See below one method of calculating page table size: First get page offset by calculating log2(page size in bytes). In your example, page size is 16 KBytes, so log2(16*2^10) is 14; that …

The frame size is 2KB. Assuming memory is byte-addressable, we need an offset into 2000 different bytes. 2000 is approximately (2^10)*2 = 2^11, so we need 11 bits for the frame offset. …

Calculate the number of bits required in the address for memory having size of 16 GB. Assume the memory is 4-byte addressable. Let ‘n’ number of bits are required. Then, Size of memory = …

We can find the page number and the page offset of a virtual address, if we know the size of pages. If virtual address v has m bits (virtual address space 2^m), and if the size of pages is …

Apr 30, 2017 · The size of each page is thus 4 KB (Kilobytes NOT kilobits). As Depaak said, we calculate the number of pages in the page table with this formula: Num_Pages_in_PgTable = …

Sep 21, 2023 · Physical Frame Number: 3 bits (8 frames require 3 bits to represent). Therefore, each page table entry requires $4$ bits + $3$ bits = $7$ bits. The size of the page table is …

Oct 27, 2014 · A paging system has the following parameters: 2^32 bytes of physical memory; page size of 2^10 bytes; 2^16 pages of logical address space. 1. How many bits are in a …

The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16 way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes.

Here the number of pages are 8 = 2 3, so you need 3 bits to represent page no field, 1024 = 2 (10) words per page hence you need 10 bits to represent offset. Now the number of offset bits …

May 31, 2017 · To determine how many bits are needed to represent the page offset in a logical address, we first need to understand the concept of page size. In this case, the page size is …