MadeEasy Test Series: Operating System - Memory Management

Question

The available main memory for loading processes is 128 MB which is divided into fixed number of partitions each of size 16 MB. If the processes of size 12 MB, and 6 MB are loaded into memory. The percentage of the internal fragmentation is ____________ (upto two decimal places)

Comments

because we use 2 partitions.

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one partition for 6mb and another partition for 12 mb tht's why dividing by 2*16 mb ( partition size)

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So it would be total unused upon size of partitions

Answer 1

for internal fragmentation we need to consider only allocated blocks

so 4+10/32=43.75%

say it there was process of  18MB then it would cause external fragmentation..it that case we consider non allocated blocks

so we get 6/8=3/4=75% external fragmentation

Comments

As u said for external fragmentation we need to consider non allocated blocks..right?

For 18 mb here no. Of  non allocated blocks is 6 out of 8 blocks.right??

Suppose we need to allocate processes of size 20mb 12 mb and 6 mb 8 mb then what will be external fragmentation??

 

For this,

internal fragmentation is (4+10+8)/48 right??

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Please explain the external fragmentation process. How is the 18mb process allocated blocks and how is it causing extrnl fragmntation