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In a database file structure, the search key field is $9$ $bytes$ long, the block size is $512$ $bytes$, a record pointer is $7$ $bytes$ and a block pointer is $6$ $bytes$. The largest possible order of a non-leaf node in a$ B+$ tree implementing this file structure is

  1. $23$
  2. $24$
  3. $34$
  4. $44$
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3 Answers

Best answer
39 votes
39 votes

Answer is (C).

From the structure of $B+$ tree we can get this equation:

$n\times p + (n-1)\times k \leq B$ ( for non leaf node)

Here, n=order, p=tree/block/index pointer,  B=size of block

I non leaf node no record pointer is there in B+ tree.

So, $n\times p + (n-1)k \leq B$

$n\times 6 + (n-1)\times 9 \leq 512$

$\implies n \leq 34.77$

Largest possible value for $n$ is $34.$

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4 votes
4 votes

option C

Because in non leaf node of the B+ tree there is not any record pointer 

so formula will be  n*p + (n-1)*(k) <= B ( for non leaf node) 

put the values and get the answer k=9 and  p=6 and B=512 

so answer is 34

–7 votes
–7 votes
option B) 24
Answer:

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