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In a database file structure, the search key field is 9 bytes long, the block size is 1024 bytes, a record pointer is 7 bytes and a block pointer is 6 bytes. The largest possible order of a leaf node in a B+ tree implementing this file structure is ________.

I am getting 63 as the answer, but in the solution, it's saying 64. Can anyone check?

63 is right ans

here it is given that we need to find the order for B+ tree leaf node

so for that we need to know the structure of B+ tree leaf node

in B+ tree leaf node, it consists of (key+record pointer) pair and a single block pointer which points to its adjacent node.

hence the formula

Order for a node is the maximum number of pointers that the node contains.

Here, in a B+ tree, leaf node has say p pairs of key and data pointers and one block pointer.

p(key + data pointer) + block pointer <= block size

p(9+7) + 6 <= 1024

p<=1018/16=63.xyz

Thus p=63...........Hence Order of leaf node= Number of Pointers( Data + Block)= 63 +1= 64

@srestha  @souravsaha  @mohitbawankar...... I think the answer 64 is correct.

Kindly correct if I am wrong.



Disk Block size = 1024 bytes

Data Record Pointer size, r = 7 bytes
Value size, v = 9 bytes
Disk Block ptr, P = 6 bytes

o(r + v) + p <= 1024
16o <= 1018
o =< 63

So Ans will be 63.

it is talking about leaf node just coz they didnt say anything and the value for record pointer is given .

so if u are suppose to consider it for and try to solve it then u will definately get the answer as 64 .