Question

Suppose we have a system with a virtual memory space of 1Gbytes and a physical memory space of 256Mbytes for paging.

a. If the page size is 4Kbytes, how many pages are available for caching?
b. How many entries are in the page table for a process?
c. How many bits of the virtual address are in the virtual page offset?
d. How many bits are in the physical page number?

Answer 1

The system has 64,000 pages available for caching, with a page table containing 262,144 entries. Virtual page offsets require 12 bits, and physical page numbers require at least 16 bits to properly address all possible pages in memory.

Step-by-step explanation:

Understanding Virtual Memory Paging

When dealing with virtual and physical memory, understanding how pages are managed is essential. In the given scenario, we have a system with a virtual memory space of 1 Gigabyte (GB) and a physical memory space of 256 Megabytes (MB) for paging.

1. To calculate the number of pages available for caching in physical memory, divide the physical memory size by the page size: 256 MB / 4 KB, which equals 64,000 pages for caching.
2. The number of entries in a page table for a process is equal to the number of pages in the virtual memory space, calculated as 1 GB / 4 KB, resulting in 262,144 entries.
3. The number of bits in the virtual page offset is based on the page size. Given a 4 KB (or 4,096 bytes) page size, we need 12 bits to represent all possible offsets within a page, as 2^12 equals 4096.
4. The number of bits in the physical page number is determined by the number of pages that can be cached in physical memory. For 64,000 pages, we need at least 16 bits since 2^16 is 65,536, which is the nearest power of two.