Suppose it takes 100 ns to access a page table and 20 ns to access associative…
2009
Suppose it takes 100 ns to access a page table and 20 ns to access associative memory with a 90% hit rate, the average access time equals :
- A.
20 ns
- B.
28 ns
- C.
90 ns
- D.
100 ns
Attempted by 1 students.
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Correct answer: B
When address translation uses an associative memory (TLB) alongside the full page table, the overall average access time is the hit-ratio-weighted combination of the fast TLB path and the slower page-table path: Average Access Time = (hit ratio x TLB access time) + (miss ratio x page-table access time).
The TLB (associative memory) access time is 20 ns, and its hit ratio is 90% (0.9).
The page-table access time is 100 ns, so the miss ratio is 10% (0.1).
Average access time = (0.9 x 20 ns) + (0.1 x 100 ns) = 18 ns + 10 ns = 28 ns.
As a check, the result must lie between the two extremes: with a 100% hit ratio the time would be 20 ns, and with a 0% hit ratio it would be 100 ns; since the hit ratio here is high (90%), a result of 28 ns, which lies close to the TLB time and far from the page-table time, is consistent with that weighting, confirming 28 ns as the average access time.