Solutions to Assignment 1

  1. The network layer (IP) binds different physical and link layer technologies under a common addressing scheme. The transport layer allows the network layer to focus on routing packets and not worry about reliability and multiplexing concerns of applications.

  2. Solution to problem 2

    1. Total Time (t1) = 2*RTT + RTT/2 + (1000KB)/(1.5 Mbps)
      = 2.5X100X10^(-3) + 1000*8*10^3/(1.5*10^6)
      = 0.25 + 5.33 = 5.58 sec
      Note: We have been sloppy above and used 1Kb = 1000 bits for the file size. If you wanted to be precise, you would use 1Kb = 1024 bits. The answer in that case would change to t1 = 5.71 sec

      2. Note: The solutions to the following parts assume that the RTT specified in the problem accounts for the propagation delay and you do not have to.

    2. Total time = t1 + (999*RTT)
      = 5.71 + 99.9 = 105.61 sec

    3. Total time = 2*RTT + ((1000/20)-0.5)RTT =51.5*RTT = 5.15 sec

    4. 2^0 + 2^1 +2^2 + ... +2^i = 1000
      2^(i+1) - 1 >= 1000 -> i= 9
      The last batch reaches 1/2RTT later at the receiver. Total time = 2*RTT + 9.5 RTT= 1.15 sec


  4. Events occur in the following order:
    t=0: start
    t=500: A finishes sending packet1, starts sending packet 2
    t=520: Packet1 finishes arriving at S
    t=555: Packet1 departs for B
    t=1000: A finishes sending packet 2
    t=1020: Packet2 finishes arriving at S
    t=1055: Packet2 departs for B
    t=1075: Packet1 arrives at B
    t=1575: Packet2 arrives at B
    Total time = 1575µsec

  5. Applying the formulae given in class, you should get the bandwidth of a satellite TV channel has 66Mbps.