Consider a logical-address space of eight pages of 1024words each, mapped onto a physical memory of 32 frames.
a. how many bits are in the logical address?
b. how many bits are in the physical address?

a)13 bits

b)15 bits

Explanation:

a)

It takes 10 bits to address within a 1024-word page because 1024= 2^10.

Since the logical address space is 8 = 2^3 pages,

We will add 10 bits for 1024 words

10 + 3 = 13 bits must be the logical addresses.

b)

There are 32 = 2^5 physical pages.

We will add 10 bits for 1024 words

Similarly, physical addresses are 5 + 10 = 15 bits long .

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