Two simple pooling test methods

We discussed two of the pooling methods, most attractive by virtue of their simplicity [Reference Mallapaty9] in this paper, with focus on the minimum number of tests required for a given infection rate. These two methods were called Method 1 and Method 2 among the four methods by Mallapaty [Reference Mallapaty9].

In Method 1, the population under test is divided into groups of equal numbers of samples (say n in each group). The n samples are pooled together and tested. If a group is tested positive, then the n samples in that group are tested individually.

Taking an example with one positive case in m = 27 samples. By using a group size of n = 9, the first round requires three tests and the second round requires nine tests, making a total of 12 vs. 27 tests without group testing.

In Method 2, the first round is the same as in Method 1, but the positive group is divided into subgroups of size k. Referring to the example of 27 samples, the positive group is divided into three subgroups each containing three samples in the second round. The samples in the positive subgroup are tested individually in the third round of testing. Thus, the total number of tests in the three rounds is 9, further reducing the number of tests.

It should be pointed that in the example above, the number 27 is used to demonstrate the methods in simple numbers. In reality, one positive case in 27 samples is a serious scenario of population infection, meaning that 3.7% of the population is infected. The advantage of pooling test is more pronounced when the infection rate is low, as was shown in the following analysis.

The value of m, that is, the average number of samples that contains one positive case, is determined by the current population infection rate θ, with m = 1/θ. In Method 1, the total number of tests L depends on the value of n. In Method 2, L depends on n and k. A problem of practical importance is the choice of n (and also k in Method 2) to make L a minimum. While the solution for Method 1 has been reported by the same authors [Reference Chow and Chow13], with n = m ^1/2 = (1/θ)^1/2, optimisation for Method 2 has not been reported. Thus, it is the aim of this short communication to show how the minimum number of tests required can be achieved via suitable choice of group size in Method 2, and to compare its efficiency with Method 1.

The two pooling test methods discussed above are summarised in the flowcharts in Figure 1.

Fig. 1.

Flowcharts summarizing the procedures in Method 1 and Method 2 in pooling test.

Optimisation

Let P be the population under test, θ be the current population infection rate. Then most probably there will be one positive case in m = 1/θ people. Further, let n be the group size in the first round and k be the subgroup size in the second round. The total number of tests L is given by

$$L = \displaystyle{P \over n} + \displaystyle{P \over m}\left({k + \displaystyle{n \over k}} \right), \;$$

First-order partial derivatives:

$$\displaystyle{{\partial L} \over {\partial k}} = \displaystyle{P \over m}\left({1-\displaystyle{n \over {k^2}}} \right), \;\displaystyle{{\partial L} \over {\partial n}} = {-}\displaystyle{P \over {n^2}} + \displaystyle{P \over {km}} = P\left({\displaystyle{1 \over {mk}}-\displaystyle{1 \over {n^2}}} \right), \;$$

First-order condition for critical points:

$$\displaystyle{{\partial L} \over {\partial k}} = \displaystyle{{\partial L} \over {\partial n}} = 0, \;$$

(1)$$\displaystyle{P \over m}\left({1-\displaystyle{n \over {k^2}}} \right) = 0, \;$$

(2)$$P\left({\displaystyle{1 \over {mk}}-\displaystyle{1 \over {n^2}}} \right) = 0.$$

Solving for (1), one has

(3)$$n = k^2.$$

Putting (3) into (2) yields

$$n^2 = mk, \;k^4-mk = 0.$$

Solving for k, one has

(4)$$k = \root 3 \of m.$$

Putting (4) into (3) yields

(5)$$n = m^{2/3}.$$

The critical point is $k = \root 3 \of m , \;n = m^{2/3}.$

Second-order partial derivatives:

(6)$$\displaystyle{{\partial ^2L} \over {\partial k^2}} = \displaystyle{{2Pn} \over {mk^3}}, \;\displaystyle{{\partial ^2L} \over {\partial n^2}} = \displaystyle{{2P} \over {n^3}}, \;\displaystyle{{\partial ^2L} \over {\partial k\partial n}} = \displaystyle{{\partial ^2L} \over {\partial n\partial k}} = {-}\displaystyle{P \over {mk^2}}.$$

Second-order condition for checking critical points:

Hessian matrix D is defined by:

(7)$$D = \left[{\matrix{ {\displaystyle{{\partial^2L} \over {\partial k^2}}} & {\displaystyle{{\partial^2L} \over {\partial k\partial n}}} \cr {\displaystyle{{\partial^2L} \over {\partial n\partial k}}} & {\displaystyle{{\partial^2L} \over {\partial n^2}}} \cr } } \right] = \left[{\matrix{ {\displaystyle{{2Pn} \over {mk^3}}} & {-\displaystyle{P \over {mk^2}}} \cr {-\displaystyle{P \over {mk^2}}} & {\displaystyle{{2P} \over {n^3}}} \cr } } \right].$$

At critical point $k = \root 3 \of m , \;n = m^{{2 \over 3}}, \;D{\rm \;}$is given by:

(8)$$D = \left[{\matrix{ {\displaystyle{{2P} \over {m^{4/3}}}} & {-\displaystyle{P \over {m^{5/3}}}} \cr {-\displaystyle{P \over {m^{5/3}}}} & {\displaystyle{{2P} \over {m^2}}} \cr } } \right].$$

Sign-checking of |D|, ∂²L/∂k ² and ∂²L/∂n ² at critical point $k = \root 3 \of m , \;n = m^{{2 \over 3}}$:

(9)$$\vert D \vert = \left\vert {\matrix{ {\displaystyle{{2P} \over {m^{4/3}}}} & {-\displaystyle{P \over {m^{5/3}}}} \cr {-\displaystyle{P \over {m^{5/3}}}} & {\displaystyle{{2P} \over {m^2}}} \cr } } \right\vert = P\left\vert {\matrix{ {\displaystyle{2 \over {m^{4/3}}}} & {-\displaystyle{1 \over {m^{5/3}}}} \cr {-\displaystyle{1 \over {m^{5/3}}}} & {\displaystyle{2 \over {m^2}}} \cr } } \right\vert = P^2\left({\displaystyle{4 \over {m^{10/3}}}-\displaystyle{1 \over {m^{10/3}}}} \right) = \displaystyle{{3P^2} \over {m^{10/3}}} > 0, \;$$

(10)$$\displaystyle{{2P} \over {m^{4/3}}} > 0, \;\displaystyle{{2P} \over {m^2}} > 0.$$

Thus, the critical point $k = \root 3 \of m , \;n = m^{{2 \over 3}}$ is a local minimum point.

The minimum required number of tests (L _min) in terms of m at the critical point $k = \root 3 \of m , \;n = m^{{2 \over 3}}$ is given by

(11)$$L_{{\rm min}} = \displaystyle{P \over {m^{2/3}}} + \displaystyle{P \over m}( {2\root 3 \of m } ) = \displaystyle{{3P} \over {m^{2/3}}}.$$

Comparison of the two methods

A comparison of the two methods is given in Table 1 (for a population of P) at a population infection rate of θ and m = 1/θ.

Table 1.

Comparison of the number of tests using different methods

Numerical values are taken with reference to the observed number of infection cases. Taking a higher population infection rate of identifying 1 AC in a building instructed to have mandatory screening of 1000 residents θ = 0.001, m = 1/θ = 1000.

Then $k = \root 3 \of m$ = 10, and n = $m^{{2 \over 3}}$ = 100. For a population P of 10 million or 10 × 10⁶, the total number of tests using Method 2 is given by Eq. (11):

$$L = \displaystyle{{3P} \over {m^{2/3}}} = \displaystyle{{3 \times 10 \times {10}^6} \over {100}} = 0.3 \times 10^6.$$

Thus the minimum number of tests required using Method 2 is only 0.03 of the total number of tests without grouping test. For comparison, when using Method 1, the minimum number of tests required is equal to 0.63 × 10⁶, which is 0.063 that without group testing. Method 2 requires less than one-half of the number of tests in Method 1.

(12)$${\rm In\ general, \;}\,\,\displaystyle{{L_{2min}} \over {L_{1min}}} = \displaystyle{3 \over 2}m^{-{1 \over 6}} = \displaystyle{3 \over 2}\theta \, ^{{1 \over 6}}.$$

As reported by the Hong Kong Special Administrative Region Government [Reference Cheung, Cheung and Lam16, 24], there are four waves of infection up to early April 2021 since early 2020. As shown in Figure 2 on the number of daily infection cases N in Hong Kong [24], the infection rate θ (related to N) could be regarded as 150 in 7 million people (or 0.0002) in the second wave. As m is related to 1/θ and θ is N/7 × 10⁶, 1/N is also plotted in each wave of infection in Figure 2.

Fig. 2.

Number of confirmed cases reported by the Hong Kong Government.

Equation (12) clearly shows that the advantage of Method 2 over Method 1 in reducing the number of tests becomes more prominent as the population infection rate decreases. That is to say, the saving in resources is larger when the infection rate is low. The dependence of ratio L _min/L_o, where L_o is the number of tests without pooling, on the infection rate θ is shown in Figure 3 for Method 1 and 2. The comparison of L _min for Methods 1 and 2 is shown in Figure 4. The curves in these figures clearly show the advantage of using Method 2 over Method 1, especially for small population infection rate θ.

Fig. 3.

Minimum number of tests required in the two pooling tests relative to that without pooling.

Fig. 4.

Comparison of the two pooling tests for different infection rates θ.

Note that in practical cases, $\root 3 \of m$ and $m^{{2 \over 3}}$ may not be integers, and the integers nearest to $\root 3 \of m$ and $m^{{2 \over 3}}$ are to be chosen as k and n.

It should also be pointed out that in the optimisation above, only the mathematical aspect is considered. In practice, other aspects have to be considered. For example, as the group size increases, the reliability of the test results might be compromised due to dilution, and this will increase the likelihood of false-negative test results [Reference Mutesa8], especially for carriers with low virus load. A balance must be made between increasing the group size and retaining test reliability. In this respect, FDA has provided detailed guidelines for employing pooled sample testing for COVID-19 [25]. The pooling test method is effective and efficient when the positivity rate is low [25].

The robustness of a pooling test method is closely related to the sensitivity and specificity of the test used. In this respect, there is a maximum allowable pool size n _max below which the pooling test may be considered as robust [Reference Bish26]. Thus to achieve robustness, the optimal pool size to be used should be determined, not only from mathematical optimisation, but in conjunction with the limiting value imposed by n _max. To minimise the influence arising from the uncertainty in prevalence or infection rate, Bish et al. [Reference Bish26] also developed a robust pooling strategy via updating pool sizes each week, for each risk group, based on prior week's testing data. This is a more sophisticated approach, though more complex in implementation.

While pooling test is an effective strategy for identifying asymptomatic carriers in mass screening test, symptomatic patients should be dealt with in a different manner in terms of testing for confirmation. In this respect, the point-of-care-testing strategy has been proposed for saving time relative to mass pooling test [Reference Chow and Chow14, Reference Zavascki27]. Point-of-care (POC) tests are diagnostic tests performed at or near the place where a specimen is collected, such as in hospitals or clinics. Thus transport time of samples to laboratory is minimal, the test is individual, and the symptomatic patient is treated on the spot (POC). Of course, as a priori of this strategy, there should be qualified personnel to initially identify the suspected patients before POC testing.