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Suppose we have variable logical records of lengths of $5$ bytes, $10$ bytes and $25$ bytes while the physical block size in disk is $15$ bytes. What is the maximum and minimum fragmentation seen in bytes?

  1. $25$ and $5$
  2. $15$ and $5$
  3. $15$ and $0$
  4. $10$ and $5$
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4 Answers

Best answer
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6 votes

Answer : option D

         please refer this link for checking solution: http://nptel.ac.in/courses/106108101/pdf/Worked_Out_Problems/Mod_2.pdf

 Question no 2.4 in the above pdf

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Now we have Logoical Records of 5B , 10 B and 25 B 

assuming that records are spanned . 

saY record 1= 5B

Recorrd 2= 10B

Record 3= 25B

Minimum amount of fragmentation will happen if i allocate record in  R1R2R3. each will give mInimum Fragmentation value OF 5

Max amount of fragmentation will happen if i allocate record in  R1R2R3OR R2R1R3. each will give mAX Fragmentation value OF 10 

hence 10 and 5  will be answer ---Option d 

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5 votes
5 votes
Block size 15 B

Maximum we can put 10B and 5B records

So, fragmentation 0

Maximum fragmentation will be 15B

So, Answer C)
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2 votes
Since the physical block size is only 15 bytes, we will require 2 physical blocks to store 25 byte data.

One block will be fully occupied and for the second block only 10 bytes will be used resulting in unused 5 bytes.

So,

Maximum fragmentation = 10 bytes

Minimum fragmentation = 5 bytes

Option D is correct.
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