CSP 214 Written Homework 3 Solutions - Fall, 1997 1. The size of the "pipe" is: 1Mbps * 270ms * 2 = 0.54Mb 0.54Mb/10000 = 54 packets. As long as the send window is larger than 54, the pipe stays full. So a window size of 1024 is large enough (indeed, its way too large). 2. If there is not transmission error (no out-of-sequence either), then both would perform equally well. One could make the argument that Go-Back-N is a simpler protocol to implement and would thus be preferred. Moreover, the additional complexity requires more CPU handling time, so if CPU cycles were in short supply, Go-Back-N might perform better. But the argument here is not very strong. 3. "Application" routines get invoked at runtime via threads. For example, asp_init gets called once at startup. This routine then schedules an event (via Event) to invoke the client and server routines. 4. To detect collision, the minimum packet size should be 10*9 b/sec * 2 * 200meters/(2*10**8 m/sec) = 2000 b 5. In a WAN environment, the delay is much higher, hence the minimum packet size has to be pretty large to be able to detect collision. The larger minimum size will significantly degrade utilization, since small packets will need to be padded with extra bytes before they can be sent. In addition, because the RTT is so large, the "contention period" during which it is possible for two stations will transmit simultaneously becomes unreasonably large, signifcantly increasing the probability that collisions will take place. 6. The longest frame it could send is 16*10**6 b/sec * 10**-2sec = 160K bits 7. Time to send a packet: 4000byte*8bits/byte/(10**8 b/sec) = .320 sec Time for last bit to come back after transmission ceases: 100km/(200000km/sec) = .500 sec. hence the channel is busy .320/(.320+.500) = 39% of the time 8. A->B SW1: SW2: SW3: SW4 2 0 1 0 3 0 0 0 0 0 3 0 C->G SW1: SW2: SW3: SW4 2 0 1 0 3 0 0 0 0 0 3 0 3 0 1 1 3 1 1 0 3 0 1 0 E->I SW1: SW2: SW3: SW4 2 0 1 0 3 0 0 0 0 0 3 0 3 0 1 1 3 1 1 0 3 0 1 0 2 0 0 1 0 1 2 0 D->B SW1: SW2: SW3: SW4 2 0 1 0 3 0 0 0 0 0 3 0 3 0 1 1 3 1 1 0 3 0 1 0 2 0 0 1 0 1 2 0 0 0 1 2 3 2 0 2 0 2 3 1 F->J SW1: SW2: SW3: SW4 2 0 1 0 3 0 0 0 0 0 3 0 3 0 1 1 3 1 1 0 3 0 1 0 2 0 0 1 0 1 2 0 0 0 1 2 3 2 0 2 0 2 3 1 1 0 0 3 0 3 1 0 2 0 3 0 H->A: SW1: SW2: SW3: SW4 2 0 1 0 3 0 0 0 0 0 3 0 3 0 1 1 3 1 1 0 3 0 1 0 2 0 0 1 0 1 2 0 0 0 1 2 3 2 0 2 0 2 3 1 1 0 0 3 0 3 1 0 2 0 3 0 1 0 2 0 1 1 3 0 0 0 3 1 9. With the new Arpanet routing algorithm, link-state updates are flooded to each node. Selective flooding is NOT used, so neighbors send back received updates on the same line on which an update was received. Updates contain a sequence number, so an IMP can distinguish new updates from old ones it has seen. Duplicate updates are discarded, and not forwarded further. Consequently, each update propagates a link twice, once in both directions. With 50 switches, each having 3 links, there are a total of 50*3/2 links, but since each link is bidirectional (i.e. the delay is measured in each direction of the link) be a total of 50*3 distinct updates would be generated every .5 seconds. Consequently, the total number of update packets per link is 150*2 = 300 updates per second. Correction 10/22/97, 10:45 AM: The last sentence should read: Consequently, the total number of update packets per link is 150*2 = 300 updates per .5 second, or 600 packets/sec. Note 2: If you assumed that the updates generated by an IMP are sent together in one packet (i.e., 1 update for 3 links rather than 3 separate links), that is OK and would reduce your answer by a factor of 3. 10. omitted here 11. Given C=8Mb, rou=1M and M=6Mbps S = C/M-rou = 1.6sec 12. When converted from analog signal to digital data, some information will always be lost. This because analog signal consists of infinite levels while digital data can only record up to a finite level of accuracy. Note that a new generations of CDs is coming out in which the CD is mastered using 22 bits per sample, with the result (somehow) converted back to 16 bits at the end of the process. The resultant CD is alleged to sound better.