Single-machine scheduling with trade-off between number of tardy jobs and compression cost

Transcription

1 Ths s the Pre-Publshed Verson. Sngle-machne schedulng wth trade-off between number of tardy jobs and compresson cost 1, 2, Yong He 1 Department of Mathematcs, Zhejang Unversty, Hangzhou , P.R. Chna 2 State Key Lab of CAD & CG, Zhejang Unversty, Hangzhou , P.R. Chna Q We 3 3 Nngbo Insttute of Technology, Zhejang Unversty, Nngbo , P.R. Chna T. C. E. Cheng 4, 4 Department of Logstcs, The Hong Kong Polytechnc Unversty, Kowloon, Hong Kong Abstract We consder a sngle-machne schedulng problem n whch the job processng tmes are controllable or compressble. The performance crtera are the compresson cost and the number of tardy jobs. For the problem where no tardy jobs are allowed and the objectve s to mnmze the total compresson cost, we present a strongly polynomal tme algorthm. For the problem to construct the trade-off curve between the number of tardy jobs and the maxmum compresson cost, we present a polynomal tme algorthm. Furthermore, we extend the problem to the case of dscrete controllable processng tmes where the processng Ths research was supported by the Teachng and Research Award Program for Outstandng Young Teachers n Hgher Educaton Insttutons of the MOE, Chna, and the Natonal Natural Scence Foundaton of Chna ( ). The thrd author was supported n part by The Hong Kong Polytechnc Unversty under a grant from the ASD n Chna Busness Servces. Emal: mathhey@zju.edu.cn Emal: weq@nt.zju.edu.cn Correspondng author; emal: LGTcheng@polyu.edu.hk 1

2 tme of a job can only take one of some gven dscrete values. We show that even some specal cases of the dscrete controllable verson wth the objectve of mnmzng the total compresson cost are NP-hard, but the general case s solvable n pseudo-polynomal tme. Moreover, we present a strongly polynomal tme algorthm to construct the trade-off curve between the number of tardy jobs and the maxmum compresson cost for the dscrete controllable verson. 1 Introducton Most classcal schedulng problems assume that the job processng tmes are fxed,.e., they are determned a pror and cannot be modfed by the scheduler. However, n real applcatons the processng of a job often requres resources, such as manpower, facltes, funds, raw materals, etc. By puttng more resources nto job processng, shorter job processng tmes may be accomplshed. Hence, t s possble to compress jobs (control the job processng tmes) by ncurrng extra costs. Schedulng problems wth controllable processng tmes have receved much attenton from researchers n the last two decades. Work n ths area was ntated by Vckson [11, 12] and Van Wassenhove and Baker [13]. For the related work on machne schedulng problems wth controllable processng tmes, the reader s referred to the survey by Nowck and Zdrzalka [9]. In ths paper we consder a sngle-machne model of jont job sequencng and resource allocaton wth the sequencng crteron beng the number of tardy jobs, whch was frst proposed by Danels and Sarn [5]. Formally, ths problem can be formulated as follows. We are gven a set J = {J 1, J 2,, J n } of ndependent jobs, whch must be scheduled on a sngle machne. The processng requrement of a job J s specfed by four non-negatve parameters p, d, u and w. Here p denotes the normal (ntal) processng tme of J, d denotes the due-date of J, u p s an upper bound on the compressblty of J, and w s the cost for unt-tme compresson of J, called the unt-compresson cost. By allocatng addtonal resources to the processng of J, the actual processng tme of J may be compressed from p to p = p x, where x [0, u ] s the amount of compresson. The cost of performng ths compresson equals w x. Let C denote the completon tme of job J under a gven schedule. Defne U = 0 f job J s early (C < d ) or on-tme (C = d ), and U = 1 f t s tardy (C > d ). We assume that all the jobs are avalable at tme zero. Moreover, let T CC = Σ n w x be the total compresson cost, and n T = Σ n U the number of tardy jobs. Then the objectve s to construct the trade-off curve between n T and T CC. We denote ths problem by P 0 : 1 contr, n T k T CC n ths paper. 2

3 For ths problem, Danels and Sarn [5] provded some theoretcal results. Cheng, Chen and L [2] proved that the problem s NP-hard, and presented a pseudo-polynomal tme algorthm. Ths paper s an extenson of the above work wth three man contrbutons. Frst, we wll resolve the specal case n T = 0 of the problem by presentng an optmal algorthm. Its tme complexty s O(n 2 ), and becomes O(n log n) f all w are the same. Ths problem, denoted by P 1 : 1 contr, n T = 0 T CC, models the followng scenaro. For some applcatons n logstcs and supply chan management, jobs denote orders from customers, and the machne denotes the manufacturer. In some cases, whle tardy jobs are allowed, havng tardy jobs may damage the goodwll of the customers, so the manufacturer s concerned wth the trade-off between n T and T CC. In other cases, customers wll accept no tardy jobs. So the manufacturer must fnd a soluton to mnmze the compresson cost wth the constrant that n T = 0. Note that whle the algorthm gven n [2] stll requres pseudo-polynomal tme for n T = 0, we present a strongly polynomal tme algorthm for ths case. Second, we consder the problem wth a new objectve. The objectve s to construct the trade-off curve between n T and the bottleneck objectve functon MCC = max,,n w x,.e., the maxmum compresson cost. Ths problem, denoted by P 2 : 1 contr, n T k MCC, models the followng stuaton: Gven that the resource s lmted, the decson maker (scheduler) seeks to fnd a soluton that balances the amount of resource used by each job under the constrant that the number of tardy jobs s no greater than a gven nonnegatve nteger k. We wll show that ths problem can be solved n O(n 2 log W ) tme, where W = max,,n w u. Thrd, we consder the problems wth dscrete controllable processng tmes. Schedulng problems wth dscrete controllable processng tmes have been studed by Chen, Lu and Tang [1], De et al. [3], [4], and Skutella [10], whch have many real-world applcatons. For ths stuaton, the allowed compresson amount x of job J s n a fnte set,.e., t must be one of some gven dscrete values, nstead of any value n the nterval [0, u ]. Hence, we assume that for each J J, the actual processng tme p = p x has l possble values a 1 = p > a 2 > > a l. Let w j be the compresson cost for the processng tme a j, = 1,, n, j = 1,, l. It s assumed that 0 = w 1 < w 2 < < w l. Ths assumpton s reasonable because to acheve smaller processng tmes requres more resources, hence ncurrng hgher costs [1]. We show that both P 3 : 1 dsc contr, n T k T CC and P 4 : 1 dsc contr, n T = 0 T CC are NP-hard even for a very specal case where all the jobs have the same due-date, all l = 2 and all unt-compresson costs w 2 /a 2, = 1,, n, are the same (.e., all the unt-compresson costs are the same for all the jobs), and present pseudo-polynomal tme algorthms based on dynamc programmng for the general case of P 3 and P 4. Moreover, we wll present a strongly polynomal tme algorthm 3

4 for the problem P 5 : 1 dsc contr, n T k MCC. The tme complexty s O(n 2 log(nl)), where l = max,,n l. Ths paper s organzed as follows: Sectons 2 and 3 consder the problems P 1 and P 2, respectvely, Secton 4 studes the problems P 3 and P 4, and Secton 5 studes the problem P 5. Fnal remarks are presented n Secton 6. 2 Strongly polynomal solvablty of the problem P 1 Defnton 2.1 For the problem P 1, a soluton s called feasble f t satsfes Σ n U = 0. Lemma 2.2 There exsts an optmal soluton such that all the jobs are scheduled n the Earlest Due-Date (EDD) order. Proof. It can be shown by applyng the standard parwse job nterchange argument. Hence, n the remander of ths secton, we assume that the jobs are re-ndexed such that d 1 d 2 d n. Snce u s the upper bound on the compressblty of job J, = 1, 2,, n, we have Corollary 2.3 The problem P 1 has a feasble soluton satsfyng the EDD order f and only f Σ j=1 (p j u j ) d, = 1, 2,, n. For each job J, = 1, 2,, n, let the actual compresson be x. Then job J s not tardy f and only f Σ j=1 (p j x j ) d,.e., Σ j=1 x j Σ j=1 p j d. Thus, the problem P 1 can be formulated as the followng lnear program: mn Σ n w x s.t. Σ j=1 x j Σ j=1 p j d, = 1, 2, n, 0 x u, = 1, 2,, n. It s clear that the lnear program can be solved n polynomal tme, so can P 1. In the followng, we present a strongly polynomal tme algorthm for ths problem. Algorthm A 1 : 1. Renumber all the jobs n the EDD order such that d 1 d 2 d n. Set x = 0, u = u, = 1, 2,, n, and k = 1. 4

5 2. Schedule the jobs n the order of J 1, J 2,, J n wth the actual processng tmes p 1 x 1, p 2 x 2,, p n x n, respectvely. If there are no tardy jobs after processng all the jobs, then go to 4; otherwse let J tk be the frst tardy job wth completon tme C tk, go to Assume that the unt-compresson weghts of the job set {J 1, J 2,, J tk } satsfy w j1 w j2 w jtk (that s to say, J j s the job wth the -th smallest unt-compresson cost n {J 1, J 2,, J tk }). Note that the current bounds on the compressblty of these jobs are u 1, u 2,, u t k, respectvely. Let s = mn{ Σ p=1 u j p > C tk d tk, 1 t k }. Then for = j 1, j 2,, j s 1, set x x +u and u 0; for = j s, set x x +C tk d tk Σ s 1 p=1 u j p and u u x. Set k k + 1, go to Output x, = 1, 2,, n, as the fnal compresson of job J, and the objectve value Σ n w x, stop. Lemma 2.4 Consder the followng lnear program: mn Σ t w y s.t. Σ t y L, (1) 0 y l, = 1, 2,, t. where L s a postve number and 0 w j1 w j2 w jt. Let s = mn{ Σ p=1 l j > L, 1 j t}. Then s an optmal soluton. l j, f 1 s 1, y j = L Σ s 1 l j, f = s, 0, f s + 1 t Proof. It s a contnuous relaxaton of the mnmzaton verson of a specal bounded Knapsack problem [8], hence the lemma holds. Theorem 2.5 Algorthm A 1 produces an optmal soluton to the problem P 1, and runs n tme O(n 2 ). Proof. Frst, t s clear that the soluton produced by algorthm A 1 s feasble. We next prove ts optmalty. 5

6 Let x k and x denote the compresson of job J, = 1, 2,, n, rght after the k-th teraton of the algorthm, and the compresson of job J n an optmal soluton satsfyng the EDD order, respectvely. We prove by nducton on k that Σ t k w x Σt k w x k, and Σt k x k s exactly the mnmum possble total compresson to assure no jobs n {J 1, J 2,, J tk } are tardy, and thus Σ t k x Σ t k x k for every k 1. It states that the algorthm yelds a soluton that has the objectve value and the total compresson no greater than those of the optmal soluton, and thus s optmal. For k = 1, n order to guarantee that job J t1 s not tardy, Σ t 1 x C t1 d t1 must hold. From Lemma 2.4, we conclude that x 1 1, x1 2,, x1 t 1 s an optmal soluton to the followng lnear program: mn Σ t 1 w y s.t. Σ t 1 y C t1 d t1, (2) 0 y u, = 1, 2,, t 1. On the other hand, x 1, x 2,, x t 1 s a feasble soluton to (2), hence Σ t 1 w x Σ t 1 w x 1. Because Σ t 1 x 1 = C t 1 d t1, we know that Σ t 1 x 1 s the mnmum possble total compresson such that job J t1 s not tardy. Hence, the result s true for k = 1. In general, suppose that the result s true for k 1, that s, Σ t k 1 w x Σt k 1 w x k 1 and Σ t k 1 x Σ t k 1 xk 1, and no jobs n {J 1, J 2,, J tk 1 } are tardy wth the processng tmes p x k 1, = 1, 2,, t k 1. Snce Σ t k 1 xk 1 s the mnmum possble total compresson that guarantees that no jobs n {J 1, J 2,, J tk 1 } are tardy, to make job J tk not tardy, the new compresson must be at least C tk d tk. From the algorthm, we know that Σ t k x k = Σ t k 1 xk 1 + C tk d tk. It mples that Σ t k x k s exactly the mnmum possble total compresson to ensure that no jobs n {J 1, J 2,, J tk } are tardy, and thus Σ t k x Σ t k x k. (3) To show that Σ t k w x Σ t k w x k, we dvde the compressons x, = 1, 2,, t k nto two parts x and x, and prove the result by verfyng that the total cost of the frst parts of all jobs s no less than that of the compressons x k 1 1, x k 1 2,, x k 1 t k 1, and the cost of the second parts of all jobs s no less than that of the new compressons n the k-th teraton. = x k 1 xk 1 1, x k 2 determned such that: x k 1. = x k 2 xk 1 2,, x k. t k = x k tk x k 1. Frst, x, = 1, 2,, t k 1 are 6 t k

7 ) If x xk 1, then x x xk 1, f x < xk 1, then x = x. ) If the processng tme of job J, = 1, 2,, t k 1 s p = p x, then the jobs J 1, J 2,, J tk 1 are all not tardy by the EDD rule, and Σ t k 1 x = Σ t k 1 xk 1. (4) Furthermore, let x = 0, = t k 1 + 1,, t k. Thus, let the second part of x 1, 2,, t k. be x = x x, = In fact, ths constructon can be performed as follows: Let z = mn{x, xk 1 }, = 1, 2,, t k 1 and T = Σ t k 1 xk 1 Σ t k 1 z. Let v = mn{ Σ j=1 (x j z j) > T, 1 t k 1 }. We then defne x, = 1, 2,, t k 1 as follows (note that x = 0 as above, = t k 1 + 1,, t k ): x, for 1 v 1, x = z + T Σ v 1 (x z ), for = v, (5) z, for v < t k 1. Obvously, x, = 1, 2,, t k 1 satsfy ) and (4). Hence, we only need to prove that no jobs n {J 1, J 2,, J tk 1 } wth processng tmes p = p x are tardy. We show ths result by contradcton. Suppose that there s a tardy job. From the algorthm, we know that job J tk 1 s an on-tme job after the k 1-th teraton n the algorthm wth processng tmes p x k 1, = 1,, t k 1, thus d tk 1 = Σ t k 1 (p x k 1 ). (6) Snce Σ t k 1 x = Σt k 1 xk 1, (6) mples d tk 1 = Σ t k 1 (p x ), and J t k 1 s not tardy wth the processng tmes p, = 1,, t k 1. Thus, let J tj, j k 2 be the frst tardy job accordng to the order from J tk 1 to J 1 wth processng tmes p, = 1,, t k 1. If t j < v, then from the constructon of x we conclude that Σt j (p x ) = Σt j (p x ) d t j, whch mples that J tj s not tardy, a contradcton. Thus, t j v. From the algorthm, we know that job J tj s an on-tme job rght after the j-th teraton wth the processng tmes p x j, = 1,, t j, thus d tj = Σ t j (p x j ). Hence, combnng ths wth (6), we have d tk 1 d tj = Σ t k 1 (p x k 1 ) Σ t j (p x j ) = Σ t k 1 =t j +1 p Σ t j (xk 1 x j ) Σt k 1 =t j +1 xk 1 Σ t k 1 =t j +1 p Σ t k 1 =t j +1 xk 1 Σ t k 1 =t j +1 p Σ t k 1 =t j +1 z (snce j < k 1 mples x k 1 x j 0 ) (snce z = mn{x, xk 1 } x k 1, = 1, 2,, t k 1 ) = Σ t k 1 =t j +1 p Σ t k 1 =t j +1 x. (by (5)) (7) 7

8 Because of the assumpton that job J tj s tardy wth the processng tmes p, = 1,, t k 1, we have Σ t j (p x ) > d t j. Combnng ths wth (7), we obtan Σ t k (p x ) > d t j + Σ t k 1 =t j +1 (p x ) > d t k 1. Thus J tk 1 s tardy wth the processng tmes p, = 1,, t k 1, a contradcton. Therefore, we conclude that no jobs n {J 1, J 2,, J tk 1 } are tardy. Now we return to the proof of the theorem. On the one hand, the compressons x 1, x 2,, x t k 1 make the jobs J 1, J 2,, J tk 1 not tardy; hence, from the nducton assumpton, we know that Σ t k 1 w x Σ t k 1 wk 1 x k 1. (8) On the other hand, subtractng (4) from (3), we get Σ t k x Σ t k x k = C tk d tk. Because satsfes ), we have x By defnton, we know 0 x = x x u x u x k 1, = 1, 2,, t k 1. 0 x = x u, = t k 1 + 1,, t k. Therefore, x, = 1, 2,, t k, s a feasble soluton to the followng lnear program: mn Σ t k w y s.t. Σ t k y C tk d tk, 0 y u x k 1, = 1, 2,, t k 1, (9) 0 y u, = t k 1 + 1, t k. From Lemma 2.4 and the algorthm, we conclude that x k, = 1, 2,, t k s an optmal soluton to the above lnear program (9). Thus, we obtan Σ t k w x k Σ t k w x. (10) Addng (8) and (10), we get Σ t k w x k Σt k w x. Thus, the soluton produced by algorthm A 1 s an optmal soluton to the problem P 1. We next study the tme complexty of algorthm A 1. It s clear that Step 1 takes O(n log n) tme. Steps 2-3 may terate at most n tmes and each teraton takes O(n) tme. Hence, algorthm A 1 runs n O(n 2 ) n the worst case and s a strongly polynomal algorthm. If all unt-compresson costs are the same,.e., w = w, = 1,, n, t can be shown smlarly that the followng smplfed verson of A 1 yelds an optmal soluton n tme O(n log n). 8

9 Algorthm A 1 : 1. Renumber the jobs n the EDD order such that d 1 d 2 d n. 2. Compute r such that Σ r p d r = max{σ j p d j j = 1, 2,, n}. 3. Let s = mn{j Σ j u > Σ r p d r, 1 j n}. Then, for 1 s 1, let x = u ; for = s, let x = Σ r p d r Σ s 1 u ; and for s + 1 n, let x = Output x 1, x 2,, x n and Σ n w x, stop. 3 Polynomal solvablty of the problem P 2 To solve the problem P 2, we assume that all w x, = 1, 2,, n are ntegers, thus the objectve value max{w x, = 1, 2,, n} s also an nteger. Ths assumpton should not be a serous lmtaton [2], snce the amount of allocaton can always be expressed n the smallest unt of resource for all practcal purposes, whch makes the costs ntegers. Let max{w u, = 1, 2,, n} = W. It s well-known that Moore s algorthm can solve the problem 1 n T n O(n 2 ) tme [7]. In the followng we present an optmal algorthm for P 2 by combnng Moore s algorthm wth the bsecton method. Defnton 3.1 Consder an nstance of the problem 1 n T wth the processng tmes {p = 1, 2,, n}. If ts optmal objectve value s n T k, then we say that {p = 1, 2,, n} s a feasble soluton to the problem P 2. The problem P 2 s called feasble f t has at least one feasble soluton. In the remander of ths secton, we use p t to denote the set consstng of processng tmes p mn{t/w, u }, = 1, 2,, n, where p s the ntal processng tme of job J. Specfcally, p 0 = {p = p = 1, 2,, n} and p W = {p = p u = 1, 2,, n}. The followng result s trval. Lemma 3.2 (1) Let {p = 1, 2,, n} be a feasble soluton to the problem P 2. If p p, = 1, 2,, n, then {p = 1, 2,, n} s a feasble soluton to the problem P 2, too. (2) The problem P 2 has a feasble soluton f and only f p W s a feasble soluton. Algorthm A 2 : 9

10 1. Invoke Moore s algorthm to solve the nstance of 1 n T wth actual processng tmes p 0. If n T k, then the current soluton s optmal wth the objectve value MCC = 0, stop; otherwse, go to Invoke Moore s algorthm to solve the nstance of 1 n T wth actual processng tmes p W. If n T > k, then output that the problem P 2 has no feasble soluton, stop; Otherwse, let t = W and t = If t t = 1, then output that the soluton p t s optmal wth the objectve value MCC = t, stop; otherwse set l = (t + t )/2, and nvoke Moore s algorthm to solve the nstance of 1 n T wth actual processng tmes p l. If n T k, then set t = l and go back to 3; otherwse, set t = l and go to 3. Theorem 3.3 If the problem P 2 s feasble, then the soluton obtaned by algorthm A 2 s an optmal soluton, and the tme complexty of algorthm A 2 s O(n 2 log W ). Proof. If the algorthm stops at Step 2, then p W s not a feasble soluton, and hence by Lemma 3.2(2), the problem s nfeasble. We now consder the case that the problem P 2 s feasble. If the algorthm stops at Step 1, then a soluton p 0 s obtaned wth the objectve value 0, whch s trvally an optmal soluton. Hence, we suppose n the followng that the optmal objectve value s not 0. Then we can clam that p W s a feasble soluton, whereas p 0 s not. So, by the bsecton procedure, algorthm A 2 can get t and t such that t t = 1, p t s feasble, but p t = p t 1 s not. Thus algorthm A 2 outputs a feasble soluton p t wth the objectve value MCC = t when t stops. Let {p = 1, 2,, n} be any feasble soluton of the problem P 2 wth the objectve value MCC = max{(p p )w = 1, 2,, n}. We now prove by contradcton that MCC = t MCC and hence p t must be the optmal soluton. Suppose MCC > MCC. As there s no nteger between t 1 and t, MCC > MCC mples that t 1 MCC. So, by Lemma 3.3(1), p t 1 s also feasble, However t = t 1, and we know that p t s not feasble, a contradcton. Thus, MCC > MCC s not true and p t s an optmal soluton to the problem P 2. Because Step 3 may repeat for at most log W tmes and the tme complexty of Moore s algorthm s O(n 2 ), we conclude that the tme complexty of algorthm A 2 s O(n 2 log W ). 10

11 4 NP-hardness of the problems P 3 and P 4 Cheng, Chen and L [2] proved that the problem P 0 : 1 contr, n T k T CC s NP-hard. Now we dscuss the problem P 3 where the possble compressons of each job are some dscrete values. Theorem 4.1 The problem P 3 s NP hard even f all the jobs have the same due-date, all l = 2, and all the unt-compresson costs are the same. Proof. Snce all the unt-compresson costs are the same, the objectve value of the problem P 3 s equvalent to the total compresson. We show the result by reducng the Partton problem [6] to ths problem. Gven an nstance I of the Partton problem wth a set of postve ntegers {h 1, h 2, h n } and 2B = Σ n h, we construct an nstance II of the problem P 4 as follows: Assocated wth each h, = 1,, n, s job J wth p = h, p = p x {h, 0}, d = B, = 1, 2,, n. In addton, we construct n more jobs J n+1,, J 2n wth p = 2B, p = p x {2B, 0}, d = B, = n + 1, n + 2,, 2n. Defne k = n and threshold UB = B. We prove that nstance I has a soluton f and only f nstance II has a soluton wth n T k and the objectve s no greater than UB. If I has a soluton, then there exst two subsets H 1 and H 2 of H such that H 1 H 2 = H, H 1 H 2 = and h H 1 h = h H 2 h = B. Construct a soluton for nstance II as follows: Let the compresson of J, H 1, be x = h for each H 1, and x = 0 for all other jobs. Then the frst n jobs can be completed early or on-tme whle the last n jobs are tardy. Hence, the number of tardy jobs s k = n, and the total compresson s exactly B. Next, suppose II has a soluton such that n T k = n and the objectve value (.e., the total compresson) s no greater than UB. If a job J, n + 1 2n, s compressed, then the total compresson s at least 2B > UB, a contradcton. Thus, none of the last n jobs can be compressed, and all of them are tardy. That s to say, the frst n jobs must be completed early or on tme. If the total compresson of the jobs J 1, J 2,, J n s less than B, then ther total actual processng tme s more than 2B B = B, hence there exsts at least a tardy job, a contradcton. Snce UB = B, the total compresson of the jobs J 1, J 2,, J n s exactly B. It follows that there exsts a subset H {1, 2,, n} such that Σ H a = B. We have shown that 1 contr, n T = 0 T CC s strongly polynomal tme solvable. However, the dscrete controllable case becomes N P -hard. 11

12 Theorem 4.2 The problem P 4 s NP hard even f all the jobs have the same due-date, all l = 2, and all the unt-compresson costs are the same. Proof. Smlarly, all the unt-compresson costs beng the same, the objectve value of the problem P 4 s equvalent to the total compresson. We agan show the result by reducng the Partton problem to ths problem. Gven an nstance I of the Partton problem wth a set of postve ntegers {h 1, h 2, h n } and 2B = Σ n h, we construct an nstance II of the problem P 3 as follows: Assocated wth each h, = 1,, n, s job J wth p = 2h, p = p x {2h, 3h /2}, d = 7B/2, = 1, 2,, n. Defne the threshold UB = B/2. It can easly be shown that nstance I has a soluton f and only f nstance II has a soluton wth n T = 0 and the objectve value s no greater than UB. Next we present a pseudo-polynomal tme algorthm that solves the general case of the problem P 3 optmally. Note that f the problem P 3 s feasble (.e., the number of tardy jobs s no greater than k), there must exst an optmal soluton such that the early and on-tme jobs are scheduled n the EDD order, and the tardy jobs are scheduled n any order followng all the early and on-tme jobs. Furthermore, the tardy jobs are not compressed snce we are to mnmze the compresson cost. Hence, we assume that d 1 d 2 d n. Algorthm A 3 : 1. Let f(, t, q) be the mnmum total cost of a partal soluton contanng the frst jobs, J 1, J 2,, J, gven that the completon tme of the early and on-tme jobs n ths partal soluton s exactly t, and the number of tardy jobs s exactly q (1 n, 0 t d n, 0 q k). 2. Recursve relatons: For = 2, 3,, n, t = 0, 1,, d n and q = 0, 1,, k: mn{f( 1, t, q 1), mn{f( 1, t a j, q) + w j, = 1, 2,, l }}, f t d 1, mn{f( 1, t a j, q) + w j, = 1, 2,, l }, f(, t, q) = f d 1 < t d,, f t > d. { p a j, f f(, t, q) = f( 1, t a j, q) + w j, x = 0, otherwse. 12

13 3. Intal values: For t = 0, 1,, d n : { w1j, f d 1 t = a 1j, f(1, t, 0) =, otherwse, x 1 = f(1, t, 1) = 0, t = 0, 1,, d n. { p1 a j, f f(1, t, 0) = w 1j, 0, otherwse. 4. An optmal soluton can be obtaned by computng mn{f(n, t, 0), f(n, t, 1), f(n, t, k) t = 0,, d n }. Remark 4.3 Let l = max,2,,n l, the tme complexty of algorthm A 2 s O(nlkd n + n log n) snce we try all possble values of ( = 1,, n), all possble values of t (t = 0, 1,, d n ), and all possble values of q (q = 0, 1,, k), and the computaton of f(, t, q) needs O(l) tme for each possble state (, t, q). In addton, sortng jobs n non-decreasng order of due-date takes O(n log n) tme. Therefore, algorthm A 3 s pseudo-polynomal tme, mplyng that the problem P 3 s only NP-hard n the ordnary sense. Remark 4.4 Algorthm A 3 can also be used to solve P 4 by keepng q = k = 0. Hence, the tme complexty of the algorthm becomes O(nld n +n log n), and P 4 s NP -hard n the ordnary sense. 5 Strongly polynomal solvablty of the problem P 5 The man dea of the algorthm for P 5 s smlar to that for P 2. However, we can construct a strongly polynomal tme algorthm by makng mnor modfcatons. Before we present the algorthm, let w 0 = 0, and re-arrange the compresson costs of all the jobs n non-ncreasng order of ther values. Then we express ther dfferent values as follows: 0 = w 0 < w 1 < w 2 < < w L. Defnton 5.1 For a gven postve number S, defne j = max{j w j S, j = 1, 2,, l } for every = 1,, n, and let p S = {p = a j = 1, 2,, n}. Specfcally, p w 0 = {p = a 1 = p = 1, 2,, n} and p w L = {p = a 1 = p = 1, 2,, n}. Algorthm A 4 : 1. Invoke Moore s algorthm to compute the objectve value of the nstance of 1 n T wth actual processng tmes p w 0. If the objectve value n T k, then output MCC = 0, stop; otherwse, go to 2. 13

14 2. Invoke Moore s algorthm to compute the objectve value of the nstance of 1 n T wth actual processng tmes p w L. If the objectve value n T > k, then output that the problem P 5 has no a feasble soluton, stop; otherwse, let t = L and t = If t t = 1, then output that the soluton p wt s optmal wth objectve value MCC = w t, stop; otherwse set l = (t + t )/2, and nvoke Moore s algorthm to compute the nstance of 1 n T wth actual processng tmes p w l. If n T k, then set t = l and go back to 3; otherwse, set t = l and go back to 3. Theorem 5.2 If the problem P 5 has feasble solutons, then the soluton obtaned by algorthm A 4 s an optmal soluton, and the tme complexty of algorthm A 4 s O(n 2 log(nl)), where l = max,2,,n l. Proof. Wth an argument analogous to the proof of Theorem 3.3, we can show that the soluton produced by algorthm A 4 s optmal. Snce the compresson costs of all the jobs have at most L nl dfferent values, Step 3 terates at most L tmes. Therefore, we conclude that the tme complexty of algorthm A 4 s O(n 2 log(nl)). 6 Conclusons In ths paper we consdered sngle-machne schedulng wth contnuously and dscretely controllable processng tmes. The goal s to construct the trade-off curve between the number of tardy jobs and the total or maxmum compresson cost. We found that the levels of dffculty between the sum objectve (.e., total compresson cost) and bottleneck objectve cases, and between the contnuous and dscrete models, are quet dfferent. For the sum objectve case, the problems 1 contr, n T k T CC, 1 dsc contr, n T k T CC and 1 dsc contr, n T = 0 T CC are NP-hard, but for the bottleneck case, both the problems 1 contr, n T k MCC and 1 dsc contr, n T k MCC are polynomal solvable. For the contnuous model, the problem 1 contr, n T = 0 T CC s strongly polynomal solvable, but for the dscrete model, even the specal case of the problem 1 dsc contr, n T = 0 T CC s NP-hard. References [1] Z.-L. Chen, Q. Lu, G.C. Tang, Sngle machne schedulng wth dscretely controllable processng tmes, Operatons Research Letters, 21(1997),

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