Consider a computer system with a 32-bit logical address and 4-KB page size. The system

supports up to 512 MB of physical memory. How many entries are there in each of the following?

a. A conventional single-level page table?

b. An inverted page table?

Respuesta :

Answer:

Conventional single-level page table  [tex]2^{20}[/tex]  pages

Inverted page table are [tex]2^{17}[/tex] frame

Explanation:

given data

logical address = 32-bit = [tex]2^{32}[/tex]  Bytes

page size = 4-KB =   [tex]2^{12}[/tex] Bytes

physical memory = 512 MB  = [tex]2^{29}[/tex]  bytes

solution

we get here number of pages that will be

number of pages = [tex]\frac{logical\ address}{page\ size}[/tex]   ..............1

put here value

number of pages =   [tex]\frac{2^{32}}{2^{12}}[/tex]

number of pages = [tex]2^{20}[/tex]  pages

and

now we get number of frames  that is

number of frames  = [tex]\frac{physical\ memory}{page\ size}}[/tex]   ............2

number of frames  = [tex]\frac{2^{29}}{2^{12}}[/tex]

number of frames  = [tex]2^{17}[/tex] frame

so

Conventional single-level page table  [tex]2^{20}[/tex]  pages

and

Inverted page table are [tex]2^{17}[/tex] frame