Question

Base machine has a 2.4GHz clock rate. There is L1 and L2 cache. L1 cache is 256K, direct mapped write through. 90% (read) hit rate without penalty, miss penalty is 4 cycles. (cost of reading L2) All writes take 1 cycle. L2 cache is 2MB, 4 way set associative write back. 95% hit rate, 60 cycle miss penalty (cost of reading memory). 30% of all instructions are reads, 10% writes. All instructions take 1 cycle - except reads which take 1 cycle if data is in L1, but otherwise have a miss penalty

a) Calculate CPI for this machine.

b) Suppose we increase L1 cache to 512K. This improves our hit rate to 95% in L1 , but slows all reads to 2 cycles. What is the CPI with this change?

c) Changing the L1 cache to 4 way set associative increases the hit rate to 94% in L1 but added complexity of hardware means we must reduce the clock rate to 2.3GHz. What is the CPI in this case?

d) Which of the 3 machines is faster for the given instruction mix?

Answer 1

I have completed this problem in very detailed manner, If you like it. Please give thumbs up.

Step 1 :- Given data is

            Base machine clock rate = 2.4GHz

            L1 cache size is =256KB

            L2 cache size is =2MB

            Hit rate of read(Hr) is =90%

            Miss penalty is =4cycles

            Write takes =1 cycle

            L2 cache is 4 way set associative

          L1 cache is direct mapped

        In L2 cache, hit rate is 95%

     Miss penalty is 60cycles

        30% read instructions,10% write instructions,60% other instructions

Step 2 := (a)

CPI of this machine =0.30(for read)*0.90(hit rate)*1cycle + 0.30(read)*0.10(miss in L1)*0.95(hit in L2)*4(miss peanlty) + 0.30(read)*0.10(miss in L1)*0.05(miss in L2)*60cycles(miss penalty) + 0.10(for write)* 1cycle + 0,60(others)*1cycles = 0.27 + 0.114 + 0.09 + 0.10 + 0.60 = 1.174 CPI

Step 3 :- (b)

Hit rate for read becomes = 95%

CPI of this machine = 0.30(for read)*0.95(hit rate)*2cycle + 0.30(read)*0.05(miss in L1)*0.95(hit in L2)*4(miss peanlty) + 0.30(read)*0.05(miss in L1)*0.05(miss in L2)*60cycles(miss penalty) + 0.10(for write)* 1cycle + 0,60(others)*1cycles = 0.57 + 0.057 + 0.045 + 0.10 + 0.60 = 1.372 CPI

Step 4 :- (c)

Hit rate for read becomes =94%

CPI of this machine =0.30(for read)*0.94(hit rate)*1cycle + 0.30(read)*0.06(miss in L1)*0.95(hit in L2)*4(miss peanlty) + 0.30(read)*0.06(miss in L1)*0.05(miss in L2)*60cycles(miss penalty) + 0.10(for write)* 1cycle + 0,60(others)*1cycles = 0.282 + 0.0684 + 0.054 + 0.10 + 0.60 = 1.104 CPI

Step 4 (d)

Among all 3 machines ,3rd machine is faster because its CPI is very less and faster in execution per instruction compares to other.