3 Consider two hosts A and B connected by a single link of rate R bps Suppose that the two hosts are separated by m meters and suppose the propagation speed along the link is s meterssec Host A is to

a. Propagation delay = distance / propagation speed = m / s b. Transmission time = packet size / link rate = L / R c. End-to-end delay = propagation delay + transmission time = m/s + L/R d. The last bit of the packet will be at distance m from host A, which is the same distance from host B. Therefore, it will arrive at host B at time t = m / s + L / R. e. If the propagation delay is greater than the transmission time, the first bit of the packet will not have been transmitted yet at time t=. Therefore, it will still be at host A. f. If the propagation delay is less than the transmission time, the first bit of the packet will have already arrived at host B at time t=. g. Substituting the given values into the expression for propagation delay: m / s = = L / R Solving for m: m = = (100 bits) / (28 kbps - 2.5·108 m/s) = 357.14 meters