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Module 4: Memory Organization

20192m

The average memory access time for a machine with a cache hit rate of 90% where the cache access time is 10 ns and the memory access time is 100 ns is

  • (i) 55 ns
  • (ii) 45 ns
  • (iii) 90 ns
  • (iv) 19 ns
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Answer: (iv) 19 ns

Average memory access time (hierarchical access: check cache first, go to memory only on miss):

Tavg=h⋅Tc+(1−h)⋅TmT_{avg} = h \cdot T_c + (1-h)\cdot T_m

Given hit rate h=90%=0.9h = 90\% = 0.9, Tc=10T_c = 10 ns, Tm=100T_m = 100 ns:

Tavg=0.9×10+0.1×100=9+10=19 nsT_{avg} = 0.9 \times 10 + 0.1 \times 100 = 9 + 10 = 19 \text{ ns}

If counting Tc+TmT_c + T_m on miss (simultaneous vs hierarchical variants), then 0.9×10+0.1×110=200.9 \times 10 + 0.1 \times 110 = 20 ns — but with given options, 19 ns is the expected answer.

Correct option: (iv) 19 ns

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