 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2004/n/n2/b/iii/solution

During phase A every item is inspected, while during phase B inspection has probability

1/ r

, independently of item quality. The long-run inspected fraction is therefore the stationary reward

I = \sum_{i < K} π_{i} + \frac{π _{K}}{r} = 1 - (1 - \frac{1}{r}) π_{K} .

This can be justified by the Markov chain ergodic theorem for state occupation and the strong law for the independent inspection coins (or the corresponding bounded martingale differences).

For the printed cumulative count this gives

I_{cumulative} = \frac{K p + q}{K p + r q} .

It does not equal the claimed expression in general. For instance

K = 2

p = 1/2

r = 2

gives

I = 3/4

, whereas the printed expression gives

4/5

For the consecutive-clearance rule, substituting

π_{K} = r q^{K} / D

instead gives

I_{consecutive} = 1 - \frac{( r - 1 ) q ^{K}}{D} = \frac{1}{1 + ( r - 1 ) ( 1 - p ) ^{K}} .

Thus the claimed formula is valid for consecutive clearance, and false for the literal cumulative count. For

p = 0

the limiting inspected fraction is

1/ r

in either model; for

p = 1

it is 1 because phase A never clears.

 New to topics? Read the docs here!