Solution HW4 (120 points) Chapter 4
Review questions: 22. (10 points) IGMP is a protocol run only between the host and its first-hop multicast router. IGMP allows a host to specify (to the first-hop multicast router) the multicast group it wants to join. It is then up to the multicast router to work with other multicast routers (i.e., run a multicast routing protocol) to ensure that the data for the host-joined multicast group is routed to the appropriate last-hop router and from there to the host. 23. (10 points) In a group-shared tree, all senders send their multicast traffic using the same routing tree. With source-based tree, the multicast datagrams from a given source are routed over a specific routing tree constructed for that source; thus each source may have a different source-based tree and a router may have to keep track of several source-based tress for a given multicast group. Problems: 21. (10 points)
23. (20 points) The center-based tree connects A to C; B to C; E to C; and F to C (all directly). In the center-based tree the total traffic intensity (upstream + downstream traffic) is 4 on links AC, BC, EC, and FC. There is one unit of traffic emanating from nodes A, B, E, F and three units of traffic coming downstream from C that was sent by the other three nodes.
When RPF is used, the source tree rooted at A has one unit of traffic on AB, AC, CE, and CF. The source tree rooted at B has one unit of traffic on AB, BD, DE, EF. The source tree rooted at E has one unit of traffic on EF, ED. DB, EC and CA. The source tree rooted at F has one unit of traffic on FE, ED, DB, FC, and CA. Summing these up, the union of the source rooted trees generates: 3 units on CA; 2 units on CF; 2 units on AB; 3 units on BD; 3 units on ED; 3 units on EF; and 2 units on CE. Note that no link has 4 units of traffic, while four of the links in the center-based tree have 4 units of traffic on them. Thus, in this example, we can say that the center-based tree tends to concentrate traffic. 24. (12 points) Since MOSPF builds source-specific trees, its state complexity is S*G. PIM SM builds single a center-based tree for the group and hence has state complexity G. PIM DM is like DVMRP and thus has a complexity of S*G.
Chapter 5
Review questions: 9. (6 points) C's adapter will process the frames, but the adapter will not pass the datagrams up the protocol stack. If the LAN broadcast address is used, then C's adapter will both process the frames and pass the datagrams up the protocol stack. 14. (7 points) After the 5th collision, the adapter chooses from {0, 1, 2, ..., 31}. The probability that it chooses 4 is 1/32. It waits 204.8 microseconds. 16. (10 points) No, there wouldn't be an advantage if the RTS and CTS frames were long. The short RTS and CTS frames enable the sender and receiver to quickly notify neighboring nodes that they shouldn't transmit for a specified period of time. If the RTS were long, then during the long RTS, a neighbor of the receiver that is hidden to the sender could start to transmit, causing a collision before the reservation is made. Problems 14. (10 points) At t = 0 A transmits. At t = 576 , A would finish transmitting. In the worst case, B begins transmitting at time t = 224 . At time t = 224 + 225 = 449 B 's first bit arrives at A . Because 449 < 576 , A aborts before completing the transmission ...