Question
This elementary problem begins to explore propagation delay and transmission 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, 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.
Answer
a) The propagation delay, dprop=m/s sec
b) The transmission time of the packet, dtrans=L/R sec
c) The end-to-end delay=(L/R+m/s) sec
d) Suppose Host A begins to transmit the packet at time t = 0. At time t = dtrans.
Then, the last bi of the packet t = dtrans
e) Suppose dprop is greater than dtrans. At time t = dtrans
Thus, the first bit of the packet is dprop> dtrans.
f)Suppose dprop is less than dtrans. At time t = dtrans,
Thus, the first bit of the packet is dprop< dtrans.
Discussions
vamshi banoth
answers for:
d)The bit is just leaving Host A
e)The first bit is in the link and has not reached Host B
f)The first bit has reached Host B
Fadi Badarni
g) For the given values:
- s = 2.5 * 10^8 meters/sec.
- L = 120 bits.
- R = 56 kbps = 56 * 10^3 bits/sec.
m = (120 * 2.5 * 10^8) / (56 * 10^3) = (3*10^7) meters ≈ 3000 Km
Post the discussion to improve the above solution.