Problem 8:

 Through the use of a set of benchmark programs a designer discovers that a two-instruction sequence (shift followed by add) is used many times in these programs. In fact, the two instruction sequence accounts for 20% of the total instructions executed. The designer decides to redesign the processor to include a single shift–add instruction that can be used in place of all the old two instruction sequences of shift followed by add. However, the new processor has a lower maximum clock frequency: freqold/freqnew = 1.1.

1. Which processor (old or new) has higher performance and by how much?

Make reasonable assumptions where necessary.

Solution:

Execution time = seconds/cycles * CPI * number of instructions

Old time = 1/ (old freq) * CPI * previous number of instructions

New time = 1/ (new freq) * CPI * new number of instructions

New freq = old freq/1.1

20% of instructions were reduced by half so that is 10%, so the new program has 90% of the original instructions.

New time = 1.1/ (old freq) * CPI * new number of instructions * 0.9

Old/new = 1/1.1/0.9 = 1/0.99 = 1.0101 so the old time is slower and the new system is faster.