2.993: Principle of Internet Computing

HW1 Solutions

1.

  1. latency(A-B) = propagation delay(A-B) + transmission time(A-B)

    2. = 1 ms + (8x8192)/10M sec.

    = 1 ms + 6.55 ms

    = 7.55 ms

  2. latency(B-C) = prop. delay(B-C) + transmit time(B-C)

    3. = 5 ms + (8x8192)/1.5 M sec.

    = 5 ms + 43.69 ms

    = 48.69 ms

    latency(A-C) = latency(A-B) + latency(B-C)

    = 7.55 ms + 48.69 ms

    = 56.24 ms

    Alternatively, latency(A-C) = prop. delay(A-C) + transmit time(A-C)

    = (1+5) ms + (8x8192)/1.3M sec.

    = 6 ms + 50.24 ms

    = 56.24 ms

    Note that if R is defined as the transmission rate (bandwidth),

    1/R(A-C) = 1/R(A-B) + 1/R(B-C)

    1/1.3M = 1/10M + 1/1.5M

    [For those who have studied circuit theory, the link rates are like capacitances. When they are connected in series, the equivalent bandwidth (capacitance) is always less than each bandwidth (capacitance) that makes up the series connection. This means that the same link performance (delay bandwidth product) can be achieved by having a point-to-point link between A and C with a bandwidth of only 1.30 Mbps.]

  3. Assume A starts transmitting at time t = 0 ms.

    4. B receives the first packet at t = 7.55 ms. It starts transmitting at t = 7.55 ms at a rate of 1.5 Mbps while packets are arriving in buffer B at 10 Mbps. At t = 7.55 ms, the buffer can support 9 more packets. The packets are accumulated in buffer B at a rate of 8.5 Mbps.

    Thus, the time to fill up buffer B is (9x8x8192)/8.5M = 69.4 ms

    So the time it takes after A starts blasting packets till B is forced to drop a packet is:

    69.4 ms + 7.55 ms = 76.9 ms

    Regardless of packet size, B will drop packets at a rate of 8.5Mbps or

    8.5M/(8x8192) = 129.7 packets/sec.

  4. Assume that the packet size of ACK is negligible, and the latency for ACK is just prop. delay(C-A) = 6 ms. Then, A stops transmission at t = (56.24+6) = 62.24 ms.

B receives the last bit sent at t = (62.24 + 1) ms = 63.24 ms.

We know that starting from t = 7.55 ms, B is accumulating packets at a rate of 8.5 Mbps. At t = 63.24 ms, the total number of accumulated packets is:

(63.24 – 7.55) ms x 8.5M/(8x8192) packets/s = 7.22 packets

Thus, the total number of packets in buffer B = (7.22 + 1) = 8.22 packets

As a result, the buffer does not overflow.

2. (Peterson & Davie, 1.12)

  1. minimum RTT = round-trip propagation delay

    2. = 2(240/186) = 2.58 sec.

  2. delay bandwidth product = 2.58 sec. x 100 Mbps = 258 Mbits = 32.3 MB.
  3. the significance is that by the time the sender receives ACK, it has already transmitted 32 MB of data. (a lot!)
  4. minimum time elapsed (from request to end of transfer)

= RTT + transmit time = 2.58 sec. + (25M x 8)/100M sec.

= 4.58 sec.

4. (Peterson & Davie, 1.14)

The answer in each part depends on the interpretation of each command. To be fair, we will not grade this problem.

 

(T. Minn)