Homework Answers
Processor without cache:-
Since main memory access time = 8 cycles
And there are 40% instruction using memory data access. Hence out of 100000 instruction, there will be 100000*0.4 = 40000 instruction which will access memory data.
Hence total execution time = (number of instructions + number of instruction accessing memory data)*(number of cycles in memory access) = (100000 + 40000)*8 = 140000*8 cycles = 1120000 cycles
Processor with cache:-
Total execution time =( (hit rate of instructions)*(number of instructions) + (hit rate of data)*(number of memory access) )*(number of cycles in cache access) + ( (miss rate of instructions)*(number of instructions) + (miss rate of data)*(number of memory access) )*(number of cycles in cache access +number of cycles in memory access)
= ( 0.9*100000 + 0.85*40000 )*1 + (0.1*100000 + 0.15*40000)*(1+8) = 124000*1 + 16000*9 = 268000 cycles
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