Made Easy Test Series : Assume a system has a 64bit VAS and 48 bit PAS. Page size is equal to 4KB. If only single level paging is used then the size of the page table (in TB)

Question

 

---

Can anybody explain why are they dividing the number of pages by page size?
According to me, it should be – 

Page table size = Number of entries * entry size

i.e PT size = #pages * entry size

                   = [(2^52 * 36 )/2^40 * 8] = 18432 TB (which is nowhere close to the answer provided)

Can anybody explain what am I missing?

---

Answer 1

When we have to find the Page table size we have a formula

                              Page Table Size = No.of Entries In The Page Table * Entry Size (in bytes)

Here point to understand is that 

                                        Entry size = no. of bits for the frame + other some extra bits like I/V bits, dirty bits etc

so if entry size is not given then we find the bits required for the frames , the that becomes our frame size 
In the above question they are doing the same (Finding the number of buts for the frames and convert it in bytes) 

I hope this explaination is sufficient

 

Comments

I did calculate the frame bits in my solution, i.e, log (2^48/ 2^12) = 36 (You can find the approach quoted below)
My doubt is in the first part of the formula, which is the number of entries in the page table
Number of entries in the page table should be equal to the number of pages of VAS, right?
 

Page table size = Number of entries * entry size

i.e PT size = #pages * entry size

                   = [(2^52) * (36 )] bits

                   = [(2^52) * (36 ) / 2^40 * 8] TB

                   = 18432 TB

---

True that,  I am getting the same , I think Question is not correctly solved

---

same answer i got 18432 TB