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The order of an internal node in a $B+$ tree index is the maximum number of children it can have. Suppose that a child pointer takes $6$ bytes, the search field value takes $14$ bytes, and the block size is $512$ bytes. What is the order of the internal node?

  1. $24$
  2. $25$
  3. $26$
  4. $27$
asked in Databases by Veteran (69k points)
edited by | 1.5k views

4 Answers

+21 votes
Best answer
Answer: C

$14$(p-$1$) + $6$p <= $512$
$20$p - $14$ <= $512$
$20$p <= $526$
Therefore, p = $26$.
answered by Veteran (36.4k points)
edited by
+13 votes

Answer: C

lets assume order of b+ tree is n.

B+ tree internal node has  n-1 keys and n node (child) pointer.

 

Given : key size: 14 bytes
 
Child pointer : 6 bytes
 
we have ,
 
block size = (n-1) * key size + n * child pointer
512>=(n-1)*14+n*6
512>=14n-14+6n
n=(512+14)/20
n=526/20
n=26.3
n=26

 

answered by Loyal (2.6k points)
–2 votes
26 ans
answered by Active (1.4k points)
–10 votes
ans d)
answered by Boss (5.1k points)
Answer:

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