GO Classes DPP | OS | VIRTUAL MEMORY

Answer 1

Virtual Address$: 48$ bits.

Page Size: $8 \mathrm{~KB}\left(2^{13}\right.$ bytes $) \rightarrow$ Offset $=13$ bits.

VPN (Virtual Page Number): $48-13=35$ bits.

Outer $($Level $1): 10$ bits

Middle $($Level $2): 10$ bits

Inner $($Level $3): 15$ bits $(35-10-10=15)$

Total Pages needed for $\mathbf{20}$ MB:
$$
\frac{20 \times 1024 \mathrm{~KB}}{8 \mathrm{~KB}}=2560 \text { pages }
$$
Level $\mathbf{3}$ (Inner): One Level $3$ table entry points to one $8$ KB page. One Level $3$ table has $2^{15}$ entries. Since $2560<2^{15}$, we only need $\mathbf{1}$ Inner Page Table.

Level $\mathbf{2}$ (Middle): One Level $2$ table entry points to one Level $3$ table. Since we only have $1$ Inner Table, we only need $\mathbf{1}$ Middle Page Table.

Level $\mathbf{1}$ (Outer): One Level $1$ table entry points to one Level $2$ table. We only need $\mathbf{1}$ Outer Page Table.

$$
\text { Total Memory }=(1 \text { Outer }+1 \text { Middle }+1 \text { Inner }) \times 8 \mathrm{~KB}=\mathbf{2 4} \mathrm{KB}
$$

Comments

It is actually a good question @GO Classes. But, Inner table can't fit in a single 8KB page size. As Inner page talbe has 2^15 entries each with size of 2^3(8) B. 
Inner page table size = (2^15) * (2^3) = 256KB. 

Please update the solution @GO Classes @Sachin Mittal 1

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@GO Classes please respected team check your answer we feel answer should be

 {8KB+8KB+256KB }= 272KB

See the below attachment this is how Sachin sir taught us

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@GO Classes please correct the solution and options of Question, it's wasted my whole day to thinking about there is something new but letter i found Option and explanation is wrong

Answer 2

Here is the simple and Detailed Explanation