Question

Consider a process with a logical address space of 4 pages of 1024 bytes per page, mapped onto a physical memory of 64 frames.

A. How many bits are there in the logical address?

B. How many bits are there in the physical address?

C. Given the following page table map: page 0 is mapped to frame 3, page 1 is mapped to frame 14, page 2 to frame 6, and page 3 to frame 33, what is the physical address of page 2 byte 256?

Answer 1

a)

Logical address space size = 4*(2^(10)) bytes = 2^(12) bytes

So log(2^(12)) = 12 bits will be required for logical address space

b)

Frame size = page size

so physical memory has 64 pages or we can say that

Physical address space size = 64*(2^(10)) bytes = 2^(16) bytes

So log(2^(16)) = 16 bits will be required for physical address space

c)

Number of bytes before 256th byte of frame 6 = 5*1024+255 = 5375

So we have to find the binary representation of 5376 to find the address of 256th byte of page 2 =

Which is 0001 0101 0000 0000

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