306 views
0 votes
0 votes

1 Answer

1 votes
1 votes

In question it is asking no. of entries but it should be no. of bits 

 

VA= 52 bits ,PA= 32 bits, Page size = 16KB , page offset bit(PO)= 14

number of bits for Frame number = PA - PO = 32-14 =18

page table entry size = additional bits + frame number = 14 + 18 = 32 = 4 bytes

No. of entries in 1st level PT = $\frac{2^{52}}{2^{14}}$ = $2^{38}$

size of 1st level PT = $2^{38} * 2^{2} B$ = $2^{40}B$

to address no. of entries in one page in PT1 bits required = $\frac{Page Size}{Page Table entry Size}$ =$ \frac{2^{14}}{2^{2}}$ = $2^{12}$ = 12 bits

----------------------------------------------------- 

No. of entries in 2nd level PT = $\frac{2^{40}}{2^{14}}$ = $2^{26}$

size of 2nd level PT = $2^{26} * 2^{2} B$ = $2^{28}B$

to address no. of entries in one page in PT2 bits required = $\frac{Page Size}{Page Table entry Size}$ =$ \frac{2^{14}}{2^{2}}$ = $2^{12}$ = 12 bits

 

----------------------------------------------------- 

No. of entries in 3rd level PT = $\frac{2^{28}}{2^{14}}$ = $2^{14}$

size of 3rd level PT = $2^{14} * 2^{2} B$ = $2^{16}B$

to address no. of entries in one page in PT3 bits required = $\frac{Page Size}{Page Table entry Size}$ = $\frac{2^{14}}{2^{2}}$ = $2^{12}$ = 12 bits

----------------------------------------------------- 

No. of entries in 4th level PT = $\frac{2^{16}}{2^{14}}$ = $2^{2}$

size of 4th level PT = $2^{2} * 2^{2} B$ = $2^{4}B$ (will fit into one page)

to address no. of entries in one page in PT4 bits required = VA - PT1 - PT2- PT3 - PO = 52- 12 - 12 -12 -14 = 2 bits

 

Answer <2,12,12,12>

edited by

Related questions

0 votes
0 votes
0 answers
1
Magma asked Aug 16, 2018
458 views