Consider a computer system with 40-bit virtual addressing and page size of…
2016
Consider a computer system with 40-bit virtual addressing and page size of sixteen kilobytes. If the computer system has a one-level page table per process and each page table entry requires 48 bits, then the size of the per-process page table is ___________ megabytes.
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Correct answer: 384
Answer: 384 megabytes.
Step-by-step reasoning:
Compute the page offset bits from the page size: 16 kilobytes = 16 × 1024 bytes = 2^14 bytes, so the offset is 14 bits.
Number of pages in the virtual address space = 2^(virtual address bits − offset bits) = 2^(40 − 14) = 2^26 pages.
Each page table entry requires 48 bits = 6 bytes.
Total size in bytes = number of pages × bytes per entry = 2^26 × 6 bytes = 6 × 2^26 bytes.
Convert to megabytes (1 MB = 2^20 bytes): (6 × 2^26) / 2^20 = 6 × 2^(26−20) = 6 × 2^6 = 6 × 64 = 384 MB.
Therefore, the per-process page table size is 384 megabytes.
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