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Calculates Page table Entry Size

Some questions come from calculating page table entry and page size.

So, we are discussing how to calculate page table entry with a method:

Let consider a question:

In a system with a 24-bit virtual address and 32KB page size, use a single-level page table. Calculate the page table entry size.

Solution:

Here, Virtual Address = Logical Address = 24 bit, so, Virtual Address Space (VAS) = Logical Address Space (LAS) = ${\mathbf{2}}^{\mathbf{24}}$ = 16MB and page size = 64KB = ${\mathbf{2}}^{\mathbf{16}}$

So, we can calculate 

Let page table entry = e.

Now, we know Page Table Size (PTS) = Number of pages × e

As we know frame size is always the same as page size i.e. page size = frame size = 4KB = ${\mathbf{2}}^{\mathbf{12}}$.

Now, in a single level paging system Page Table Size (PTS) is always same or less than a frame size otherwise we cannot load this page table into a single frame of main memory.

So, PTS < frame Size (FS) or Page Size (PS) 

Then 

Another Example:

Question:

A computer system supports 32-bit virtual addresses as well as 256KB page size with the use of a single-level page table. Page table entry size is less or equal to

(A) 16 B          (B) 32B          (C)  64B          (D)  256B

Solution:

Here, Virtual Address = Logical Address = 32 bit, so, Virtual Address Space (VAS) = Logical Address Space (LAS) = ${\mathbf{2}}^{\mathbf{32}}$ = 4GB and page size = 256KB = ${\mathbf{2}}^{\mathbf{18}}$

So, we can calculate 

Let page table entry = e.

Now, we know Page Table Size (PTS) = Number of pages × e

As we know frame size is always the same as page size i.e. page size = frame size = 4KB = ${\mathbf{2}}^{\mathbf{18}}$.

Now, in a single level paging system Page Table Size (PTS) is always the same or less than a frame size otherwise we cannot load this page table into a single frame of main memory.

So, PTS < frame Size (FS) or Page Size (PS)

Then

So, option (B) is correct.