By Christian Schmidt
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Example text
K}, set n ← 0, τ ← n0 , β ← 1 − (1 − α∗ )1/(k−1) . 3. WHILE | I | > 1 DO another stage: (a) Observe τ additional samples from system i, for all i ∈ I. Set n ← n + τ . Set τ ← ξ. (b) Update: Set η ← 21 (2β)−2/(n−1) − 1 and h2 ← 2η(n − 1). For all i ∈ I, update the sample statistics x¯i and s2i . (c) Screen: For all i, j ∈ I and i > j, set dij ← x¯j − x¯i and h2 (s2i +s2j ) δ∗ − n . If dij > ij then I ← I\{i}. If ij ← max 0, 2n δ∗ 2 dij < − ij then I ← I\{j}. 4. Return remaining system, system D, as best.
Inoue, Chick, and Chen 1999; Chen, Y¨ ucesan, Dai, and Chen 2005]). The variations involve different approximations for PCSBayes , and different thought experiments for how additional samples might improve the probability of correct selection. Here we specify the idea behind the OCBA and the variations used for this thesis. The OCBA assumes that if an additional τ replications are allocated for system i, but none are allocated for the other systems, then the standard error is scaled back accordingly.
The ‘winner’ then received τ = 1 34 2. 2 The Procedures replication. Usually, at most 3 doublings (τ = 8) were sufficient to select a winner. If there was no clearly defined best because two or more systems whose EVI or EAPCSi − APCS had overlapping intervals but the intervals did not contain 0, then we allocated τ = 1 replication to the system with the highest upper bound for the interval. The performance of 0-11 and LL1 does not appear to be significantly hurt by the collisions, as the curves in the (E[N ], log(1 − PCSiz )) and (E[N ], log EOCiz ) planes appear relatively straight.