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Table 1 Choosing the optimal value for the number of pools per layer, q

From: A new pooling strategy for high-throughput screening: the Shifted Transversal Design

q Γ k v gain
≤ 13 ≥ 3 ≥ 16 k > q+1, can't use these values
17 3 16 272 36.8
19 3 16 304 32.9
23 2 11 253 39.5
29 2 11 319 31.3
... 2 11 ... ...
97 2 11 1067 9.4
101 1 6 606 16.5
  1. This table shows the gains obtained with various q values, when the total number of variables to be tested is n = 10000 and the number of expected positives is t = 5, in a noiseless experiment (E = 0). Γ is the compression power (i.e. logarithm of n in base q, see Preliminaries in Results(1) section), k is the number of layers, v is the number of pools (i.e. k·q), and the gain is defined as n/v. By construction, STD requires k ≤ q+1; and to guarantee the identification of t positives while correcting E errors, section 3.3 showed that we must choose k = t·Γ+2·E+1; in this example, k = 5Γ+1. Often, the smallest useable q (i.e., satisfying k ≤ q+1), qmin, yields the highest gain, but this is not always the case. In this example, qmin = 17, but q = 23 (smallest q such that Γ = 2) yields the highest gain: 39.5.