Answered

Consider two hosts, A and B, connected by a single link of rate R bps. Suppose thatthe two hosts are separated by m meters, and suppose the propagation speed along thelink is s meters/sec. Host A is to send a packet of size L bits to Host B.(a) Express the propagation delay, dprop, in terms of m and s.(b) Determine the transmission time of the packet, dtrans, in terms of L and R. (c) Ignoring processing and queueing delays, obtain an expression for the end-to-end delay.(d) Suppose Host A begins to transmit the packet at time t = 0. At time t = dtrans, where is the last bit of the packet?(e) Suppose dprop is greater than dtrans. At time t = dtrans, where is the first bit of the packet?(f) Suppose dprop is less than dtrans. At time t = dtrans, where is the first bit of the packet?(g) Suppose s = 2.5×108 , L = 100 bits, and R = 28 kbps. Find the distance m so that d(prop) equals d(trans).

Answer :

ofentsemaphalla

Answer:

A) d(prop)=m / s

(B) d(trans)=L / R

C) certain delays are overlooked, so, d(total)= amount of the illustration above.

D(total)=(mR+Ls)/Rs D) The bit only exits Host A.

E) The bit has been in the link and hasn't arrived until Host B.

F) Host B has received the bit.

G) m= L / R s= 536 km= 120/56x 10 ^ 3 x (2.5x 10 ^ 8)

Other Questions