6) Execution ofprogram P on computer X entails 4-10, fixed-point instructions an

Question

6) Execution ofprogram P on computer X entails 4-10, fixed-point instructions and 2·10, floating- t). The CP point instructions (you do not need to be concerned about what these instructions do ye for fixed-point instructions on X is 3.0 and that for floating-point instructions for X is 2 GHz is 9.0. The clock rate a) What is the average CPI for the instructions of P? b) How long does it take P to execute? e) What would be the overall speed up of an improvement reducing the floating-point CPI to 6?

Explanation / Answer

Solution:

a)

Total number of instructions = 4*10^9 + 2*10^9 = 6*10^9

The ratio of X is 4/6 = 0.6667 and for floating point 2/6 = 0.333

Average CPI = 0.66667*3 + 0.3333*9 = 5

b)

Total execution time = CPI*number of instructions/clock frequency

= (5 * 6 * 10^9)/(2*10^9) = 15 seconds

c)

then the average CPI = 0.66667*3 + 0.3333*6 = 4

Overall speedup = 4/5 = 0.8

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