INDEXING

Question

Comments

what is the size of index entry.??

we have to take  SSN +7 as index entry as they said primary index is to be build using SSN value .????

Answer 1

A record size is (30+9+9+40+9+8+1+4+4)=114 and 1 B for deletion marker 1 so 115 B record size . No. of record 30000.

block size =512 B  , blocking factor floor (512/115) =4 so no. of block required celin (30000/4 ) = 7500 .

SSN= key = 6 B

KEY+block pointer = 9+ 6 = 15 B

No.of index per record   = (512/15) = 34.13 = 34

1st level :

 7500/34=220.58=221 block required for 1st level

2nd level:

221/34=6.5=7 block required for 2nd level

3rd level

1 block required (its straightforward )

and avg block access is 4 (1+1+1+1).

Comments

in spanned u r calculating like this 30000*115/512 =6738.28 =6739 but in above answer .u first calclulate to know block factor than after that calculate the no of blocks...??

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blocking factor 512/115=4.4521739130434782608695652173913 for a large amount of records the fraction part will be important . so for spanned it is better to calculate that way .

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thanku :)