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ISRO | ISRO CS 2014 | Question 35
• Last Updated : 16 May, 2018

Using the page table shown below, translate the physical address 25 to virtual address. The address length is 16 bits and page size is 2048 words while the size of the physical memory is four frames.

```Page   Present(1-In 0-out)   Frame
0      1                     3
1      1                     2
2      1                     0
3      0                     -```

(A) 25
(B) 6169
(C) 2073
(D) 4121

Explanation: Given, virtual address size is 16 bits and page size is 211 bytes. Therefore,
Number of pages = 216/211 = 25

So, physical address = (number of frames) * (size of each frame) = 4 * 211 = 213

Physical address (25)10 = (0000000011001)2 in 13 bits
Here first two bits represent frame number and another 11 bits represent the offset within the page: (00 00000011001)2

According to the page table: frame 00 is mapped on with page number 2
Page # 2 = (00010)2
Offset within the page = (00000011001)2
Therefore, 16 bit virtual address = (00010 00000011001)2 = (4121)10

So, option (D) is correct.

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