Details of a paging system for memory management are as follows:
Logical address space: 32 KB
Page Size: 4 KB
Physical Memory size: 64 KB
The number of pages in the logical address space and number of page frames in physical memory, respectively, are:
Logical address space: 32 KB
Page Size: 4 KB
Physical Memory size: 64 KB
The number of pages in the logical address space and number of page frames in physical memory, respectively, are:
Show Hint
- 4, 16
- 4, 12
- 8, 16
- 3, 16
The Correct Option is A
Solution and Explanation
Step 1: Calculate the number of pages in logical address space.
The logical address space is 32 KB and the page size is 4 KB. The number of pages is:
\[
\frac{32 \, \text{KB}}{4 \, \text{KB}} = 8 \, \text{pages}.
\]
Step 2: Calculate the number of page frames in physical memory.
The physical memory size is 64 KB and the page size is 4 KB. The number of page frames is:
\[
\frac{64 \, \text{KB}}{4 \, \text{KB}} = 16 \, \text{page frames}.
\]
Step 3: Conclusion.
Thus, the number of pages in the logical address space is 8, and the number of page frames in physical memory is 16. Hence, the correct answer is **(3) 8, 16**.
Top Questions on Paging
- Consider a page reference string as: 7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2. Assume that there are 3-page frames available. Calculate the total number of page faults for the above reference string if LRU policy is used for page replacement.
Consider a demand paging system with three frames, and the following page reference string: \[ 1\ 2\ 3\ 4\ 5\ 4\ 1\ 6\ 4\ 5\ 1\ 3\ 2. \] The contents of the frames are as follows initially and after each reference (from left to right):
The *-marked references cause page replacements. Which one or more of the following could be the page replacement policy/policies in use?
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