# GATE | GATE-CS-2017 (Set 2) | Question 41

In a two-level cache system, the access times of L1 and L2 1 and 8 clock cycles, respectively. The miss penalty from the L2 cache to main memory is 18 clock cycles. The miss rate of L1 cache is twice that of L2. The average memory access time(AMAT) of this cache system is 2 cycles. The miss rates of L1 and L2 respectively are:
(A) 0.111 and 0.056
(B) 0.056 and 0.111
(C) 0.0892 and 0.1784
(D) 0.1784 and 0.0892

Explanation:

• Access time of L1 = 1
• Access Time of L2=8
• miss penalty  L1 cache (2*L2)  = 18*2 = 2*a
• miss penalty  L2 cache say a = 18
• AMAT (average memory access time) =2

AMAT = Access time of L1 + (MissRate L1 * miss penalty  L1)  where miss penalty L1 = Access time of L2 + (MissRate L2 * miss penalty L2)

2 = 1+ 2*a *(8 + a* 18)

Solving the equation,

a=0.111

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