= Solution
For $0\leq q<1$, the total number of forward <channel> uses is $T_n=\sum_{j=1}^nM_j$, so the random <code rate> of a message with $n$ <bits> is
$$
R_n=\frac{n}{T_n}=\left(\frac1n\sum_{j=1}^nM_j\right)^{-1}.
$$
The <independent and identically distributed random variables> $M_j$ have finite <expectation> $1/(1-q)$. By the <strong law of large numbers>,
$$
\boxed{R_n\longrightarrow1-q\quad\text{almost surely}.}
$$
Thus the protocol reaches the <binary erasure channel> <channel capacity> in its asymptotic forward throughput while guaranteeing zero error on every completed message. Its random transmission length and use of <channel feedback> distinguish it from the fixed-length <channel codes> in part (b). Independent erasure indicators suffice for this calculation; the message <bits> themselves need not be <independent> or uniformly distributed. At $q=1$, the algorithm never produces a completed transmission, so $R_n$ as a completed-code rate is undefined; the physical throughput is zero.
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