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Posted on Jul 31, 2023 in Computers

11. (a) What are two reasons for using layered protocols? (b) What are the two possible disadvantages of using layered protocols?        (a) Among other reasons for using layered protocols, using them leads to,(1) breaking up the design problem into smaller, more manageable pieces, and(2) layering means that protocols can be changed without affecting higher or lower ones.   b) Disadvantage of using layered protocols: (1) More overhead, due to added information or operation done in each layer. 2) Information for one layer which could be useful for other layer (say, application might care whether it is running over wired or wireless) could be hidden or, not available, or, may require added computational overhead for retrieval.
12. Provide a list of reasons for why the response time of a client may be larger than the bestcase delay.       One reason is request or response messages may get corrupted or lost during transmission, which will necessitate a retransmission delay. Another reason is the processing unit in the client may get overloaded processing several requests at once. Yet another reason is that the request or response messages may be queued in the network along with other messages.
13. A factor in the delay of a store-and-forward packet-switching system is how long it takes to store and forward a packet through a switch. If switching time is 10 μsec, is this likely to be a major factor in the response of a client-server system where the client is in New York and the server is in California, separated by 5000 km? Assume the propagation speed in copper and fiber to be 2/3 the speed of light in vacuum (299,792,458 m/s in the vacuum, but use 300,000 km/s to keep it simple). Assume, 50 switches along the way.                The speed of propagation is (2/3 × 300,000 k/s) or, 200,000 km/sec or, 200 meters/μ sec. In 10 μsec the signal travels 2 km. Thus, each switch adds the equivalent of 2 km of extra cable. If the client and server are separated by 5000 km, traversing even 50 switches adds only 100 km to the total path, which is only 2%. Thus, switching delay is not a major factor under these circumstances.
14. Five routers are to be connected in a point-to-point subnet. Between each pair of routers, the designers may put a high-speed line, a medium-speed line, a low-speed line, or no line. If it takes 100 ms of computer time to generate and inspect each topology, how long will it take to inspect all of them?           Call the routers #1, #2, #3, #4 and #5.  Then, ten potential lines: 1-2, 1-3,1-4, 1-5, 2-3, 2-4, 2-5, 3-4, 3-5, 4-5. Each of these lines has four possibilities (three lines and one no line), so the total number of topologies is 4 10 = 1,048,576. At 100 ms each, it takes 104,857.6 sec (1 sec = 1000ms), or 29.127 hours to inspect them all.
15. A disadvantage of a broadcast subnet is the capacity wasted when multiple hosts attempt to access the channel at the same time. As a simplistic example, suppose that time is divided into discrete slots, with each of the n hosts attempting to use the channel with probability p during each slot. What fraction of the slots will be wasted due to collisions?      We can compute the probabilities from the case where no collision has happened, as: i) Case 1: a host has transmitted with probability p and can have no collision if, (n-1) hosts do not transmit with probability (1-p) each; thus the probability of this event per host is:  p^1 × [(1-p) × (1-p) × …{(n-1) times}] or, p (1-p)^(n-1) This can be true for n different host in n different ways, thus finally, it is: n×p×(1-p)^(n-1).
16. The subnet of Figure (below) was designed to withstand a nuclear war. How many bombs would it take to partition the nodes into two disconnected sets? Assume that any bomb wipes out a node and all of the links connected to it.       There does not appear to be a group of nodes that are connected to the rest of the network by fewer than three other nodes, so we conclude that three bombs are needed to partition the network. For example, the two nodes in the upper-right corner can be disconnected from the rest by three bombs knocking out the three nodes to which they are connected. The system can withstand the loss of any two nodes.
17. Consider the traceroute following traceroute below:
1 cs-gw (128.119.240.254) 1.009 ms 2 128.119.3.154 (128.119.3.154) 0.931 ms 3 border4-rt-gi-1-3.gw.umass.edu (128.119.2.194) 1.032 ms 4 acr1-ge-2-1-0.Boston.cw.net (208.172.51.129) 10.006 ms 5 agr4-loopback.NewYork.cw.net (206.24.194.104) 12.272 ms 6 acr2-loopback.NewYork.cw.net (206.24.194.62) 13.225 ms 7 pos10-2.core2.NewYork1.Level3.net (209.244.160.133) 12.218 ms 8 gige9-1-52.hsipaccess1.NewYork1.Level3.net (64.159.17.39) 13.081 ms 9 p0-0.polyu.bbnplanet.net (4.25.109.122) 12.716 ms 10 cis.poly.edu (128.238.32.126) 14.080 ms. Which link has the largest delay? Which link has the second largest delay? Where are these links located?       The roundtrip to router 8 is 13.1 msec and the roundtrip delay to router 9 is 12.7. Two separate roundtrip probes were sent to routers 8 and 9 at two (slightly) different times. During these times, congestion in the links along the path changed. Apparently there was more congestion in at least one of the first eight links during the first probe than in the second probe; hence, the roundtrip delay to router 8 is greater than the roundtrip delay to router 9. The two largest delays occur between routers 3 and 4 and between routers 4 and 5. Routers 3 and 4 appear to be located in Amherst Massachusetts (UMass) and Boston. Routers 4 and 5 appear to be located in Boston and New York City