UGC NET 2021 | Find number of pages required for
| 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. Find number of pages required for the given virtual address space? |
| i➥ 210 |
| ii➥ 212 |
| iii➥ 220 |
| iv➥ 230 |
Answer: Option III
Solution:
Given,
Logical Address Space = 32 bits,
Page size = 4KB = 212.
Formula used,
Page offset = ⌈log2(page size)⌉,
Total number of pages = 2page number.
Calculation,
Page offset = ⌈log2(212)⌉
= ⌈12*log22⌉
= ⌈12⌉
= 12 bits
Note: log22 = 1
Page number = 32-12 = 20 bits.
Therefore, total number of pages = 220
So, Option(III)is correct.
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