NIELIT 2017 OCT Scientific Assistant A (CS) - Section B: 4

Question

If a processor has $32$-bit virtual address, $28$-bit physical address, $2$ kb pages. How many bits are required for the virtual, physical page number?

• A. $17,21$
• B. $21,17$
• C. $6,10$
• D. None

Comments

option B

Answer 1

Number of bits required to address a virtual page is:  $log_{2}\left ( \frac{VirtualAddressSpace}{PageSize} \right )$

Number of bits required to a address a physical page is :   $log_{2}\left ( \frac{PhysicalAddressSpace}{PageSize} \right )$

Here Virtual Address Space $=$  $2^{32}B$ and the physical address space  $=$  $2^{28}B$,  PageSize $= 2^{11}B$

$\therefore$ Bits required for virtual page number  $=$  $log_{2}\left ( \frac{2^{32}}{2^{11}} \right )=21$  $bits$

and  Bits required for physical page number  $=$  $log_{2}\left ( \frac{2^{28}}{2^{11}} \right )=17$  $bits$

Option B is correct.

Answer 2

using logical and physical address concept …..