Question Number 55: Route Aggregation Master paper question 55 Figure 1: Question Description screenshot from master paper GATE answer key: B, D My opinion: Marks to all. As question can be solved only with certain assumptions. Explanation 255=11111111 254=11111110 GATE 2022 Page 1 of 8 252=11111100 255.255.254.0=/23 255.255.252.0=/22 Network1: 12.20.164.0/22 = 12.20.10100100.0 Network2: 12.20.170.0/23 = 12.20.10101010.0 Network3: 12.20.168.0/23 = 12.20.10101000.0 Network4: 12.20.166.0/23 = 12.20.10100110.0 Below Table1 has details about all four networks. Slash notation of subnet mask has been computed using decimal dotted notation given in question. Subnet mask Range Number of IPs Network 1 12.20.164.0/22 12.20.164.0 - 12.20.167.255 210 = 1024 Network 2 12.20.170.0/23 12.20.170.0 - 12.20.171.255 29 = 512 Network 3 12.20.168.0/23 12.20.168.0 - 12.20.169.255 29 = 512 Network 4 12.20.166.0/23 12.20.166.0 - 12.20.167.255 29 = 512 Table 1: All networks with subnet mask, range and number of IP addresses Figure2 depicts IP spectrum range of all networks. Visual representation in Figure2 helps to understand allocation of IP addresses among all netwroks GATE 2022 Page 2 of 8 Figure 2: Visual representation of IP ranges on number line Please note that there are some of the extra IPs in the range 12.20.172.0 - 12.20.175.255 which are not part of any network given. Visual representation for range of all options along with range of all networks is shown in Figure3. Figure 3: Visual representation of Options and network ranges on number line It can be visualized from Figure3 that we have a couple of choices to aggregate these four networks. GATE 2022 Page 3 of 8 Interestingly, all choices are subject to assumptions in the topology of the router. • Choice 1: Option A Option A standalone covers all networks but it also cover some extra IPs. • Choice 2: Options B,D Options B,D collectively covers all networks and also does not cover some extra IPs. • Choice 3: Options C,D Options C,D collectively covers all networks but also cover some extra IPs Let us talk about each choice one by one. And we will show that each of these choices are possible only when we assume something or other. Choice 1: Options B, D This cannot be the correct choice because we must somehow assume that router R1 is connected via interface 1 (Figure4). But there could be a topology where router R1 and interface 1 are different (Figure5). Figure 4: R1 is connected via interface1 and R2 is connected via interface0 GATE 2022 Page 4 of 8 Figure 5: R1 and R2 are not connected via interface1 and interface0 respectively Please note that topologies represented in Figure4 and Figure5 are valid with respect to routing table given in the question. Figure5 shows the possibility where first half of Network1 is connected via R1 and second half of Network1 (which is Network4) is connected via interface1. This is valid since any packet destined to Network4 will be sent based on longest mask match between first and fourth entry of routing table given in question. Finally, We conclude that Choice 1 is correct only if we assume topology given by Figure 4. In case of Figure 5 Choice 1 is not correct. In fact, In case of Figure 5 aggregation is not at all possible and there can not be any answer to this question. This can also be seen in Reference [1] which is an answer on Network Engineering Stack Exchange website. Answer in Reference [1] states that, “Separate the entries by next hop. You have to summarize them separately”. Which means that the very first step in aggregation is to group by “Next hop”. Therefore, if there is no common next hop then aggregation is not possible. Now talking about Choice 2 and 3. Choice 2 and Choice 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. In case extra IPs in Figure 3 are already allocated to some hosts then Choice 2 and Choice GATE 2022 Page 5 of 8 3 are not possible. (Assumption) Now onward, we assume that extra IPs in Figure 3 are unallocated. Choice 2: Option A Option A is possible only when Interfaces 0 and 1 are the same names. Similarly router names R1 and R2 shoule be synonyms for each other. Figure 6 below illustrates the topology. Figure 6: 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 4. Final remarks: 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. Appendices Consider Figure 7 of Tanenbaum’s book [2] from page [2, p. 448] GATE 2022 Page 6 of 8 Figure 7: Router New York aggregate three networks even though there is one block of extra IPs (missing IPs) in red is present As we see in the above figure 7, we can aggregate IPs even when some IPs are missing or extra. Again, 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 8 is taken from same book page number [2, p. 449] Figure 8: Here Author has allocated extra IPs to different geographical locations than London. Table 2 and Table 3 respectively shows the routing table after aggregation for router New York in Figure 7 and in Figure 8. Subnet Number Subnet mask Next Hop Subnet Number Subnet mask Next Hop 192.24.0.0 /19 London 192.24.0.0 /19 London 192.24.12.0 /22 San Francisco ... ... ... Table 2: Aggregated Routing table for New York in ... ... ... Figure 7 Table 3: Aggregated Routing table for New York in Figure 8 GATE 2022 Page 7 of 8 Routing Table for Figure 8 has two entries and conflict will be resolved based on the longest mask match Next, please consider below Figure 9 and Figure 10 which are exercise problem and solution of same book. Where the author has mentioned missing or extra IPs case. Figure 9: Exercise problem of Tanenbaum given at Pg492. Figure 10: Solution of above problem In solution(figure 10), it is mentioned clearly that we just need to add one extra entry in future when unallocated IPs get assign. The same problem could also be seen at [3]. Where they also specifically mention that we can aggregate extra unallocated IPs. And in future when IPs are allocated then just add new entry in routing table and longest mask matching will take care of routing. References [1] R. T. (https://networkengineering.stackexchange.com/users/2142/ron trunk), “How to aggregate a routing table given ip addresses and their subnet masks.” Network Engineer- ing Stack Exchange. URL:https://networkengineering.stackexchange.com/a/7238/82339. [2] A. S. Tanenbaum and D. J. Wetherall, Computer Networks. USA: Prentice Hall Press, 5th ed., 2010. [3] https://cis.temple.edu/ latecki/Courses/CIS617-04/FinalQuestions/Yu.doc. GATE 2022 Page 8 of 8
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