 Solution

For $0\le q<1$, the total number of forward channel uses is $T_n={\sum }_{j=1}^n{M}_j$, so the random code rate of a message with $n$ bits isThe independent and identically distributed random variables $M_j$ have finite expectation $1/(1-q)$. By the strong law of large numbers,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.