Question

P6. This elementary problem begins to explore propagation delay and transmis- sion delay, two central concepts in data networking. 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 meters/sec. Host A is to send a packet of size L bits to Host B a. Express the propagation delay, dprops in terms of m and s b. Determine the transmission time of the packet, drans, in terms of L and R. c. Ignoring processing and queuing delays, obtain an expression for the end- d. Suppose Host A begins to transmit the packet at time t = 0. At time t = dtrans' e. Suppose drop is greater than dran . At time t = d, ans, where is the first bit of f Suppose dprop is less than dtrans. At time t = dtrans, where is the first bit of g. Suppose s = 2.5-108, L = 120 bits, and R = 56 kbps. Find the distance m to-end delay where is the last bit of the packet? the packet? the packet?o so that dprop equals drans

Answer 1

Answer:

P6.a)Propagation delay d_prop is equal to d/s where d is the distance and s is the wave propagation speed.

here m is the distance between 2 hosts.so d_prop=m/s

b)Packet transmission time is obtained as,packet size/bit rate

here packet size is L and bit rate is R bps.

so d_trans=L/R

c)Ignoring processing and queuing delay an expression of end-to-end delay will be=m/s+L/R or d_prop+d_trans

d)At time t=0 and t=d_trans then the last bit of a packet is leaving node A.

e)If d_prop>d_trans then at t=d_trans first bit of packet leaves from node A at t=0 and it is in the link and not reached to node B yet.

f)If d_prop<d_trans then at t=d_trans first bit of packet has reached to node B.

g)s=2.5x10⁸  L=120bits R=56kbps then m=LR/s=120x56/2.5x10⁵km/s=0.02688km/s.