Two-phase inspection model ($I=(Kp+q)/(Kp+rq),q=1-p$)

A two-phase inspection model alternates between inspecting every item and inspecting each item with probability $1/r$, where $r>1$. Suppose item defects are independent with probability $p\in (0,1)$. The first phase ends after $K\ge 1$ good items have been accumulated, while the second ends upon detecting a defect. With state $i<K$ equal to the cumulative number of good items found during the current full-inspection phase, the Markov chain has transitions $P_{ii}=p$, $P_{i,i+1}=q=1-p$, $P_{K0}=p/r$, $P_{KK}=1-p/r$.

Its stationary distribution is ${\pi }_i=p/(Kp+rq)$ for $i<K$ and ${\pi }_K=rq/(Kp+rq)$. Indeed, balance gives equal mass to every phase-A state and $p{\pi }_K/r=q{\pi }_{K-1}$; normalization fixes the common mass. The finite chain is irreducible and aperiodic, so the Markov chain ergodic theorem identifies these probabilities with long-run occupation fractions. The long-run inspected fraction is $1-(1-1/r){\pi }_K=(Kp+q)/(Kp+rq)$. The fraction of all production consisting of undetected defective items is $p(1-1/r){\pi }_K=p(r-1)q/(Kp+rq)$.

An important modeling choice is whether defects during full inspection preserve accumulated good items or reset the count. Resetting produces the distinct consecutive-clearance inspection model and changes the long-run fractions.

In the consecutive-clearance inspection model, a run of $K$ consecutive good items is required to leave full inspection. With $q=1-p$, $0<p<1$, its Markov chain has transitions $P_{i0}=p$, $P_{i,i+1}=q$ for $i<K$, and $P_{K0}=p/r$, $P_{KK}=1-p/r$. Thus every defect in the clearance phase resets the count, rather than preserving the cumulative total.

Writing $D=1+(r-1)q^K$, the stationary distribution isTo prove this, balance at $i=1,\dots ,K-1$ gives ${\pi }_i=q^i{\pi }_0$, while balance at $K$ gives $(p/r){\pi }_K=q^K{\pi }_0$. Summing the geometric series and normalizing gives ${\pi }_0=p/D$; balance at zero follows as well. The inspected fraction is $1-(1-1/r){\pi }_K=1/D$. The fraction of all production that is both defective and uninspected is $p(r-1)q^K/D$; the fraction among defective items alone is $(r-1)q^K/D$. All these long-run limits follow from the Markov chain ergodic theorem and the independent inspection coins. The distinction from cumulative clearance matters already at $K=2$.