max{O, d - tj - c + pi},

since pj + tj > pi + ti.

If m # m’, then either AZ, < 0 or AzR < 0. This implies that S* is not optimal, which is a contradic- tion. Thus, there must be some job in S* which is completed at one end of its delivery window.

All the cases and subcases have now been exhausted. We conclude that it is preferable to have job Ji come first if pi + ti 2 pj + tj, and to have job Jj come first if pi + ti < pj + tj except when

pi > pj and d > tj + c + pj - pi,

in which case job Ji should come first. Considering the fact that pi 2 pj was assumed at the outset, we can restate this result: It is preferable to have job Jk come first except when

If m = m’, then AzL = AzR = 0. We can keep shifting S* to the right or the left until some job is completed at one end of its delivery window. Since the cost does not change by such shifting, the re- sulting schedule must be also optimal. The proof is thus complete. 0

4. Problem P is NP-complete

Pk + tk > Pr + tr3 PI > Pk and d > C + tk i- pk - PI,

in which case job Jr should come first, where k = arg ITlaX{pi + hi, Pj + tj> and r = arg

min{pi + ti, pj + tj}. 0

In this section, we prove that Problem P is NP-complete by reducing in polynomial time the even-odd partition problem, which is known to be NP-complete [16], to the recognition version of Problem P.

Theorem 4. Let Ji and Jj be two adjacent jobs in an Even-odd Partition problem. Given a set of 2n posi- arbitrary schedule S and s denote the time at which tive integers B = { bI, b2, . . . , b2,,} with bi > bi _ 1 for job Ji or job Jj can be started. Ifs > d, then it is each i = 2,3, . . . , 2n, is there a partition of B into two

222 M.X. Weng, J.A. Ventura Jlnt. J. Production Economics 42 (I 995) 217-227

subsets B1 and B2 such that

C hi = C hi =A, h,EB, h,sB,

und B1 contains exactly] one qf {hzi- 1, hzi) for each i = 1, 2, . . . . n?

Recognition version of Problem P. Does there exist

a schedule with totul earliness and tardiness penalty no greater than some threshold value yO?

Given an arbitrary instance of the Even-Odd Partition problem, construct an instance (called Instance I) of Problem P as follows:

Number of jobs: 2n + 3. Processing times: PO = 3A,

pi = hi, i = 1, 2, . . , 2n,

p212+ I = 3A, p2,,+2 = 4A.

Common due date: d = 10A. Due date tolerances: ti = 0, i = 0, 1, 2, . . . ,2n,

t ,_,,+I = 4A, t 2,,+2 = 4A.

Threshold value: .YO=Cl <i<n i(p2i- I + p2i).

For Instance I, we have ti < tj if pi < pj, for all i andj. Thus, Theorem 2 implies that jobs in sets, E and T of an optimal schedule must be in LPT and SPT orders, respectively.

The 2n jobs Ji, J2, . . . , Jzn as defined above are called partition jobs. Consider a partitioning of the set of partition jobs (J1, Jz, , J2,,} into two subsets B1 = {JI1, J21, . . . . J,l} and B1 = {J12, Jz2, . , J,,z}, where (Jil, Ji2} = {J2i_ 1, Jzi}, for each i = 1, 2, . . . , n.

Lemma 1. If there exists a solution,for the Even-Odd

Partition problem, then the cost of schedule So con-

strutted as shown in Fig. 2 is equal to the threshold

v&e yo.

Proof. As implied by Theorem 2, jobs in subsets B1 and Bz are sequenced in LPT and SPT orders, respectively. The Lemma then follows from straightforward computation. 0

Conversely, we need to show that there is a schedule with total cost z < jlo only if the Even-Odd Partition problem has a solution. We first define another instance of Problem P as follows.

Instance I’.

Number of jobs: 2n + 1. Processing times: ~0 = 3A and pi = hi,

i= 1, 2 ,..., 2n. Common due date: d = 10A. Due date tolerances: ti = 0, i = 0, 1, 2, . , 2n.

Proposition 1. Schedules constructed us shown in Fig. 3 are optimal for Instance I’, with the optimal total cost z*(F) equal to y,. There are 2” dijtkent

optimal schedules corresponding to the 2” d#erent purtitioning of set B into subsets B1 and B2, and any other schedule has a cost strictly greater than yo.

Proof. ti = 0, for all i, implies that Instance I’ is also an instance of the MAD problem. Since the common due date d > co 4 i G znpi (makespan), I’ is unrestricted. Then, the proposition follows immediately from Kanet [9]. 0

Lemma 2. If there is an optimal schedule for Instance I with a total cost z*(I) < yo, then there must exist a solution ,for the Even-Odd Partition problem.

0 d

Fig. 2. Schedule S,,.

M./Y Weng, J.A. Ventura/lnt. J. Production Economics 42 (1995) 217-227 223

1 Jo B, B*

0 d

Fig. 3. Optimal schedules for I’.

Proof. It is easy to observe that z*(I) > z*(r) = y,. Since Instance I’ is obtained from Instance I by removing jobs Jzn + 1 and Jzn + 2, Proposition 1 implies that if z*(Z) d y,, Ez,+ 1 = T2,,+ 1 = Ezn+ 2 = T 2n+ 2 = 0. This happens only if there is an

optimal schedule which has the same structure as the schedule So shown in Fig. 2. That is, if z*(Z) d y,,, then there must exist a solution for the Even-Odd Partition problem. !Zl

Theorem 6. Given a set of jobs and a nonnegative integer y, the problem of determining whether there exists a schedule with total cost z < y is NP-complete.

Proof. For any instance of the Even-Odd Partition problem, construct a set of jobs as described above and set y = y,. This transformation requires polynomial time. Theorem 6 then follows immediately from Lemmas 1 and 2. 0

5. Algorithms

In this section, a polynomial time algorithm is presented to find a heuristic solution to Problem P. A dynamic programming algorithm with pseudo- polynomial time bound is also proposed to find an optimal schedule to Problem P if pi < pj implies ti d tj, for all i and j.

Assume that jobs are numbered such that pi < pj if i <j. Then, it is well known [3] that for the unrestricted MAD problem (i.e., ti = 0, for all i), the schedule (n, n - 2, . . . , 1, . . . , n - 3, n - 1) is optimal, where job 1 (job 2) is completed on d if n is odd (even). This schedule is called Kanet’s schedule.

Note that the Kanet’s schedule may not be optimal to the unrestricted MAD problem if jobs are numbered in a different manner.

Algorithm H (Heuristic) Step 1: Find an initial schedule by resequencing

jobs in set E and set T of the Kanet’s schedule, respectively, in nonincreasing order of pi + ti and in SDDT (smallest due date tolerance) order.

Step 2: Apply Theorems 3 and 4 to adjacent jobs in set E and set T of the initial schedule until no improvement is possible.

Step 3: If the number of early jobs (nE) in set E is equal to the number of tardy jobs (nT) in set T, stop. Otherwise, shift the current best schedule left (right) if nE < nT (nE > nT), until no improvement is possible.

If jobs are numbered differently, Algorithm H will apparently generate different heuristic solutions. We propose the following three strategies to number jobs.

Strategy 1: Jobs are numbered in SPT order, where a tie is broken by the SDDT rule.

Strategy 2: Jobs are numbered in SDDT order, where a tie is broken by the SPT rule.

Strategy 3: Jobs are numbered such that pi + ti d pj + tj, if i <,j, where a tie is broken by numbering jobs in SPT order.

An example The following example is provided to illustrate

the proposed method: n = 8. Job processing times and corresponding tolerances are (1,5; 3,2; 4,l; 5,lO; 7,9; 9,3; 10,2; 12,l) with the common due date d = cpi = 51.

224 M.X. Weng, J.A. Ventura/M. J. Production Economics 42 (1995) 217-227

Strategy 1: Jobs are numbered in SPT order as Strategy 3: Jobs are numbered in nondecreasing follows: order of pi + ti as follows:

i 12 3 45 6 7 8 i 12 3 4 5 6 78

pi 1 3 4 5 7 9 10 12 pi 3 4 1 9 10 12 5 7

ti 5 2 1 109 3 2 1 ti 2 1 5 3 2 1 10 9

Then, Kanet’s schedule in this case is shown in Fig. 4. Algorithm H generates the schedule with a total cost 48, shown in Fig. 5.

Strategy 2: Jobs are numbered in SDDT order as follows:

The Kanet’s schedule for this case is shown in Fig. 8. Algorithm H then generates the schedule with a total cost 41, as shown in Fig. 9.

The schedule shown in Fig. 9 is the actual optimal.

i 1 23 45 6 7 8

Pi 4 12 3 109 17 5 A dynamic programming algorithm

ti 1 1 2 2 3 5 9 10 We now propose a dynamic programming algo-

rithm (called Algorithm DP) to find an optimal Kanet’s schedule in this case is shown in Fig. 6. schedule for the special case that pi < pj implies

Algorithm H then generates the schedule with a ti d tj, for all i and j. Assume that jobs have been total cost 47, as shown in Fig. 7. numbered by strategy 3. In Algorithm DP, we first

d I

I I I I II I I I

12 9 5 314 7 10 I I

0 22 34 43 48 51 52 56 63 73

Fig. 4. Kanet’s schedule for strategy 1

d

I I I I I I I I 12 5 9 341 7 10

I 0 22 34 39 48 51 55 56 63 73

Fig. 5. Schedule generated by Algorithm H.

d

I I I I I I I I

51 10 12 4 3 9 7 I

0 23 28 29 39 51 55 58 67 74

Fig. 6. Kanet’s schedule for strategy 2.

d

I ' I I I I I

12 5'

10 1 4 3'

9 7 I

0 20 32 37 47 48 52 55 64 71


M.X. Weng, J.A. VenturalInt. J. Production Economics 42 (I 99s) 217-227 225

d

I I I I II I I

7 12 9 4 31 10 5 1

0 19 26 38 47 51 54 55 65 70

Fig. 8. Kanet’s schedule for strategy 3.

d I

I I I I I I 1 12 7 9 4 31 5 10

I

0 20 32 39 48 52 55 56 61 71


choose some job that may be started before and completed after d, and schedule the rest of the jobs one at a time, in the increasing order of job indices. In addition, Si denotes the start time of job J,, for all i.

Algorithm DP Optimal value function:

f(r, k, s) = the minimum total cost if job J, and the k jobs with the smallest index numbers (excluding Y) have been scheduled, given that the first job is started at s.

Recursive relation: For r = 1,2, . . . , n and all k # r,

f(r, k, s) = min

i

f(r, k - 1, .s - pk) + max(0, d - tk - s - pk},(2)

f(r, k - 1, s) + max(0, s + Q - d - tk}, (3)

where Q is the total processing time of all scheduled jobs (including job I). Ties are broken arbitrarily. Optimal policy function:

1 P(r, k, s) =

if k is scheduled in the front,

0 otherwise. (4)

Boundary conditions:

f(r, u, s) = max{O, S, + p, - d - t,>,

for s,=d-p,,d-p,-l,..., d,

f(r, u, s) = CC , if s, < d - p, or s, > d,

(5)

(6)

where u = 1 if r = 1, and u = 0 if r > 1.

Optimal solution:

min{f(r, n’, s): r = 1, 2, . . , n and

S = S,inr S,in + 1, ... > d}, (7)

where S,in = min(0, d-MS}, n’ = n if r < n and n’ = n - 1, otherwise, and MS = Ci pi.

Theorem 7. Zf the agreement condition (i.e., pi < pj implies ti < tj) holds, the schedule found by Algo- rithm DP is optimal.

Proof. Algorithm DP first schedules job J, which may be started before and completed after d. The- orem 2 implies that job Jk (the next job to be scheduled) must be scheduled to immediately precede or succeed the partial schedule from the previous stage. If job Jk is scheduled to precede the partial schedule, the new cost is given by (2). If job Jk is scheduled to succeed the partial schedule, the new cost is given by (3). By Theorem 5, there is an optimal schedule in which some job (say job k) completes at d - tk or d + tk. Under the assumption that the due date, job processing times and

226 M.X. Weng, J.A. Venturaflnt. J. Production Economics 42 (1995) 217-227

I

d I

I I I I II I I I 12 10 7 413 5 9 I

0 19 31 41 48 52 53 56 61 70

Fig. 10. An optimal schedule for the new problem.

tolerances are all integers, it is only necessary to consider integer start times for any job since it is assumed that d and ti are integer. Because r = 1, 2,..., n and s = S,in, s,rn + l,..., d have exhausted all possible cases, (7) yields an optimal schedule. 0

Since for every r = 1, 2, . , n, Algorithm DP considers all integers s, min{O, d-MS} < s < d, its running time is bounded by O(n2min{d, MS}).

Now, we propose a method to generate a lower bound for Problem P. Given any instance of Prob- lem P, let all jobs be numbered in SPT order (ties are broken by numbering jobs in SDDT order). Define another instance of Problem P as follows:

Pi = Pi3 i = 1, 2, . . . , n,

and

tj = max{t,, t2, . . . , ti}, i = 1, 2, . . , n.

Since tf > ti, for all i, the optimal objective value of the new problem is a lower bound for the original problem. However, the agreement condition holds for the new problem and, consequently, an optimal solution to the new problem can be easily found by Algorithm DP. Let us take the previous example to illustrate the proposed method to generate a lower bound. The new problem will be as follows (remem- ber that the jobs in the original problem are numbered in SPT order): i 1 2 3 4 5 6 7 8 p; 1 3 4 5 7 9 10 12 t; 5 5 5 10 10 10 10 10

Then an optimal schedule generated by Algo- rithm DP is shown in Fig. 10 and its total cost 39 is a lower bound to the optimal solution of the original problem which is 41.

Table 1

Computational results for the proposed methods

n Heuristic solution/z* (optimal solution) Lower

bound 12’

Strategy I Strategy 2 Strategy 3

8 1.145 1.161 1.071 0.943

9 I.158 1.153 1.074 0.928

IO 1.129 1.140 1.059 0.93 1

Optimal solutions z* were obtained by total enumeration.

6. Computational study

This section presents the computational results for some randomly generated small problems in order to investigate the performance of the proposed heuristic (Algorithm H) and the quality of the dynamic programming method to generate a lower bound.

Problems with n = 8, 9 and 10 are considered. For each n, 30 problems are tested. The processing times and tolerances of all test problems are randomly generated from the uniform distribution [ 1,201. The common due date of each test problem is set equal to its makespan MS. The programs are coded in Borland C ++ and run on a Gateway 2000 486/33 MHz PC.

The computational results are given in Table 1. The CPU time for every test problem is less than 1 s. For the heuristic algorithm, Strategy 3 performs better than Strategies 1 and 2 and produces fairly good heuristic solutions to the test problems. The computational results also show that the lower bounds generated by the DP algorithm are close to the optimal solutions and, therefore, can be used to facilitate a search for optimal solutions to problems of large sizes.

M.X. Wag, J.A. VenturalInt. J. Production Economics 42 (1995) 2177227 227

7. Conclusion

In this paper, we have considered the problem of minimizing the mean absolute deviation (MAD) of job completion times about a given common due date with different sizes of tolerance in an n-job, single-machine scheduling environment. Some dominance properties have been presented and the problem has been proven to be NP-complete. A heuristic algorithm is proposed to find an approximate solution to the problem. A dynamic programming algorithm with computational complexity 0(n2 min{d, MS}) is also proposed to optimally solve the problem if the agreement condition defined in the paper holds. The dynamic programming algorithm can also be used to find a lower bound for the problem by adjusting tolerances of some jobs. The computational results indi- cate that the heuristic algorithm produces fairly good solutions with Strategy 3 performing better than Strategies 1 and 2. The computational results also show that the lower bounds generated by the DP algorithm are close to the optimal solutions and, therefore, can be used to facilitate a search for optimal solutions of large-size problems.

Acknowledgement

This research was partially supported by the National Science Foundation under Grant DDM 90-57066.

References Cl51

[l] Raghavachari, M., 1988. Scheduling problems with non-

regular penalty functions A review. Opsearch, 25: 141-164.

Cl61

[2] Baker, K.R. and Scudder, G.D., 1990. Sequencing with

earliness and tardiness penalties: A review. Oper. Res., 38:

22236.

[3] Kanet, J.J., 1981. Minimizing the average deviation ofjob completion times about a common due date. Naval Res.

Logist. Quart., 28: 643-651.

[4] Sundararaghavan, P. and Ahmed, M., 1984. Minimizing

the sum of lateness in single-machine and multimachine

scheduling. Naval Res. Logist. Quart., 31: 3255333.

[S] Panwalker, S.S., Smith, M.L. and Seidmann, A., 1982.

Common due date assignment to minimize total penalty

for the one machine sequencing problem. Oper. Res., 30:

C61

c71

PI

c91

Cl01

Cl11

Cl21

[I31

Cl41

391-399. Raghavachari, M., 1986. A V-shaped property of optimal

schedule of jobs about a common due date. Eur. Oper.

Res., 23: 401402.

Hall, N.G., 1986. Single and multi-processor models for

minimizing completion time variance. Naval Res. Logist.

Quart., 33: 49-54.

Bagchi, U., Sullivan, R. and Chang, Y., 1986. Minimizing

mean absolute deviation ofcompletion times about a common due date. Naval Res. Logist. Quart., 33: 227-240.

Hall, N.G., Kubiak, W. and Sethi, S.P., 1991. Earli-

ness-tardiness scheduling problems II: Deviation of com-

pletion times about a restrictive common due date. Oper. Res., 39: 8477856.

Hoogeveen, J.A. and Van De Velde, S.L., 1991. Scheduling

around a small common due date. Eur. Oper. Res., 55: 237-242.

Hall, N.G. and Posner, M.E., 1991. Earlinessstardiness

scheduling problems I: Weighted deviation of completion

times about a common due date. Oper. Res., 39: 836-846.

Cheng, T.C.E., 1988. Optimal common due date with limited completion time deviation. Comput. Oper. Res., 15:

91-96.

Dickman, B., Wilamowsky, Y. and Epstein, S., 1991.

A note on optimal common due date with limited comple-

tion time. Comput. Oper. Res., 15: 125-127.

Weng, M.X. and Ventura, J.A., 1994. Scheduling about

a large common due date with tolerance to minimize mean

absolute deviation of completion times. Naval Res. Logist., 14: 8433851.

Baker, K.R., 1974. Introduction to Sequencing and Sched-

uling. Wiley, New York.

Carey, M.R., Tarjan, R.E. and Wilfong, G.T., 1988. One-

processor scheduling with earliness and tardiness penalties. Math. Oper. Res., 13: 33&348.

Documents

Single-machine earliness-tardiness scheduling about a common due date with tolerances