GATE | GATE-CS-2015 (Set 3) | Question 56
Consider B+ tree in which the search key is 12 bytes long, block size is 1024 bytes, record pointer is 10 bytes long and block pointer is 8 bytes long. The maximum number of keys that can be accommodated in each non-leaf node of the tree is
Let m be the order of B+ tree m(8)+(m-1)12 <= 1024 [Note that record pointer is not needed in non-leaf nodes] m <= 51 Since maximum order is 51, maximum number of keys is 50.
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