TLB with fault rate.

Question

consider a paging system uses 1 level page table residing in RAM and TLB. RAM access time is 55 ns while TLB lookup takes 17 ns. Each page transfer takes upto 4700 ns to/from swap space. Miss ratio of TLB is 95% and Page fault ratio is 33%. Calculate the EMAT when TLB update time is negligible.

Answer 1

EMAT=H[ tlb + p(pfstm)+(1-p)(tm)]      +  (1-H)[tlb + tm+ p(pfstm)+(1-p)(tm)]

H=TLB Hit ratio = 0.05

tlb= time to access TLB = 17ns

p=page fault ratio =0.33

pfstm=page fault service time = 4700ns

tm=Main meneory access time= 55ns

Answer 2

 

When page fault doesn’t occur

TLB hit: time to access (TLB) + time to access (M/M)

TLB miss: time to access (TLB) + time to access (page table) + time to access (M/M)

Effective access time (EAT) = P x (TLB hit) + (1-P) x (TLB miss) = 0.05 x (17 + 55) + 0.95 x (17 + 55 + 55) = 124.25 ns

where page table resides in main memory (i.e. time to access page table = time to access main memory). And, P is the hit ratio, which is 5%, and (1-p) is the miss ratio is 95%.

 

When page faults occur

In this case, the page needs to swap in from the hard disk.

TLB hit: time to access (TLB) + time to access (M/M) + swap in from (H/D)

TLB miss: time to access (TLB) + time to access (page table) + time to access (M/M) + swap in from (H/D)

Effective access time (EAT) = P x (TLB hit) + (1-P) x (TLB miss) = 0.05 x (17 + 55 + 4700) + 0.95 x (17 + 55 + 55 + 4700) =  4824.25 ns

 

Effective Memory Access Time

= Page hit ratio x EAT + Page fault ratio x EAT = 0.67 x (124.25) + 0.33 x (4824.25) = 1675.25 ns

 

Reference: https://www.youtube.com/watch?v=Z4vzWxCcDCY&t=11s [lang: hindi]