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12 November, 11:40

A system has a level 1 cache and a level 2 cache. The hit rate of the level 1 cache is 95%, and the subsequent hit rate of the level 2 cache is 80%. An access to level 1 cache requires one cycle, and access to level 2 cache requires 3 cycles, and an access to main memory requires 80 cycles. What is the average access time in clock cycles?

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  1. 12 November, 15:31
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    Step-by-step explanation:

    H1 = Hit rate of level 1 cache = 0.95

    T1 = Access time for level 1 cache = 1

    H2 = Hit rate of level 2 cache = 0.8

    T2 = Access time for level 2 cache = 3

    Hm = Hit rate of Main Memory = 1

    Tm = Access time for Main Memory = 80

    Average access time = [H1*T1]+[ (1-H1) * H2*T2]+[ (1-H1) (1-H2) * Hm*Tm]

    = [0.95*1]+[1-0.95]*0.8*3]+[ (1-0.95) (1-0.8) * 1*80]

    = [0.95+0.05*0.8*3]+[0.05*0.2*1.80]

    = 2.4+0.018 = 2.418

    Average access time = 2.418
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