Size of address spaces (logical and physical)

Question

Suppose a system in which addresses (physical and logical) occupy 32 bits, page size is 1024 bytes (2¹⁰), and physical memory is of size 32MB. How many frames are in physical memory? Is the logical address space larger than physical memory? Explain.

The solution given is:

2³² byte logical address space (assuming byte-addressable machine)

2²⁵ byte physical memory

2¹⁰ byte pages

At one page per frame, there are 2²⁵/2¹⁰ = 2¹⁵ pages in physical memory. The logical memory address space is larger than physical memory.

Can someone explain how to arrive at this solution?

Answer 1

The size of a frame is the same as that of a page, so the size of a frame is 1024 bytes (2¹⁰ bytes).

If the physical memory is 32MB (2²⁵ bytes), the number of frames is 2²⁵ / 2¹⁰ = 2¹⁵ and this is also the maximum number of pages that can be present in memory at the same time.

The logical address space is larger than the physical address space, a process can address 2³² memory locations that is 2³² / 2¹⁰ = 2²² pages.

Answer 2

Since the page size is 2^10 , in the virtual address, 10 bits are allocated as offset bits. Since it has 32 bits total, the remaining 22 bits represents the Frame number. So there are 2^22 logical address frames.

The physical memory is 2^25 bytes and the frame size is 2^10 bytes. So the number of frames in the physical address space is (2^25)/(2^10) = 2^15.

Each process has its own logical address space. That may be smaller than the physical address space. But at a given time there are more than one process running. So altogether the logical address space exceeds the physical address space.