# Operating Systems | Memory Management | Question 9

A computer uses 46â€“bit virtual address, 32â€“bit physical address, and a threeâ€“level paged page table organization. The page table base register stores the base address of the firstâ€“level table (T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the secondâ€“level table (T2). Each entry of T2 stores the base address of a page of the thirdâ€“level table (T3). Each entry of T3 stores a page table entry (PTE). The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16 way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes. What is the size of a page in KB in this computer?

(A)

2

(B)

4

(C)

8

(D)

16

Explanation:

```Let the page size is of \'x\' bits

Size of T1 = 2 ^ x bytes

(This is because T1 occupies exactly one page)

Now, number of entries in T1 = (2^x) / 4

(This is because each page table entry is 32 bits
or 4 bytes in size)

Number of entries in T1 = Number of second level
page tables

(Because each I-level page table entry stores the
base address of page of II-level page table)

Total size of second level page tables = ((2^x) / 4) * (2^x)

Similarly, number of entries in II-level page tables = Number
of III level page tables = ((2^x) / 4) * ((2^x) / 4)

Total size of third level page tables = ((2^x) / 4) *
((2^x) / 4) * (2^x)

Similarly, total number of entries (pages) in all III-level
page tables = ((2^x) / 4) * ((2^x) / 4) * ((2^x) / 4)
= 2^(3x - 6)

Size of virtual memory = 2^46

Number of pages in virtual memory = (2^46) / (2^x) = 2^(46 - x)

Total number the pages in the III-level page tables =
Number of pages in virtual memory

2^(3x - 6) = 2^(46 - x)

3x - 6 = 46 - x

4x = 52
x = 13

That means, page size is of 13 bits
or Page size = 2^13 bytes = 8 KB ```

Quiz of this Question
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