Question

Suppose you have a 16-bit machine with a page size of 16B. Assume that any unsigned 16-bit integer can be a memory address. Also, assume that the machine can support up to only 32KB of physical memory. (1K = 210)
Required:
For paging-based memory allocation, how many page table entries do you need to map the virtual address space of a process, with pure paging alone?

Answer 1

Answer:

2^11

Explanation:

Physical Memory Size = 32 KB = 32 x 2^10 B

Virtual Address space = 216 B

Page size is always equal to frame size.

Page size = 16 B. Therefore, Frame size = 16 B

If there is a restriction, the number of bits is calculated like this:  

number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]

where

physical memory size = 32KB  which is the restriction

n bit machine = frame size = 16

Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11