A Note on Scheduling Tall/Small Multiprocessor Tasks with Unit Processing Time to Minimize Maximum Tardiness

Transcription

1 A Note on Scheduling Tall/Sall Multiprocessor Tasks with Unit Processing Tie to Miniize Maxiu Tardiness Philippe Baptiste and Baruch Schieber IBM T.J. Watson Research Center P.O. Box 218, Yorktown Heights, NY February 8,

2 Scheduling Tall/Sall Multiprocessor Tasks 2 Abstract We study the scheduling situation where n tasks, subjected to release dates and due dates, have to be scheduled on parallel processors. We show that, when tasks have unit processing ties and either require 1 or processors siultaneously, the iniu axial tardiness can be coputed in polynoial tie. Two algoriths are described. The first one is based on a linear prograing forulation of the proble while the second one is a cobinatorial algorith. The coplexity status of this tall/sall task scheduling proble P r i,p i = 1,size i {1,} T ax was unknown before, even for 2 processors. Keywords: Multiprocessor Task Scheduling, Linear Prograing, Unit Processing Ties

3 Scheduling Tall/Sall Multiprocessor Tasks 3 1 Introduction We study the following scheduling proble. The input consists of n unit processing tie tasks that have to be scheduled on parallel identical processors. Each task i is associated with a release date r i before which it cannot start and a due date d i by which it needs to be copleted. Each task requires siultaneously a fixed nuber size i of processors, yet the processors required are not specified. In the tall/sall proble, there are two possible sizes, either 1 (sall tasks) or (tall tasks), while in the arbitrary size proble, size i {1,..., }. In the following, T 1 and T denote the sets of tasks of size 1 and, respectively. Throughout this paper, we consider the axiu tardiness objective function. The tardiness of task i is defined as T i = ax{0, C i d i }, where C i is the copletion tie of i, and the axiu tardiness is T ax = ax i T i. Following the classical scheduling notation (see for instance [1, 2]), this proble is referred to as P r i, p i = 1,size i {1, } T ax. When all release dates are equal, the arbitrary size proble for any fixed nuber of processors, denoted P p i = 1,size i T ax, can be solved in polynoial tie [3]. Siilarly, the proble for any fixed nuber of sizes, denoted P p i = 1, {size i } = c T ax, can be solved in polynoial tie. Both proble can be solved using dynaic prograing. However, the arbitrary size proble is NP-Hard in the strong sense [4]. When preeption is allowed, even with arbitrary processing ties and release dates, the proble is easy to solve. Existing algoriths are based on a Linear Prograing forulation where a variable is associated to each subset of tasks whose total resource requireent is no ore than (see for instance [1]). Unfortunately, there are soe instances for which the non-preeptive axiu tardiness is strictly larger than the preeptive axiu tardiness. In the preeptive schedule of Figure 1, T ax = 0, while the value of T ax is at least 1 for any non-preeptive schedule. Processor 1 Processor 2 A B D E X B C E F A, B, C: r= 0, d = 2 X: r = 1, d = 3 D, E, F: r = 2, d = 4 Figure 1: An optial preeptive schedule. On two processors, the tall/sall proble and the arbitrary size proble are exactly the sae and the coplexity status of P2 r i, p i = 1,size i T ax has been open. In

4 Scheduling Tall/Sall Multiprocessor Tasks 4 this paper we show that P r i, p i = 1,size i {1, } T ax can be solved in polynoial tie. Without loss of generality, we can assue that that ax n i=1{d i } in n i=1{r i } is O(n). Indeed, consider any schedule built chronologically as follows: schedule at each tie t an unscheduled task aong {i r i t}. In such a schedule, at ost one task is scheduled in each tie slot and it is easy to see that if no task is scheduled at t then the tasks can be split in two subsets {i : r i t} and {i : r i > t} that can be scheduled independently. Hence we can assue that ax n i=1{r i } in n i=1{r i } n. We can also assue that the largest due date is not larger than the largest release date plus n. Hence ax n i=1 {d i} in n i=1 {r i} is O(n). In the following, we focus on the decision variant of the axial tardiness proble. For a fixed value of T ax, it is easy to copute a deadline δ i = d i +T ax for each task i and a schedule is said to be feasible if tasks are copleted by their deadlines. To copute the inial axial tardiness, one can find the sallest value of T ax for which there is a feasible schedule. Since one can easily build a feasible schedule with T ax = n, there are no ore than n values to test. A dichotoic search could be used to reduce the nuber of iterations. So, the T ax proble can be solved in polynoial tie provided that the deadline scheduling proble is solvable in polynoial tie. We describe two independent techniques to solve the tall/sall proble. In Section 2, we introduce a linear prograing forulation of the proble and in Section 3 we describe a cobinatorial algorith to solve the proble in O(n 4 ). Finally, we draw soe conclusions in Section 4. 2 An LP Forulation In Section 2.1 we describe soe linear constraints that ust be et by any feasible schedule. In Section 2.2 we show that if these constraints hold, we can build a preeptive schedule of tall tasks that iplicitly takes into account the sall tasks. This schedule is transfored in Section 2.3 into a non-preeptive schedule of both sall and tall tasks. 2.1 Necessary Conditions To siplify the presentation of the algorith we will introduce soe tie indexed variables. Since ax n i=1 {d i} in n i=1 {r i} is O(n), there are few relevant tie points and the total nuber of variables is linear in n. In the following, unless precisely stated, tie points and tie slots are integral.

5 Scheduling Tall/Sall Multiprocessor Tasks 5 Consider a feasible schedule and, for each tall task i and for each integer tie point, let t, x t i [0, 1] be the total tie during which i executes in the tie-slot [t, t + 1). Each tall task has to be scheduled soewhere between its release date and its deadline so, δ i 1 i T, x t i = 1. (1) t=r i On top of that, the total tie during which tall tasks are processed in a single tie slot does not exceed the size of the tie slot, i.e., t, 1. (2) i T x t i Now let us focus on sall tasks. For any tie interval [t 1, t 2 ), and any size u {1, }, let T u (t 1, t 2 ) be the set of sall (u = 1) or tall (u = ) tasks that have to execute in the tie interval [t 1, t 2 ), i.e., T u (t 1, t 2 ) = {i T u : t 1 r i < δ i t 2 }. Note that in a non-preeptive schedule, q sall tasks cannot be scheduled in less than q tie units. Since in a tie interval [t 1, t 2 ) there are only t 2 t 1 t2 1 i T t=t 1 x t i tie units available to schedule tall tasks, we get [t 1, t 2 ), t 2 1 x t i + T1 (t 1, t 2 ) t 2 t 1. (3) i T t=t 1 Hence, if there is feasible schedule, there is a feasible solution of the Linear Progra (4). δ i 1 i T, x t i = 1 t=r i t, x t i 1 i T (4) [t 1, t 2 ), t 2 1 x t i + T1 (t 1, t 2 ) t 2 t 1 i T t=t 1 i T, t, x t i 0 In the following, we show that a feasible schedule exists if there is a feasible solution of (4).

6 Scheduling Tall/Sall Multiprocessor Tasks Preeptive Schedule of Tall Tasks Fro now on, we assue that tasks 1,..., T are the tall ones and that they are sorted in non-decreasing order of due dates, i.e., d 1... d T. A solution x of (4), specifies the duration x t i a tall task i is scheduled in [t, t + 1). To precisely build a preeptive schedule of tall tasks, it reains to decide how pieces of tall tasks are scheduled inside each tie slot [t, t + 1). Let S(x) be the schedule where, in each tie slot, pieces of tall tasks are scheduled fro left to right according to their initial nubering (i.e., in non-decreasing order of deadlines). Now let us consider the solution x that lexicographically iniizes the vector of average copletion ties (C 1,...,C T ), where C i = δ i 1 t=r i t x t i (i.e., C 1 is iniu, C 2 is iniu given the previous value of C 1,..., and finally C n is iniu given the previous values of C 1,...,C n 1 ). The next proposition shows that in any such solution x t i is either 0 or 1. Proposition 1. In S( x) tall tasks are not preepted and they start at integer tie points. Proof. Let k be the first task for which the proposition does not hold (all tasks with saller indices are not preepted and start at integer tie points). Let [t, t + 1) and [t, t + 1) be the tie slots in which k respectively starts and is copleted in S( x). First, we show that in [t, t + 1), k is the only tall task to execute. Indeed, if there was another tall task l that executes there, we would have l > k (because of our assuption on k). Therefore, we could exchange a sall piece of l that executes in [t, t + 1) with a sall piece of k that executes in [t, t + 1). In other ters, we could build a vector ˆx that equals x except for the following values: ˆx t k = xt k + ε ˆx t k = xt k ε, ˆx t l = x t l + ε ˆx t l = xt l ε where ε = in{ x t k, xt l } > 0. In the resulting schedule S(ˆx) the average copletion tie of k is saller than its average copletion tie in S( x) and task l is copleted before its deadline (because δ l δ k ). Moreover, the load of each tie slot [τ, τ +1) is the sae in both schedule, i.e., τ, x τ i = ˆx τ i. i T i T

7 Scheduling Tall/Sall Multiprocessor Tasks 7 It follows that ˆx is a feasible solution of (4) and it is saller lexicographically than x. This contradicts our hypothesis on x. Next, we exaine what prevents us fro obtaining a lower feasible solution in the defined lexicographical order by increasing x t k by ε and decreasing xt k by ε. The reason for this ust be soe constraints of (4) that are tight for x. Constraints (1) do not prevent us fro aking this exchange. Neither do Constraints (2) since k is the only tall task to execute in tie slot t. Now, let us exaine Constraints (3). We are going to show that a slightly ore coplex exchange is possible, hence leading to a contradiction. In the following, the notation I(t 1, t 2 ) refers to the constraint (3) over the interval [t 1, t 2 ). Let Ω be the set of constraints (3) with t 1 t and t < t 2 t that are tight for x. It is easy to see that Constraints (3) that do not belong to Ω do not prevent us fro increasing x t k. Aong the constraints of Ω let us pick one with axiu t 1. Since the constraint is tight, we have t 2 1 x t i = t T1 (t 1, t 2 ) 2 t 1. t=t 1 i T Hence, i T t2 1 t=t 1 x t i takes an integer value and consequently, there is another tall task u that is partially executed between t 1 and t 2. If u were partially executed between t and t 2 then we could exchange a piece of it with the last piece of k that executes in [t, t + 1) (we have δ u δ k because u is preepted and so the exchange is feasible). We would have decreased the average copletion tie of k; which would contradict our initial hypothesis. So u is partially executed in a tie slot [τ, τ + 1) between t 1 and t. Let x be the vector that equals x except of the following values: x τ u = xτ u ε x t u = x t u + ε x t k = xt k ε, x t k = xt k + ε for a sall positive value ε. Note that a piece of u can be scheduled at t because δ u δ k. Moreover, the only constraints that could be violated by the exchange are those in the set Ω (the value of x t k is consistent with (2) because k is the only tall task that executes in [t, t + 1)). Let I(t 1, t 2 ) be a violated constraint of Ω. Because of our hypothesis on t 1, we have t 1 t 1. Hence, it is easy to verify that the load induced by the tall tasks between t 1 and t 2 does not increase. Therefore, the resulting schedule S( x) is lower than S( x) in the lexicographical order. This contradicts our hypothesis.

8 Scheduling Tall/Sall Multiprocessor Tasks Fro Preeptive to Non-Preeptive Schedules In Section 2.2, we have shown that there is a solution x of (4) such that in S( x), tall tasks are not preepted and start at integer tie points. In Proposition 2 we show that sall tasks can be scheduled in S( x) too. Proposition 2. Sall tasks can be scheduled in S( x). Proof. Let us sort sall tasks in non-decreasing order of deadlines and let us add the one after the other into S( x). Each tie, the current task starts at the first tie point after its release date where one processor is available. Note that because tall tasks are not preepted and start at integer tie points, sall tasks are not preepted either and also start at integer tie points. Suppose that we do not obtain a feasible schedule. Let then k be the first sall task that is copleted after its deadline and let t be the earliest tie point such that all processors are full between t and δ k. Note that since k is not copleted by its deadline t r k. Let Ψ be the set of sall tasks that are scheduled in [t, δ k ). Tasks in Ψ have a release date greater than or equal to t (otherwise they would be scheduled in [t 1, t)) and a deadline saller than or equal to δ k (because tasks are sorted in non-decreasing order of deadlines). Since all processors are full, there are exactly (δ k t δk 1 i T t =t x t i ) tasks in Ψ. On top of that, Ψ T 1(t, δ k ) and k / Ψ but k T 1 (t, δ k ). Hence, This contradicts (3). T 1 (t, δ k ) > (δ k t δ k 1 i T t =t x t i ). 3 A Cobinatorial Algorith For each tie interval [t 1, t 2 ), we define recursively the slack of the interval, denoted (t 1, t 2 ). We show later that there exists a feasible schedule if and only if all the intervals have non-negative slack. We also show how to use the slack to actually construct a schedule is such exists. Intuitively, (t 1, t 2 ) is an upper bound on the nuber of free tie slots in [t 1, t 2 ) in any schedule of the tasks in T (t 1, t 2 ). (For a given schedule a free tie slot is a tie unit in which all the processors are idle.) To define the slack we define static slack and dynaic slack. In the recursive definition the coputation the dynaic slack (and thus the slack) of an interval [t 1, t 2 ) uses the values of the slack of intervals contained in [t 1, t 2 ).

9 Scheduling Tall/Sall Multiprocessor Tasks 9 Definition 1. For any interval [t 1, t 2 ), with t 2 > t 1 + 1, the slack (t 1, t 2 ) is the iniu of the static slack s (t 1, t 2 ) and of the dynaic slack d (t 1, t 2 ). For any interval [t 1, t 1 + 1), (t 1, t 1 + 1) = s (t 1, t 1 + 1). Definition 2. The static slack over [t 1, t 2 ), s (t 1, t 2 ), equals ( ) T1 (t 1, t 2 ) t 2 t 1 T (t 1, t 2 ) + Definition 3. The dynaic slack over [t 1, t 2 ), with t 2 > t 1 + 1, d (t 1, t 2 ), is the iniu of (t 1, t 2 ) + (t 1, t 2) T (t 1, t 2 ) \ (T (t 1, t 2 ) T (t 1, t 2)) over t 1, t 2 with t 1 < t 1 t 2 < t 2. We clai that all slacks can be coputed in O(n 4 ). Indeed, there are O(n) tie points to consider and thus there are O(n 2 ) values (t 1, t 2 ) to copute. We observe that all the values of T 1 (t 1, t 2 ) and T (t 1, t 2 ) can be coputed inductively in O(n 2 ) tie. To see this, note that all the O(n) values T 1 (t 1, t 1 + 1) and T (t 1, t 1 + 1) can be coputed in linear tie. Now, for any x > 1, and for i {1, }, T i (t 1, t 1 + x) = T i (t 1, t 1 + x 1) + T i (t 1 + 1, t 1 + x) T i (t 1 + 1, t 1 + x 1). Thus, given all values T 1 (t 1, t 1 +y) and T (t 1, t 1 +y), for y < x, T 1 (t 1, t 1 +x) and T (t 1, t 1 + x) can be coputed in linear tie. Clearly, this iplies that all static slacks can be coputed in O(n 2 ) tie. For each of the O(n 2 ) coputations of the dynaic slack we need to find the iniu of O(n 2 ) quantities. Since T (t 1, t 2 ) \ (T (t 1, t 2 ) T (t 1, t 2)) = T (t 1, t 2 ) T (t 1, t 2 ) T (t 1, t 2) + T (t 1, t 2 ), each of these O(n 2 ) quantities can be coputed in constant tie totalling to O(n 4 ) tie for the whole coputation. The next proposition shows that if there is a feasible schedule, then all intervals have non-negative slack. Proposition 3. There are at ost (t 1, t 2 ) idle tie slots between t 1 and t 2 in any feasible schedule of T (t 1, t 2 ).

10 Scheduling Tall/Sall Multiprocessor Tasks 10 Proof. Consider an interval [t 1, t 2 ) such that there is a feasible schedule of T (t 1, t 2 ) with ore than (t 1, t 2 ) idle tie slots and assue that t 2 t 1 is inial aong such intervals. ( There are T ) (t 1, t 2 ) tall tasks and T 1 (t 1, t 2 ) sall tasks and at least T (t 1, t 2 ) + T1 (t 1,t 2 ) tie units are required to schedule these tasks. It follows that the nuber of idle tie slots in a feasible schedule is no ore than the static slack s (t 1, t 2 ). Hence the slack equals the dynaic slack; that is, (t 1, t 2 ) = d (t 1, t 2 ). Suppose that d (t 1, t 2 ) is deterined by the two tie points t 1 and t 2 ; that is, (t 1, t 2 ) = (t 1, t 2 )+ (t 1, t 2) T (t 1, t 2 ) \ (T (t 1, t 2 ) T (t 1, t 2)). Consider the feasible schedule S of T (t 1, t 2 ) and let S r and S l be the restriction of S to the tasks of T (t 1, t 2) and to the tasks of T (t 1, t 2 ). The nuber of idle tie slots in S is less than or equal to the nuber of those in S r plus the nuber of those in S l. Given our hypothesis on t 2 t 1 this is lower than or equal to (t 1, t 2 ) + (t 1, t 2). On top of that, large tasks in T (t 1, t 2 ) \ (T (t 1, t 2) T (t 1, t 2 )) are not scheduled in S l nor in S r so the nuber of idle tie slots in S is less than or equal to (t 1, t 2 ) + (t 1, t 2) T (t 1, t 2 ) \ (T (t 1, t 2 ) T (t 1, t 2)), which equals (t 1, t 2 ). This contradicts our initial hypothesis. The ore coplicated part is to show that if there is no feasible schedule then there ust be at least one interval with negative slack. In order to prove this we first prove two propositions. In the proofs we consider several instances siultaneously. Given an instance I, (I, t 1, t 2 ), T (I, t 1, t 2 ), T 1 (I, t 1, t 2 ) and T (I, t 1, t 2 ) denote the slack and the sets associated to I for a given interval [t 1, t 2 ). Fro now on, to siplify notation we also assue that the sallest release tie is 0. Proposition 4. Consider an instance I, and let J be and instance given by reoving a tall task i, with r i = 0, if such exists. For all t < δ i, (J, 0, t) = (I, 0, t), and for all t δ i, (J, 0, t) = (I, 0, t) + 1. Proof. The first part follows directly fro the definition of slack. Consider the second part and to obtain a contradiction assue that the equality does not hold. Let t be the iniu tie such that that (J, 0, t) (I, 0, t) + 1. We first show that (J, 0, t) (I, 0, t) + 1. We consider two cases. Case 1: (I, 0, t) = s (I, 0, t). Since s (J, 0, t) = s (I, 0, t)+1, we have (J, 0, t) (I, 0, t) + 1. Case 2: (I, 0, t) = d (I, 0, t). Let t 1, t 2 be two tie points that deterine d (I, 0, t). Note that (I, t 1, t) = (J, t 1, t) (because 0 < t 1 ). If δ i t 2 then i

11 Scheduling Tall/Sall Multiprocessor Tasks 11 does not belong to T (I, 0, t) \(T (I, 0, t 2 ) T (I, t 1, t)). Hence, the slack of J is not larger than (J, 0, t 2 ) + (I, t 1, t) T (I, 0, t) \ (T (I, 0, t 2 ) T (I, t 1, t)). Since t 2 < t, by our assuption (J, 0, t 2 ) = (I, 0, t 2 ) + 1. Hence, (J, 0, t) (I, 0, t) + 1. If δ i > t 2 then (J, 0, t) (I, 0, t 2 ) + (I, t 1, t) ( T (I, 0, t) \ (T (I, 0, t 2 ) T (I, t 1, t)) 1) = (I, 0, t) + 1. Siilarly, it can be shown that (I, 0, t) (J, 0, t) 1, by considering the two cases (J, 0, t) = s (J, 0, t) and (J, 0, t) = d (J, 0, t). Proposition 5. Consider an instance I. Let k be the nuber of sall tasks in I with release date 0. Let J be the instance derived fro I by reoving in{k, } sall tasks with the sallest deadline aong tasks with release date 0, and by changing the release date of the rest of the tasks with release date 0 (if any) to 1. Let x be a tall task in I with the earliest deadline aong tall tasks with release date 0 (if such exists), and let t be δ x if x is exists and otherwise. For all 1 < t 2 < t, in { (I, 0, t 2 ), (I, 1, t 2 )} (J, 1, t 2 ). Proof. To obtain a contradiction assue that t 2 is the iniu tie for which the inequality does not hold. If k then since no tall task in I has release date 0 and deadline at ost t 2, we have (I, 1, t 2 ) = (J, 1, t 2 ); a contradiction. Suppose that k >. Let δ be the earliest deadline of a task in I with release date 0 that was not reoved. If t 2 < δ then again T (J, 1, t 2 ) does not contain any task with release tie 0 in I, and thus, (I, 1, t 2 ) = (J, 1, t 2 ); a contradiction. The only reaining case is t 2 δ. We consider two cases. Case 1: (J, 1, t 2 ) = s (J, 1, t 2 ). Since t 2 δ T 1 (J, 1, t 2 ) = T 1 (I, 0, t 2 ) and thus T1 (J, 1, t 2 ) T1 (I, 0, t 2 ) = 1. Recall also that since no tall task in I has release date 0 and deadline at ost t 2 we have for all 1 < s t 2, T (I, 0, s) = T (J, 1, s). We get T1 (J, 1, t 2 ) (J, 1, t 2 ) = t 2 1 ( T (J, 1, t 2 ) + T1 (I, 0, t 2 ) = t 2 ( T (I, 0, t 2 ) + ) = s (I, 0, t 2 ) (I, 0, t 2 ). )

12 Scheduling Tall/Sall Multiprocessor Tasks 12 Case 2: (J, 1, t 2 ) = d (J, 1, t 2 ). Let t 1, t 2 with 1 < t 1 t 2 < t 2 be tie points that deterine the dynaic slack. By our assuption in { (I, 0, t 2), (I, 1, t 2)} (J, 1, t 2 ). Since t 1 > 1, (I, t 1, t 2) = (J, t 1, t 2). We get (J, 1, t 2 ) = (J, 1, t 2 ) + (J, t 1, t 2) T (J, 1, t 2 ) \ (T (J, 1, t 2 ) T (J, t 1, t 2)) in { (I, 0, t 2), (I, 1, t 2)} + (I, t 1, t 2 ) T (I, 0, t 2 ) \ (T (I, 0, t 2 ) T (I, t 1, t 2)) in { (I, 0, t 2 ), (I, 1, t 2 )}. Proposition 6. If all tie intervals have non-negative slack then there is a feasible schedule. Proof. We prove that if there is no feasible schedule then there ust be an interval with a negative slack. Define the size of an instance I as T δ i r i. We prove it by induction on the size of the instance. The proposition clearly holds for instances of size 1 since such instances consist of one task. Suppose that the proposition holds for all instances of size at ost s. We prove it for instances of size s + 1. Consider an instance I of size s+1 with no feasible schedule. If I has both a tall task and a sall task with release date 0 and deadline 1 or ore than sall tasks with release tie 0 and deadline 1 then (I, 0, 1) < 0. Suppose that this is not the case. Let k be the nuber of sall tasks in I with release date 0. Consider two instances J S and J T defined as follows. The instance J S is derived fro I by reoving in{k, } sall tasks with the sallest deadline aong tasks with release date 0, and by changing the release date of the rest of the tasks with release date 0 (if any) to 1. The instance J T is derived fro I by reoving a tall task x with the sallest deadline aong tall tasks with release date 0 (assuing that the task x exists), and by changing the release date of the rest of the tasks with release date 0 (if any) to 1. Note that if any of J T and J S has a feasible schedule then this schedule can be extended to a feasible schedule for I. We conclude that both J S and J T have no feasible schedule. Since the size of both J S and J T is at ost s, there is an interval with negative slack in both J S and J T. If for J {J S, J T }, there exists 1 < t 1 < t 2 such that (J, t 1, t 2 ) < 0 then since (I, t 1, t 2 ) = (J, t 1, t 2 ) we are done. Suppose that this is not the case. Hence, there are t S, t T > 1 such that (J S, 1, t S ) < 0 and (J T, 1, t T ) < 0. We consider several cases.

13 Scheduling Tall/Sall Multiprocessor Tasks 13 Case 1: t T δ x. Let I be the instance given by reoving task x fro I. Since the sets of tasks in J T and I are the sae (I, 0, t T ) (J T, 1, t T )+1. By Proposition 4 (I, 0, t T ) = (I, 0, t T ) 1. Hence, (I, 0, t T ) (J T, 1, t T ) < 0. Case 2: t S < δ x. By Proposition 5 in { (I, 0, t S ), (I, 1, t S )} (J S, 1, t S ) < 0. Case 3: The reaining case is t S δ x and t T < δ x. Let l be the nuber of large tasks in I with release date 0 and deadline at ost t S. Let J S be the instance given by reoving these l tasks fro J S. By proposition 4 (J S, 1, t S) = (J S, 1, t S ) + l < l. Let I be the instance given by reoving these l tasks fro I. Note that (I, 1, t S ) = (I, 1, t S ). By Proposition 5 in { (I, 0, t S ), (I, 1, t S )} (J S, 1, t S) < l. If (I, 0, t S ) < l then by Proposition 4 (I, 0, t S ) = (I, 0, t S ) l < 0. Otherwise, (I, 1, t S ) < l. We get that (I, 0, t S ) (I, 0, t T ) + (I, 1, t S ) T (I, 0, t S ) \ (T (I, 0, t T ) T (I, 1, t S )) < 0 + l l = 0 To conclude the proof, we just have to consider the situation where the task x does not exist (i.e., there is no tall task with release date 0). In such a case, the sae reasoning as for Case 2 applies. Finally, note that the proof of Proposition 6 iplies also an algorith for constructing a schedule. Given an instance I we deterine the schedule at tie 0 by testing the feasibility of J S and J T, and continue in the sae anner to construct the rest of the schedule. 4 Conclusion We have presented two algoriths for scheduling tall/sall ultiprocessor tasks with unit processing tie to iniize axiu tardiness. The first one relies on linear prograing and the second one is purely cobinatorial. An interesting open question is whether one of the could be extended to solve a larger class of proble such as P r i, p i = 1,size i T ax. Acknowledgent The authors would like to thank Professor Peter Brucker who introduced the first author to the sall/tall proble in 1998.

14 Scheduling Tall/Sall Multiprocessor Tasks 14 References [1] J. Blazewicz, K. H. Ecker, E. Pesch, G. Schidt and J. Weglarz. Scheduling Coputer and Manufacturing Processes. Springer, [2] P. Brucker. Scheduling Algoriths, 3rd edition. Springer, [3] P. Brucker and A. Kräer. Polynoial algoriths for resource-constrained and ultiprocessor task scheduling probles. European Journal of Operational Research, 90: , [4] E. L. Lloyd. Concurrent task systes. Operations Research 29: , 1981.