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Consider a computer with 64 MB physical memory and 32 bit virtual address space. If the page size is 4KB , then the size of page table in MB is _____________    (Assume 2 bits per page table entry for extra information other than that for addressing)

### 1 comment

Sir Why are we not doing multilevel paging here as page table is not fitting in a single page

## 1 Answer

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Answer is 2MB.

We need a page table entry for each possible page (virtual) means for $2^{32}/4 KB = 2^{20}$ possible pages.

Each page table entry must address a physical page whose address can be of $\log (64 MB / 4KB) = 14$bits.

We are mentioned 2 extra bits per page table entry- so 16 bits = 2 bytes per PTE.

So, size of page table = $2 \times 2^{20} = 2 \; MB.$

### 4 Comments

Got confused with bits and converted KB to MB. Anyways thanks.
@bikram sir, What is physical and virtual page here? Is it like the pages in physical memory are called physical pages and those in virtual memory are called virtual pages??

Virtual page is like the virtual address which a program generate.

Physical page is like the actual page that what physical memory store.

In page table this virtual to physical address translation is done.

Virtual address translation refers to the process of finding out which physical page maps to which virtual page.

Read this :

https://www.bottomupcs.com/virtual_addresses.xhtml

In this question 32 bit is virtual address, not virtual address space Am I right? @Bikram sir
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