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James F. Kurose, Keith W. Ross
Computer Networks And The Internet
Question:6 | ISBN:9780132856201 | Edition: 6


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.



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 = 0. At time dtrans.

Then, the last bi of the packet dtrans

e) Suppose dprop is greater than dtrans. At time dtrans

Thus, the first bit of the packet is dpropdtrans.

f)Suppose dprop is less than dtrans. At time dtrans,

Thus, the first bit of the packet is dpropdtrans.

0 0


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

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