Project ForgeOperating Systems
Memory Management
GATE 2015
Medium
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
- A1
- B2
- 4
- D8
Explanation
4 KB page size needs 12 offset bits (2^12 = 4096), leaving 32−12 = 20 bits for the page number, i.e. 2^20 page table entries. Each entry is 4 bytes, so page table size = 2^20 × 4 bytes = 2^22 bytes = 4 MB.
Source: GATE CS 2015 (Session 1 / 7th Feb Shift 1) (2015)
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