Question

Question 4 - [25 Points] Part (a) - Average Access Time (AMAT) The average memory access time for a microprocessor with One (1) level (L1) of cache is 2.4 clock cycles - If data is present and valid in the cache, it can be found in 1 clock cycle If data is not found in the cache, 80 clock cycles are needed to get it from off- chip memory Designers are trying to improve the average memory access time to 1.4545 clock cycles and are considering adding a 2nd level (L2) of cache on-chip. This second level of cache could be accessed in 6 clock cycles The addition of this cache does not affect the first level cache's access patterns or hit times Off-chip accesses would still require 80 additional clock cycles. To obtain the desired speedup, how often must data be found in the 2nd level cache (L2 hit rate)?

Answer 1

Formula of AMAT with hit ratio(h1) of One level cache( with access time L1):

amat = h1 * L1 + (1-h1)*M (M is the memory access time)
As per question, L1 = 1 clock cycle, M = 80 clock cycle, h1=?(we need to find out and amat = 2.4 clock cycles.
==> 2.4 = h1*1 + (1-h1)*80
==> 2.4 = h1 - 80h + 80
==> h1= 77.6/79 = 0.98227
==> h1 =  0.98(approx) hit ratio of L1 cache h1

Formula of AMAT with two level cache,
L1 = access time for level 1 cache, h1 = hit ratio of level 1 cache, L2 = access time for Level 2 cache, h2 = hit ratio of level 2 cache, M = memory access time
amat = h1*L1 + (1-h1)(h2 * L2 + (1-h2)*M)
As per question, amat = 1.4545, h2=?(we need to find out) , L2 = 6 clock cycles and M = 80 clock cylces ( here M stands for off chip access)

==> 1.4545 = 0.98*1 + (1-0.98)( h2*6 + (1-h2)*80 )
==> 1.4545 - 0.98 = 0.02*(6*h2 + 80 - 80*h2)
==> h2 = 0.76 (approx.)