INVESTIGADORES
FERRARI pablo Augusto
artículos
Título:
Group testing with nested pools
Autor/es:
ARMENDARIZ INES; FERRARI, PABLO A.; FRAIMAN, DANIEL; MARTÍNEZ, JOSÉ MARIO; PONCE DAWSON, SILVINA
Revista:
IEEE TRANSACTIONS ON INFORMATION THEORY
Editorial:
IEEE-INST ELECTRICAL ELECTRONICS ENGINEERS INC
Referencias:
Lugar: Piscataway; Año: 2021
ISSN:
0018-9448
Resumen:
In order to identify the infected individuals of a population, their samples are divided inequally sized groups called pools and a single laboratory test is applied to each pool. Individualswhose samples belong to pools that test negative are declared healthy, while each pool that testspositive is divided into smaller, equally sized pools which are tested in the next stage. Thisscheme is called adaptive, because the composition of the pools at each stage depends on resultsfrom previous stages, and nested because each pool is a subset of a pool of the previous stage.Is the infection probability p is not smaller than 1=3 it is best to test each sample (nopooling). If p < 1=3, we compute the mean Dk(m; p) and the variance of the number oftests per individual as a function of the pool sizes m = (m1; ...;mk) in the rst k stages; in the(k + 1)-th stage all remaining samples are tested. The case k = 1 was proposed by Dorfman inhis seminal paper in 1943. The goal is to minimize Dk(m; p), which is called the cost associatedto m. We show that for p the optimal choice is one of four possible schemes,which are explicitly described. For p > 2^-51 we show overwhelming numerical evidence that thebest choice is (3^k or 3^(k -1)4; 3^k-1; : : : ; 3^2; 3), with a precise description of the range of p´s whereeach holds. We then focus on schemes of the type (3k; : : : ; 3), and estimate that the cost of thebest scheme of this type for p, determined by the choice of k = k_3(p), is of order O(p log(1-p)).This is the same order as that of the cost of the optimal scheme, and the dierence of these costsis explicitly bounded. As an example, for p = 0:02 the optimal choice is k = 3, m = (27; 9; 3),with cost 0:20; that is, the mean number of tests required to screen 100 individuals is 20.