0 votes 0 votes A B+ tree index is built on the key ‘CS’ attribute of the relation ME16. The attribute ‘CS’ is of length 32 bytes, disk blocks are of size 2048 bytes and index pointers are of size 8 bytes. The number of pointers per node is _______. thor asked Jan 10, 2017 thor 524 views answer comment Share Follow See all 10 Comments See all 10 10 Comments reply Show 7 previous comments Sushant Gokhale commented Jan 14, 2017 reply Follow Share let #keys per block = x $\therefore$ 32(x) + 8(x+1) <= 2048 $\therefore$ 40x + 8 <= 2048 $\therefore$ 40x <= 2040 $\therefore$ x = 51 (without any doubt) 1 votes 1 votes braindead commented Jan 12, 2018 reply Follow Share Eg taken from - https://courses.cs.washington.edu/courses/cse326/08sp/lectures/11-b-trees.pdf So, node structure is like : <index-pointer, key, index-pointer, key,............................, index-pointer> and not like : <key,index pointer> Thus 52 is correct, and the solution given is also correct 0 votes 0 votes braindead commented Jan 12, 2018 reply Follow Share @Sushant, you have thought of 'x' as no of keys, thus there are 51 keys, but there are 52 pointers(per node) 0 votes 0 votes Please log in or register to add a comment.