Performance
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Need help on problems 4.19, 4.20, 4.21, and 4.22.
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The solution assists with the given questions on performance.
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4.19.
Execution time on floating point before enhancement = 10/2 = 5 seconds
Execution time on floating point after enhancement = 5 * 1/5 = 1 second
Execution time after improvement = 1 + (10 - 5) = 6 seconds
Speedup = Execution time before improvement / Execution time after improvement = 10/6 = 5/3 = 1.67
4.20 Let me assume the new floating-point unit is as described in Exercise 4.19, i.e., it has a performance improvement of 5 times faster, and let x be the percentage of the initial execution time floating-point instructions have to account for, then
Execution time after improvement = 100 * x * 1/5 + 100 * (1 -x) = 100 - 80 *x
So the speed up = Execution time ...
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