Authors:

James F. Kurose, Keith W. Ross

Chapter:

Computer Networks And The Internet

Exercise:

Problems

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, *d*_{prop},
in terms of *m
*and
*s*.

b.
Determine the transmission time of the packet, *d*_{trans},
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
*=
*d*_{trans},where
is the last bit of the packet?

e.
Suppose *d*_{prop
}is
greater than *d*_{trans.}
At time *t
*=
*d*_{trans},
where is the first bit of the packet?

f.
Suppose *d*prop
is
less than *d*_{trans}.
At time *t
*=
*d*_{trans},
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 *d*_{prop}
equals *d*_{trans.}

a) The propagation delay, *d*_{prop}=m/s sec

b) The transmission time of the packet, *d*_{trans}=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 *= *d*_{trans}.

Then, the last bi of the packet *t *= *d*_{trans}

e) Suppose *d*_{prop }is greater than *d*_{trans.} At time *t *= *d*_{trans}

Thus, the first bit of the packet is *d*_{prop}> *d*_{trans}.

f)Suppose *d*prop is less than *d*_{trans}. At time *t *= *d*_{trans},

Thus, the first bit of the packet is *d*_{prop}< *d*_{trans}.

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

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.**