An approach for joint optimization of probabilistic group test based on cost and time value: taking nucleic acid detection of COVID-19 as an example

Abstract

In recent years, the world has encountered many epidemic impacts caused by various viruses, COVID-19 has spread and mutated globally since its outbreak in 2019, causing global impact. Nucleic acid detection is an important means for the prevention and control of infectious diseases. Aiming at people who are susceptible to sudden and infectious diseases, considering the control of viral nucleic acid detection cost and completion time, a probabilistic group test optimization method based on the cost and time value is proposed. Firstly, different cost functions to express the pooling and testing costs are used, a probability group test optimization model that considers the pooling and testing costs is established, the optimal combination number of samples for nucleic acid testing is obtained, and the positive probability and the cost functions of the group testing on the optimization result are explored. Secondly, considering the impact of the detection completion time on epidemic control, the sampling ability and detection ability were incorporated into the optimization objective function, then a probability group testing optimization model based on time value is established. Finally, taking COVID-19 nucleic acid detection as an example, the applicability of the model is verified, and the Pareto optimal curve under the minimum cost and shortest detection completion time is obtained. The results show that under normal circumstances, the optimal combination number of samples for nucleic acid detection is about 10. Generally, 10 is used to calculate for the convenience of organization, arrangement and statistics, except for cases where there are special requirements for testing cost and detection completion time.

Introduction

The group testing method was first used in the USA during World War II. To ensure the safety and effectiveness of military personnel, medical institutions conducted rigorous medical examinations to prevent carriers of deadly infectious diseases from joining the army. In order to reduce cost and time consumption, this method was proposed by the medical staff team (e.g., Dorfman 1943). The blood sample is divided into two parts: one part is added to a group for mixing test, and the other part is used as standby. If the test result in this group is positive, the spare blood samples of each individual in this group need to be tested separately. In the later research process, the group test method has not only been applied to the screening of various infectious diseases, but also can be used to detect battery necrosis, detect regional water pollution and determine potato planting density (e.g., Aprahamian 2018; Chiang et al. 2010). In recent years, regional public health incidents have occurred frequently, directly threatening human life and hindering social and economic development. In 2020, COVID-19 wreaked havoc around the world, with significant impacts. Take China as an example. At the beginning of the outbreak, in order to ensure that the virus does not spread widely, the whole country implemented closed management. Most people were required to quarantine at home, which directly led to a severe slowdown in economic growth. With the epidemic has been under regular prevention and control, measures of early detection, early reporting, early isolation, and early treatment have been implemented, targeted prevention and control in communities have been strengthened, and the scope of detection has been expanded. For sporadic cases and clusters of epidemics in Beijing, Qingdao, Dalian, Xinjiang, and other places, we should find out and extinguish them together, and carry out large-scale nucleic acid testing in areas where confirmed cases have occurred.

With the development of group testing methods, there are probabilistic group tests, combinatorial group testing, and other types. Dorfman's infectious disease screening problem uses the classical probabilistic group test (e.g., Dorfman 1943). Many scholars have already explained the probability group testing; they used multi-level thinking and dichotomy to perfect the group testing model (e.g., Chen 2000; Yin 2001; Zheng 2009; Chen 2001). The method of pooling cost and testing cost is proposed. These two costs are of practical significance in the real-life group testing process and are more maneuverable (e.g., Turner and Danny 2009). In a specific situation, a multistage group test procedure is proposed. When the procedure is used in low-prevalence populations with lower accuracy kits, the sensitivity of detection can be improved (Kennedy 2004).

In addition, there are other types of group testing. Consider the optimal configuration of the square array group testing algorithm to reduce the number of tests, and illustrate the results through the HIV-infected specimens of Marvela (e.g., Hudgens 2011). A design is proposed to accommodate the error of any prespecified nominal test sensitivity and specificity values (e.g., Huang et al. 2020).

The COVID-19 nucleic acid test is very different from the US military's virus carrying test during World War II. The COVID-19 nucleic acid test has a larger scale, and there are certain requirements for the completion time of the test. The faster the epidemic controlled, the faster the regional economy and residents' production and life can recover. Therefore, on the basis of considering the cost control and time value, the optimization problem of the group test of the nucleic acid detection probability of COVID-19 is studied (e.g., Zhu et al. 2020; Tao et al. 2021; Melin et al. 2021). These studies focus on the optimization of the group test method process in order to reduce the time of tests under the premise of accurate testing. These methods are widely used in tests which do not pay attention to timeliness. However, in an epidemic caused by a rapidly spreading virus, without considering the detection time in the process, it may not be able to control the spread of the virus in time and achieve the purpose of controlling the spread of the epidemic.

When a highly infectious virus appears in a certain region, the flow of people greatly increases the risk of the spread of the virus (e.g., Konstantinos et al. 2021; Nikolay et al. 2021; Amaro et al. 2021; Fanelli and Piazza 2020), which threatens the lives of residents in the region and impedes the economic and social development of the whole region. Based on the anti-epidemic experience of Wuhan (e.g., He et al. 2021), in China, once the infected person of the COVID-19 appears in a certain area, the infected person will be sent to the designated hospital for treatment promptly, and the affected area will be closed. Then immediately start the virus detection and conduct a carpet search for virus-infected people to avoid a larger spread. In addition, under normal circumstances, large-scale mixed detection of viral nucleic acid is usually free. So, it is of great significance for government public health management to study how to reduce the cost of viral nucleic acid detection while improving detection efficiency. Using the probabilistic group test method that considers the cost control and time value, several samples are tested in groups according to scientific and reasonable quantities. The time and cost of average single sample in virus nucleic acid detection are optimized. In this process, it is known that the detection of viral nucleic acid on the combined sample has no effect on the sensitivity of the detection.

The cost of group testing includes the cost of pooling and testing, the cost of preliminary preparation for viral nucleic acid detection, and the cost incurred during the sampling process are combined costs. It corresponds to the first three modules in Fig. 1. The cost incurred during the detection process is the cost of detection. It corresponds to the fourth module in Fig. 1.

Fig. 1.

Process of nucleic acid detection of COVID-19

The principle of the conditional probability model is to use the mutual calculation relationship between the newly diagnosed number and the newly infected number, and the maximum likelihood estimation method is used to obtain the estimated value of the expected value of the daily newly infected number. Introduce the Bootstrap method to give the corresponding confidence interval, to further calculate the number of existing infections (undiagnosed) and predict the trend of the number of newly diagnosed people. Use the method to estimate the number of positive samples, estimate the probability of positive samples, and calculate the group test of the best combination number (e.g., Ding et al. 2020).

There are two methods for group testing. One is to sample several people separately at the sampling point and put them in the same sampling tube. This mode is called “mixed sampling”; the other is to test in laboratory testing, the fixed combined testing samples are mixed together in the same volume. This mode is called “mixed testing.” This study does not retain backups and uses the “mixed sampling” method.

In summary, on the basis of considering the regional risk level, detection cost and completion time, study how to achieve the shortest detection completion time in higher-risk areas and how to achieve the lowest detection cost in lower-risk areas. The ultimate goal is to effectively allocate resources, restore social order as soon as possible, resume work, production, and school, and restore the regional economy. In order to establish a better model, the following hypotheses are proposed: (1) Regardless of human and resource constraints, sampling and testing can be performed continuously for 24 h, and the two tasks can be performed simultaneously; (2) the number of combined samples is not considered, however; (3) in the same area, the sampling ability of sampling points per unit time is the same; and (4) the same virus nucleic acid detection tool is used in the same area, and medical expenses and testing expenses are fixed in the same area.

The model

Symbol definition

N ≡ the total number of samples for nucleic acid testing; k ≡ number of combined samples tested in group test; I (N, k) ≡ N divided by the integer part of k; L ≡ L = N−k*I(N, k), the remaining part after grouping can also be expressed as N mod k; C₁(k) ≡ pooling cost of group test; C₂(k) ≡ cost of viral nucleic acid detection for group testing; C₃ ≡ the cost of virus nucleic acid detection for a single sample, C₃ = a; C₁(L) ≡ the pooling cost of the remaining group; C₂(L) ≡ cost of viral nucleic acid detection for the remaining group; A₁ ≡ sampling capability, it indicates the number of samples per unit time at the sampling point; A₂ ≡ detection capability, it indicates the number of samples tested by the testing agency per unit time; p ≡ randomly select the probability of a certain group of measuring units; q ≡ q = 1−p.

Optimal model considering pooling cost and group testing cost control

Large-scale viral nucleic acid detection often uses a mixed sampling method without backing up the samples. Once a mixed sample is tested positive, it is necessary to resample all the combined samples of this group of mixed samples and then perform viral nucleic acid testing separately. When the probability of a positive test is a too large relative to the total number of test samples, more samples will be resampled and retested, and it is not suitable to use a probability group test method.

The total cost of viral nucleic acid testing includes the combined cost and testing cost of the integer group and the remaining group, plus the cost of retesting. The integer group is the integer part of the total number of test samples divided by the number of combined samples. The pooling cost of the integer group refers to the various labor costs, resource costs, and medical expenses in the process of gathering this part of the sample. The testing cost of the integer group refers to the cost of testing this group of samples. The remaining group is the group formed when the total number of test samples divided by the number of combined samples has a remainder. The combined cost of the remaining group is the various expenses incurred when combining the remaining groups. The test cost of the remaining group refers to the test cost of the remaining group. The following explains the optimization study of probabilistic group trials considering combination cost and mixed detection cost control.

When k = 1, it is equivalent to a single sample of viral nucleic acid detection, and the total cost is the cost of a single test multiplied by the total number of samples tested:

$$C=C_3·Nk=1$$ 1

When k ≥ 1, the calculation cost includes two stages. In the first stage, all the tested samples are negative and the normal group is tested. In the second stage, if there are positive samples in the combined sample, the combined sample will be tested again. Probability is involved in the second stage process. Let p be the probability of a specific event occurring, and q the probability that the event will not occur. Now define this event as a positive is found of viral nucleic acid test. The probability of a single sample being negative is q¹, the probability of a combined sample being all negative is q^k, and the probability that all samples in the remaining group are negative is q^{N−kI (N, k)}. Therefore, a probabilistic group test model considering combination cost and mixed detection cost control is obtained:

$$C_{\text{pooling}}\equiv \left\{\begin{array}{cc}I(N,k)C_1(k)& ifN/k=I(N,k)\\ I(N,k)C_1(k)+C_1(L)& ifN/k>I(N,k)\end{array}\right)$$ 2

$$C_{\text{testing}}\equiv \left\{\begin{array}{cc}I(N,k)C_1(k)& \text{if}N/k=I(N,k)\\ I(N,k)C_2(k)+C_2(L)& \text{if}N/k>I(N,k)\end{array}\right)$$ 3

If the samples are all negative, Eq. (2) represents the pooling cost of the integer group and the remaining group, and Eq. (3) represent the testing cost of the integer group and the remaining group. The sum of the above two formulas is the final viral nucleic acid detection cost, which are:

$$C_I=I(N,k)\left(C_1(k)+C_2(k)\right)N/k=I(N,k)$$ 4

$$C_L=I(N,k)\left(C_1(k)+C_2(k)+C_1(L)+C_2(L)\right)N/k>I(N,k)$$ 5

Equation (4) is the final viral nucleic acid detection cost obtained when all samples in the integer group are negative, and Eq. (5) is the final viral nucleic acid detection cost obtained when all samples in the remaining group are negative.

After the test result shows a positive sample, in order to quickly screen out other positive infections in the area, a large-scale virus nucleic acid detection is performed. The probability model is used to obtain the estimated number of people who will be retested individually, so as to calculate the cost of retesting the positive pooling sample in the second stage:

$$E_I(C_a)=\left(1-q^k\right)I(N,k)k{C}_{3}N/k=I(N,k)$$ 6

$$E_L(C_a)=\left(1-q^k\right)I(N,k)k{C}_3+\left(1-q^L\right)L{C}_{3}N/k>I(N,k)$$ 7

Equation (6) is the expected cost of retesting for integer groups, and Eq. (7) is the expected cost of retesting for remaining group. The sum of the cost of group testing and the cost of retesting is the final expected cost:

$$E_I(C)=I(N,k)\left(C_1(k)+C_2(k)\right)+\left(1-q^k\right)I(N,k)k{C}_{3}N/k=I(N,k)$$ 8

$$\begin{array}{cc}\hfill E_L(C)=& I(N,k)\left(C_1(k)+C_2(k)\right)+C_1(L)+C_2(L)+\left(1-q^k\right)I(N,k)k{C}_3\hfill \\ \hfill & +\left(1-q^L\right)L{C}_{3}N/k>I(N,k)\hfill \\ \hfill \end{array}$$ 9

Equation (8) is the expected cost of the integer group, and Eq. (9) is the expected cost of the remaining group. Use the average cost E (Total Cost) / N to select the number of combined sample k corresponding to the lowest cost. If the average cost under the combined sample number k is less than the single test cost C₃, then the group test is meaningful. We can simplify the formula in order to calculation convenience. Let C₁(k) + C₂(k) = C(k), let C₁(L) + C₂(L) = C(L), C₃ = a. The average cost of the group test is:

$$E=\frac{I(N,k)C(k)+C(L)+(1-q^k)I(N,k)ka+(1-q^L)La}{N}$$ 10

The above formula must meet the constraint of 1 ≤ k ≤ N. Equation (10) can be expressed as:

$$f(k)=a+\frac{I(N,k)C(k)+C(L)+\left(1-q^k\right)I(N,k)ka+\left(1-q^L\right)La}{N}$$ 11

Let Eqs. (11) be:

$$F(k)\equiv \left\{\begin{array}{cc}0,& \text{if}k=1\\ I(N,k)C(k)+C(L)-a\left[kI(N,k)q^k+q^{L}L\right],& \text{if}1,k<N\end{array}\right)$$ 12

Then proceed to the proof of probabilistic group test optimization considering combination cost and mixed detection cost control:

The first thing to determine is that $\text{k}\text{I}\text{(}N,\text{k})$ is positive and $\text{N}-\text{k}\text{I}\text{(N},\text{k})$ is also positive. For the convenience of proof, the following settings are made.

$$A_1(k)=kI(N,k)q^k+q^{L}L>0$$

$$A_2(k)=I(N,k)C(k)+C(L)\ge 0$$

Proposition 1

The optimal group size is k^∗  > 1 if and only if there exists an integer k > 1 such that F(k) > 0.

Proof

This follows immediately from Eq. (12) because the relative expected cost is a when k = 1.

Proposition 2.

Suppose that the cost function C(k) is nondecreasing for 2 ≤ k ≤ N. If there is no integer k > 1 such that aA₁(k) > C (2), then the optimal group size is k ∗  = 1. Moreover, if there exists an integer k’ > 1 such that aA₁(k’) > (1 + N/2) C(N), then the optimal group size is k^∗  > 1.

Proof

First, observe that C(k) ≥ C(L) because k > N-kI (N, k). Also, we have that1 ≤ I (N, k) ≤ I(N, 2) because 2 ≤ k ≤ N. Therefore, I (N, k) C(k) ≤ A₂(k) ≤ (1 + I (N, k)) C(k). Moreover, C(2) ≤ C(k) ≤ I(N, k) C(k) and (1 + I(N, k))C(k) ≤ (1 + N/2)C(N) yield that C(2) ≤ A₂(k) ≤ (1 + N/2)C(N), from which we obtain C(2)-aA₁(k) ≤ F(k) = A₂(k)- aA₁(k) ≤ (1 + N/2)C(N)-aA₁(k). The result follows from the last sequence of inequalities.

Proposition 3

Suppose C(k)is nonincreasing for 2 ≤ k ≤ N. If aA₁(k) > C(N)fails to hold for all k > 1, then the optimal group size is k^∗  = 1, and if there exists an integer k’ > 1 for which aA₁(k’) > (1 + N/2) C (2), then the optimal group size is k^∗  > 1.

Proof

Note from the definition of A₂(k) that I (N, k) C(k) ≤ A₂(k). When k ≥ 2, then C(N) ≤ C(k) ≤ I (N, k) C(k) and C(L) ≤ C (2), from which we have that C(N) ≤ A₂(k) ≤ (1 + N/2) C (2). This fact implies that C(N)-aA₁(k) ≤ F(k) = A₂(k)-aA₁(k) ≤ (1 + N/2) C(N)-aA₁(k). The proposition follows immediately from the last set of inequalities.

Proposition 4

Suppose m ≤ C(k) ≤ M for 2 ≤ k ≤ N, where m and M are positive constants. If there is no integer k > 1 such that aA₁(k) > m, then the optimal group size is k^∗  = 1, and if there does exist an integer k’ > 1 such that aA₁(k’) > (1 + N/2) M, then the optimal group size is k ∗  > 1.

Proof

The result follows by noting that m- aA₁(k) ≤ F(k) = A₂(k)- aA₁(k) ≤ (1 + N/2) M-aA₁(k).

Two-objective optimization of probabilistic group test based on group test cost and time value

The epidemic has a huge impact on economic and social development. With the rapid response of relevant departments and scientific prevention and control, the epidemic has been gradually brought under control. In order to restore regional economic and social life as soon as possible, large-scale viral nucleic acid testing is carried out in areas where confirmed cases or asymptomatic infections have occurred. In this process, not only cost control is required, but also the testing completion time is required to be shortened as much as possible. Completing viral nucleic acid testing for all personnel in the relevant area in the shortest possible time will help prevent the spread of the epidemic and restore regional development and people’s daily life. Completing viral nucleic acid testing for all personnel in relevant areas with as little cost as possible will help save government expenditures and reduce financial burdens. Therefore, the completion time of the test and the cost of the group test should be the smallest under the norms of viral nucleic acid detection, and the number of combined samples at which the two targets are both the smallest is the optimal number of combined samples. The following first gives the derivation process of the establishment of the probability group test model considering the time value, and then combined with the probabilistic group test model considering the detection cost, a dual-objective programming model considering the cost and time value is established.

Use the sampling ability and detection ability to calculate the detection time of virus nucleic acid. The unit of these two indicators is per hour. Divide the total sample number detected by the sampling capability to get the sampling time, and divide the total group number by the detection capability to get the detection time. The completion time of the integer group is the sampling time plus the detection time; the completion time of the remaining group is based on the integer group plus the detection time of the remaining group. Thus, the first stage time in the two states is obtained.

$$T_{I(N,k)}=\frac{N}{A_1}+\frac{I(N,k)}{A_2}$$ 13

$$T_L=\frac{N}{A_1}+\frac{I(N,\text{k})+N-kI(N,\text{k})}{A_2}$$ 14

Equation (13) is the completion time when the test results of the integer group are all negative, and Eq. (14) is the completion time when the test results of the remaining groups are all negative. Because the mixed sampling mode is adopted, there is no backup of the test samples, so the positive combination sample must be retested in the second stage. The first stage time plus the time of the positive combination sample resampling and testing is the final test completion time.

$$T_{I\text{final}}=\frac{N+kI(N,k)\left(1-q^k\right)}{A_1}+\frac{I(N,k)+kI(N,k)\left(1-q^k\right)}{A_2}$$ 15

$$\begin{array}{c}{T}_{L\text{final}}=\frac{N+kI(N,\text{k})\left(1-q^k\right)+N-kI(N,\text{k})\left(1-q^{N-kI(N,k)}\right)}{A_1}+\\ \frac{I(N,\text{k})+N-kI(N,\text{k})+kI(N,\text{k})\left(1-q^k\right)+N-kI(N,\text{k})\left(1-q^{N-kI(N,\text{k})}\right)}{A_2}\\ \end{array}$$ 16

Equation (15) is the final detection completion time of the integer group, and Eq. (16) is the final detection completion time of the remaining group.

The purpose of management activities pursued by management is to make full use of limited human, material, and financial resources to optimize the allocation of resources and strive to achieve the greatest efficiency and benefit with the smallest cost. The idea of the Pareto optimal solution is the same. The Pareto curve is a situation where one objective A optimizes another objective B degradation, and Pareto optimal is the solution when there is no way to optimize another objective B without damaging another objective A. The goal of time optimization requires the goal of damage average inspection cost. On the contrary, if you want to optimize the goal of average detection cost, you need the goal of damage completion time.

$$\begin{array}{c}minT=\frac{N+kI(N,k)\left(1-q^k\right)}{A_1}+\frac{I(N,k)+kI(N,k)\left(1-q^k\right)}{A_2}+\\ \left[\frac{L+L\left(1-q^L\right)}{A_2}+\frac{L\left(1-q^L\right)}{A_1}\right]x_0\\ \end{array}$$ 17

$$minE=I(N,k)C(k)+akI(N,k)\left(1-q^k\right)+\left[L\left(1-q^L\right)a+C(L)\right]x_0$$ 18

s.t.:

$$N/k\ge I(N,k)$$ 19

$$1\le k\le N$$ 20

In the formula, a: single virus nucleic acid detection cost; C(k): sum of combined cost and mixed detection cost; C(L): the sum of the remaining group combination cost and the mixed detection cost.

$$x_0=\left\{\begin{array}{cc}1,& N/k\ne 0,\text{There}\text{are}\text{remaining}\text{groups};\\ 0,& \text{other}\end{array}\right)$$

In the above model, Eq. (17) is the optimization goal for the shortest detection completion time, Eq. (18) is the optimization goal for the lowest total cost of detection, and Eq. (19) is the constraint condition. This constraint exists in both cases where there is integer group and there are remaining groups. Equation (20) is the model constraint, that is the maximum number of combined samples in the model can be the total number of test samples. The function of the parameter x₀ introduced in the model is to distinguish whether there is a remaining group in the actual situation where the total number of people is uncertain, x₀ is 1 means that N/k has remaining samples to form the remaining group, and x₀ is 0 means that N/k has no remaining samples.

Results and analysis

When the average cost of the group test is lower than the cost of the single test, the group test is meaningful. If there is no problem of probability and the upper limit of detection, the number of combined samples will be larger. What affects the average cost of group testing is the number of combined samples and the probability of positive samples in the combination. The optimization research of probabilistic group test considering combination cost and mixed detection cost control is applied to real life. This method is generally applicable to areas where large-scale viral nucleic acid detection is performed.

In the entire model, the total number of detected samples, the probability of positive samples, and the cost of virus nucleic acid detection for a single sample are known values in the application scenario and can be adjusted according to the actual situation. The pooling cost is a function that changes with the number of combined samples. When the number of combined samples increases, the cost of the combination increases. The cost function selects a monotonically increasing function, and the increase cannot be too large.

When calculating the cost, convert the actual cost into a unitless number and use different cost functions to calculate, let $C(k)=1,C\left(k\right)=lnk,C\left(k\right)=\sqrt{k}$, the remaining component cost C(L) takes $C(L)=1,C\left(L\right)=lnL,C\left(L\right)=\sqrt{L}·C\left(k\right)=1$ is a comparative cost function, this function is a fixed value. It has no effect on the trend, and it can be more directly observed how the average test cost of the group test is affected by the positive probability of the sample. Because of the detection level and medical limitations, the maximum number of combined samples will not exceed 100, so the maximum number of combined samples is 100. The optimal combined sample number is obtained from the cost function; under this combined sample number the cost is the lowest. Figure 2 shows the optimal combination sample sizes obtained under different cost functions. The solution results of the combined sample sizes of three different cost functions are put together to facilitate the observation of the difference in the optimal combination sample number among the three. Let the total number of people in the test N = 300, the probability p = 0.03, and the single test cost C₃ = 0.5. The three cost functions are solved separately with the programming language C +  + , and the solution results under these three cost functions are obtained:

Fig. 2.

Comparison of solving results of three types of group test cost functions

From the results in Fig. 2, we can see that under different cost functions, the number of optimal combination samples is different. When C $\left(\text{k}\right)\text{=}\sqrt{\text{k}}$, the number of optimal combination samples is 15, and the average cost is 0.44175. Saving (0.5–0.44175) × 300 = 17.475; when C $\left(\text{k}\right)=\text{lnk}$, the optimal combination sample number is 12, and the average cost is 0.3601. The total saving is (0.5–0.3601) × 300 = 41.97; C $\left(\text{k}\right)=\text{1}$, the optimal combination sample number is 10, and the average cost is 0.2313, which is a total saving of (0.5–0.2313) × 300 = 80.61.

Different cost functions have different effects on the average test cost in the group test state, but the trend is basically the same. When the number of combined samples is relatively low, the average test cost of the group test is also larger, but as the number of combined samples becomes larger, the average detection cost gradually decreases. When the number of combined samples is a certain value, the smallest average detection cost of the group test is obtained. After that, because the number of combined samples becomes larger, the probability of positive samples in the combined sample becomes higher, which makes the sample for retesting become more, the average inspection cost of the group test will increase accordingly.

Let the total number of detection N = 500,000, the probability p = 0.001, and the single test cost C₃ = 0.5. The three cost functions are solved by the programming language C +  + , and the solution results under these three cost functions are obtained:

It can be seen from the results in Fig. 3 that the sample positive probability and the total number of tested samples affect the number of combined samples. When the total number of detected samples is small and the relative probability of positive samples is greater, there are an optimal number of combined samples; when the total number of test samples increases, the probability of positive samples is low, and there is no restriction on the number of combined samples. Then the optimal number of combined samples will increase as the limit on the number of combined samples is released, and the average combined test cost gradually tends to a fixed value.

Fig. 3.

Change in the estimated probability of sample positive and the average test cost of the group test after changing the total test sample

Figure 4 shows the change of average detection cost after changing the total number of detection personnel. It can be seen that under different cost utility functions, the optimal number of mixed samples is different. When the total number of people tested changes, the impact on the optimal number of mixed samples is small, but the cost savings will also increase as the number of people increases.

Fig. 4.

Change in the total number of people tested and the average detection cost after changing the total number of detection personnel

Viral nucleic acid detection needs to be completed within a specified time, and high-risk areas often have higher requirements for testing time. However, the sampling capability and detection capability of different regions are different. The sampling capabilities and detection capabilities of regions with developed medical levels are obviously higher than those in regions with underdeveloped medical levels, and the detection time will be faster.

Using actual data in a certain area, the shortest completion time and the optimal number of combined samples under the three types of estimated sample positive probabilities are obtained. The total number of tested samples N = 245,000, sampling capability A₁ = 2000, detection capability A₂ = 980, and the estimated positive probability of the sample is taken as p = 0.03, p = 0.003, and p = 0.0003, respectively, using the programming language C +  + to solve the results. The results are as follows:

In the above conditions in Fig. 5, when the estimated sample positive probability is p = 0.03, there is an optimal combined sample number k = 5, and the shortest completion time is 224 h. When the estimated sample positive probability is p = 0.003, there are five combined samples with the same shortest completion time. They are 14, 15, 16, 17, 18, and the shortest completion time is 154 h. When the estimated sample positive probability is p = 0.0003, there are seven combined samples with the same shortest completion time, which are 42, 43, 44, 51, 52, 53, 54, and the shortest completion time is 131 h.

Fig. 5.

Completion time under different estimated probabilities of being tested positive

In order to get more accurate results in the face of practical problems, take hour as the time unit in the above calculation. In the previous problem description, it is proposed that the sampling and testing can be carried out 24 h without considering the human and resource constraints. However, in actual situations, only in very urgent situations, around-the-clock non-stop virus nucleic acid detection activity will be carried out. Therefore, when using the model in different regions, the actual working time per day plus the extra part of the assumed working time should be used in the final total time. For example, according to the above model, it takes 131 h to complete the test, but the working time in this area is 12 h; 12 h should be added to every 12 h of the 131 h of completion time, so the actual completion time should be 251 h.

The total time used for sampling cannot be optimized by the number of combined samples. The total number of tested samples N (people) is divided by the sampling capability A₁ (people/h) = 245,000/2000 = 122.5 (h), the 122.5 is the total number of tested samples and the base value of the sampling capability. In other regions where the total number of tested samples and sampling capability are different, this value will change. The number of samples in the optimal combination is also changing under the influence of the positive probability of different samples. As the positive probability of the sample decreases, the number of samples in the optimal combination gradually increases, and the change in completion time gradually decreases. There are multiple optimal combination samples. In this case, if there is no cost requirement, you can choose any combination of samples. If there is a cost requirement, the time function and cost function should be formed into a dual-objective optimization and the optimal solution should be selected.

In the previous article, we set different cost utility functions to reach the conclusion that the change trend of detection cost is consistent. To observe the test completion time under different estimated positive probabilities proves that there is an optimal value of time. Under realistic conditions, both of them are important factors that cannot be ignored in epidemic prevention and control. Therefore, we need to consider both of them together to draw a model for our final research. The dual-objective optimization plan based on the cost and time value of the probability group test is a method to minimize the test cost under the shortest test completion time. Use the corresponding number of combined samples in the probability group test model when the time value is considered separately, to find the minimum cost in the probability group test based on the cost of the group test. This cost model uses the cost function C $\left(\text{k}\right)\text{=}\sqrt{\text{k}},$ to solve the optimal solution. Because the number of combined samples is an integer, you can list the points, and the resulting data generate a Pareto curve graph.

As shown in Fig. 6, The shadow covers a total of eight points, namely (224, 0.51785), (225, 0.49176), (228, 0.47433), (233, 0.46296), (238, 0.4549), (244, 0.44752), (250, 0.44563), (256, 0.44175), the corresponding combined sample number k is 5, 6, 7, 8, 9, 10, 11, 12. These eight points are all Pareto optimal solutions. When a certain area has specific requirements for the test completion time, first select the items that meet the test completion time requirements, and then select the point with the lowest average test cost among the eligible items, and finally get the optimal sample mixture number. For example, it is now required to complete all viral nucleic acid testing within 240 h. From the eight Pareto optimal solutions covered by the shadow, select the point that meets the completion time requirements and the average testing cost is the smallest. This point should be (238, 0.4549), the best number of good samples is 9.

Fig. 6.

Pareto curve of average test cost under completion time and group test state

In addition, when the combined sample number k is greater than 25, the completion time and the average detection cost curve in the group test state are both increasing due to the influence of the positive probability of the sample. The curve after the corresponding combination sample number k is 25 is an increasing trend, that is, the average test cost under the group test state increases as the completion time increases.

Because the combined sample number k is an integer, the detection cost and completion time obtained by the combined sample number k are both discrete. Therefore, it is easier to obtain the data for calculation of the virus nucleic acid detection in different regions, and then plot the Pareto curve, select the Pareto optimal solution, there may be multiple Pareto optimal solutions, and these optimal solutions can all be k*.

Conclusion

To solve the problem of virus nucleic acid detection, a probabilistic group test model based on the cost of group testing and a probabilistic group test model considering the value of time have been established. Use the probability group test method considering the time value and cost control to solve the dual-objective optimal combination sample number. Under the optimal number of combined samples, the minimum cost and the shortest completion time for viral nucleic acid detection are obtained. It provides specific suggestions for the government's public health management to achieve cost reduction and efficiency increase under the normal epidemic situation. At present, the global epidemic prevention and control situation is constantly changing, and COVID-19 virus itself is also changing, so our research cannot be limited in the past. Because of the common characteristics of similar epidemic situation, such as rapid transmission and strong ability, the model established in this article still has certain reference value for the study of variant COVID-19 virus now and even in the future.

There are still some shortcomings. Firstly, when using sampling and detection capabilities to establish probability group test optimization based on time value, for the convenience of calculation, it is assumed that the sampling process and the detection process are carried out without interruption, which is actually not continuous but does not affect for optimization results. An independent function can be considered in subsequent studies to express the sampling completion time that changes with the number of combined samples so that the optimization goal of the model is closer to the real situation. Secondly, the study only involves one round of viral nucleic acid testing, but from a practical point of view, the results of one round of viral nucleic acid testing are not completely reliable. Asymptomatic patients have been tested negative many times, and the impact of asymptomatic samples on combined testing should be studied next. In addition, the number of samples to be mixed has an upper limit. The next step should be to fully investigate the influence of medical equipment and testing reagents on the upper limit of the number of combined samples.

Appendix

The cost function is used to solve the average cost code under each combination sample in the 2.2 part.

Funding

This work was supported by the Social Science Planning Fund of Liaoning Province for Youths (L21CGL006); the National Natural Science Foundation of China (72204034), Dalian Maritime University Talent Research Startup Fund (CN) (02502324).

Data availability

The data used to support the findings of this study are available from the corresponding author upon request.

Declarations

Conflict of interest

The authors declare that there is no conflict of interests regarding the publication of this paper.

Ethical approval

This article does not contain any studies with human participants or animals performed by any of the authors.