Routing Equal-Size Messages on a Slotted Ring

Transcription

1 Journal of Scheduling 15(4) (2012) Routing Equal-Size Messages on a Slotted Ring Dariusz Dereniowski Department of Algorithms and System Modeling Gdańsk University of Technology Gdańsk, Poland deren@eti.pg.gda.pl Wiesław Kubiak Faculty of Business Administration Memorial University St. John s, Canada wkubiak@mun.ca Abstract We deal with the problem of routing messages on a slotted ring network in this paper. We study the computational complexity and algorithms for this routing by means of the results known in the literature for the multi-slot just-in-time scheduling problem. We consider two criteria for the routing problem: makespan, or minimum routing time, and diagonal makespan. A diagonal is simply a schedule of ring links i = 0,..., q 1 in q consecutive time slots respectively. The number of diagonals between the earliest and the latest diagonals with non-empty packets is referred to as the diagonal makespan. For the former, we show that the optimal routing of messages of size k, is NP-hard in the strong sense, while an optimal routing when k = q can be computed in O(n 2 log 2 n) time. We also give an O(n log n)-time constant factor approximation algorithm for unit size messages. For the latter, we prove that the optimal routing of messages of size k, where k divides the size of the ring q, is NP-hard in the strong sense even for any fixed k 1, while an optimal routing when k = q can be computed in O(n log n) time. We also give an O(n log n)-time approximation algorithm with an absolute error 2q k. Keywords: just-in-time scheduling, multi-slot scheduling, optical network, routing, slotted ring 1 Introduction According to the Websters online dictionary a slotted ring is A LAN architecture that continually carries a constant number of fixed length packets or slots round the ring. The nodes then use, by replacement, empty slots as they pass through, to transmit data. All the nodes have the ability to recognise empty slots or slots addressed to them. A regular topology of the ring and no buffering in the slotted ring are two desirable features of modern high speed networks according to a recent survey, Li et al. [2007]. Consequently, slotted rings have found numerous real-life implementations. Although slotted rings with variable packet sizes have been proposed, the asynchronous network model with variable size packets have several disadvantages, see e.g. Bianco et al. [2004], Yang et al. [2004]. Therefore the DAVID Metropolitan Area Network described in Bianco et al. [2004] is dealing with variable packet sizes by breaking them into fixed size packets, transmitting these through the network and then assembling them at the destination once the transmission is complete. The topology of the DAVID network is a Wide Area Network consisting of several optical Metropolitan Area Networks (MAN). Each MAN contains several optical slotted ring networks, see Bianco et al. [2002, 2004]. The new French slotted ring ECOFRAME is described in Chiaroni [2008]. For other examples of fixed size slotted ring networks models and their analysis see e.g. Carena et al. [2002],Chaitou et al. [2007], Kang et al. [1995], King and Mitrani [1987], Marsan et al. [1997]. In this paper we consider a slotted ring network with one wavelength. Though in practice several wavelengths may be available, their number is usually much smaller than the possible number of nodes attached to the network. This leads to similar algorithmic and complexity problems as the ones considered in this paper. One important consequence of this assumption is the fact that the transmission of packets happens in one direction only in the slotted ring. The complexity and related results concerning the ring loading problem for the rings with a possibility of two-way transmission are given in Lee and Chang 1

2 [1997], Myung and Kim [2004], Schrijver et al. [1998], Wang [2005]. A survey of various wavelengthdivision multiplexing (WDM) metropolitan ring networks and protocols can be found in Herzog et al. [2004]. Despite studies concerning the network efficiency for slotted and un-slotted models even with more general than ring network topologies, see Yao et al. [2001], very little is known about computational complexity and algorithms for routing slotted rings. To our knowledge the work of Barth et al. [2004] is the first and only paper dealing with these issues. The main motivation of this paper is to fill in this void. One of the contributions of this paper is a characterization of a connection between the optimization of routing on a slotted ring and the optimization of multi-slot just-in-time schedules on parallel machines. In the latter problem a set of n jobs and an integer L > 0 are given. Each job is characterized by its relative due-date, which is an integer not greater than L, and its integer processing time. The time is divided into time slots, each of length L. In a feasible schedule no machine processes two or more jobs at the same time, and each job has to be executed without preemption to complete exactly at its due-date in one of the time slots. The due-dates are relative to the beginning of slots, not to the beginning of schedule. The problem consists in minimizing either the number of slots or the makespan of the schedule. The latter objective is more general for a schedule minimizing makespan minimizes the number of slots at the same time but not the other way round. The minimization of the number of slots of a just-in-time schedule is strongly NP-hard even if the processing time of each job is at most L, however there exists an approximation algorithm with absolute error of one slot, Sung et al. [2007b], for this case. If the processing time of each job does not exceed its due-date, then a minimum makespan solution can be found in time O(n log n), see Dereniowski and Kubiak [2010]. An optimal schedule for a single machine and any set of jobs can be found in O(n log n)- time, Sung et al. [2007b], and O(n log 2 n) time, Dereniowski and Kubiak [2010], for the minimum number of slots and the makespan objectives, respectively. Other results on multi-slot just-in-time scheduling can be found in Čepek and Sung [2004]. For general scheduling models and notation see Tanaev et al. [1994] and Błażewicz et al. [1996], and for just-in-time scheduling models see Józefowska [2007], and Kubiak [2009]. For the terminology and definitions concerning algorithms and complexity see e.g. Garey and Johnson [1979], Hromkovič [2001], Vazirani [2001]. The paper s main result proves that the optimization of routing on a slotted ring is NP-hard in the strong sense. This proof improves the result of Barth et al. [2004] in two ways. First, by holding for the messages of equal size. Second, by dropping the initial staircase-pattern machine availability assumption, see Błażewicz et al. [2000] for the scheduling with machine availability constraints. This assumption is a key assumption needed in the NP-hardness proof given in Barth et al. [2004]. Finally, the paper proposes several off-line optimal and approximation algorithms for the optimization of routing on a slotted ring. The paper is organized as follows. The next section introduces the routing on a slotted ring and defines two optimization criteria: the diagonal makespan, or the number of diagonals, and the makespan. The section also defines the notation used thorough the paper. Section 3 is devoted to the problem of minimizing the diagonal makespan, its main contribution is a mapping between the ring routing problem and the problem of multi-slot just-in-time scheduling discussed in Sections 3.1 and 3.2. Section 3.3 applies this mapping to prove that the problem of minimizing diagonal makespan is strongly NP-hard even for messages of size 1, and to derive an efficient approximation algorithm with an absolute error of 2q k for the problem, where k is the message size. Section 4 deals with the makespan minimization problem in slotted rings. Section 4.1 shows that this problem is NP-hard in the strong sense even for equal size messages. However, if the size equals the size of the ring, then the problem is solvable in polynomial time, which is shown in Section 4.2. Sections 4.3 and 4.4 give algorithms for two special cases of the problem: routing on a path and short routing with a bottleneck link. These algorithms are then used as subroutines in an O(n log n)-time constant approximation ratio algorithm for unit size messages in Section 4.5. We summarize the results, and list some open problems in Section 5 which concludes the paper. 2

3 2 Routing slotted ring The network with the ring topology contains q nodes, numbered 0,..., q 1, and q links, (0, 1), (1, 2),..., (q 1, 0). We refer to the link (l, (l + 1) mod q) simply as link l. The set of messages is denoted by P = {P 1,..., P n }. Each message P i is characterized by its size, size(p i ) - the number of packets the message P i contains. Its source node, source(p i ), is the node in the ring where the broadcast of P i originates from, and its destination node, dest(p i ), is the node in the ring where P i is destined to. The time is divided into steps of equal duration sufficient for a link to carry a single packet. At each step s a link l carries a packet p P i if and only if either link (l 1) mod q carried p in step s 1 and dest(p i ) l, or source(p i ) = l and the broadcast of P i begins in step s. Each node can receive a packet and send another packet in the same step. Without loss of generality we assume that the ring carries packets clockwise. A routing is defined by assigning each message P i an integer b i - the step when the broadcast of P i begins in the source(p i ). Figure 1(a) depicts a ring network with 9 nodes numbered 0,..., 8. The three distinguished nodes 1, 5 and 8 are both broadcasting and receiving packets in this example. Suppose that a set P = {P 1,..., P 5 } of 5 unit size messages is to be transmitted, where the message source and destination nodes are defined as follows. Message source(p i ) dest(p i ) P P P P P (a) (b) time (c) 8 P P 1 2 P3 3 4 P 2 Figure 1: (a) a ring with 9 nodes; (b) a matrix t for routing R; (c) the state of the ring in the third step Figure 1(b) gives a possible routing R for this instance. The routing R is presented by a matrix t where t l,s = i means that link l carries a packet of message P i in step s, and t l,s being blank means that link l is empty, carries no packet, in step s. Figure 1(c) shows the state of the ring in step 3 of R. The step when the transmission of the last packet of P i begins is b i +size(p i ) 1, while the transmission of P i ends in step e i = b i + size(p i ) 1 + λ i, where λ i is the number of links that each packet of P i visits on its way from the source to the destination, formally λ i = { dest(pi ) source(p i ) if dest(p i ) > source(p i ), q (source(p i ) dest(p i )) if dest(p i ) < source(p i ). (1) The makespan of a routing R is ms(r) = max{e i : i = 1,..., n}. In the example from Figure 1 we have that ms(r) = 7. Following the notation in Barth et al. [2004], we use the MS-FSP (Makespan criterion - Fixed Size Packet) to denote the problem of minimizing makespan for the routing problem with all messages of size 1. We extend this notation to MS-FSP k to denote the problem of minimizing makespan for the routing problem with all messages of size k 1. The k may or may not be a part of the input, whichever is the case will be clear from the context. In particular we have MS-FSP = MS-FSP 1. 3

4 Given a routing R, let t l,s = i if link l, l = 0,..., q 1, in step s, s = 1,..., ms(r), carries a packet of message P i, or t l,s = otherwise. If s 0 or s > ms(r) then t l,s =. For r 1 define a vector I r : {0,..., q 1} P { } I r = (t 0,r q+1, t 1,r q+2,..., t h,r q+1+h,..., t q 1,r ). The vector I r is called diagonal r of routing R. We also use a notation I r (a, b) = (t a,r q+1+a,..., t b,r q+1+b ), 0 a b < q, to denote a sub-vector of I r which includes all coordinates of I r between a and b inclusive. We say that I r or I r (a, b) is empty if all their coordinates are. Note that by definition we have that for each routing R, the sub-vectors I r (0, q 1 r) and I ms(r)+r (q r, q 1) are empty for r = 1,..., q 1. The example in Figure 1 has b 1 = 1 so that t 8,1 = P 1. In step 2 the only packet of P 1 is carried by link 0 so that t 0,2 = 1. The broadcast of message P 1 is thus contained in diagonals I 1 and I 10 of R. Given a routing R for P, define its diagonal makespan, diag(r), as the number of diagonals I r between the earliest and the latest nonempty diagonals inclusive, diag(r) = r max r min + 1, (2) where r max = max{r : I r (,..., )} and r min = min{r : I r (,..., )}. The I(R) will denote the set of all diagonals I r of R such that r min r r max. Thus, diag(r) = I(R). We reserve the DIAG-FSP k to denote the problem of minimizing diagonal makespan for the routing problem with all messages of size k 1. For the routing R in Figure 1(b) we have diag(r) = 13. We conclude this section with a theorem that allows us to assume without loss of generality that if k q, that is if k divides q, then each schedule for the problem DIAG-FSP k starts broadcasting each message P i P in some diagonal I jk+1 for some j 0. In other words no message starts broadcasting on the in-between diagonals I jk+2,..., I j(k+1). Theorem 1 Let P and q be given. If R is a routing for P and k q, then there exists a routing R such that diag(r ) diag(r) and R has the following property: for j = 0, 1,.... I jk+1 = I jk+2 = = I (j+1)k, (3) Proof: Given a feasible R. If I r, r > 1, is the first nonempty diagonal of R, then subtract r 1 from all the diagonal indices in I(R). Thus, without loss of generality I 1 is the first nonempty diagonal of R. Let us consider routing R with P i starting in diagonal I jk+1 provided that P i starts in diagonal I jk+t for some j 0 and 1 t k. Clearly, diag(r ) diag(r). We now prove that (3) holds. Consider diagonals I jk+1,..., I (j+1)k. By the construction of R and since k q, these diagonals may only contain packets of messages P i that start their broadcasts either in I jk+1 or in I jk+1 q = I jk+1 (q/k)k = I (j q/k)k+1. The latter holds since if a packet of message P with source(p ) > dest(p ) passes the link q 1 in step s, that is, it completes the diagonal I s, then it continues on link 0 in step s + 1 which is the start of diagonal s + q. Consequently, the condition (3) is satisfied, because size(p i ) = k. We argue, by contradiction, that R is feasible. Suppose that packets p P i and p P m, for different messages P i and P m, share a link in the same step in R. For each P P, we have that size(p ) = k and the broadcast of P starts in a diagonal I jk+1 in R for some j 0, thus without loss of generality we may assume that p and p are the first packets of P i and P m, respectively. However, since b i b i b i + k 1 and b m b m b m + k 1, then a packet of P i and a packet of P m must share a link in the same step in R as well because size(p i ) = size(p m ) = k and (b i mod k) (b m mod k) < k. This implies the infeasibility of R and leads to a contradiction. 3 Diagonal makespan is just-in-time scheduling This section focuses on the DIAG-FSP k problem. Our approach transforms DIAG-FSP k to the multislot just-in-time scheduling problem. The transformation is linear in P and not only does it give a one-to-one mapping between the two problem s instances but also a one-to-one mapping between their 4

5 schedules and routings. The latter, however, holds as long as the size of the ring q is divisible by the size of messages k, that is k q, (4) which we assume throughout this section. The advantage of this transformation based approach to solving DIAG-FSP k is that a number of results already obtained for the multi-slot just-in-time scheduling problem, see Sung et al. [2007a] and Dereniowski and Kubiak [2010], almost immediately translates in the results for the routing problem DIAG-FSP k and thus no new complexity proofs or algorithms are required for the latter. We begin with introducing the multi-slot just-in-time scheduling problem. 3.1 Just-in-time scheduling We follow terminology of Sung et al. [2007a] and Dereniowski and Kubiak [2010] in our introduction of the multi-slot just-in-time scheduling here. The time is divided into slots of equal size L. A set of jobs J = {J 1,..., J n } is given, where each job J j has integer processing time p j and integer due-date d j. The execution of job J j must end exactly at d j in one of the slots. The jobs are executed on a set of identical machines M = {M 1,..., M m }. No preemptions are allowed. A schedule S for J is feasible if no machine M performs more than one job at the same time and the execution of each job J j ends at its due-date. We define s j for each J j J to denote the start of J j relative to the beginning of a slot, formally { dj p s j = j if p j d j, (5) L p j + d j if p j > d j. Note that s j depends on J j but not on schedule. However, the absolute start, S j = s j + a j L, of J j depends on schedule since the schedule defines the slot a j where J j starts. The completion time of J j is then C j = S j + p j. Let S be a schedule of J on machines in M. Then S M, M M, is simply S restricted to a single machine schedule on machine M. If S is a single machine schedule, then slots(s) denotes the number of slots used in S. If S is a schedule on machines in M, M > 1, then we define slots(s) = max{slots(s M ) : M M}. For any schedule S let len(s) = max{c j : J j J }. 3.2 The transformation The transformation from an instance of the routing problem to an instance of the just-in-time scheduling problem is as follows. Set L = q and m = q/k, (6) and for each message P j P create a corresponding job J j J with processing time p j = λ j, where λ j is defined in (1), due date d j = dest(p j ), and s j = source(p j ). Note that the size k of all messages affects only the number of machines. Moreover, by (1) the outcome of the transformation contains only jobs J i shorter than L. That is, a job either ends in the same slot it has started or in the next one in a multi-slot just-in-time schedule. We now claim that there exists a routing R for P with diag(r) bq if and only if there exists a just-in-time schedule S for J with slots(s) b, for an integer b 0. Therefore, since q is fixed in any instance of the routing problem, the diagonal makespan minimization is related to the number of slots minimization. First, let us observe that there exists a one-to-one mapping between the slots of schedule S and the diagonals I r of routing R; simply slot p on machine M f M is mapped into the set of diagonals {I k(f 1)+1+(p 1)q,..., I kf+(p 1)q }, (7) in this mapping. However, by Theorem 1, all these diagonals are equal. Thus, we may limit ourselves to diagonals I ki+jq for j 0 and i = 1,..., q/k in the mapping. Moreover, without loss of generality we 5

6 may assume that the first nonempty diagonal in R is I 1. Thus, the schedule S Mf is actually mapped into the set {I kf, I kf+q,..., I kf+(slots(s) 1)q }, (8) of diagonals; since km = q, any two different machines are thus mapped into disjoint sets of diagonals. The schedule S on M f is simply the concatenation of I kf, I kf+q,..., I kf+(slots(s) 1)q in that order. Next, observe that the unit time interval [l, l + 1], for some integer l {0,..., q 1}, in slot p of S Mf corresponds to the coordinate l of diagonal I k(f 1)+(p 1)q. Therefore, any time interval [a, b], for integers a b, b a < L, entirely contained in slot p of S Mf in S, corresponds to the sub-diagonal I kf+(p 1)q (a (p 1)L 1, b (p 1)L), (9) and any time interval [a, b] intersecting two slots, p and p + 1, on M f, corresponds to the concatenation of two sub-diagonals I kf+(p 1)q (a (p 1)L 1, L 1) and I kf+pq (0, b pl). (10) This proves that a job J i is executed in the interval [a, b] in S Mf if and only if the last packet of the corresponding message P i is scheduled in either (9) or (10) in R. Therefore, there is a one-to-one mapping between feasible routings R for P and feasible schedules S for J. Finally, by (8) we have that slots(s) b if and only if diag(r) bmk = bq for an integer b. We thus proved the following theorem. Theorem 2 Let size(p i ) = k for each P i P, L = q, m = q/k, where k q. There exists a schedule S for J satisfying slots(s) b if and only if there exists a routing R for P with diag(r) bmk = bq. Let us give an example of this transformation for q = 5 and k = 1. The set P contains the following 8 messages: Message source(p i ) dest(p i ) P P P P P P P P Figure 2(a) gives a multi-slot just-in-time schedule for J = {J 1,..., J 8 }. This schedule is executed within two slots on each of the 5 machines. The corresponding routing for P is given in Figure 2(b). (a) M 1 J 1 M 2 J 2 J 3 J 4 M 3 M 4 M 5 J 5 J 8 J 6 J 7 0 (b) P 2 P 1 P 3 P 4 P P 2 P 5 P 3 P 7 2 P 1 P 2 P 5 P 8 P 7 3 P 1 P 3 P 8 P 7 4 P 1 P 3 P 4 P 6 P 8 Figure 2: (a) a just-in-time schedule for J ; (b) the corresponding routing for P 3.3 Applications of the transformation We are now ready to prove that DIAG-FSP k is NP-complete in the strong sense. We begin by recalling the following result. 6

7 Theorem 3 (Sung et al. [2007b]) Given an integer b 1 and a set of jobs J satisfying p j < L for each J j J, and J p j J j = bml, the problem of deciding whether there exists a schedule S with slots(s) b is NP-complete in the strong sense. In the following we show that we can restrict the inputs of the multi-slot just-in-time scheduling problem in Theorem 3 to the ones with L = km, for any fixed integer k 1, and this problem still remains NP-complete in the strong sense. First of all, by multiplying L, d j s and p j s by 2k, we may assume without loss of generality that L = kl for some even integer l > 1, and that there are two odd integers s, s {0,..., L 1} such that s j, d j / {s, s } for each job J j. If L < km, then we set the slot size to L = km, and the number of machines to m = m. Each due date increases by km L. Thus, a new due-date is at the same relative distance from L, as the old one was from L. Moreover, if s j d j for J j, then we set p j = p j, otherwise we set p j = p j + km L. In the former case the job is entirely processed in a single slot of a just-in-time schedule, in the latter it is processed in two adjacent slots. Observe that after the change no job starts or ends in time interval (0, km L) which means that we can readily obtain a schedule for one instance from a schedule for the other by simply executing each job in the same slot and on the same machine as in the original schedule. If L = kl > km, then we add l m long jobs to J each with due date L and processing time bl. We increase the number of machines to m = L/k = l, and set L = L. Thus, each new long job must occupy a separate machine by itself in any just-in-time schedule S with slots(s) b. We then split each new long job into the maximum number of jobs shorter than L and with due-dates s and s. Observe that the total processing time equals bm L for the resulting instance, provided that J j J p j = bml for the initial instance. Thus, the new shorter than L jobs may not share a machine with the original jobs in J ; the sharing would create at least one unit idle time which is not allowed as long as slots(s) b. Thus, we obtain instances with L = km and p j < L, j = 1,..., n, (11) in both cases. To prove that the transformation is polynomial observe that {s j, d j : J j J } 2n for an input J, L to the multi-slot just-in-time scheduling problem. If u, v {s j, d j : J j J }, u < v, meet the following condition s j, d j / (u, v) for all J j J, then we can we convert (u, v) into a unit time interval. This can be done by setting the slot size to L v + u + 1, and for each J j with s j u and d j v setting their new values s j v + u + 1 and d j v + u + 1, respectively. (Note that by changing s j and/or d j we also modify p j so that (5) holds.) This allows us to assume that L 2n. We may also restrict ourselves to nontrivial instances where m n. Finally, b 2n, because each job intersects at most two consecutive slots in each schedule, so we can schedule each job exclusively within two adjacent slots and thus obtain a schedule using at most 2n slots. This leads to the following corollary. Corollary 1 Given integers L, b and k, a set of jobs J satisfying p j < L for each J j J, and a set of m machines M, where k M = L, the problem of finding a just-in-time schedule S satisfying slots(s) b is NP-complete in the strong sense. This corollary and Theorem 2 immediately imply the following theorem. Theorem 4 The routing problem DIAG-FSP k is NP-hard in the strong sense for each fixed k 1. This transformation readily leads to an approximation algorithm for DIAG-FSP k which we now show. To begin with, we recall the O(n log n)-time multi-slot just-in-time approximation scheduling algorithm given by Sung et al. [2007b]. Theorem 5 (Sung et al. [2007b]) There exists an O(n log n)-time algorithm, which finds a just-in-time schedule S for J and M with slots(s) slots(s ) + 1, where S is an optimal schedule for J and M. Though this theorem is a point of departure for the approximation algorithm for DIAG-FSP k, we also need the following relative lower bound on the number of diagonals in the routing problem resulting from the transformation. 7

8 Lemma 1 Let J, L, m and P, q = L, k = q/m be the corresponding instances of the multi-slot just-intime scheduling and the routing problems, respectively. If an optimal just-in-time schedule for J uses b slots on some machine, then for each routing R for P, diag(r) (b 1)L + k. Proof: By contradiction. Let R be a routing for P with diag(r) < (b 1)L + k. By Theorem 1, diag(r) (b 1)L. This, by Theorem 2, implies that there exists a just-in-time schedule S for J of length slots(s) b 1 a contradiction. We are now ready to give the approximation algorithm for DIAG-FSP k. The algorithm constructs the corresponding instance of the multi-slot just-in-time scheduling problem, applies the algorithm from Theorem 5 to find a just-in-time schedule, and finally the schedule is transformed into a routing as described earlier in this section. Theorem 6 If k q, then there exists an approximation O(n log n)-time algorithm with an absolute error of 2q k for the problem DIAG-FSP k. Proof: Let an optimal schedule S for J results into slots(s ) slots. Let R, S and R be an optimal routing for the DIAG-FSP k problem, a schedule for J found by the algorithm from Theorem 5, and the routing for P corresponding to S in the transformation, respectively. We have that slots(s) slots(s ) + 1 by Theorem 5. By Theorem 2, diag(r) (slots(s ) + 1)q. Finally, by Lemma 1, diag(r ) (slots(s ) 1)q + k. Thus, diag(r) diag(r ) 2q k and the theorem holds. We close this section pointing another benefit of the transformation. Namely, whenever an instance of the just-in-time scheduling problem belongs to DIAG-FSP q, that is k = q, then by Lemma 1 and Theorem 2, an optimal schedule for the DIAG-FSP q problem is simply an optimal single machine multislot just-in-time schedule for J. The latter can be found in O(n log n) time Sung et al. [2007a,b] as has been earlier mentioned. Thus, the problem DIAG-FSP q is solvable in O(n log n) time. It is somewhat interesting that DIAG-FSP k is NP-hard in the strong sense for any fixed k = 1, 2, 3,... by Theorem 4, yet whenever k = q the problem is efficiently solvable. 4 Makespan of the routing problem This section deals with the makespan minimization problem MS-FSP k. We begin by showing that the problem is NP-hard in the strong sense in Section 4.1. Then, in the subsequent sections we give polynomial time optimal and approximation algorithms for some special cases of the problem. We begin in Section 4.2 with MS-FSP q where all messages have the same size equal the size of the ring q. We then focus on the problem MS-FSP 1 where all messages are of unit size, k = 1. Section 4.3 considers the case of a ring with a link never used by any message in P. We refer to this case as routing unit size messages on a path. We show a 2-approximation algorithm running in O(n log n) time for this case. Section 4.4 addresses a somewhat complementary case of a ring with a bottleneck link used by all messages in P, and with an additional requirements that the makespan of an optimal routing is at most q. We show that this case reduces to finding maximum matchings in convex bipartite graphs (we give a definition in Section 4.4). Finally, we give an approximation algorithm with a constant approximation ratio for any set of messages P of unit size in Section 4.5. The approximation relies on the two special cases studied in Sections 4.3 and 4.4 and a reduction to the multi-slot just-in-time scheduling problem. 4.1 Makespan minimization is NP-hard The reduction is from the circular-arc graph coloring problem we now briefly introduce. A simple graph G = (V, E) is a circular-arc graph if each vertex v V can be represented by an arc on a circle and for any two vertices u, v V, {u, v} E iff the arcs corresponding to u and v intersect. For each v V let l(v) and r(v) be respectively the left and right endpoint of the arc corresponding to v. We may without loss of generality assume that the endpoints of all vertices are pairwise different. For more details on circular-arc graph see Brandstädt et al. [1999]. For a positive integer c and a graph G, we say that a function f : V {1,..., c} is a c-coloring of the vertices of G if for each {u, v} E we have f(u) f(v). The smallest number c, denoted by χ(g), 8

9 for which there exists a c-coloring of G is referred to as the chromatic number of G. Given c and a circular arc graph G, the problem of deciding whether there exists a c-coloring of the vertices of G is NP-complete in the strong sense, see Garey et al. [1980]. We now reduce this problem to the following decision version of the routing problem. Input: a ring of size q, and a set of packets P of size k each. Question: is there a routing with makespan at most b for P? Assume that a circular-arc graph G = (V, E) is given along with the required number of colors c. We compute a mapping π : {1,..., 2 V } {l(v), r(v) : v V (G)} satisfying π(i) < π(j) for 1 i < j 2n. Let q = 2 V. For each v V include into P a message P v such that source(p v ) = i, where π(i) = l(v), dest(p v ) = j, where π(j) = r(v), and k = size(p v ) = (c + 1)q. Finally, let b = c(k + q). In order to prove that the mapping we just defined is a required reduction, we introduce a useful relation. Given a routing R for the routing problem, define a relation on P as follows: first, begin with a relation such that P 1 P 2 if there exists such a link l {0,..., q 1} that each packet of P 1 appears on l earlier than each packet of P 2, i.e. if t l,j = P 1 and t l,i = P 2 then j < i in R. Then, set to be the reflexive and transitive closure of. Figure 3(a) depicts a routing of a set of messages which results in being a partial order. The corresponding Hasse diagram is given in Figure 3(b). Observe that is not a partial order generally. (a) 0 1 P P 7 P 4 P 2 (b) P 6 P 4 P 8 P 7 P P3 6 P 2 P 5 P P 8 5 P 7 P 3 10 P 1 Figure 3: (a) a routing for P; (b) a Hasse diagram corresponding to (P, ) Lemma 2 If there exists a routing of makespan c(k + q) for the routing problem with all messages of size k = (c + 1)q on a ring with 2 V nodes, then there exists a c-coloring f of the circular-arc graph G. Proof: Given a routing R for P, observe that the corresponding relation is a partial order which is guaranteed by k = (c + 1)q. Given the relation, we partition the set P into P 1,..., P h, in such a way that P i is the set of messages on the ith level of a Hasse diagram of, where level 1 is the one containing only minimal elements. Given the partial order (P, ) and a partition P = P 1 P h, let f(v) = i if P v P i. To prove that f is a coloring of G assume that {u, v} E. This means that P u and P v belong to different sets P i, which implies f(u) f(v). Now we prove by contradiction that f uses at most c colors. Suppose h > c. Clearly, if P u P i for some P u P then e u i k. Thus, for P u P c+1 we have e u (c + 1)k. Since, k = (c + 1)q we obtain ms(s) e u = (c + 1)k = c(k + q) + q > c(k + q) = b which leads to a contradiction. Moreover, we have: Lemma 3 If there exists a c-coloring f of the circular-arc graph G, then there exists a routing of makespan c(k + q) for the corresponding routing problem with all messages of size k = (c + 1)q on a ring with 2 V nodes. Proof: Let the transmission of P v P begin at b v = (i 1)(k + q) + 1, where f(v) = i. It is easy to check that all these transmissions are then finished by b. It remains to prove that this routing is feasible. Let P u, P v be two different messages. If f(u) < f(v) then b u = (f(u) 1)(k + q) + 1 and b v = (f(u) 1)(k + q) + (f(v) f(u))(k + q) + 1, which implies that the transmission P u ends before the transmission of P v begins. The case f(u) > f(v) is similar. Finally, consider f(u) = f(v). Then 9

10 {u, v} / E. Thus, P u and P v use disjoint sets of links and consequently the transmissions of P u and P v never interfere. Since the problem of coloring circular-arc graphs is strongly NP-hard, Garey et al. [1980], the following theorem is immediately implied by Lemmas 2 and 3. Theorem 7 The problem of finding an optimal solution to the routing problem with all messages of size pq, where p is a part of input to the problem, is NP-hard in the strong sense. 4.2 Routing messages of size q The MS-FSP q problem can be solved to optimality in polynomial time. The solution relies on the transformation given in Section 3.2, and on an O(n log 2 n) time algorithm minimizing the makespan of a single machine just-in-time schedule given by Dereniowski and Kubiak [2010]. The latter algorithm is based on the well-known Gilmore-Gomory algorithm for a special case of the traveling salesman problem, given in Gilmore and Gomory [1964]. The Gilmore-Gomory algorithm computes a complete weighted graph for the input to the just-in-time scheduling problem. The vertices of the graph correspond to the jobs in J, while the weight of a directed edge between two vertices depends on the idle time between the two corresponding jobs when they are being executed one immediately after the other. Then, finding a Hamiltonian cycle in the graph gives a permutation of the jobs in a schedule. Moreover, a dummy job J 0 with due-date d and processing time d, for a given d < L, ensures producing a minimum makespan just-in-time schedule with idleness in [0, d] in its first slot. This job replaces the job J 0 in the algorithm of Dereniowski and Kubiak [2010] (Section 4.2). The choice of d can be limited to d {s j : J j J } which leads to the following algorithm: Step 1: For given P and q, let J, L = q be its corresponding instance of the just-in-time scheduling problem in the mapping given in Section 3.2 for k = q. Set b := +. Step 2: For each d {s j : J j J } run steps 3-5. Step 3: Find a minimum makespan just-in-time schedule S with idle time [0, d] in the first slot using the algorithm given in Dereniowski and Kubiak [2010]. Step 4: Map S into a routing R for P and q, see Section 3.2. Step 5: If ms(r) < b, then set b := ms(r). Step 6: Return the routing R corresponding to the minimum value of b. We thus have the following theorem. Theorem 8 There exists an O(n 2 log 2 n) time algorithm for the MS-FSP q routing problem, where n = J and q is the size of the ring. Proof: The optimality follows from the preceding discussion. The instance J, L = q in step 1 can be found in time linear in n = P = J, and so can be the routing R in step 4. Step 3 can be carried out in time O(n log 2 n), see Dereniowski and Kubiak [2010]. Since the steps 3-5 are executed n times, the running time of the algorithm is O(n 2 log 2 n). 4.3 Routing unit size messages on a path We now consider a special case of the MS-FSP 1 problem, where no message uses a selected link. Without loss of generality we assume that this selected link is q 1. Therefore in principle we can cut this link out of the ring obtaining a simple path which motivates the title of this section. The condition that no message uses link q 1 is equivalent to the following one: source(p i ) < dest(p i ) for each P i P. (12) 10

11 Dereniowski and Kubiak [2010] show that the makespan of multi-slot just-in-time multiple machine schedule can be minimized in time O(n log n) if p j d j for all jobs, which is equivalent to (12) in the transformation of Section 3.2. These results will now be used to obtain a 2-approximate solution for routing messages satisfying (12). We start by proving the following key technical lemma. Lemma 4 Given P satisfying (12), if there exists a routing R with ms(r) b for P, then there exists a routing R for P such that diag(r ) b. Proof: Given R with ms(r) b, we create R as follows. For each P j P, let i j {0, b, 2b, 3b,...} be a translation such that t source(pj),b j i j belongs to one of the diagonals I 1,..., I b in R. Moreover, if P j is contained in one of these diagonals in R, then i j = 0. Otherwise, if P j is contained in one of the diagonals I b+1,..., I b+q 1 in R, then i j is a positive integer multiple of b. It can be checked that the translation with the two required properties always exists; one simply extends the matrix t for R to the left, negative side to fully cover the empty prefixes of diagonals I 1,..., I b, and then one moves to the left along the row source(p j ) with the step b starting at t source(pj),b j until one of the diagonals I 1,..., I b is finally reached. The number of steps multiplied by b equals the required translation. Finally, set the transmission of P j to start in step b j = b j i j in R. In other words, the transmission of P j starts exactly i j steps earlier in R than in R. Note that b j b j 0. However, a negative starting point of P j does not cause any problem when counting the number of diagonals in R. By the definition of the translation, all messages are routed within diagonals I 1,..., I b in R. We prove by contradiction that no packets of two different messages P j, P k go through the same link in the same step in R. Suppose that a packet of P j uses a link l in R at b j + α j for some 0 α j λ j b and a packet of P k uses the same link l in R for some 0 α k λ k b, and in R. The last equality implies b k + α k b j + α j = b k + α k b j + α j b k α k = i j i k. However, the absolute value of the left hand side of this inequality is positive and less than b for R is a feasible routing with ms(r) b. The absolute value of the right hand side belongs to {0, b, 2b, 3b,...}, which gives that i j = i k. This leads to a contradiction and proves that no packets of two different messages P j, P k go through the same link in the same step in R. Finally, R can be easily translated to start at 0 and thus a feasible routing with diag(r ) b. This lemma implies the following lower bound for the routing makespan. Corollary 2 Let P satisfying (12) be given. If R is a routing with minimum diag(r ) then for each routing R it holds ms(r) diag(r ). We are now ready to give the 2-approximation algorithm for routing unit size messages on a path. Let the jobs J correspond to the messages P, as in the transformation of Section 3.2. First, calculate a routing R with the minimum number of b diagonals. Second convert R into a routing R of makespan at most 2b. Finding a routing R with the minimum number of diagonals for P satisfying (12) is equivalent to finding a single machine multi-slot just-in-time schedule for J with minimum number of slots. The two numbers are equal, say b. For the latter problem we use the O(n log n)-time (more general) algorithm given by Dereniowski and Kubiak [2010] for J satisfying p j d j for each J j J. The slot i of the optimal single machine multi-slot just-in-time schedule for J corresponds to the diagonal I i in an optimal routing R for P, i = 1,..., b. By Corollary 2, the makespan of each routing for P is at least b. We convert R to R so that ms(r) 2b. To that end let us set the transmission of P j to start in step b j = (b j mod b) + 1 in R, where b j is the step when the transmission of P j starts in R. 11

12 We prove that R is feasible by contradiction. Suppose that a packet of P j uses a link l in R at b j + α j for some 0 α j λ j b and a packet of P k uses the same link l in R at b k + α k for some 0 α k λ k b, and in R. The last equality implies b j + α j = b k + α k with α k α j < b. However, (b j mod b) (b k mod b) = α k α j b j b k α k α j since R is feasible. This leads to a contradiction which proves that R is feasible. Finally, b j + λ j 2b for each massage P j which implies ms(r) 2b. Thus, by Corollary 2 we just proved the following theorem. Theorem 9 There exists an O(n log n)-time 2-approximation algorithm for finding routings for the set of messages P satisfying source(p i ) < dest(p i ) for each P i P. We conclude this section with a remark about optimal routing of unit size messages on a path. Though somewhat surprisingly the computational complexity of this optimal routing remains open for instances with makespan less than q, the optimal routing for instances with makespan b q, can be found in polynomial time as follows. First observe that in order to directly translate an optimal multi-slot just-in-time schedule for J (as described in Section 3.2) into an optimal routing for the corresponding P, one needs to ensure that each machine has some predefined idle intervals at the beginning of the first and at the end of the last slot in the multi-slot just-in-time schedule. These constraints are typical staircase-pattern machine availability constraints used in scheduling, see Błażewicz et al. [2000] for scheduling with machine availability constraints, and roughly speaking they result from the translation of the horizontal slots of the just-in-time schedules in the 45 diagonals of the routings. To enforce these staircase-pattern availability constraints, we add jobs J j, J j, j = 1,..., q 1, where d j = p j = j and d j = q, p j = j to mimic the constraints. A schedule for J {J j, J j : j = 1,..., q 1} uses the same number of slots on each machine as the corresponding schedule for J with availability constraints for the additional jobs fit exactly the predefined idle intervals. The latter schedule then directly translates into an optimal routing for P. There is no longer need for the availability constraints in the new instance. However, we still need to ensure that an optimal multi-slot just in time schedule for J {J j, J j : j = 1,..., q 1} does not make two jobs in {J j, J j : j = 1,..., q 1} to share a slot. To that end we transform the multi-slot just-in-time scheduling problem to a interval graph coloring with pre-coloring problem. Each job J j corresponds to the interval (s j, d j ) in this transformation. By (12), s j < d j for each J j J, which proves the correctness of the transformation. Moreover, the jobs in {J j, J j : j = 1,..., q 1} are forced not to share slots by pre-coloring their corresponding nodes in the interval graph with 2(q 1) distinct pre-colors. The problem of finding optimal interval graph coloring with pre-coloring, provided that each pre-color is pre-assigned to exactly one node can be solved in polynomial time Biró et al. [1992]. This implies the existence of a polynomial-time algorithm for finding long optimal routing of unit size messages on a path. For more on solving the multi-slot just-in-time scheduling problem via the interval graph coloring see Dereniowski and Kubiak [2010]. Sotskov et al. [2002], Ries and de Werra [2008] provide results on mixed graph coloring and scheduling problems. 12

13 4.4 Short routing of unit size messages with a bottleneck link Here we consider the routing problem MS-FSP 1, where each message in P uses a selected bottleneck link. Without loss of generality we may assume that this bottleneck link is q 1. This assumption can be re-stated as follows source(p i ) > dest(p i ) for each P i P. (13) We consider only the instances of MS-FSP 1 with short routings, that is, we assume b q throughout this subsection. In any such routing, each message P i is transmitted at the end of diagonal I s and the beginning of diagonal I s+q whenever dest(p i ) > 0, and at the end of diagonal I s only whenever dest(p i ) = 0, for some s 1. In the former case, no other message is transmitted in I s+q. In the latter case the diagonal I s+q must be empty. Otherwise I s+q (q 1) which implies b > q in both cases. This allows us to formulate the following simple observation. Lemma 5 If R solves the MS-FSP 1 problem with ms(r) b q, then each message exclusively, i.e. without sharing with other messages, uses two diagonals in R. This observation gives raise to an algorithm for the routing problem based on a transformation to the maximum matching problem in simple (unweighted) bipartite graphs. The two partitions of the vertex set of the graph will be denoted by V 1 and V 2. The vertices in V 1 correspond to the messages in P and are denoted by v(p i ), P i P, while the vertices in V 2 = {v 1,..., v b } correspond to the pairs (I j, I j+q ), j = 1,..., b, of diagonals. The set of edges E of the graph G = (V 1 V 2, E) is defined as follows {v(p i ), v j } E iff source(p i ) q j dest(p i ) b j, (14) for each v(p i ) V 1 and v j V 2. In other words, there is an edge {v(p i ), v j } in G if a message P i can be transmitted in the coordinates q j,..., q 1 of I j and in the coordinates 0,..., b j of I j+q. We have the following lemma. Lemma 6 There exists a matching M E satisfying M = P in G if and only if there exists a routing R with ms(r) b q for P. Proof: Let M E be a matching of size P. Clearly, for each v(p i ) V 1 there exists an edge {v(p i ), v j } M. By (14), the message P i can be transmitted in diagonals I j and I j+q. Since j + q > b, the pairs of diagonals where different messages in V 2 are transmitted do not overlap. Also, M = P ensures that each message appears in the routing. Finally, (14) ensures that the makespan of the routing is at most b. Now, let R be a routing with ms(r) b q for the set P. Then, by Lemma 5, a message P i P is transmitted exclusively within two diagonals I j, I j+q for some 1 j b. Observe that the diagonal I j+q may be empty if dest(p i ) = 0. It holds source(p i ) q j and j + dest(p i ) ms(r) b for the routing finishes by b and does not start before 0. Thus, by (14) there is an edge between v(p i ) and v j in G. Make this edge belong to M. Since every message is transmitted in R, then M = P and the lemma holds. Using Lemma 6 we readily obtain an optimal schedule for the short routing of unit size messages with a bottleneck link problem. Given b and P, we simply transform this problem to the problem of finding maximum cardinality matching in G. If a matching M of size P exists, then we have the corresponding routing R with ms(r) b. Otherwise no such routing exists. We can then use a binary search on b {1,..., q} to find a minimum value of makespan. Fortunately, a maximum cardinality matching in G can be found rather efficiently for G is a convex bipartite graph. Precisely, a bipartite graph G is V 2 -convex if there exists an ordering of vertices in V 2 such that for each u V 1 if {u, u i1 } E and {u, u i2 } E, i 1 < i 2, then for each i 3, i 1 i 3 i 2, {u, u i3 } E. It is easy to see that graphs G created in our algorithm are V 2 -convex, and the corresponding ordering of the vertices in V 2 is simply v 1,..., v b. Hence, the set of neighbors of v(p i ) V 1 can be represented by two integers f i, g i such that v(p i ) is adjacent to exactly the vertices v fi, v fi+1,..., v gi. By (14), f i = q source(p i ) and g i = b dest(p i ). This means that G can be created in linear time and a maximum cardinality matching for G can be obtained in O( V 1 )-time by the results of Lipski Jr. and Preparata [1981] and Gabow and Tarjan [1985], see also Steiner and Yeomans [1996]. 13

14 Theorem 10 There exists an O(n)-time algorithm which, for the given integer b q and set of unit size messages P satisfying (13), finds a routing R with ms(r) b or decides that no such routing exists. We end this section with an example of the execution of this algorithm for q = b = 5. Assume that P = {P 1,..., P 4 }, where source(p 1 ) = 4, dest(p 1 ) = 2, source(p 2 ) = 4, dest(p 2 ) = 3, source(p 3 ) = 2, dest(p 3 ) = 1, source(p 4 ) = 3, dest(p 4 ) = 2. Figure 4(a) shows the pairs of diagonals corresponding to the vertices in V 2. The bipartite graph G is given in Figure 4(b). Consider the following matching M (a) v 5 v 1 v 2 v 3 v 4 1 v 4 v 5 v 1 v 2 v 3 2 v 3 v 4 v 5 v 1 v 2 3 v 2 v 3 v 4 v 5 v 1 4 v 1 v 2 v 3 v 4 v 5 (b) v 1 v(p (c) 1 ) P 2 P 1 P 4 P v 3 2 v(p 1 P 2 P 1 P 2 ) 4 v 3 2 P 3 P 2 v 4 v(p 3 P 4 P 3 ) 3 4 P 2 P 1 P 4 P 3 v 5 v(p 4 ) Figure 4: (a) the diagonals corresponding to the vertices v 1,..., v 5 ; (b) the bipartite graph G; (c) the routing corresponding to the matching M M = {{v 1, v(p 2 )}, {v 2, v(p 1 )}, {v 4, v(p 3 )}, {v 3, v(p 4 )}}. The final routing obtained from M is depicted in Figure 4(c). 4.5 Makespan approximation algorithm for unit size messages We begin by observing that the makespan and the number of diagonals are within a distance of q 1 for any R. The two following lemmas hold for any routing R and any P, i.e. even with messages of variable sizes. Lemma 7 If R is a routing for P, then diag(r) ms(r) + q 1. Proof: Consider the diagonals I 1,..., I ms(r)+q 1. Again, without loss of generality we assume that I 1 is the first nonempty diagonal in R. If s {1,..., ms(r)} and l {0,..., q 1} then t s,l must belong to exactly one of these diagonals. Thus, the lemma holds. Lemma 8 If R is a routing for P, then ms(r) diag(r) + q 1. Proof: By the definition there are diag(r) consecutive indices r such that I(R) defines R. The lemma then follows from the fact that each diagonal I r intersects exactly q consecutive steps in a routing. We are ready to give an O(n log n)-time approximation algorithm for the MS-FSP 1 routing problem. In the following R D and R M denote optimal routings for P for the diagonal DIAG-FSP 1 and the makespan MS-FSP 1 minimization problems, respectively. By Theorem 6 we can find a routing R for P with diag(r) diag(r D ) + 2q 1 in time O(n log n). Clearly, diag(r D ) diag(r M ). By Lemma 8, ms(r) diag(r) + q 1. Thus, ms(r) diag(r M ) + 3q 2. (15) By Lemma 7, diag(r M ) ms(r M ) + q 1. This, by (15), gives ms(r) ms(r D ) + 4q 3 which implies the theorem. Theorem 11 There exists an O(n log n)-time approximation algorithm for finding a routing R of length ms(r) ms(r M ) + 4q 3, where R M is an optimal routing for MS-FSP 1. 14