[Solution] UGC NET 2021 | process of size 34KB is to be executed

UGC NET 2021 | process of size 34KB is to be executed

Comprehension:
Read the following and answer the questions:
Consider a machine with 16 GB main memory and 32-bits virtual address space, with page size as 4KB. Frame size and page size is same for the given machine.
If a process of size 34KB is to be executed on this machine, then what will be the size of internal fragmentation for this process?

NTA UGC NET June 2021

Operating System

Memory Management

i➥ 1 KB

ii➥ 2 KB

iii➥ 4 KB

iv➥ Zero

Answer: Option II 

Solution:
Internal Fragmentation
Internal fragmentation occurs when memory is partition into fixed size. If memory block size assign to process is bigger than size of that process. Due to this, some part of the memory is left unused and this cause internal fragmentation.

For example: Suppose memory block size is 4KB and process size is 3KB, then if 4KB memory block size is assign to 3KB of process, then 1KB of memory is left unused. In this case, size of internal fragmentation is 1KB.


Now, let's come to solution

Given,
Page size = 4KB,
Process size = 34KB.

Calculation,


Therefore, out of 34KB of process size, 32KB (8*4KB) of space is occupied by 8 pages having no internal fragmentation. Remaining 2 KB of space is left unused and this 2KB is the size of internal fragmentation.
                            

So, Option(II)is correct.