Consider a virtual memory system with a $16$GB virtual address space with $8$KB pages. There are $4$GB of physical memory. Answer the following questions:

Answer the following questions about the virtual memory management system $($using the same information as given above$).$ Unless the question specifies bits, be sure to back convert the answer back to bytes.

How many bits are required to hold the virtual address?How many bits are required to hold the physical memory address?How many page table entries are there in each page table?How many frames are there in physical memory?How wide is each page table entry? Remember, this depends on how many bits are required to hold the frame number $($see previous question$).$ Also, you have to round this number up to $8,16,$ or $32$ bits, representing $1,2,$ or $4$ bytes, respectively.How big is each page table?

Let's add a second level page table so we can greatly reduce the size of the page tables in memory. Again, give each answer in bytes unless otherwise specified.

How many frames does it take to hold a single page table? $($i$.$e$.,$ how many contiguous frames would need to be allocated for the primary page table, if the entire table had to be allocated?$)$How many page table entries are in each frame?How much physical memory can a single frame of page table entries cover?Assume some process has program, data, and heap size of $320$MB and a stack of $550$MB$.$ How many pages of the original page table would need to be allocated to cover these areas?How many pages are needed in the second level page table to cover the primary page table?What is the total number of pages that needs to be allocated for the new two$-$layered paging system?

Consider a main memory access time of $32$ps$.$

If we used a two$-$layered paging system, how long would it take to convert a virtual address and access the memory location?Let's add a really fast cache of $8$ps that caches the frame number for a virtual page. What would be the average access time for the memory if the cache hit rate were $87\%?$ Remember that you still have to go to the actual frame to get the data!What cache hit rate would we need if we wanted the average time to be $56$ps$?$ Again, remember that the actual data$/$code memory access of $32$ps is included in this figure. Hint: This is an algebra problem. Plug what you have into the formula and solve for what you don't have.