testbook test series Database

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Suppose the search key value of a B-tree is V=8B long, the disk block size is B=512B, a record(data) pointer is P=6B long and  a block pointer is i=5B long.Calculate the order of B-tree and also in each node of B tree is 69 percent full.Find the average number of keys per node of the above tree.

Answer 1

See here  search key value of a B-tree is k=8B
 disk block size is B=512B,
 record(data) pointer is P=6B
 block pointer is i=5B  so as per our knowledge we know 
n*i+(n-1)*(k+P_r)<= disk block size    
so 
n*5+(n-1)*(8+6)<=512

 => n  <=27.68
=> n will give 27
1ans        means no. of node will be 27

and so no. of key will be 26  // ^{here sometimes we forget to do  that (no. of key = node no. - 1)}
now since each node of B tree is 69 percent full
2ans therefore  the average number of keys per node of the above tree
= 0.69 ˟ 26 =17.94  and approx we will take it 17

hope it is clear

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