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GATE | GATE-CS-2015 (Set 2) | Question 65

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A Computer system implements 8 kilobyte pages and a 32-bit physical address space. Each page table entry contains a valid bit, a dirty bit three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is _______________ bits

(A)

36

(B)

32

(C)

28

(D)

40


Answer: (A)

Explanation:

Max size of virtual address can be calculated by 
calculating maximum number of page table entries.

Maximum Number of page table entries can be calculated 
using given maximum page table size and size of a page 
table entry.

Given maximum page table size = 24 MB

Let us calculate size of a page table entry.

A page table entry has following number of bits.
1 (valid bit) + 
1 (dirty bit) + 
3 (permission bits) + 
x bits to store physical address space of a page.

Value of x = (Total bits in physical address) - 
             (Total bits for addressing within a page)
Since size of a page is 8 kilobytes, total bits needed within
a page is 13.
So value of x = 32 - 13 = 19

Putting value of x, we get size of a page table entry =
                                   1 + 1 + 3  + 19 = 24bits.

Number of page table entries 
           = (Page Table Size) / (An entry size)
           =  (24 megabytes / 24 bits)                             
           =  223

Virtual address Size 
             = (Number of page table entries) * (Page Size)
             =  223 * 8 kilobits
             = 236  
Therefore, length of virtual address space = 36


Quiz of this Question
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Last Updated : 28 Jun, 2021
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