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Consider a paging system with a page size of 4 KB. If a process of size 38 KB is in logical address space, find the internal fragmentation in Kbytes.
- a)4
- b)5
- c)2
- d)7
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Consider a paging system with a page size of 4 KB. If a process of siz...
Calculation of Internal Fragmentation in Paging System
Given:
Page size = 4 KB
Process size = 38 KB
Solution:
To calculate the internal fragmentation in paging system, follow the below steps:
Step 1: Calculate the number of pages required for the process
Number of pages = Process size / Page size
= 38 KB / 4 KB
= 9.5 pages
≈ 10 pages (as we can't have fractional pages)
Step 2: Calculate the total memory allocated to the process
Total memory allocated = Number of pages * Page size
= 10 * 4 KB
= 40 KB
Step 3: Calculate the internal fragmentation
Internal fragmentation = Total memory allocated - Process size
= 40 KB - 38 KB
= 2 KB
Therefore, the internal fragmentation in the paging system is 2 KB.
Hence, option (c) is the correct answer.
Given:
Page size = 4 KB
Process size = 38 KB
Solution:
To calculate the internal fragmentation in paging system, follow the below steps:
Step 1: Calculate the number of pages required for the process
Number of pages = Process size / Page size
= 38 KB / 4 KB
= 9.5 pages
≈ 10 pages (as we can't have fractional pages)
Step 2: Calculate the total memory allocated to the process
Total memory allocated = Number of pages * Page size
= 10 * 4 KB
= 40 KB
Step 3: Calculate the internal fragmentation
Internal fragmentation = Total memory allocated - Process size
= 40 KB - 38 KB
= 2 KB
Therefore, the internal fragmentation in the paging system is 2 KB.
Hence, option (c) is the correct answer.
Community Answer
Consider a paging system with a page size of 4 KB. If a process of siz...
Concept:
Internal fragmentation:
Internal fragmentation occurs when memory is divided into fixed-sized blocks. Internal fragmentation occurs when the memory assigned to the process is somewhat larger than the memory requested.
Internal fragmentation:
Internal fragmentation occurs when memory is divided into fixed-sized blocks. Internal fragmentation occurs when the memory assigned to the process is somewhat larger than the memory requested.
The given data,
Size of page = 4 KB
Size of process = 38 KB
The number of pages required by process = (38 KB)/(4 KB) = 9.5
So, the number of pages required by the process is 10.
But required = 38 KB
Here 10 x 4 = 40 KB
40 – 38 = 2 KB is the internal fragmentation.
Hence the correct answer is 2.
Size of page = 4 KB
Size of process = 38 KB
The number of pages required by process = (38 KB)/(4 KB) = 9.5
So, the number of pages required by the process is 10.
But required = 38 KB
Here 10 x 4 = 40 KB
40 – 38 = 2 KB is the internal fragmentation.
Hence the correct answer is 2.
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Consider a paging system with a page size of 4 KB. If a process of size 38 KB is in logical address space, find the internal fragmentation in Kbytes.a)4b)5c)2d)7Correct answer is option 'C'. Can you explain this answer?
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