In: Computer Science
Consider the “index calculation” problem we have studied in the class (see lecture note), where we have a relation of 8,000,000 tuples. Again, make the following assumptions:
described by databaseSystems-71-indexing, page 2? [16 points]
No of records = 800000
Disk Block Size = 4000B
Key = 40B
Page address size = 10B
Swap time = 10ms
1)Use Dense Indexing (level 1)
No of entry in index = no of rewards (dense)
No of entries in index table = 800000
size of each entry in index table = key + address size
= 40 + 10 = 50B
Number of blocks to search in index table =
= 22.9315 units = 23 units
2)Using level 2 index table with sparse indexing
no of records in level 1 of index = 800000
size of records in level 1 of index = 50B
no of entries in index which can be stored in 1 block = Block size / record size of level 1 index
= 4000 / 50 = 80
no of blocks needed for level 1 index = 800000 / 80 = 100000
For level2 indexing, 100000 entries are required since number of blocks at level1 are 100000 &
we use sparse index here.
No of blocks to search in level2 =
= 16.6
17
Total no of blocks to search = 17+1+1 = 19
Total time = 19*10ms = 190ms.