GATE CSE 2008 | Question: 70

Question

Consider a file of $16384$ records. Each record is $32\;\text{bytes}$ long and its key field is of size $6\;\text{bytes}$. The file is ordered on a non-key field, and the file organization is unspanned. The file is stored in a file system with block size $1024\;\text{bytes}$, and the size of a block pointer is $10\;\text{bytes}$. If the secondary index is built on the key field of the  file, and a multi-level index scheme is used to store the secondary index, the  number  of  first-level  and second-level  blocks  in  the  multi-level  index  are respectively

• A. $8$ and $0$
• B. $128$ and $6$
• C. $256$ and $4$
• D. $512$ and $5$

Comments

How can we decide whether level is from left hand side or rhs as in OS in multilevel paging first level means 0th level?

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                                  $1^{st} level$                   $2^{nd} level$

$\fbox{16384:records}$   $\fbox{16384:indexes}$         $\fbox{256:indexes}$

DB                      $\left \lceil{\dfrac{16384}{64}}\right \rceil=256$     $\left \lceil{\dfrac{256}{64}}\right \rceil=4$

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key point to solve this question  index file only contain: key value+pointer(address to required block)

Answer 1

it's very simple..

Block size is 2¹⁰ bytes and if you see option C .. it's 256 and 4 which is 2⁸ and 2² 

So only option C satisfies ​​​^​​​​​​^​​​​​​.

Comments

is it any shortcut?

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As it is dense index, we need to use record pointer. But in the question block pointer is given ie 10B. Why are using block pointer here when we should be using record pointer?

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Finally ,someone think , why block ptr is used in place of record ptr.

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if block ptr or record ptr any of them is given then we have to consider it for both.