Consider a 32-bit machine where four-level paging scheme is used. If the hit ratio to TLB is 98%, and it takes 20 nanosecond to search the TLB and 100 nanoseconds to access the main memory what is effective memory access time in nanoseconds?
Explanation: For 4 level Paging scheme, Effective Memory access Time, EAT =
HTLB * TTLB + (1 - HTLB)[ TTLB + 4*Tm] + Tm] where, HTLB = hit ratio of TLB TTLB = search time of TLB Tm = Memory access time
Applying the formula:
EAT = (0.98 *20) + 0.02(20 + 400) + 100 EAT = 19.6 + 8.4 + 100 EAT = 128 ns
Option (B) is correct.
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