55 55 votes Consider a network with $6$ routers $R1$ to $R6$ connected with links having weights as shown in the following diagram. Suppose the weights of all unused links are changed to $2$ and the distance vector algorithm is used again until all routing tables stabilize. How many links will now remain unused? $0$ $1$ $2$ $3$ Computer Networks gatecse-2010 computer-networks routing distance-vector-routing normal + – go_editor 20.7k views answer comment Share Follow Print See all 3 Comments 3 3 Comments reply ritiksri8 commented Jun 1, 2024 reply Follow flag Then R5-R6 will remain unused. 0 0 replyShare Redlex69 commented Jan 21, 2025 reply Follow flag I hope this is correct 3 3 replyShare js__ commented Nov 13, 2025 reply Follow flag If there is a shortest link bw any two Routers then that link will definitely be there for example bw R3 and R4 paths can be R3-R1-R2-R4 = 3+2+7 = 12 and R3-R2-R4 = 2+7 = 9 so , R3-R2 will definitely be present 0 0 replyShare Please log in or register to add a comment.
Best answer 56 56 votes First we need to find which are the unused links in the graph For that we need not make distance vector tables, We can do this by simply looking into the graph or else DVT can also give the answer. So, $R1-R2$ and $R4-R6$ will remain unused. Now If We changed the unused links to value $2$. $R5-R6$ will Now remain unused. So, the correct answer is option B). bad_engineer answered Apr 21, 2016 • edited Jun 20, 2018 by Milicevic3306 bad_engineer comment Share Follow See all 6 Comments 6 6 Comments reply Show 3 previous comments kp1 commented Oct 11, 2019 2 flags: ✌ Edit necessary (js__)✌ Edit necessary (under_50) reply Follow flag Answer would be zero instead of 1. –1 –1 replyShare rohith1001 commented Jan 3, 2020 reply Follow flag @kp1 No! R5---R6 edge is never present in any shortest path between any two pairs of nodes. Hence B)1 is the correct answer. 1 1 replyShare kp1 commented Jan 6, 2020 reply Follow flag @rohith1001 yes u r right. Actually in previous year book there is a misprint ( it is given cost of R5-- R6 link is 3) so if u take 3 then answer will be zero.. But in the actual question R5--R6 link is 4. Thus answer will be 1. 3 3 replyShare Please log in or register to add a comment.
29 29 votes Only one link is not used pC answered Jul 12, 2016 pC comment Share Follow See all 2 Comments 2 2 Comments reply ꧁༒☬ĿọŗԀ 🆂🅷🅸🆅🅰☬༒꧂ commented Aug 7, 2024 reply Follow flag @pC $R3 $ to $R6$ distance should be $11$ 0 0 replyShare Arnav Singh_01 commented Oct 21, 2024 reply Follow flag Yes R3->R2->R4->R6 0 0 replyShare Please log in or register to add a comment.
8 8 votes The links R1-R2 and R4-R6 will never be used for data transfer because there are shorter paths available in any case.If those two link weights are changed to 2, now only one link ie R5-R6 will never be used. Hunaif answered Apr 28, 2016 Hunaif comment Share Follow 0 reply Please log in or register to add a comment.
8 8 votes Option-B SImple reason for this.Use your intuition like below way Find all the shortest path from each node to other and mark your visited edge .Then automatically in first searching u will find that only two edge are unused example edge R1---->R2 as well as R4---->R6. When u change it into 2,Then again apply your intuition to find out shortest path from each node to other in this satuation u again find on link is unused. Paras Nath answered Sep 14, 2016 Paras Nath comment Share Follow 0 reply Please log in or register to add a comment.
2 2 votes Ans is B now OO7 answered Nov 28, 2018 • edited Nov 28, 2018 by OO7 OO7 comment Share Follow 0 reply Please log in or register to add a comment.
2 2 votes Please correct the question in your text book who thinks answer should be 0 Distance between R5 and R6 is 4 and not 3 as asked in gate 2010 please correct the question Ayush3007 answered Dec 8, 2019 Ayush3007 comment Share Follow 0 reply Please log in or register to add a comment.