Network Delay Calculation: Propagation, Transmission, and End-to-End Delay

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=dprop, where is the last bit of the packet?

e. Suppose dprop is greater than dtrans. At time t=dprop, where is the first bit of the packet?

f. Suppose dprop is less than dtrans. At time t=dprop, where is the first bit of the packet?

g. Suppose s=2.5*10^8, L=100 bits, and R=28 kbps. Find the distance m so that dprop equals dtrans.

Solution:

a. Propagation delay = distance / propagation speed = m / s

b. Transmission time = packet size / link rate = L / R

c. End-to-end delay = propagation delay + transmission time = m/s + L/R

d. The last bit of the packet will be at distance m from host A, which is the same distance from host B. Therefore, it will arrive at host B at time t = m / s + L / R.

e. If the propagation delay is greater than the transmission time, the first bit of the packet will not have been transmitted yet at time t=dprop. Therefore, it will still be at host A.

f. If the propagation delay is less than the transmission time, the first bit of the packet will have already arrived at host B at time t=dprop.

g. Substituting the given values into the expression for propagation delay:

m / s = dprop = L / R

Solving for m:

m = dprop * s = (100 bits) / (28 kbps / 2.5*10^8 m/s) = 357.14 meters.