Consider the computer organization with following specifications: The computer h
ID: 3804888 • Letter: C
Question
Consider the computer organization with following specifications: The computer has 1 level cache The average memory access time for a microprocessor is 2.4 clock cycle 1 clock cycle is required to find whether the data is present and valid in the cache or not. If the data is not found in the cache, it needs 80 clock cycles to get it from the off-chip memory. You need to improve the average memory access time by 65% and need to add 2nd level of cache such that: This new 2nd level of cache can be accessed in 6 clock cycles. This new 2nd level cache should not affect 1st level cache access time. If the data is not even found in this 2nd level cache, it should take 80 clock cycles to get it from the off-chip memory. To obtain this improvement in access time, how often must the data be found in this 2nd level cache?Explanation / Answer
Let us assume the memory access time of processor with L1 cache is 2.4
L1 cache hit is 1 clock cycle
L1 cache miss is 80 clock cycles
memory access time =L1cache hit xL1 cache access time + L1cache missx (L1cache access time +MMaccess time)
2.4=1xh1+(1-h1)x81
h1=0.9825
L2 cache hit rate is 6.L2 cache miss rate is 80 clock cycles.
From the problem it is given that the memory access improvement is 65 percent.Or in other words the rate is 1.65.
or 2.4/x=1.65
new access time x=1.4545
The memory access time for L2 cache can be given as
1.4545=h1x1+(1-h1)xh2x(6+1)+(1-h1)(1-h2)(80+6+1)
=0.9825+0.0175xh2x6+1+0.0175(1-h2)87
solving this eqation we get h2=0.75,L2 cache hit rate.