ISRO2014-66

Question

Consider a system where each file is associated with a 16 bit number. For each file, each user should have the read and write capability. How much memory is needed to store each user's access data?

• A. 16 KB
• B. 32 KB
• C. 64 KB
• D. 128 KB

Comments

i think need to know about total number of files and no. of users?

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everybody is thinking only.. plz any1 tell the correct way...

Answer 1

I think answer is A .

Question is that how much memory is needed to store each user's access data so its only the memory reqd for read/write information  which is asked. For 2^16 files, there are 4 different combinations of read(R) write(W). Read, write, no read, no write.

So bits reqd for 1 file to store access info=2

Bits reqd for 2^16 files=(2^16)×2 bits=128Kb

Converting this bits into bytes we get 128Kb/(8bits/Byte)

=16KB

Answer 2

Files are associated with 16 bit number.

=> File ID ranges from 0000 0000 0000 0000 to 1111 1111 1111 1111.

=> Total File IDs possible = Total files possible = $2^{16}$

 

For each user, we need 2 bits. 1 bit for read (0 => can't read; 1 => can read)

and 1 bit for write (0=> can't write; 1 => can write)

 

So, $2^{16}*2bits$  =  $2^{16}*2/8bytes$  =  $2^{14} bytes$  =  $16KB$

Answer 3

i think answer will be a. plz post the answer if correct. total number of files by 16 bit number will be 2^16 and assuming 2 bit for read and write as permission can be no read no write no read but write .... so on 4 cases . 2 bit for that . and the user will have these 2 byte for all the files so 2^16*2=128Kbit

128/8=16kB

Comments

according to key answer is correct that is option (a)...

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i missed the bit  . its asked in kB. so i missed the divide

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@Tendua  I didn't understand this part "and the user will have these 2 byte for all the files" of your sentence. Can you please elaborate on this?