Size of address spaces (logical and physical)

Problem Detail: 

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?

Asked By : Greg Miller

Answered By : Giorgio Bar

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.

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Question Source : http://cs.stackexchange.com/questions/41931