On the relationship between the distribution of maximal queue length in the M/G/1 queue and the mean busy period in the M/G/1/n queue

Takács has shown that, in the M/G/1 queue, the probability P(k | i) that the maximum number of customers present simultaneously during a busy period that begins with i customers present is P(k | i) = Q_k–i/Q_k, where the Q's are easily calculated by recurrence in terms of an arbitrary Q₀ ≠ 0. We augment Takács's theorem by showing that P(k | i) = b_k–i/b_k, where b_n is the mean busy period in the M/G/1 queue with finite waiting room of size n; that is, if we take Q₀ equal to the mean service time, then Q_n =b_n.