ISRO2015-18

Question

Given a block can hold either $3$ records or $10$ key pointers. A database contains n records, then how many blocks do we need to hold the data file and the dense index

• A. $13n/30$
• B. $n/3$
• C. $n/10$
• D. $n/30$

Comments

please post the question

Answer 1

total number of block required by data file = n/3

total number of records needed by pointer = n/10. ( key pointer = no of keys)

total blocks = n/3+n/10 =13n / 30

Comments

for sparse index wt will be ans.??

---

then it will be 11n/30 because the number of pointer will be n/3/10 = n/30 blocks will be required for index file. while for record file it will be n/30+n/3= 11n/30

---

if we find no of levels then ?

---

Follow the topic ‘Dense Index’ in the below link and try to make an analogy with what is given in the question to the diagram:

https://www.guru99.com/indexing-in-database.html#3

As per diagram, there are total 6 blocks (2 for storing pointers and 4 for storing actual records).
Total no. of records = 8.

Total no. of blocks = Total no. of records/2 + Total no. of records/4

Answer 2

This is a question on simple indexing and so we should not bother about higher level of indexing..

So no of records given  =  n

 no of records per data block given = 3

So no of data blocks    =  n / 3

Now we know in dense index ,

No of <key,pointer> pairs in dense index = No of data records 

So

No of <key,pointer> pairs in 1st level index here = n

So no of  index blocks  = n/10 [ As there are 10 such pairs per index block]

So total no of data blocks and index blocks required  =  n / 3 + n / 10

                                                                             =  13 n / 30

Hence A) should be the correct answer..

Comments

how many levels ?

---

no level is here

---

Given a block can hold either 3 records or 10 key pointers.

10 key pointers means ==> 10 ( key+block pointer ) 

right ??

Answer 3

For $n$ records, blocks required = $\frac{n}{3}$

For $n$ indices in the index file, blocks required = $\frac{n}{10}$

Total = $\frac{n}{3}+\frac{n}{10}$ = $\frac{10n}{30}+\frac{3n}{30}$ = $\frac{13n}{30}$

Option A

---

Bonus:-

In case of sparse indexing:

For $n$ records, blocks required = $\frac{n}{3}$

For $\frac{n}{3}$ indices in the index file, blockes required = $\frac{n}{30}$

Total = $\frac{n}{3}+\frac{n}{30}$ = $\frac{10n}{30}+\frac{n}{30}$ = $\frac{11n}{30}$

---

For dense indexing, we have a pointer (index in the index file) for each record.

For sparse indexing, we have a pointer for each block. (More specifically for every first record of the block)