Suppose size of block 1000 bytes search key of 12 bytes, pointer size 8 bytes. How many maximum records in DB file which can index by dense B+ tree of 2 levels?
a:) 1200
b:) 1250
c:)15000
d:) 2499
Doubt:- Now here if we solve for P,we get P=50
Given answer is 2499 If we index 2499 records ,then at the last level we need to point to 2499 records.
At last level we will have max. 49 keys/ptr so we can need Ceil(2499/49)=51 nodes.
At second las level we will have Ceil(51/50) => 2 nodes.
And then we need one more 3rd level. So how can 2499 is the answer?
Isnt 1250 correct?
