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sel sh load balancing in .NET Creator Code 39 in .NET sel sh load balancing

sel sh load balancing Using Barcode generator for VS .NET Control to generate, create ANSI/AIM Code 39 image in .NET framework applications. bar code of each ser Code 39 Full ASCII for Visual Studio .NET ver in Lk is at least k. Analogously to the analysis in the proof of Theorem 20.

7, one shows the recurrence . Lk (k + 1) . Lk+1 , for 0 k c 2, and Lc 1 1. Solving the recurrence yields L0 . (c 1)! = (c). Thus, . L0 . = m implies c 1 (m) = (ln m/ ln ln m). Now let C = max c + 1, ln m ln ln m = ln m . ln ln m In the rest of the proof, we show that the expected makespan of the equilibrium assignment can exceed C at most by a factor of order ln ln m/ ln ln ln m so that the expected makespan is O(ln m/ ln ln ln m), which proves the theorem as we assume opt(G) = 1. As the next step, we prove a tail bound on j , for any xed j [m] and, afterward, we use this tail bound to derive an upper bound on the expected makespan. j For a machine j [m], let Tj(1) denote the set of tasks i with pi 1 and Tj(2) 4 j the set of tasks i with pi (0, 1 ).

Let j(1) and j(2) denote random variables that 4 describe the load on link j only taking into account the tasks in Tj(1) and Tj(2) , respectively. Observe that j = j(1) + j(2) . For the tasks in Tj(1) , we immediately obtain.

(1) j i Tj(1). wi 4 sj i Tj(1). wi pi = 4 E[ sj (1) j ]. 4C.. (20.2). To prove an Code 39 Full ASCII for Visual Studio .NET upper bound on j(2) , we use the weighted Chernoff bound from Lemma 20.14.

This bound requires an upper bound on the maximum weight. As a rst step to bound the weights, we prove a result about the relationship between the speeds of the machines in the different groups that are de ned by the pre xes. For 0 k c 2, let Gk = Lk \ Lk+1 , and let Gc 1 = Lc 1 .

For 0 k c 1, let s(k) denote the speed of the fastest machine in Gk . Clearly, s(c 1) s(c 2) s(1) s(0). We claim that this sequence is, in fact, geometrically decreasing.

Lemma 20.16 For 0 k c 4, s(k + 2) 2 s(k)..

proof To pr ove the claim, we rst observe that there exists a task j with wj s(k + 2) that has positive probability on a machine in Lk+3 . This is because an optimal assignment strategy has to move some of the expected load from the machines in Lk+3 to machines in L \ Lk+3 and it can only assign those tasks to machines in L \ Lk+3 whose weights are not larger than the maximum speed among this set of machines, which is s(k + 2). Now suppose s(k) > 1 s(k + 2).

2 The expected load of the fastest machine in Gk = Lk \ Lk+1 is at most k + 1. Thus the expected cost of j on the fastest machine in Gk is at most k+1+ 2wj wj <k+1+ k + 3. s(k) s(k + 2).

mixed equilibria on uniformly related machines This contra dicts that the expected cost of j in the considered Nash equilibrium is at least k + 3 as it has positive probability on a machine in Lk+3 . Thus, Lemma 20.16 is shown.

Now we apply Lemma 20.16 to prove an upper bound on the weights of the tasks in the set Tj(2) . Lemma 20.

17 For every j [m] and i Tj(2) , wi 12 sj .. proof Let i .NET framework Code39 be a task from Tj(2) , i.e.

, pi (0, 1 ). Let j Gk , for 0 k c 1. 4 The expected cost of i on j is ci = E[ j ] + 1 pi.

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