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Consider a computer system with 32-bit virtual addressing and 44-bit physical addressing and page size is 4 KB. Each page table entry contains 2 valid bits, 3 protection bits and 2 permission bits. Size of the page table when virtual memory uses single level paging ________ (MB).
Correct answer is '5'. Can you explain this answer?
Verified Answer
Consider a computer system with 32-bit virtual addressing and 44-bit ...
Number of pages = = 232 / 212 = 220
Number of frame = 244 / 212 = 232
Number of bits in page table entry =32+2 valid bit +3 protection bit +2 permission bit =39 bit = 5B
Page table size = 220 x 5B = 5MB
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Most Upvoted Answer
Consider a computer system with 32-bit virtual addressing and 44-bit ...
Given Information:

- Virtual addressing: 32-bit
- Physical addressing: 44-bit
- Page size: 4 KB
- Page table entry:
- Valid bits: 2
- Protection bits: 3
- Permission bits: 2

Calculating the Size of the Page Table:

To calculate the size of the page table, we need to determine the number of entries in the page table and then multiply it with the size of each entry.

Number of Entries in the Page Table:

The virtual address space is 32-bit, which means there can be a maximum of 2^32 = 4,294,967,296 virtual addresses.

Since the page size is 4 KB (2^12), each page table entry maps to a single page. Therefore, the number of entries in the page table is equal to the number of pages in the virtual address space.

Number of pages = (2^32) / (2^12) = 2^20

Size of Each Page Table Entry:

Each page table entry contains 2 valid bits, 3 protection bits, and 2 permission bits.

Size of each entry = (2 valid bits + 3 protection bits + 2 permission bits) / 8 (to convert bits to bytes)

Size of each entry = 7 bits / 8 = 0.875 bytes

Size of the Page Table:

The size of the page table is calculated by multiplying the number of entries in the page table with the size of each entry.

Size of the page table = Number of entries × Size of each entry

Size of the page table = 2^20 × 0.875 bytes

To convert bytes to MB, we divide by 2^20.

Size of the page table = (2^20 × 0.875) / (2^20) MB

Size of the page table = 0.875 MB

Therefore, the size of the page table when virtual memory uses single-level paging is approximately 0.875 MB.
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Consider a computer system with 32-bit virtual addressing and 44-bit physical addressing and page size is 4 KB. Each page table entry contains 2 valid bits, 3 protection bits and 2 permission bits. Size of the page table when virtual memory uses single level paging ________ (MB).Correct answer is '5'. Can you explain this answer?
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