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Suppose a B tree is used for indexing a database file. Consider the following information:size of the search key field $=10$ bytes, block size $=1024$ bytes, size of the record pointer $=9$ bytes, size of the block pointer $=8$ bytes.Let $K$ be the order of internal node and $L$ be the order of leaf node of B tree, then $(K,L)=$ ______.





Solution

For internal node, if order is $K$, then it has $K$ block pointers and $K-1$ keys.

So,

$8K+10(K-1)\le 1024$

$8K+10K-10\le 1024$

$18K\le 1034$

$K\le 57.44$

So, $K=57$.

For leaf node, each entry has search key and record pointer.

One extra block pointer is also needed.

So,

$L(10+9)+8\le 1024$

$19L+8\le 1024$

$19L\le 1016$

$L\le 53.47$

So, $L=53$.

Therefore, $(K,L)=(57,53)$.


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