memory management

Question

Comments

1 + 2⁸ + 2¹⁶

Answer 1

In such question , we should continue finding the page table size till its size becomes same as that of page size.

So 1st level page table size = (Logical address space * page table entry size)/page size

                                        = 2³⁴ * 2² / 2¹⁰ = 2²⁶ bytes which is again large and > 2¹⁰ bytes(1 KB as given for a page size)

So in 2nd level page table size = 2²⁶ * 2² / 2¹⁰ = 2¹⁸ bytes

In 3rd level page table size   =   2^{18 *} 2² / 2¹⁰ = 2¹⁰ bytes = 1 kB

So it is at the 3rd  level of paging that we are getting the page table size = page size

Hence , no of page tables required = 3

Comments

If size of entry in a page table is 4B then number of entries in a page will be 1KB/4B= 256

total no of page table entries are 2^24

then no of pages required = 2^24/256

---

and why u added 1and 2⁸ then?

---

2^16 pages of first level...

So 2^16 entries in 2nd level page table..so 2^8 pages in 2nd level..

1 page in 3rd level corresponding to 2^8 page table entries

Answer 2

The splitting for Physical address would be PA = 24|10

Now, #entries per page = 2⁸     which means 8 bits are required for offset within each page of page table.

So, for 1st level paging, the splitting would be as follows:  PA = 16| 8| 10

For 2nd level, PA = 8| 8| 8| 10

So, for 3rd level, its 8.

So, 3 levels of paging + 10 bits for accesing each frame.

Comments

SImply go on splitting using the offsets.