INFS 612 Computer Networking

Assignment1 of chapter1: Computer Networks and the Internet

2.

  1. I think a circuit-switched network will be more appropriate for the application that

the question mentioned.

Because in circuit-switched networks, the resources needed along a path to provide for communication between the end systems are reserved for the duration of the session. So when the network establishes the circuit, it also reserves a constant transaction rate in the network’s links for the duration of the connection. This reservation allows the sender to transfer the data to the receiver at the guaranteed constant rate.

b. In the 2.b part, I think the congestion control will NOT be needed.

If the network is reserved for this application and the sum of this application data rates is less than the capacities of each link, there will not be any congestion. In this situation, even we use “packet-switching” network, we will not need to worry about congestion problem.

3. We use L: bits/packet, M: packets, Q: Links:

a. Time for each packet: (L + h) / R;

Answer T = ts + Q * ((L + h) / R) + (M - 1) * ((L + h) / R).

  1. Answer T = Q * ((L + 2h) / R) + (M - 1) * ((L + 2h) / R).
  1. Answer T = Q * ((M * L + 2h) / R).
  1. Answer T = ts + (M * L + h) / R. (Note: For a circuit-switched network, the

transmission time is independent of the number of links.)

4. I did the experiment with the message-switching Java applet in this chapter. I found

the delays in the applet correspond to the delays in question 3. And I found out that the

link propagation delay have the SAME effect on the overall end-to-end delay for

packet switching and for message switching.

5. The total delay should be: (ceiling (F / S) + 1) * ((S + 40) / R)

Using calculus:

We will say when S = (40 * F)1/2, the delay of moving the packet from Host A to Host

B will be the minimum.

6.

  1. dprop = m / s;
  1. dtrans = L / R;
  1. d = m / s + L / R; (Note: there are only one link between these two Hosts.)
  1. The last bit will be in Host A.
  1. The first bit will be in somewhere in the link.
  1. The first bit will be in Host B.
  1. Because: m /s = L / R, so m = L * s / R =8.9 * 105 (meters).

7. The time will be:

Each packet has: 48 * 8 = 384 bits;

Time = 384 / (64 * 103) + 384 / (1 * 106) + 2 * 10-3 + 384/ (64 * 103)

= 6 * 10-3 + 2 * 10-3 + 6 * 10-3 + 0.384 * 10-3

= 14.384 * 10-3 (seconds)

9. For the N packets arrive the buffer at the same time every (L / R) * N seconds: The

first packet transmitted has no queuing delay; the second packet transmitted has a

queuing delay of L / R seconds; and more generally, the nth packet transmitted has a

queuing delay of ((n - 1) * L / R) seconds.

Total queuing delay = 0 + L / R + 2 * (L / R) + ………. + ((n - 1) * L / R)

= (N * (0 + ((n - 1) * L / R))) / 2

= (n * (n - 1) * (L / R)) / 2

Average queuing delay time = Total delay / n = ((n -1) * (L / R)) / 2.

10.

  1. The formula for the total delay = L / R + (I * L) / (R * (1 - I));
  1. If we use L / R to show this formula, it will be: Let x = L / R, so delay = x + (a * x * x) / (1 – a * x) = x / (1 – a * x).

11. Links: Q; Routers: Q –1; L: packet size.

a. (1). Processing delay: DTproc = dproc1 + dproc2 + …….+ dprocQ

(2). Transmission delay: DTtrans = L / R1 + L / R2 + …….+ L / RQ

(3). Propagation delay: DTprop = dprop1 + dprop2 + …….+ dpropQ

So Dend-end = DTproc + DTtrans + DTprop

b. if we have queuing delay:

The Dend-end = DTproc + DTtrans + DTprop + (Q – 1) * dqueue