simple paging system

Question

I have this question and I am not sure how to solve it:

Consider a simple paging system with the following parameters:

1. 2^32 bytes of physical memory
2. page size of 2^10 bytes
3. 2^16 pages of logical address space

How many bits are there in logical memory?

How many bytes in a frame?

Answer 1

How many bits are there in logical memory (you mean address)?

• 16 (2^16 is the number of pages) + 10 (2^10 is the size of page table) = 26 bits

How many bytes in a frame?

• A frame is the same size as a page, so 2^10 bytes are required.

Answer 2

1)size of 1 page is 2^10 bytes and there are 2^16 pages of logical address space so if we multiply both we calculate total bytes of pages in logical address;

2^16 * 2^10 = 2^26 bytes 1 byte = 8 bits so (2^26)*(2^3) is your answer.

2) page is 2^10 and 2^16 pages of logical space so again if we multiply both we find how many bytes in a frame 2^16 * 2^10 = 2^26 bytes