Consider a machine with 16 GB of main memory and a 32-bit virtual address space, with page size 4 KB (frame size equals page size). If a process of size 34 KB is to be executed on this machine, what will be the size of internal fragmentation for this process?
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Correct answer: B. 2 KB
Explanation
Page size = 4 KB = 4096 bytes, and the process size is 34 KB = 34×1024 = 34816 bytes. The number of pages the process needs is ⌈34816/4096⌉ = ⌈8.5⌉ = 9 pages, so the OS actually allocates 9×4096 = 36864 bytes = 36 KB to it. Internal fragmentation is the unused space in the last allocated page: 36 KB − 34 KB = 2 KB, matching option (ii).More Programming Languages and Computer Graphics questions
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