Consider two hosts A and B connected by a single link of rat

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, 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 queuing 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 = 120 bits, and R = 56 kbps. Find the distance m so that dprop equals dtrans

Solution

Answer:

A) d(prop)=m/s

(B) d(trans)=L/R

C) other delays are neglected, so, d(total)= sum of above expression.

d(total)=(mR+Ls)/Rs

D) The bit is just leaving the Host A.

E) The bit is in the link and has not reached till Host B.

F) The bit has reached host B.

G)

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