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Calculation of Logical address bit and Number of pages

Calculation of the number of bits required for Logical Address:

• # Logical Address Space (LAS) = Size of process
• # Logical Address Space (LAS) is divided into an equal number of pages
• # Page size is always a power of 2.

Let Logical Address Space (LAS) = L bytes 

Example:

LAS = 128 MB = ${\mathbf{2}}^{\mathbf{27}}$ Byte then

So, number of bits requirement for logical address = 27 bits

We know Logical address contains two parts: page number and Offset

Calculation of the number of pages in Logical Address Space:

LAS (process size) = L bytes = ${\mathbf{2}}^{\mathbf{m}}$ bytes (We take as a power of 2)

Page size = P bytes = ${\mathbf{2}}^{\mathbf{n}}$ bytes (We take as a power of 2)

And the number of bits for the page number in LAS = m - n

Example:

LAS = 256KB = ${\mathbf{2}}^{\mathbf{18}}$ bytes and page size = 16KB = ${\mathbf{2}}^{\mathbf{14}}$  bytes 

Number of bits required for the page number in LAS = 4

What we know from above

Now,

And  

So, the number of bits for the page number in LAS = m - n bits  = k bits.

We can also calculate page offset directly because we consider 

Page size = P bytes = ${\mathbf{2}}^{\mathbf{n}}$ bytes

So, page offset = n bits

So, here w - k bits = n bits