What is the effective average instruction execution time in a 2-level paging scheme?

Consider a system with a two-level paging scheme in which a regular memory access takes 150 nanoseconds, and servicing a page fault takes 8 milliseconds. An average instruction takes 100 nanoseconds of CPU time, and two memory accesses. The TLB hit ratio is 90%, and the page fault rate is one in every 10,000 instructions. What is the effective average instruction execution time?

  1. 645 nanoseconds
  2. 1050 nanoseconds
  3. 1215 nanoseconds
  4. 1230 nanoseconds
3Comments
shivani @shivani1234
2 Sep 2017 08:22 pm

All options are wrong
Reason being following:
effective average instruction execution time = cpu time +emat
emat(effective memory access time) = address translation time + (total time in accessing data from memory)

= t+(1-pt)*k*m + m +pfr*pfst
=[0+ (1-09)*2*150 +150 +800/1000000 ]ns = 980ns

So, effective average instruction execution time = 100 + 2*980 2060ns

set2018 @setgate
2 Sep 2017 09:16 pm

@shivani1234 Ma'am

Why we are not  considering this?

tlb hit rate (tlb access+ memory access)+tlb miss rate (tlb access+2level page access+memory acess)

after that it may be chance that instruction is not present in memory (page fault)

then

page fault rate*page service time +page hit rate *memory access

shivani @shivani1234
3 Sep 2017 11:57 pm
since, tlb hit rate +tlb miss rate =1
tlb hit rate (tlb access+ memory access)+tlb miss rate (tlb access+2level page access+memory acess)
so. it can be written as tlb access+ memory access+tlb miss rate (2level page access)
try to rearrange your expression , then u will find my exp and yours are same