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Unformatted text preview: interesting thought problem. It helps you reason about the range of possible times that can lead to a given interval count. The following diagram illustrates the two cases: B.9. MEASURING PROGRAM PERFORMANCE 729
Minimum Maximum A A
0 10 20 30 40 50 60 70 80 For the minimum case, the segment started just before the interrupt at time 10 and ﬁnished right as the interrupt at time 70 occurred, giving a total time of just over 60ms. For the maximum case, the segment started right after the interrupt at time 0 and continued until just before the interrupt at time 80, giving a total time of just under 80ms. Problem 9.5 Solution: [Pg. 457] This problem requires thinking about how well the accounting scheme works. The seven timer interrupts occur while the process is active. This would give a user time of 70ms and a system time of 0ms. In the actual trace, the process ran for 63.7ms in user mode and 3.3ms in kernel mode. The counter overestimated the true execution time by ¼ ´ ¿ · ¿ ¿µ ½ ¼ X. Problem 9.6 Solution: [Pg. 465] This problem requires reasoning about the different sources of delay in a program and under what conditions these sources will apply. From these measurements we get:
· · ¿ ½¿¿ Ô ¦ ½ ¿½ From this we conclude that ½¼¼ , ¿¿ ,Ô ¾½ , and Ñ . Problem 9.7 Solution: [Pg. 475] This problem requires applying probability theory to a simple model of process scheduling. It demonstrates that obtaining accurate measurements beco...
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This note was uploaded on 09/02/2010 for the course ELECTRICAL 360 taught by Professor Schultz during the Spring '10 term at BYU.
- Spring '10
- The American