Question

Consider a logical address with a page size of 8 kb. how many bits must be used to represent the page offset in the logical address?

Answer 1

8 kilobytes have addresses from hex 0x0000 to 0x1fff (=8191). You would need 13 bits for that. 2^13-1 = 8191. If you write the highest address in binary, you get:

0001 1111 1111 1111. Just count the number of 1's.

If you find this difficult to visualise, write down all logical addresses of a small page size of e.g. 8 bytes. You can see you can make all addresses with 3 bits.

Answer 2

Bits and bytes are the smallest scale of information represented on a computer.

The number of bits to represent the page offset in the logical address is 13

The page size is given as:

$\mathbf{Size = 8kb}$

Convert to bytes

$\mathbf{Size = 8\times 1024b}$

Express as exponents of base 2

$\mathbf{Size = 2^3 \times 2^{10}b}$

Apply law of indices

$\mathbf{Size = 2^{3+10}b}$

$\mathbf{Size = 2^{13}b}$

The number of bits (n) is calculated as follows:

$\mathbf{2^n = Size}$

So, we have:

$\mathbf{2^n = 2^{13}}$

Cancel out the base (2)

$\mathbf{n = 13}$

Hence, the number of bits is 13

Read more about bits and bytes at:

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