fragmentation

Question

If holes are half as large as processes, the fraction of memory wasted in holes is 1/5 or 1/3

Comments

If total size of processes = P bytes ,   total space occupied by holes =P + P/2 bytes( holes are half as large as processes)

total memory = P+P+P/2 = (5P)/2

fraction of memory wasted in holes =( 3P/2 ) / (5P/2)=3/5

Please tell if something wrong.

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@ gari

but the given ans is 1/5

can u plzz upload any document or link regarding your ans.

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Hi ... I think earlier i misunderstood the question so i have made changes to it ...and i m still not able to get 1/5 according to my understanding.

@Bikram sir please help!!

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Total memory occupied by process=x.
 

Holes are half as large memory occupied by process : x/2
Total mem=3x/2
Memory wasted : x/2 / 3x/2=1/3


 

Answer 1

let there are n processes of size x then hole size will be x/2 

space occupied by processes= nx

space occupied by holes= nx/2

total memory= nx + nx/2 => 3nx/2

fraction of memory wasted by holes = space occupied by holes / total memory

= (nx/2) / (3nx/2)

=1/3 ans

Answer 2

I would think of memory as linear, not planar, because of the way it is addressed. Then you have as many holes as processes, so the fraction of holes is 1/2÷1/2+1=1/3

at somewhere this solution is given what it mean?