Question: Let γ be the expected number of transmitted frames from A to B per successfully accepted packet at B. Let β be the expected number of transmitted frames from A to B between the transmission of a given frame and the reception of feedback about that frame (including the frame in transmission when the feedback arrives). Let p be the probability that a frame arriving at B contains errors (with successive frames assumed independent). Assume that A is always busy transmitting frames, that n is large enough that A never goes back in the absence of feedback, and that A always goes back on the next frame after hearing that the awaited frame contained errors. Show that γ satisfies γ = 1 + p(β + γ). Define the efficiency η as 1/ γ, and find η as a function of β and p.