0 votes 0 votes What will be the order of B+ tree with a database of 5,00,000 records of 200 bytes each and the search key is 15 bytes?Assume tree and data pointers are are 5 bytes each and the index node is 1024 bytes? neha singh asked Mar 18, 2017 neha singh 1.7k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Akriti sood commented Mar 19, 2017 reply Follow Share internal node - n*(5) + (n-1) * 15 <= 1024 leaf node - n*(5 + 5) <=1024 pls correct me and what has to be done with number of records?? 0 votes 0 votes akash.dinkar12 commented Mar 19, 2017 reply Follow Share now what has to do?? 0 votes 0 votes neha singh commented Mar 20, 2017 reply Follow Share As nothing is mention in the question so we can take internal node so answer here is 51 0 votes 0 votes Akriti sood commented Mar 20, 2017 reply Follow Share but what has to be done with the number of reccords given? 1 votes 1 votes akashjeet commented Nov 16, 2018 reply Follow Share While Solving for internal nodes 15 * (n-1) should be = 15n - 15 but you took 15n - 5 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes the formula nPb+(n-1)(k+Pd) <= Block size in given picture k+Pd= 15 put, how? whereas Pb and Pd =5B k+Pd =15+5 = 20 should there? deovrat57 answered Aug 14, 2018 • edited Aug 14, 2018 by deovrat57 deovrat57 comment Share Follow See 1 comment See all 1 1 comment reply arvin commented Aug 15, 2018 reply Follow Share https://gateoverflow.in/232929/b-tree-b-tree?show=232938#a232938 u can see the same question. and i think u are confused between b and b+ tree and between leaf node and internal node of b+ tree. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes for linternal node:- P*(B.P.)+(P-1)K <=Block size 5(P)+15(P-1)<=1024 20P-15<=1024 P=51.9 P= ~51 for leaf node: B.P.+(P-1)(K+R)<=1024 5+(P-1)(20)<=1024 20P-15<=1024 P=51.9 P=~51 Kaushal22 answered Oct 3, 2020 Kaushal22 comment Share Follow See all 0 reply Please log in or register to add a comment.