A paging scheme uses a Translation Look-aside Buffer (TLB). A TLB access takes…
2008
A paging scheme uses a Translation Look-aside Buffer (TLB). A TLB access takes 10 ns and a main memory access takes 50 ns. What is the effective access time (in ns) if the TLB hit ratio is 90% and there is no page fault?
Answer: C. 65 — Concept: When a system uses a TLB, every memory reference either hits the TLB or misses it. The effective access time (EAT) is the probability-weighted…
- A.
54
- B.
60
- C.
65
- D.
75
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Show answer & explanation
Correct answer: C
Concept: When a system uses a TLB, every memory reference either hits the TLB or misses it. The effective access time (EAT) is the probability-weighted average of the access time on a hit and the access time on a miss: EAT = p × (hit time) + (1 − p) × (miss time), where p is the TLB hit ratio.
Application: Apply the concept to the given values.
TLB access time = 10 ns; main memory access time = 50 ns.
On a TLB hit, the entry is already found in the TLB, so only one memory access is needed to fetch the data: hit time = 10 + 50 = 60 ns.
On a TLB miss (no page fault), the page table must first be looked up in main memory (one memory access), and then the data is fetched (another memory access): miss time = 10 + 50 + 50 = 110 ns.
With hit ratio p = 0.9 and miss ratio 1 − p = 0.1: EAT = 0.9 × 60 + 0.1 × 110 = 54 + 11 = 65 ns.
Cross-check: Equivalently, EAT = TLB access + p × (memory access) + (1 − p) × (2 × memory access) = 10 + 0.9 × 50 + 0.1 × 100 = 10 + 45 + 10 = 65 ns — matching the value above.
Answer: 65 ns.
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