Doubt

Question

Consider a system having 4K entries. LA = 29 bits. Calculate page size(PS)?

Solution: I know that ,

PS = LAS/No. of pages ...............so,

PS = 2^29 (bits)/2^12 = 2^17 KB

How it comes as KB. I guess it should be 2^14 KB( dividing by 8 to convert it in bytes)

Please tell me whether conversion is required or not. In some places i see the conversion and some places no conversion.

Comments

Given LA= 29 bits then How size - 2²⁹ bits or 2²⁹ Bytes?

By default take Byte addressable so 2²⁹​​​​​​​ Bytes.

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Post (or) give link of the questions where you see conversion.

Answer 1

LA $= 29$ bits, it means that $29$ bits are required to address complete process.

So, Process Size $= 2^{29}$ Bytes.

So process Size $= 2^{29}B$ and number of pages $= 2^{12}$

So, Page size $= \frac{2^{29}}{2^{12}} = 2^{17}B = 128KB$

I think you may be talking about conversion in case when Page table entry is given as  $14$ bit. then You need to convert it to Bytes, but it's better if you give specific question.

Answer 2

Imagine LAS as an array. Each cell of the array has an index. Each cell size = 1B(1 word = 1B is general assumption).

To point to specific index of this array we need 29 bits.

Therefore no of cells in the array are 2^29.

Each cell is of size 1B, Hence LAS = 2^29 * 1B = 2^29 B

Therefore PS = 2^29B / 2^12 = 2^17B = 128KB

Hope this clears the doubt.