Question

We want to set up a virtual memory system with 4 GiB of virtual memory, mapped to 4 MiB of physical memory In order to set this up, we decide to use a page size of 4 KiB, with a single layer page table.

How many bits long is the page offset?

Answer 1

The page offset for a virtual memory system using a 4 KiB page size is 12 bits long, as 4 KiB equates to 4096 bytes and 12 bits are needed to uniquely address each byte within the page.

Step-by-step explanation:

The question asks about setting up a virtual memory system and specifically how many bits the page offset would be when using a 4 KiB page size. The page offset is used to locate a specific byte within a page. Since we’re using 4 KiB pages, we need to determine the number of bits required to address all bytes in a single page.

4 KiB equates to 4 * 1024 bytes, which is 4096 bytes. To find out how many bits are needed to represent 4096 unique values, we need to calculate the logarithm base 2 of 4096, which equals 12. Therefore, the page offset must be 12 bits long.

To find the number of bits for the page offset, we need to determine the page size. We are given that the page size is 4 KiB. Since 1 KiB is equal to 1024 bytes, the page size is 4 * 1024 = 4096 bytes. To represent any byte within a page, we need to have unique combinations for the bits in the page offset. Since 2 raised to the power of x is equal to the number of unique combinations, we can find x. In this case, 2 raised to the power of x is equal to 4096, so solving for x, we get x = log2(4096) = 12. Therefore, the page offset is 12 bits long.