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GO Classes CS Test Series | Operating Systems | Subject Wise Test 3 | Question: 20

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1 1 vote
A unix file system has $2$ KB blocks and $4$ byte disk addresses. Each i-node contains $10$ direct entries, one singly-indirect entry and one doubly-indirect entry. Suppose half of the files are exactly $1.5$ KB and the other half of all files are exactly $2\mathrm{KB}$, what percentage fraction of disk space would be wasted?

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2 2 votes
A block is a minimum allocation unit of data. So, the Block can not be shared between the two files.

Half of the block takes $2$ KB space wasting $0$ disk storage.

Half of the block takes $1.5$ KB space wasting $0.5$ KB disk storage.

$\%$ waste of total $=(0.5 \mathrm{~KB} / 2 \mathrm{~KB}) 50 \%+(0 K B / 2 K B) * 50 \%=12.5 \%$
1 1 vote

@GO Classes Support @GO Classes Please, check this.

We know that every file is assoiated with it's own i-node. Then why don't you take the vacancy of i-node?

We have the i-node size=(10 direct address+ 1 singly-indirect + 1 double indirect) * 4B=48 B

We need two disk blocks to store the entire file, in which one disk block wastes 0.5 KB and another disk block wastes 0 KB.

So, there is two direct entries for those two disk blocks which are used for storing the actual file data and remainning address entries are wasted. So, wasted space in i-node=(10*4 B)=40 B.

So, percentage wasted space should be= (0.5 KB + 40 B)/( 2 * 2 KB + 48 B)=(512+40) B/(4096+ 48) B = 0.1332 = 13.32 %

 

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