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- It has become evident that identifying persons in a population who have the virus before they have symptoms is critical to controlling an outbreak. There is a simple way to drastically reduce the number of tests required: test in groups of G where ##G = 1/\sqrt{p}## and p is the probability that a randomly chosen person will test positive.
Germany has been virus testing in groups of 10. In a high risk population where p=1/100 this requires, on average, 196 tests to test 1000 people. In a moderate risk population, it requires 109 tests to test 1000 people. But if they were to test in groups of 32 in such a population one would need only 63 tests.
The tests are done as follows:
1) swabs containing biological material that would contain the virus if it is present in an individual are taken from each of G individuals
2) the biological material in these swabs, G in number, is extracted from the swabs and all mixed together to create a single uniform composite test sample
3) the composite test sample is tested for the virus
4) if it is positive, all G individuals are tested individually to determine who in the group is positive
5) if it is negative, all G individuals are eliminated as carriers
The number of tests is determined as follows:
Let T = number of tests; p = probability that an individual test will be positive; G=number of individuals in the group; N= number of individuals in the population you want to test; Pgroup = probability that a group of G will test positive
(1) T = N/G + (N/G) x Pgroup x G = N(1/G + Pgroup)
So, for example in a population of 1,000,000 people in groups of 10 the number of tests would be:
T = 1,000,000 (1/10 + Pgroup) = 100,000 + 1,000,000 x Pgroup
Now the probability that a group will test positive is 1 - the probability that all individuals in the group will be negative:
##P_{group} = 1- (1-p)^G##
So (1) becomes:
(2) ##T = N(1/G + 1-(1-p)^G)##
Since p is very small the higher order terms in the expansion of ##(1-p)^G## can be ignored so that to a very close approximation:
##(1-p)^G = (1-p)(1-p)...(1-p) \approx 1 - Gp ##
So (2) becomes:
(3) ##T = N(1/G + Gp)##
T is minimum where dT/dG = 0
##dT/dG = N((-1/G^2) + p)##
So T is minimal where:
(4) ##G = \frac{1}{\sqrt{p}}##
So, if, for example, p = 1/1000, optimal G would be the closest integer to ##\sqrt{1000}## which is 32. Substituting into (3) results in T = 1000/32+32 = 63
I have to thank my mathematician brother Dave for his help in working this out.
AM
The tests are done as follows:
1) swabs containing biological material that would contain the virus if it is present in an individual are taken from each of G individuals
2) the biological material in these swabs, G in number, is extracted from the swabs and all mixed together to create a single uniform composite test sample
3) the composite test sample is tested for the virus
4) if it is positive, all G individuals are tested individually to determine who in the group is positive
5) if it is negative, all G individuals are eliminated as carriers
The number of tests is determined as follows:
Let T = number of tests; p = probability that an individual test will be positive; G=number of individuals in the group; N= number of individuals in the population you want to test; Pgroup = probability that a group of G will test positive
(1) T = N/G + (N/G) x Pgroup x G = N(1/G + Pgroup)
So, for example in a population of 1,000,000 people in groups of 10 the number of tests would be:
T = 1,000,000 (1/10 + Pgroup) = 100,000 + 1,000,000 x Pgroup
Now the probability that a group will test positive is 1 - the probability that all individuals in the group will be negative:
##P_{group} = 1- (1-p)^G##
So (1) becomes:
(2) ##T = N(1/G + 1-(1-p)^G)##
Since p is very small the higher order terms in the expansion of ##(1-p)^G## can be ignored so that to a very close approximation:
##(1-p)^G = (1-p)(1-p)...(1-p) \approx 1 - Gp ##
So (2) becomes:
(3) ##T = N(1/G + Gp)##
T is minimum where dT/dG = 0
##dT/dG = N((-1/G^2) + p)##
So T is minimal where:
(4) ##G = \frac{1}{\sqrt{p}}##
So, if, for example, p = 1/1000, optimal G would be the closest integer to ##\sqrt{1000}## which is 32. Substituting into (3) results in T = 1000/32+32 = 63
I have to thank my mathematician brother Dave for his help in working this out.
AM
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