Question

Assume that the: Clock rate is 2 GHz, L1 access time is 1 cycle, L2 access time is 10 cycles, Memory access time is 100 cycles, L1 hit rate is 60%, L2 hit rate is 70%. What is the average memory access time? (4 marks)

Answer 1

Given clock rate =2 GHz

=2 x 10⁶  Hz

=1/(2 x 10⁶ ) s

=0.5 x 10^-6 s

=5 x 10^-7s

L1 access time = 1 cycle

=5 x 10^-7s

L1 hit rate =60%=0.6

L2 access time = 10 cycles

=10 x 5 x 10^-7s

L2 hit rate=70%=0.7

Memory access time=100 cycles

=100 x 5 x 10^-7s

average memory access time is given by

=H1*(L1)+(1-H1)(H2)(L1+L2)+(1-H1)(1-H2)MM

where H1=hit rate of L1 cache

L1=L1 cache access time

H2=hit rate of L2 cache

L2= L2 cache access time

MM=main memory access time

So,

=H1*(L1)+(1-H1)(H2)(L1+L2)+(1-H1)(1-H2)MM

=0.6 * (5 x 10^-7s)+0.4 * 0.7 * (5 x 10^-7s+10 x 5 x 10^-7s)+0.4 * 0.3 * ( 100 x 5 x 10^-7s)

=3 x 10^-7s + 2.8 * ( 55 x 10^-7s)+ 1.2 * (500 x10^-7s)

=3 x 10^-7s + 154 x 10^-7s+ 600 x10^-7s

=757 x10^-7s

=7.57 x10^-5s

=7.57x10^-2 ms (ans)