1, Host A and Host B are directly connected by a 10 × 106 bps link. The distance

Question

1, Host A and Host B are directly connected by a 10 × 106 bps link. The distance between the two hosts is 1,000 kilometers. Assume that the propagation speed over the link is 2 × 108 m/s, and that all packets are 125 bytes long. a. What is the propagation delay for a single packet? (4 points) b. What is the transmission delay between the two hosts? (4 points) c. If Host A begins to send a single packet at time t-O, at what time does Host B receive the FIRST bit of the packet? (4 points) LAST bit of the packet? (4 points) packet? (4 points) d. If Host A begins to send a single packet at time t 0, at what time does Host B receive the e. If the packet size is increased to 1 million bytes, what is the propagation delay for a single

Explanation / Answer

a) Propagation delay means how much time a bit (most significant bit) takes to
   reach from host A to host B.It only depends on the diatance between the host
   and the propagation speed and it has nothing to do with the length of the packet.

   In this case prpagation delay = (Distance/Speed) = (1000 * 1000)/2 * 10^8)
                                                    = 2 * 10^(-2) seconds
                                                    = 20 ms
b) Tramsmission delay = time to push out the packet by the sender = (Packet length)/(link speed)
                      = (125*8)/(10 *10^6) = 1000 * 10^(-7) = 10^(-4) = 0.1 ms

c) Time required for get the first bit by host B = time requred by A to put the packet on wire
                                                   + one bit to travel to B + time taken by B to pick one bit up

                                                 = (125*8)/(10 *10^6) + propagation delay + 1/(link speed)
                                                 = 0.1 ms + 20 ms + 10^(-4) ms
                                                 = 20.1001 ms

d) Time required for get the first bit by host B = time requred by A to put the packet on wire
                                                   + bits to travel to B + time taken by B to pick all bits up

                                                 = (125*8)/(10 *10^6) + propagation delay + 125/(link speed)
                                                 = 0.1 ms + 20 ms + 125 * 10^(-4) ms
                                                 = 0.1 ms + 20 ms + 0.0125 ms
                                                 = 20.1125 ms

e) Propagetion delay has notghing to do with the length of the packet. So in this case also
   we have propagation delay = (Distance/Speed) = (1000 * 1000)/2 * 10^8)
                                                    = 2 * 10^(-2) seconds
                                                    = 20 ms

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