Question

Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by d 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) (3 points) Define the propagation delay, d_prop (i.e., explain what is propagation delay and how to compute it).

b) (3 points) Define the transmission time of the packet, d_trans? (i.e., explain what is transmission delay and how to compute it).

c) (4 points) Suppose d_prop is greater than d_trans. At time t = d_trans, where is the first bit of the packet and where is the last bit of the packet?

Answer 1

Solution:

Given,

=>Bandwidth = R bps

=>Distance = d meters

=>Propagation speed = s meters/sec

=>Packet length = L bits

(a)

Explanation:

Calculating propagation delay:

=>Propgation delay is the ration of distance and propagation speed.

=>Propagation delay = distance/propagation speed

=>Propagation delay = d meters/s meters/sec

=>Propagation delay = (d/s) sec

Calculating dprop:

=>dprop is the ration of distance and propagation speed.

=>dprop = distance/propagation speed

=>dprop = d meters/s meters/sec

=>dprop = (d/s) sec

(b)

Explanation:

Transmission delay:

=>Tranmission delay is the time taken to put all the bits of packet into the channel/link

Calculating transmission delay(dtrans):

=>Transmission delay(dtrans) = packet length/bandwidth

=>Transmission delay(dtrans) = L bits/R bps

=>Transmission delay(dtrans) = (L/R) sec

(c)

Explanation:

=>At t = dtrans, first bit wil be at s*dtrans means at d*(L/R) meters from the host.

=>At t = dtrans last bit of packet will be leaving the host.

I have explained each and every part with the help of statements attached to it.