I have found this exercise on the Internet but Im having problems with sections 4 and 5, because I dont understand where do the solutions appear from.Example: Mapping VAs to PAsSuppose - Virtual memory of 2^32 bytes - Physical memory of 2^24 bytes - Page size is 2^10 (1K) bytes  1. How many pages can be stored in physical memory at once?2^(24-10) = 2^14  2. How many entries are ther in the page table?2^22  3. How many bits are necessary per entry in the page table? (Assume each entry has PPN, residentbit, dirty bit)16  4. How many pages does the table require?2^23 bytes = 2^13 pages  5. What is the largest fraction of VM that might be resident?1/(2^8)6. A portion of the page table is given to the left. What is the physical address for virtual address 0x1804?VPN = 6 --> PPN = 2 --> PA = 0x804$$\begin{array}{c|c|c|} \text{VPN} & \text{D} & \text{R} & \text{PPN} \\ \hline\text{0} & 0 & 0 & 7 \\ \hline\text{1} & 1 & 1 & 9 \\ \hline\text{2} & 1 & 0 & 0 \\ \hline\text{3} & 0 & 0 & 5 \\ \hline\text{4} & 1 & 0 & 5 \\ \hline\text{5} & 0 & 0 & 3 \\ \hline\text{6} & 1 & 1 & 2 \\ \hline\text{7} & 1 & 0 & 4 \\ \hline\text{8} & 1 & 0 & 1 \\ \hline\text{...} & ... & ... & ... \\ \hline\end{array}$$Could you help me with this two exercises?Thank you in advance
How many pages does the page table require?
operating systems;virtual memory
Here some of the answers seem incorrect to me.3] Number of bits required for a page table entry.Page table Entry=>   R==D==PPNR ==>1bitD ==>1bitNumber of frames =2^14 ==>14 bitsNumber of bits in a page table entry= 1+1+14=164] Number of pages required by page tableAs the page table itself will be stored in main memory it will require some pages.Amount of memory required for page table =size of an entry * Number of entries                                     =16 bits * 2^22                                     =67108864 bits =8388608 bytes                                     =8192= 2^13 pages (: 1 page=1024 bytes)5] What is the largest fraction of virtual memory resident:i.e. the amount of virtual memory that can reside in RAM.i.e Total Physical memory / Total Virtual memory                                     =2^24 /2^32                                     =1 / 2^8                                     =0.39% of virtual memory can be in RAM ant an instant of time.Please go through Virtual memory from Charles C. Lin.I hope this may help....