Question

Considering that the TLB takes 5 ns access time and main memory takes 200ns, compute the Effective Access Time (EAT) for the following case: Number of hits to the TLB is 95%.

Answer 1

Page Fault Rate 0 ≤ p ≤ 1 if p = 0 no page faults if p = 1

Effective Access Time (EAT) EAT = (1 – p) x memory access + p (page fault overhead + [swap page out ] + swap page in + restart overhead)

Lets say memory access (including TLB and extra page table accesses) = 20 nsecs

overhead at both ends say 10000 instructions ≈ 10000nsecs

swap page in = 4 msecs = 4 000 000 nsecs

95% of time have to swap page out = 2 000 000 nsecs So EAT = (1 – p) x 20 + p(10000 + 6 000 000) ≈ 20 + 6 000 000p nsecs