Virtual memory 32G Physical memory 8G Page size 512 bytes Fo

Virtual memory: 32G Physical memory: 8G Page size: 512 bytes For such a system, answer the following questions: 1. Assuming simple one-level paging How is the virtual address partitioned into page number and offset? Identify how many bits are used for each. How is the physical address partitioned into frame number and offset? What is the minimum size of page table entry? Note that the page table entry must include present and modified bit. Assuming that page table entry must be 1, 2, 4, 8, 16 bytes, what is the size of the page table entry you would have? How many entries are in one page table? What is the size of the one page table in bytes? What is the size of the one page table in frames? Assuming that you have 1024 processes, what is the size of ALL page tables?

Solution

a .How is the virtual address partitioned into page number and offset? Identify how many bits are used for each.

Answer :-

The size of the page table hierarchical grows with the size of the virtual address space. If we have a large virtual address space (such as in a 64 bit architecture)

e .How many entries are in one page table?

Answer : -

From page size we get offset =10bits. For number of page, we have - 2^32/2^10 = 2^22 pages.

Since, we have two level paging system, we will page the pages,

i.e. 2^22 pages will constitute page table entry for next paging so,

So total number of entries in one page =2^10 / 2^2 = 2^8. (2^2 size of one entry)
Now, number of pages in outer page table= 2^22/2^8= 2^14
Therefore, we have 14 bits in outer page table and 22-14=8 bits for inner page table and 10 bits for offset.

F. What is the size of the one page table in bytes?

Answer :-

The total size of the page table is 2^20 entries * 2^2 bytes/entry = 2^22 bytes = 4MB.

g. What is the size of the one page table in frames?

Answer :-

Instead of making a very large sparse array, we can instead use a hash table with one entry per frame. The hash table maps page numbers to frame numbers.