Answer :
The access time for L1 is 91.75 clock cycles.
Given: The L1 hit time is 2 clock cycles and the L1 hit rate is 75%.
Let the access time for L1 be t.
Let the L2 access time be t2.
The L2 hit time is 5 clock cycles and the L2 hit rate is 80%.
The L2 miss penalty is 120 clock cycles.
Now, the L1 hit rate is 75%, which means 75% of the time the data will be found in L1 itself and 25% of the time it will not be found in L1 and will need to be accessed from L2.
Hence, the average time taken to access data from L1 is: t1= (L1 hit time) + (L2 hit rate * L2 access time) + (L2 miss rate * miss penalty)
Substituting the given values,
we get,t1 = (2) + (0.25 * 5) + (0.75 * 120) = 91.75
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