Q10.Marks: +2.0UGC NET Paper 2: Computer Science 26th Nov 2021
📄 Passage
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?
1.4KB
2.Zero
3.1KB
4.2KB✓ Correct
Solution
The correct answer is 2 KB.
Key Points
According to the question, the total size of the main memory is 16 GB. The size of the virtual address space is 32 bits. The page size is 4KB. It is also given that the process size is 34 KB.
So, the internal fragmentation can be calculated by dividing the process size by page size. If there is no reminder then there is no internal fragmentation. But if there is a remainder, then the remainder is known as internal fragmentation.
So, the internal fragmentation is = 34KB/ 4KB. After the division, the result is 8 and the remainder is 2. So, the internal fragmentation is 2KB.
Thus the correct answer is2KB.
Additional Information
Internal fragmentation happens when the partition of the memory divides into a fixed memory size.
The page is referred to as a continuous memory block with a fixed memory size in an operating system. But it is virtual memory. The page is the smallest data set unit in the virtual memory.
The frame in the operating system is referred to as a fixed length of the physical block. The frame is often found on the RAM block.
