Question

Suppose two hosts, Host A and B, are separated by 10,000 kilometers and are connected by a direct link of R-1 Mbps. Suppose the propagation speed over the link is 2.5 10 meters/sec. The propagation delay of the link is dprop (a) Calculate the bandwidth-delay product, R dpop (b) Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximum number of bits that will be in the link at any given time? Provide an interpretation of the bandwidth-delay product. What is the width (in meters) of a bit in the link? Derive a general expression for the width of a bit in terms of the propagation speed s, the transmission rate R, and the length of the link m (c) (d) (e)

Answer 1

#1 : Bandwidth-delay product = Data rate * (end-end delay)

          (bits)                                   (bps)                 (s)

#2 : End - end delay for hosts A and B = (Data link length) m

                                                          ------------------------

                                                          (propogation speed) m/s

#3 : Data rate R = 1Mbps = 10^6 bps (approx)

#4 : Distance between A and B = Link length = 10000 km =10 ^7 m

#5 : Propogation speed = 2.5*10^8m/s

=>Bandwidth-delay product = (10^6 * 10 ^7)/( 2.5*10^8)

                                         = 40000 bits