FoundationNumeric answerFree solution

An L1 cache reference costs about 11 ns and a main-memory reference about 100100 ns. By what factor does a cache miss slow that access?

Answer with a number. Fractions, powers and expressions like 23/6 or C(52,5) are read correctly in practice.

Show the answer and worked solution

Answer

100100

The ratio is simply 100 ns/1 ns=100100\,\text{ns}/1\,\text{ns} = 100. It is worth carrying alongside the rest of the ladder: L1 about a nanosecond, L2 a few, L3 around ten, main memory a hundred, an SSD read tens of microseconds, a network round trip within a datacentre tens of microseconds. That hundredfold step is why hot-path work is counted in cache misses rather than in operations, and why a data structure that touches memory contiguously beats one with better asymptotics but a pointer chase per element.

Worked solution

  1. Formula
    tRAMtL1\frac{t_{\text{RAM}}}{t_{\text{L1}}}
  2. Substitute
    =100 ns1 ns= \frac{100\,\text{ns}}{1\,\text{ns}}
  3. Solve
    =100= 100
  4. Answer
    100×100\times

Takeaway: Main memory is about a hundred times slower than L1.

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An L1 cache reference costs about 1 ns and a main-memory reference about 100 ns. · QuantMax