retagged by
356 views
1 votes
1 votes
In a B-Tree, the block size is $512$ Bytes, search key is $4$ Bytes, block pointer size is $8$ Bytes and record pointer size is $8$ Bytes. Every node contains $n-1$ record pointers, $n-1$ search keys and n block pointers (where ‘$n$’ is the order of the B-tree).

The maximum number of search keys that can be accommodated in level $0$ of the above B-tree (where root is at level $0$) is _________.
retagged by

1 Answer

Best answer
3 votes
3 votes
Let n be the order of B Tree.

8(n) + (n-1) * (4+8) ≤ 512

=> n ≤ 26

Number of search keys is one less than the order = 26 - 1 = 25
selected by
Answer:

Related questions

1 votes
1 votes
1 answer
1
Bikram asked Aug 26, 2017
399 views
Assume that a B-Tree is used as an index for a large database table which has six levels (including the root node). If a new key is inserted into this index, then the max...
2 votes
2 votes
1 answer
3
Bikram asked Aug 26, 2017
431 views
The above ER diagram depicts a book rental scheme.If this ER diagram is mapped to a relational model, to correctly depict this above scenario, the total number of relatio...
3 votes
3 votes
2 answers
4
Bikram asked Aug 26, 2017
409 views
$\sigma_{A=B \text{ and } B=C \text{ and } C=A} \bigg( \Pi_A (R) \times \Pi_B (R) \times \Pi_C (R) \bigg)$The number of rows returned by the above relational algebraic ex...