Thread: Expected number of blood tests.

It is known that 5% of the members of a population have disease A, which can be discovered by a blood test. Suppose that N (a large number) people are to be tested. This can be done in two ways: (1) Each person is tested separately, or (2) the blood samples of k people are pooled together and analyzed. (Assume that N = nk, with n an integer.) If the test is negative, all of them are healthy (that is, just this one test is needed). If the test is positive, each of the k persons must be tested separately (that is, a total of k + 1 tests are needed).

a For fixed k, what is the expected number of tests needed in option 2?
b Find the k that will minimize the expected number of tests in option 2.
c If k is selected as in part (b), on the average how many tests does option 2 save in comparison with option 1?

i have solve part a...answer is E[X] = n [1 + k(1-(0.95^k)) ]
but i cant solve part b....anyone can help me???

Differentiate that expression of E[X] wrt 'k' and find the root of the differentiated equation. That will give you the required value of 'k'. Here goes the calculation in Wolframalpha using n=10 ( I have taken an integral value, since Wolframalpha won't give the result properly otherwise. You can take a general 'n' while calculating in hand. Observe the steps here. )
Solution

k=0 solves the equation. Hence E[X] will be minimized if k=0.

Originally Posted by Sambit

Differentiate that expression of E[X] wrt 'k' and find the root of the differentiated equation. That will give you the required value of 'k'. Here goes the calculation in Wolframalpha using n=10 ( I have taken an integral value, since Wolframalpha won't give the result properly otherwise. You can take a general 'n' while calculating in hand. Observe the steps here. )
Solution

k=0 solves the equation. Hence E[X] will be minimized if k=0.

If k = 0, how does there exist n such that nk = N?

At the very least you need to replace n in the equation with N/k. I guess you can use calculus, but you have to be careful because you need to choose k so that k divides N.

Well...in that case k=5 (approxly) gives the minimum value.

i got the prove of k=5
although i not really sure izzit correct...
i am just using simple differentiation,
since n is not fixed, so i chang it to N/k which N is a constant to make my differentiation easier...
i get k=4.47, thn i approximate to 5 wo...

Originally Posted by Sambit

Well...in that case k=5 (approxly) gives the minimum value.

And what happens if N is prime?

Originally Posted by theodds

And what happens if N is prime?

According to the problem N can not be prime.

Originally Posted by Sambit

According to the problem N can not be prime.

Of course it can. If N is prime, then n = N and k = 1 is the only possibility, so the k that minimizes the expected number of blood tests is k = 1. N is fixed at the outset of the problem, and we have to find a k such that nk = N for some natural number n and k is chosen to minimize the expected number of blood tests.

Originally Posted by theodds

Of course it can. If N is prime, then n = N and k = 1 is the only possibility, so the k that minimizes the expected number of blood tests is k = 1. N is fixed at the outset of the problem, and we have to find a k such that nk = N for some natural number n and k is chosen to minimize the expected number of blood tests.

Theoretically that's possible. But it is useless to write N=nk when N is prime since in that case k does not play any role.

The point is that you can't just use calculus on the expression given and maximize it in k. You have to make sure that k divides N. You can't decide k = 5 is optimal, because what if 5 doesn't divide N? Maybe N is still composite and some other k other than 1 will do. Maybe the optimal k is the divisor of N that is closest to 5, or something like that, but you need to do further work to find out what is going on.

Originally Posted by theodds

You can't decide k = 5 is optimal...

There is no reason for not choosing k=5 as optimal. Calculus gives the optimal value as 4.47, which means if k=4.47 is taken, it will satisfy the desired criteria for the sample. We then choose k as 5 since k is an integer. You might have asked "why don't you take k=4 as optimal?" The fact is: though 4.47 is closer with k=4 than with k=5, the option of taking k as 4 is rejected since it stands below the optimality criterion that is 4.47.

Originally Posted by theodds

You have to make sure that k divides N

Once you use calculus, there is no way you can put both k and N (or n) in the same equation.

Originally Posted by theodds

The point is that you can't just use calculus on the expression given and maximize it in k....you need to do further work to find out what is going on.............

What else do you suggest?

N is a fixed large number. You can't choose k = 5 if your large number N is 7^5, because you can't divide 7^5 people into groups of 5.

The obvious thing to do is look at the behavior of the function around its minimum and see if you can find a way to extend the result to the values of k you can choose e.g. you might find that the value of k that minimizes the expected number of blood tests is the divisor of N that is closest to 4.47 (this is assuming OP is correct in the derivation of the expected number of tests).

You have to find either n or k. In any case, the value of N changes; that is, the value of N depends on the optimization-- which does not keep N a fixed quantity.

Originally Posted by Sambit

You have to find either n or k. In any case, the value of N changes; that is, the value of N depends on the optimization-- which does not keep N a fixed quantity.

This is not at all in the spirit of the problem...you don't get to choose how many people need to have their blood tested!

Can't get you. What is your method and what does it leave you with?