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Unformatted text preview: ¢ ½¼¾ ¢ ½¼ ½¼ , i.e., ½¼± of the total cycles the processor can supply. Problem 9.2 Solution: [Pg. 454] This problem requires careful study of the trace, and an anticipation of the type of pattern that will arise. A. They occur every 9.98–9.99ms: 358.93, 368.91, 378.89, 388.88, 398.86, 408.85, 418.83, 428.81. Note that the ones that are not italicized were determined by adding 9.98 to the preceding time. B. The italicized times shown above. They caused a new period of inactivity. C. The inactive times include the time spent servicing two interrupts in addition to the time spent executing the other process. D. Our process is active for around 9.5ms every 20.0 ms, i.e., 47.5% of the time. Problem 9.3 Solution: [Pg. 457] This problem involves simply labeling the execution sequence according to the process that is executing, and determining whether the process is in user or kernel mode. A B A B A A 100u + 40s B 80u + 30s Au Au As Bu Bu Bu Bu Bs Bu As Au Au Au Au Bs Bu Bu Bu Bs Au As Au Au Au As Problem 9.4 Solution: [Pg. 457] This is an...
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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