GATE answer key: B, D My opinion: Marks to all as question can be solved only with certain assumptions/ 255=11111111 254=11111110 252=11111100 255.255.254.0=/23 255.255.252.0=/22 Range Number of IPs Network 1 10 188.8.131.52/22 184.108.40.206 - 220.127.116.11 2 = 1024 addresses Network 2 9 18.104.22.168/23 22.214.171.124 - 126.96.36.199 2 = 512 addresses Network 3 9 188.8.131.52/23 184.108.40.206 - 220.127.116.11 2 = 512 addresses Network 4 9 18.104.22.168/23 22.214.171.124 - 126.96.36.199 2 = 512 addresses Network1: 188.8.131.52/22=12.20.10100100.0 Network2: 184.108.40.206/23=12.20.10101010.0 Network3: 220.127.116.11/23=12.20.10101000.0 Network4: 18.104.22.168/23=12.20.10100110.0 The lowest IP address in all four networks is 22.214.171.124 and the highest IP address is 126.96.36.199. IP spectrum range is shown below with diagram – Figure A: IP range spectrum of all four networks Please note that there are some of the extra IPs in the range 188.8.131.52 - 184.108.40.206 which are not part of any network given. Figure B: Range of each option As it can be visualized from Figure B, we have a couple of choices to aggregate these networks. Interestingly, all choices are subject to assumptions in the topology of the router. Choice 1: Options B, D Options B, D collectively covers all networks and does not cover some extra IPs. Choice 2: Option A Option A standalone covers all networks, but it also covers some extra IPs. Choice 3: Options C, D Options C, D collectively covers all networks but also cover some extra IPs Choice 1: Options B, D This cannot be the correct choice because we must somehow assume here that router R1 is connected via interface 0. (Figure 1) But there could be a topology where router R1 and interface 0 are different. (Figure 2) Figure 1 Figure 2 Moreover, the very first step in aggregation is to group by “Next hop”. This is explained in Ref https://networkengineering.stackexchange.com/questions/7234/how-to-aggregate-a- routing-table-given-ip-addresses-and-their-subnet-masks On networkengineering.stackexchange, it is clearly specified that, “separate the entries by next hop. You have to summarize them separately”. Unless we assume figure 1, we can not aggregate networks. Hence B,D is correct subject to assumptions. Choice 2 and 3 are possible only when we are allowed to cover extra IPs Covering extra IPs - In the Appendix, we have given a reference to Tanenbaum’s book that says we can cover extra IPs if extra IPs are unallocated. (Assumption) Suppose extra IPs are unallocated and we are free to aggregate them. For Choice 2 and 3, we are going to follow the same assumption from now onwards. Choice2: Option A Option A is possible only when Interfaces 0 and 1 are the same names. Similarly router names R1 and R2 are synonyms for each other. Figure 3 below illustrates the topology. Figure 3: Assumption in Topology and naming convention required for Choice 2 Choice 3: Options C, D Options C, D are only possible assuming topology and naming in Figure 1. As we see that every possible choice must have some assumptions to follow. Hence there cannot be a single answer unless assumptions are specified in the question. I request you to kindly give marks to all for this question. Appendix Consider figure 4 from Tanenbaum’s book (Book name: ) Figure 4: Router New York aggregate three networks even though there is one block of extra ips (missing ips) is present As we see in the above figure, we can aggregate IPs even when some IPs are missing or extra. This aggregation can be done only when extra IPs are unallocated. Author Tanenbaum has further extended the above example and allocated extra IPs to hosts. Figure 5 is taken from same book page number – Figure 5: Here Author has allocated extra IPs to different geographical locations than London. Table 1 and 2 respectively shows the routing table after aggregation for router New York in Figure 4 and in Figure 5. Routing Table for Figure 5 has two entries and conflict will be resolved based on the longest prefix match Subnet No Subnet Next Hop Subnet No Subnet Next Hop Mask Mask 220.127.116.11 /19 London 18.104.22.168 /19 London 22.214.171.124 /22 San Francisco Table 1: Aggregated Routing table for New Table 2: Aggregated Routing table for New York in Figure 4 York in Figure 5 Next, please consider below Figure 6.a and Figure 6.b which are exercise problem and solution of same book. Where the author has mentioned missing or extra IPs case. Figure 6.a: Exercise problem of Tanenbaum given at Pg492. Figure 6.b: Solution of above problem from solutions of Tanenbaum 5Th Ed In solution, it is mentioned clearly that we just need to add one extra entry once unallocated IPs get assign and longest matching rule will take care of correct routing. This same problem could also be seen here: Link: https://cis.temple.edu/~latecki/Courses/CIS617-04/FinalQuestions/Yu.doc They specifically say when got allocated then we add that entry till that point no entry required.