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Calculation of Physical address bit and number of frames

Calculation of the number of bits required for Physical Address:

• # Physical Address Space (PAS) =Size of main memory.
• # Physical Address Space (PAS) is divided into an equal number of frames.
• # Page size = frame size which is always a power of 2.

Let Physical Address Space (PAS) = P bytes 

Example:

PAS = 256 KB = ${\mathbf{2}}^{\mathbf{18}}$ Byte then,

So, number of bits requirement for physical address = 18 bits

Physical address contains two parts:  frame number and Offset

Calculation of the number of frames in Physical Address space:

PAS (Main memory size) = P bytes = ${\mathbf{2}}^{\mathbf{x}}$  bytes (Take it as a power of 2)

Frame size = F bytes = ${\mathbf{2}}^{\mathbf{y}}$ bytes (Take it as a power of 2)

And the number of bits for frame number in PAS = x-y

Example:

PAS = 64 MB = ${\mathbf{2}}^{\mathbf{26}}$ bytes and page size = frame size = 16KB = ${\mathbf{2}}^{\mathbf{14}}$ bytes 

Number of bits required for frame number in PAS = 12

What we know from above

And  

So, the number of bits required for frame number in PAS = x - y bits = k bits.

We can also calculate page offset directly because we consider Frame size P bytes = ${\mathbf{2}}^{\mathbf{y}}$ bytes

So, frame offset = y bits

So, here m - k bits = y bits