401 views
1 votes
1 votes

A database relation has 5000 records block can hold either 10 records or 15 keys and pointer pairs. If sparse index is used at 1st level and multilevel indexing is used in system, then the number of disk block required to store relation and index is _______

2 Answers

1 votes
1 votes
Here total 5000 records and each block can take at most 10 records So.

No of block in the for relation will be = 5000/10 => 500

No of block is required for 1st level index will be 500/15 => 33.33 which will be 34

#block at second level = 34/15 => 2.26 => 3

#bloc at level 4 because in multi level index we try to put last index in one block but in upper case here 3 block are in use so ceil(3/15) => 1

So total #blocks will be 500+34+3 +1=538

This question i have calculated for dense Index.
0 votes
0 votes
i think

to store relation 500 blocks req

and for index 34

No related questions found