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Consider a system using paging and segmentation. The virtual address space consist of up to 8 segments and each segment is 229 bytes long. The hardware pages each segment into 28 byte pages.
How many bits in the virtual address specify the offset within page?
  • a)
    8 bits
  • b)
    16 bits
  • c)
    32 bits
  • d)
    64 bits
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Consider a system using paging and segmentation. The virtual address s...
Virtual address space = 8 x 229 bytes
= 232 bytes
No. of pages within each segment

Offset within the page = 8 bits (for 28 bytes)
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Most Upvoted Answer
Consider a system using paging and segmentation. The virtual address s...
Explanation:


The virtual address space consists of up to 8 segments, and each segment is 229 bytes long. The hardware pages each segment into 28-byte pages. So, we need to find out how many bits in the virtual address specify the offset within page.

Step 1: Calculate the number of pages per segment


Page size = 28 bytes


Segment size = 229 bytes


Number of pages per segment = 229 bytes / 28 bytes = 8.1785 ≈ 8 pages

Step 2: Calculate the number of bits required to specify the page number


Number of pages per segment = 8 pages = 23 pages


So, we need 3 bits to specify the page number (23 = 8)

Step 3: Calculate the number of bits required to specify the offset within page


Page size = 28 bytes


So, we need log2(28) = 4.8074 ≈ 4 bits to specify the offset within page

Step 4: Calculate the total number of bits required for the virtual address


Total number of bits = Number of bits for segment number + Number of bits for page number + Number of bits for offset within page


Total number of bits = log2(8) + 3 + 4 = 3 + 3 + 4 = 10 bits

Therefore, the correct answer is option A, 8 bits.
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Consider a system using paging and segmentation. The virtual address space consist of up to 8 segments and each segment is 229 bytes long. The hardware pages each segment into 28 byte pages.How many bits in the virtual address specify the offset within page?a)8 bitsb)16 bitsc)32 bitsd)64 bitsCorrect answer is option 'A'. Can you explain this answer?
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