Test by Bikram | Operating Systems | Test 2 | Question: 12

Question

A certain computer provides its users with a virtual-memory space of $2^{32}$ bytes. The computer has $2^{18}$ bytes of physical memory. The virtual memory is implemented by paging, and the page size is $4,096$ bytes or $4K$ bytes.  A user process generates the virtual address $11123456$ ( in Base $16$). What is displacement (page offset) in the page?

• A. $0001 \ 0001 \ 0001$
• B. $0001 \ 0001 \ 0001 \ 0010 \ 0011$
• C. $0010 \ 0011 \ 0100 \ 0101 \ 0110$
• D. $0100 \ 0101 \ 0110$

Answer 1

The user process generated a virtual address in binary form is

             0001   0001   0001   0010   0011  0100   0101  0110

                1         1       1        2        3        4          5      6 

$$\begin{array}{|c|c|c|} \hline 0001 & 0001 & 0001 & 0010 & 0011 & 0100 & 0101 & 0110 \\ \hline 1 & 1 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline  \end{array}$$

Now it is given  the page size is $2^{12}$ , 

we take lower order 12 bits  from above binary form of virtual address,  that is  0100  0101  0110 are used as the displacement into the page .

The page table size is $2^{20}$, so remaining 20 bits ( 0001 0001  0001 0010  0011 ) are used as the displacement in the page table. 

Comments

@Bikram Sir, Here displacement means the page no in this virtual address ?

---

sir its not clear in the ques that the virtual address given is in what format? Your explanation matches only it in hexadecimal format..

---

@ parulk  

When processor generate address it generate in Hexadecimal , we assume it .. but you are correct it need to mention , i add that info now.

Thanks.

@ saxena0612

Displacement means  page offset .