A computer spends 80% of its time executing user?code and 20% of its time performing input/output operations. If you regard I/O as an unnecessary overhead, the efficiency of the computer is 80/(80 + 20) = 80%. Suppose that CPU performance is increasing at a compound rate of 40% per year and that the performance of I/O systems is increasing at a compound rate of 10% per year. In another two years, a new computer can be bought with both faster processing power and I/O performance. How efficient would the new computer be in two years’ time?
What will be an ideal response?
The new time to execute the same code will be 0.8T × 0.6 × 0.6, and the new time to perform the I/O will be 0.2T × 0.9 × 0.9. The new efficiency will be (80 × .6 × .6)/(80 × .6 × .6 + 20 × .9 × .9) = 28.8/(28.8 + 15.2) = 0.64 = 64%
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