Question 7.15: Consider a manufacturing process that sequentially produces ......
Consider a manufacturing process that sequentially produces items, each of which is either defective or acceptable. The following type of sampling scheme is often employed in an attempt to detect and eliminate most of the defective items. Initially, each item is inspected and this continues until there are k consecutive items that are acceptable. At this point 100% inspection ends and each successive item is independently inspected with probability α. This partial inspection continues until a defective item is encountered, at which time 100% inspection is reinstituted, and the process begins anew. If each item is, independently, defective with probability q,
(a) what proportion of items are inspected?
(b) if defective items are removed when detected, what proportion of the remaining items are defective?
Remark Before starting our analysis, note that the preceding inspection scheme was devised for situations in which the probability of producing a defective item changed over time. It was hoped that 100% inspection would correlate with times at which the defect probability was large and partial inspection when it was small. However, it is still important to see how such a scheme would work in the extreme case where the defect probability remains constant throughout.
We begin our analysis by noting that we can treat the preceding as a renewal reward process with a new cycle starting each time 100% inspection is instituted. We then have
proportion of items inspected =\frac{E[number inspected in a cycle]}{E[number produced in a cycle]}
Let N_{k} denote the number of items inspected until there are k consecutive acceptable items. Once partial inspection begins—that is, after N_{k} items have been produced—since each inspected item will be defective with probability q, it follows that the expected number that will have to be inspected to find a defective item is 1/q. Hence,
E[number inspected in a cycle] =E[N_{k}] +\frac{1}{q}
In addition, since at partial inspection each item produced will, independently, be inspected and found to be defective with probability αq, it follows that the number of items produced until one is inspected and found to be defective is 1/αq, and so
E[number inspected in a cycle] =E[N_{k}] +\frac{1}{αq}
Also, as E[N_{k}] is the expected number of trials needed to obtain k acceptable items in a row when each item is acceptable with probability p=1 − q, it follows from Example 3.14 that
E[N_{k}]=\frac{1}{p} +\frac{1}{p^{2}}+… +\frac{1}{p^{k}} =\frac{(1/p)^{k} − 1}{q}Hence we obtain
P_{I} ≡ proportion of items that are inspected =\frac{(1/p)^{k}}{(1/p)^{k}-1+1/α}To answer (b), note first that since each item produced is defective with probability q it follows that the proportion of items that are both inspected and found to be defective is qP_{I}. Hence, for N large, out of the first N items produced there will be (approximately) NqP_{I} that are discovered to be defective and thus removed. As the first N items will contain (approximately) Nq defective items, it follows that there will be Nq − NqP_{I} defective items not discovered. Hence,
proportion of the nonremoved items that are defective ≈\frac{N_{q}(1−P_{I})}{N(1 −qP_{I})}
As the approximation becomes exact as N→∞, we see that
proportion of the nonremoved items that are defective =\frac{q(1−P_{I})}{(1 −qP_{I})}