To conclude the basic de nitions, the sequential time of a task graph is de ned in .NET framework

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To conclude the basic de nitions, the sequential time of a task graph is de ned
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De nition 412 (Sequential Time) Let G = (V, E, w, c) be a task graph G s sequential time is seq(G) =
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which corresponds to G s execution time on one processor only (remember, local communication is cost free) 421 Schedule Example The above de nitions and conditions are illustrated by examining a sample schedule Figure 41(a) depicts a Gantt chart of a schedule for the sample task graph of Figure 315 on three processors A Gantt chart is an intuitive and common graphical representation of a schedule, in which each scheduled object (ie, node) is drawn as a rectangle The node s position (ie, the position of its rectangle) in the coordinate system spanned by the time and space axis (ie, processor axis) is determined by the node s allocated processor and its start time, while the size of the rectangle re ects the node s execution time Some examples from this gure illustrate the foregoing de nitions and conditions The start time of node a executed on processor P1 is ts (a) = 0 (ie, ts (a, P1 ) = 0) and, with a computation cost of w(a) = 2, its nish time is tf (a) = tf (a, P1 ) = 2
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Figure 41 The Gantt chart (b) of a schedule for the sample task graph (a) of Figure 315 on three processors
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Node d begins at ts (d) = 5 and nishes at tf (d) = ts (d) + w(d) = 5 + 5 = 10 Its start time ts (d) = 5 is the earliest possible, because it has to wait for the data from node a, tf (ead , P1 , P3 ) = tf (a) + c(ead ) = 2 + 3 = 5 So in this case, the communication is remote and therefore causes a delay of 3 time units corresponding to the weight of edge ead Node b, on the other hand, also receives data from node a, but as the communication is local both a and b are on the same processor it takes no time: ts (b) = tf (eab , P1 , P1 ) = tf (a) = 2 A good example for the in uence of various precedence constraints on the start time of a node is node h The two entering communications edh and eeh result in a data ready time of tdr (h, P2 ) = 14 Responsible is the communication from node d, since d nishes at tf (d) = 10 and the communication of edh lasts 4 time units, while eeh arrives at P2 at tf (eeh , P1 , P2 ) = 10 + 2 = 12 So node h starts at its DRT In contrast, node e starts at ts (e) = 6, which is later than its DRT (tdr (e, P1 ) = tf (eae , P1 , P1 ) = tf (a) = 2), because processor P1 is busy with node b until that time The length of the schedule is sl = tf (k) = 24, that is, the nish time of the last node k, as the rst node a starts at time unit 0
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422 Scheduling Complexity The process of creating a schedule S for a task graph G on a set of processors P is called scheduling It should be obvious from the example in Figure 41 that there is generally more than one possible schedule for a given graph and set of processors (eg, a could be executed on processor P2 , which of course would have consequences for the scheduling of the other nodes) Since the usual purpose in employing a parallel system is the fast execution of a program, the aim of scheduling is to produce a schedule of minimal length De nition 413 (Scheduling Problem) Let G = (V, E, w, c) be a task graph and P a parallel system The scheduling problem is to determine a feasible schedule S of minimal length sl for G on P Unfortunately, nding a schedule of minimal length (ie, an optimal schedule) is in general a dif cult problem This becomes intuitively clear as one realizes that an optimal schedule is a trade-off between high parallelism and low interprocessor communication On the one hand, nodes should be distributed among the processors in order to balance the workload On the other hand, the more the nodes are distributed, the more interprocessor communications, which are expensive, are performed In fact, the general decision problem (a decision problem is one whose answer is either yes or no ) associated with the scheduling problem is NP-complete The complexity class of NP-complete problems (Cook [41], Karp [99]) has the unpleasant property that for none of its members has an algorithm been found that runs in polynomial time It is unknown if such an algorithm exists, but it is improbable, given that if any NP-complete problem is polynomial-time solvable then all NP-complete problems are polynomial-time solvable (ie, P = NP) Introductions to NP-completeness, its proof techniques, and the discussion of many
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