Question

Consider a system using the simple paging technique with a logical address of 32 bits, a page size of 2 KB, and a page table entry of 32 bits. What must be the page size in bytes such that the page table of the process should exactly fit in the frame size?

(a) 1 KB
(b) 2 KB
(c) 4 KB
(d) 8 KB

Answer 1

The page size in bytes that allows the page table to fit exactly in the frame size is 2048 bytes, which is equivalent to 2 KB.

Step-by-step explanation:

In order to determine the page size in bytes that allows the page table to fit exactly in the frame size, we need to calculate the maximum number of entries that can fit in the frame size. The frame size is given by the page size, which is 2 KB, and the page table entry size, which is 32 bits or 4 bytes. So the maximum number of entries would be frame size / entry size. Substituting the given values, we have 2 KB / 4 bytes = 512 entries.

Since the page table needs to exactly fit in the frame size, the number of entries must be equal to the number of frames. Therefore, the frame size would also be 512 entries. To convert this to bytes, we multiply the number of entries by the page table entry size, i.e. 512 entries * 4 bytes/entry = 2048 bytes.