(Solved): Consider an operating system that uses 48-bit virtual addresses and 16KB pages. The system uses a mu ...

Consider an operating system that uses 48-bit virtual addresses and 16KB pages. The system uses a multi-level page table design to store all the page table entries of a process, and each page table entry is 4 bytes in size. What is the total number of page that are required to store the page table entries of a process, across all levels of the page table? You may follow the hint below or finish from scratch to fill the blanks. Please show your calculations to get partial points like 2^10/2^4=2^6. 1. We need to calculate the total number of page table entries needed for a process (i.e., the total number of pages for a process) 2. We need to calculate how many entries each page can store 3. With 1 and 2, we can calculate how many pages needed for the lowest (innermost) level 4. Each page from 3 requires an entry (pointer) in the upper (next) level. We need to calculate how many pages are required to store this next level entires (please note the entry size is always 4 bytes. i.e., the number of entries that can be stored in each page is alway the number from 2) 5. So on and so forth until one directory page can hold all entries pointing to its inner level. Now, we can calculate the total number of pages required to store all page table entries Show transcribed image text Consider an operating system that uses 48-bit virtual addresses and 16KB pages. The system uses a multi-level page table design to store all the page table entries of a process, and each page table entry is 4 bytes in size. What is the total number of page that are required to store the page table entries of a process, across all levels of the page table? You may follow the hint below or finish from scratch to fill the blanks. Please show your calculations to get partial points like 2^10/2^4=2^6. 1. We need to calculate the total number of page table entries needed for a process (i.e., the total number of pages for a process) 2. We need to calculate how many entries each page can store 3. With 1 and 2, we can calculate how many pages needed for the lowest (innermost) level 4. Each page from 3 requires an entry (pointer) in the upper (next) level. We need to calculate how many pages are required to store this next level entires (please note the entry size is always 4 bytes. i.e., the number of entries that can be stored in each page is alway the number from 2) 5. So on and so forth until one directory page can hold all entries pointing to its inner level. Now, we can calculate the total number of pages required to store all page table entries