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Consider a system with byte-addressable memory, 32 bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the system in megabytes is ___________
  • a)
    2
  • b)
    4
  • c)
    8
  • d)
    16
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Consider a system with byte-addressable memory, 32 bit logical address...
Number of entries in page table = 232 / 4Kbyte
= 232 / 212
= 220
Size of page table = (No. page table entries)*(Size of an entry)
= 220 * 4 bytes
= 222 = 4 Megabytes
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Most Upvoted Answer
Consider a system with byte-addressable memory, 32 bit logical address...
To determine the size of the page table in megabytes, we need to calculate the number of entries in the page table and then convert it to megabytes.

Given information:
- Byte-addressable memory
- 32-bit logical addresses
- 4-kilobyte page size
- Page table entries of 4 bytes each

Calculating the number of entries in the page table:
Since the page size is 4 kilobytes, each page can hold 4 * 1024 bytes.
Since the logical addresses are 32 bits, the page table needs to have enough entries to cover the entire logical address space.

The total number of pages in the logical address space can be calculated as follows:
Total number of pages = (2^32) / (4 * 1024)

Calculating the size of the page table in bytes:
The size of each page table entry is given as 4 bytes. So, the total size of the page table can be calculated as follows:
Size of page table = (Total number of pages) * (Size of each page table entry)
= [(2^32) / (4 * 1024)] * 4

Converting the size to megabytes:
The size of the page table is currently in bytes, but we need to convert it to megabytes. Since 1 megabyte is equal to 1024 kilobytes and 1 kilobyte is equal to 1024 bytes, we can divide the size of the page table by (1024 * 1024) to convert it to megabytes.

Size of page table in megabytes = [(Total number of pages) * (Size of each page table entry)] / (1024 * 1024)

Substituting the values:
Size of page table in megabytes = [((2^32) / (4 * 1024)) * 4] / (1024 * 1024)
= [2^32] / (1024 * 1024)

Simplifying this expression, we find that it is equal to 4 megabytes.

Therefore, the size of the page table in the given system is 4 megabytes, which corresponds to option B.
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Consider a system with byte-addressable memory, 32 bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the system in megabytes is ___________a)2b)4c)8d)16Correct answer is option 'B'. Can you explain this answer?
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