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Some Important Example 3:

Question:

Consider the system which has Logical Address Space (LAS) 512KB, number of pages 1KB, number of frames 256 Byte, and page offset 10. Memory is byte-addressable then calculates Physical Address Space (PAS), Page size, and page table size.

You should know:

Before seeing the solution please follow Calculation of offset for logical and physical address, Calculation of logical address bit and number of pages, Calculation of physical address bit and number of frames, Calculation of page table size. 

Solution:

Given data

LAS = 512KB = ${\mathbf{2}}^{\mathbf{19}}$, number of pages = 1KB = ${\mathbf{2}}^{\mathbf{10}}$, offset = 10 and number of frames = 256B

So, the number of bits required for Logical Address (LA) = 19 bits.

Page Size:

Now offset = 10 is given.

[For details see: Calculation of offset for logical and physical address]

Physical Address Space (PAS):

So, now we can calculate

So,  the  number of bits required for Physical Address (PA) = 18 bits.

Page Table Size (PTS): 

Now, if we elaborate the address fields of Logical Address (LA):

Now, how bits require for page number = Total size of LA - offset

                                 =19  - 10 = 9 bits                      

Now Page Table Size (PTS) = Number of pages × Page entry size.

But page entry is not given here and as we know if page entry is not given then we take frame number (number of bits for frame number) as page entry. 

So, number of bits for frame number at PA =  8 bits