GATE answer key : B, D My opinion : Marks to all as question can be solved only with cert ai n assumptions/ 255 = 11111111 254 = 11111110 252 = 11111100 255.255.254.0 =/ 23 255.255.252.0 =/ 22 Range Number of IPs Network 1 12.20.164.0 / 22 12.20.164.0 - 12.20.16 7 255 2 10 = 1024 addresses Networ k 2 12.20.170.0 / 23 12.20.170.0 - 12.20.17 1 255 2 9 = 512 addresses N etwork 3 12.20.1 68 .0 / 23 12.20.168.0 - 12.20.16 9 255 2 9 = 5 1 2 addresses N etwork 4 12.20.166.0 / 23 12.20.16 7 .0 - 12.20.16 7 255 2 9 = 5 1 2 addresses Network1: 12.20.164.0 / 22 = 12.20.1010 01 00 0 Network2: 12.20.170.0 / 23 = 12.20. 1010 101 0 0 Network3: 12.20.1 68 .0 / 23 = 12.20.1010 100 0. 0 Network4: 12.20.166.0 / 23 = 12.20.1010 011 0.0 The lowest IP address in all four networks is 12.20.164.0 and the highest IP a ddress is 12.20.17 1 255 IP spectrum range is shown below with diagram – Figure A : IP range spectrum of all fo ur networks P lease n ote that there are some of the extra I Ps in the range 1 2. 20. 1 72 .0 - 1 2. 20. 1 7 5 255 which are not part of any network given. Figure B : Range of ea ch option As it can be visualized from F igure B , we have a couple of choices to aggregate these networks. I nterestingly , all choices are subject to ass umption s in the top olog y of the router Ch oice 1 : O ption s B, D O ption s B, D collectively covers all networ ks and does not cover some extra IPs Ch oice 2 : O ption A Option A stand alone covers all networks, but it also covers some extra IPs Ch oice 3 : O ption s C , D O ption s C , D collectively covers all networ ks but also cover some extra IPs Choice 1 : O ption s B, D This cannot be the correct choice because we must s omehow assume here that router R1 is connected via interface 0 ( Figure 1 ) B ut there could be a topol ogy where router R1 and interfa ce 0 are different ( Figure 2 ) Figure 1 Figure 2 Moreover, the very first step in aggregation is to group by “ Next hop ” T his 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 sp ecified that , “ separate the entries by next hop. You have to summarize them separatel y ” Unle ss we assume figure 1, we can not aggregate networks. Hence B, D is correct subject to assumptions. Choice 2 and 3 a re possible only when we are allowe d to cover e xtra IPs Covering extra IPs - I n the Appendix , we have given a reference to Tanenbaum’s book that says we can cover extra IPs if extra IPs are una llocated. ( Assu mption ) Suppose extra IP s are u nallocated and we are free to aggregate the m. For C hoice 2 and 3, we are going to follow the same assumption from now onwards Choice2: O ption A Option A is possible only when I nterfaces 0 and 1 are the sam e names. Similarly router names R 1 and R2 are synon yms for each other Figure 3 below i llustrates the top olo gy. Figure 3 : Ass umption in Topology and naming convention requi red for C hoice 2 Ch oice 3 : O ption s C , D Options C , D are only possible assuming top olog y and naming in Fi gure 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 yo u to kindly give marks to all for this question Appendix C onside r figure 4 from Tanenbaum’s book ( Book name: ) Figure 4 : Router New York aggregate three networks even th ough there is one block of extra ips ( missing ips ) is present As we see in t he above figure, we can aggregate IP s even when some IP s are missing or extra. This aggregation can be done only when ext ra IP s are unallocated. Author Tanenbaum has further extended the above example and allocated extra IPs to hosts. Figure 5 is ta ken from same book page number – Figure 5: Here Au thor has allocated extra IPs to different geographical locations than London. Table 1 and 2 respectively show s the routing table after aggregation for router New York in Figure 4 and in F igure 5. Routing Table for F igure 5 has two entries and conflict will be resolved based on the longest prefix match Subnet No Subnet M ask Next Hop 192. 24.0.0 /19 L ond on Subnet No Subnet M ask Next Hop 192. 24.0.0 /19 L ond on 192. 24. 1 2 .0 / 22 San Francisco Table 1: Aggregated Rou ting table for Ne w York in F igure 4 Table 2 : Aggregated Rou ting table for Ne w York in F igure 5 Next, please consider below F igure 6 .a and F igure 6.b whi ch are ex erci se problem and solution o f same book. W here the author has mentioned missing or extra IP s case Figure 6.a : Exerci se problem of Tanenbaum given at Pg492 Figure 6. b : Solution of above problem f rom solutions of Tanenbaum 5 Th Ed In solution, it is mentioned clearly that we just need to add one extra entry once un allocate d 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.