先来先服务调度和最短作业优先调度算法实验报告

1、实验概述：【实验目的及要求】理解并掌握处理机调度算法【实验原理】基于先来先服务调度和最短作业优先调度算法思想用C语言编程实现【实验环境】（使用的软件）Visual C+6.0实验内容：本实验模拟在单处理机情况下处理机调度，用C语言编程实现先来先服务和最短作业优先调度算法。【实验方案设计】FCFS流程图：SJF流程图：试验总设计流程图：输入选择数以选择调度算法报错继续输数先按提交时间排序按照提交时间排序1FCFS算法2SJF算法其他数进程等待时间进程开始运行时间进程提交时间(即进程处于就绪态时间) 输入作业个数、作业号、提交时间、运行时间进程周转时间进程结束时间进程提交时间 再按最短运行时间排序

2、计算各作业参数：开始时间，结束时间等待时间，周转时间 并输出 并输出计算各作业参数：开始时间，结束时间等待时间，周转时间 并输出【实验过程】（实验步骤、记录、数据、分析）测试用例1：屏幕显示：Please input the total number of jobs输入：4 <回车>屏幕显示：Please input job number, submit time and run time 输入：1 9.0 0.2<回车>2 8.5 0.5<回车>3 8.0 1.0<回车>4 9.1 0.1<回车>屏幕显示：What kind of

3、algorithm do you want? Please input 1 to select FCFS, or 2 to select SJF.输入：3 <回车>屏幕显示：You have input a wrong number, please input again.输入：1<回车>屏幕输出结果： submit run starting final wait turnaround3 8.0 1.0 8.0 9.0 0.0 1.02 8.5 0.5 9.0 9.5 0.5 1.01 9.0 0.2 9.5 9.7 0.5 0.74 9.1 0.1 9.7 9.8 0

4、.6 0.7屏幕显示：The average turnaround time is 0.85What kind of algorithm do you want? Please input 1 to select FCFS, or 2 to select SJF, or 0 to exit.测试数据二： submit run 1 6 0.52 5 0.93 6.3 0.1FCFS和SJF算法结果一样：submit run starting final wait turnaround2 5 0.9 5 5.9 0 0.91 6 0.5 6 6.5 0 0.53 6.3 0.1 6.5 6.6 0

5、.2 0.3The average turnaround time is 0.567测试数据三： submit run1 5 0.22 4.2 0.33 5.1 0.34 5.2 0.1FCFS：submit run starting final wait turnaround2 4.2 0.3 4.2 4.5 0 0.31 5 0.2 5 5.2 0 0.23 5.1 0.3 5.2 5.5 0.1 0.44 5.2 0.1 5.5 5.6 0.3 0.4The average turnaround time is 0.325SJF:submit run starting final wai

6、t turnaround2 4.2 0.3 4.2 4.5 0 0.31 5 0.2 5 5.2 0 0.24 5.2 0.1 5.2 5.3 0 0.13 5.1 0.3 5.3 5.6 0.2 0.5The average turnaround time is 0.275源程序：#include "stdio.h"#include"stdlib.h"#include"string.h"#define M 50struct sjfint jobnumber;float submittime;float runtime;float s

7、tarttime;float finishtime;float waittime;float turnaroundtime;temp;static struct sjf stM;void input(struct sjf *p,int N)int i;printf("Please input the job number,submit time and run time:nFor exmple:1 8.5 2.0n");for(i=0;i<N;i+) scanf("%d%f%f",&pi.jobnumber,&pi.submitti

8、me,&pi.runtime); void print(struct sjf *p,int N)int k;float h,g;printf("run order:");printf("%d",p0.jobnumber); for(k=1;k<N;k+) printf("->%d",pk.jobnumber); printf("nThe process's information:n"); printf("njobnumtsubmittruntstarttfinaltwaitt

9、turnaroundn"); for(k=0;k<N;k+) h+=pk.turnaroundtime; printf("%dt%-.1ft%-.1ft%-.1ft%-.1ft%-.1ft%-.1ftn",pk.jobnumber,pk.submittime,pk.runtime,pk.starttime,pk.finishtime,pk.waittime,pk.turnaroundtime); g=h/N; printf("nThe average turnaround time is %-.2fn",g); /*按提交时间从小到大排序

10、*/void sort1(struct sjf *p,int N) int i,j; for(i=0;i<N;i+) for(j=0;j<=i;j+) if(pi.submittime<pj.submittime) temp=pi; pi=pj; pj=temp; /*运行*/void deal(struct sjf *p,int N)int k;for(k=0;k<N;k+)if(k=0) pk.starttime=pk.submittime; pk.finishtime=pk.submittime+pk.runtime; else if(pk.submittime&

11、gt;pk-1.finishtime) pk.starttime=pk.submittime; pk.finishtime=pk.submittime+pk.runtime;else pk.starttime=pk-1.finishtime; pk.finishtime=pk-1.finishtime+pk.runtime; for(k=0;k<N;k+)pk.turnaroundtime=pk.finishtime-pk.submittime; pk.waittime=pk.starttime-pk.submittime; void sort2(struct sjf *p,int N)

12、int next,m,n,k,i;float min;sort1(p,N);for(m=0;m<N;m+) i=0;if(m=0)pm.finishtime=pm.submittime+pm.runtime; else if(pm.submittime>pm-1.finishtime) pm.finishtime=pm.submittime+pm.runtime;else pm.finishtime=pm-1.finishtime+pm.runtime; for(n=m+1;n<N;n+) if(pn.submittime<=pm.finishtime)/*判断内存中每

13、次完成之后又多少到达的进程*/ i+; min=pm+1.runtime; next=m+1; for(k=m+1;k<m+i;k+)/*找出到达后的进程中运行时间最小的进程*/ if(pk+1.runtime<min) min=pk+1.runtime; next=k+1; temp=pm+1; pm+1=pnext; pnext=temp; deal(p,N); print(p,N); void main()int N,i;printf("Please input the total number of jobs:");scanf("%d"

14、;,&N);input(st,N);loop2:printf("What kind of algorithm do you want? Please input 1 to select FCFS, or 2 to select SJF or 0 to exit:");loop:scanf("%d",&i);if(i=0)exit(1);else if(i=1)sort1(st,N);deal(st,N);print(st,N);goto loop2;else if(i=2) sort2(st,N);goto loop2; else pri

15、ntf("You have input a wrong number, please input again:"); goto loop; 【结论】（结果）测试1：测试二：测试三：【小结】实验中产生的错误及原因分析：测试用例1的结果：错误1：错误解决方式：主要是子函数sort2()中出的错：i的作用域，程序修改：将原来：int next,m,n,k,i=0;float min;sort1(p,N);for(m=0;m<N;m+) 改为：int next,m,n,k,i;float min;sort1(p,N);for(m=0;m<N;m+) i=0;测试用例2的

16、结果：错误1：错误原因：未考虑到“提交时间（submittime）大于上个进程的结束时间”的情况：解决方法：将原来的：else pm.finishtime=pm-1.finishtime+pm.runtime; 修改为：else if(pm.submittime>pm-1.finishtime) pm.finishtime=pm.submittime+pm.runtime;else pm.finishtime=pm-1.finishtime+pm.runtime; 测试用例3的结果：错误1：错误分析：同2 一样，未在子函数sort()中未考虑到“提交时间（submittime）大于上个进

17、程的结束时间”的情况：解决方法：将原来的：else pm.finishtime=pm-1.finishtime+pm.runtime; 修改为：else if(pm.submittime>pm-1.finishtime) pm.finishtime=pm.submittime+pm.runtime;else pm.finishtime=pm-1.finishtime+pm.runtime; 实验的体会及收获：通过这次试验，我对处理机的调度算法特别是FCFS和SJF有了更深的理解，而且锻炼了我的思维能力，使我能更全面地思考问题，以后还需要多做些这方面的练习。试验不足之处：试验未考虑同一时间

18、提交多个进程的情况，如：测试数据： submit run1 7 0.22 7.2 0.53 7 0.1结果应该是：FCFS：submit run start final wait turnaround1 7 0.2 7.0 7.5 0.0 0.53 7 0.1 7.5 7.6 0.5 0.62 7.2 0.5 7.6 7.8 0.4 0.6The average turnaround time is 0.57.SJF:Jobnum submit run starting final wait turnaround3 7 0.1 7.0 7.1 0.0 0.11 7 0.5 7.1 7.6 0.1 0.62 7.2 0.2 7.6 7.8 0.6 0.6The average turnaround time is 0.43.而程