TLB paging

Question

Consider a system with 2-levels of paging and a TLB with hit rate of 95% and TLB access time of 1ns. Find the effective memory access time if there’s a data cache whose hit rate is 85% and cache access time is 1ns, and main memory access time is 100ns.

100ns.

1.   27ns
2.  ​​​​​​​ 25ns
3.  ​​​​​​​ 30ns
4.  ​​​​​​​ 20ns

Answer 1

method "1"

EMAT=TLB HIT( Cache Hit( TLB A.T + Cache A.T) + Cache Miss( TLB A.T + Cache A.T + M/M A.T)
            +TLB Miss( Cache Hit( TLB A.T + 2*M/M A.T + Cache A.T) + Cache Miss(TLB A.T + 2*M/M A.T + Cache A.T + M/MAT 

 =.95[.85*(1+1)+.15(1+1+100)] + .05[.85(1+200+1) +.15(1+200+1+100)

= 26.99 =27

Method "2"

EMAT=.95[1+.85(1)+.15(1+100)]  + .05[1+200+.85(1)+.15(1+100) ]

=27

similar example :-

https://gateoverflow.in/100831/operating-system-cache?show=100831#q100831

Comments

i can see that you have solved the question in assumption that page table is not in cache, only pages are in the cache and page table is in main memory..

since it is two level paging wouldn’t it better to have page tables in cache and pages in memory for better access time.

can anyone clear my doubt?

Answer 2

Is it 17.35 nsec?

Answer 3

Answer 1. 27ns