Gate2020 cs Q47

47Consider a paging system that uses a 1-level page table residing in main memory and a TLB for address translation. Each main memory access takes 100 ns. TLB lookup takes 20 ns. Each page transfer to/from the disk takes 5000 ns. Assume that the TLB hit ratio is 95%, page fault rate is 10%. Assume that for 20% of the total page faults, a dirty page has to be written back to disk before required page is read in from disk. TLB update time is negligible. The average memory access time is ns (round off to 1 decimal places) is ______.

Note – Numerical Type question

Option : D
Explanation :
Main memory access time, T_m = 100 ns
TLB lookup, T_TLB = 20ns
Page transfer time, T_PT = 5000 ns
TLB hit ratio, x = 0.95 (95%)
page fault rate, p = 0.10 (10%)

We know,
EMAT for multilevel paging,
EMAT = x (T_c + T_m) + (1 – x) (T_c + (n + 1) T_m)
EMAT, when there is a page fault, S → is service time
EMAT = (1 – P) T_m + Ps
Here, we are using TLB, and page fault occurs whenever there is a miss in TLB, So the required EMAT is ,
EMAT = x(T_tlb + T_m) + (1 – x) [(1 – P) (T_tlb + T_m + T_m) + p(% dirty (T_tlb + T_m + 2T_PT) + % clean (T_tlb + T_m + T_PT)
∴ EMAT = 0.95 (20 + 100) + 0.05 (0.9 (20 + 100 + 100) + 0.1 (0.2 (20 + 100 + 2(5000)) + 0.8 (20 + 100 + 5000))
= 154.5 ns

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