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By C Georgiou; Alex Allister Shvartsman

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N · κ⌋ · κ⌋ . . · κ⌋, and i i times (b) pi ≥ (1 − κ) p0 . Proof. Part (a) is immediate from the definition of A. 1 with the following definitions: Let m = ui−1 , and let a1 , . . , am be the quantities of processors assigned to each task, sorted in ascending order. , a1 is the least number of processors assigned to a task, a2 is the next least quantity of processors, etc. (In other words, j a1 ≤ a2 ≤ . . ) Let j = ui . Thus the adversary stops exactly i=1 ai processors. At the beginning of iteration i, the number of processors pi−1 = m m i=1 ai , therefore, the number of surviving processors pi = i=j+1 ai .

We now present an upper bound considering moderate number of crashes (f ≤ p/ log p). 13. The Do-All O AC(n, p, f ) problem can be solved with f ≤ p/ log p using work S = O n + p log fp p . Proof. 1, let ∆f denote the number of processor crashes in this iteration. ) We set p b = b(p, f ) = 2f , and we define S(n, p, f ) to be the work required to solve DoAll O (n, p, f ). Our goal is to show that for all u, p and f , the work S(u, p, f ) AC is no more than 16p + u + p log 2fp (min(u, p)), where u ≤ n denotes the number of undone tasks.

We show that this maximum is attained at f1 = f2 = . . = fr . For simplicity, treat fi as a continuous parameter and consider the factor in the single round work expression (given above) that depends on fi : c/ log( fpi ), where c is the constant hidden by the O(·) notation. p ∂ = c/fi (log p − log fi )2 , c/log The first derivative over fi is ∂fi fi ∂2 p and its second derivative is c/log = 2c/fi2 (log p − log fi )3 − 2 ∂fi fi c/fi2 (log p − log fi )2 . Observe that the second derivative is negative in the domain considered (assuming p > 16).

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