Group testing with nested pools

In order to identify the infected individuals of a population, their samples are divided in equally sized groups called pools and a single laboratory test is applied to each pool. Individuals whose samples belong to pools that test negative are declared healthy, while each pool that tests positive is divided into smaller, equally sized pools which are tested in the next stage. This scheme is called adaptive, because the composition of the pools at each stage depends on results from 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 $1-3^{-1/3}$ it is best to test each sample (no pooling). If $p<1-3^{-1/3}$, we compute the mean $D_k(m,p)$ and the variance of the number of tests per individual as a function of the pool sizes $m=(m_1,dots,m_k)$ in the first $k$ stages; in the $(k+1)$-th stage all remaining samples are tested. The case $k=1$ was proposed by Dorfman in his seminal paper in 1943. The goal is to minimize $D_k(m,p)$, which is called the cost associated to~$m$. We show that for $pin (0, 1-3^{-1/3})$ the optimal choice is one of four possible schemes, which are explicitly described. For $p>2^{-51}$ we show overwhelming numerical evidence that the best choice is $(3^ktext{ or }3^{k-1}4,3^{k-1},dots,3^2,3 )$, with a precise description of the range of $p$s where each holds. We then focus on schemes of the type $(3^k,dots,3)$, and estimate that the cost of the best scheme of this type for $p$, determined by the choice of $k=k_3(p)$, is of order $Obig(plog(1/p)big)$. This is the same order as that of the cost of the optimal scheme, and the difference of these costs is 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.