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SSD6卡耐基梅陇大学联系6答案 绝对正确 SSD6数据结构 是一门很重要的课程 希望对大家有帮助

1,Using the timestamp program, what is the estimated clock rate of your processor in cycles per second?
  Answer: 2264017000 cycles per second

2,What is the approximate timer interval of your operating system? Report your answer in both cycles and in seconds.
  Answer:241092844-205693166=35399678
	It is 35399678 cycles per second!And that is 35399678/2264017000=0.0156s

3,What are the minimum and maximum lengths of the timer interrupts that you observed? Do not count timer interrupts    that switched to a different thread or different process.
  Answer:The maximum is 963526 cycles.
	 The minimum is 103309 cycles.
4,How many total interrupts did you record?
  Answer:19 interrupts.

5,How many of these were timer interrupts?
  Answer:12 interrupts.

6,How many timer interrupts resulted in context switches to another thread that belongs to the timestamp program? 
  Answer:2 interrupts.

7,Did a timer interrupt resulted in a context switch to a different process? If so, how many times did this occur,     and how many timer intervals did these other processes run? 
  Answer:Yes!This occurs for 4 times.Other processes run for more than 3 timer intervals.

8,Did a thread finish executing before the end of a timer interval? If so, how many times did this happen and what   did the operating system schedule for the rest of that timer interval? Was it one of the other timestamp threads,   or another process?
  Answer:Yes.This happens for 3 times.They are thread 0,thread 1 and thread 2.When they finish executing ,the rest          is for another process.

9,
 1),How many timer intervals elapse from the time your program started to the time your program ended? 
  Answer:1,18 timer intervals elapse from the time program started to the time program ended
 2),How many of those intervals did your program not run? 
  Answer:2,3 timer intervals.

10,Expressed as a hyphen delimited string of the label numbers, what was the execution order of the threads?
  Answer: The execution order is 0-1-0-0-1-1-2-2-2-3-0-1-2-3-3-3