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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; P

(1) T = N/G + (N/G) x P

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 + P

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; P

_{group}= probability that a group of G will test positive(1) T = N/G + (N/G) x P

_{group}x G = N(1/G + P_{group})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 + P

_{group}) = 100,000 + 1,000,000 x P_{group}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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