先来先服务调度和最短作业优先调度算法实验分析报告

欢迎共阅欢迎共阅欢迎共阅实验概述：【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）VisualC++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算法其他数输入作业个数、输入作业个数、作业号、提交时间、运行时间再按最短运行时间排序计算各作业参数：开始时间，结束时间再按最短运行时间排序计算各作业参数：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：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：Pleaseinputthetotalnumberofjobs输入：4<回车>屏幕显示：Pleaseinputjobnumber,submittimeandruntime输入：19.00.2<回车>28.50.5<回车>38.01.0<回车>49.10.1<回车>屏幕显示：Whatkindofalgorithmdoyouwant?Pleaseinput1toselectFCFS,or2toselectSJF.输入：3<回车>屏幕显示：Youhaveinputawrongnumber,pleaseinputagain.输入：1<回车>屏幕输出结果：submitrunstartingfinalwaitturnaround38.01.08.09.00.01.028.50.59.09.50.51.019.00.29.59.70.50.749.10.19.79.80.60.7屏幕显示：Theaverageturnaroundtimeis0.85Whatkindofalgorithmdoyouwant?Pleaseinput1toselectFCFS,or2toselectSJF,or0toexit.测试数据二：submitrun160.5250.936.30.1FCFS和SJF算法结果一样：submitrunstartingfinalwaitturnaround250.955.900.9160.566.500.536.30.16.56.60.20.3Theaverageturnaroundtimeis0.567测试数据三：submitrun150.224.20.335.10.345.20.1FCFS：submitrunstartingfinalwaitturnaround24.20.34.24.500.3150.255.200.235.10.35.25.50.10.445.20.15.55.60.30.4Theaverageturnaroundtimeis0.325SJF:submitrunstartingfinalwaitturnaround24.20.34.24.500.3150.255.200.245.20.15.25.300.135.10.35.35.60.20.5Theaverageturnaroundtimeis0.275源程序：#include"stdio.h"#include"stdlib.h"#include"string.h"#defineM50structsjf{ intjobnumber; floatsubmittime; floatruntime; floatstarttime; floatfinishtime; floatwaittime; floatturnaroundtime;}temp;staticstructsjfst[M];voidinput(structsjf*p,intN){ inti; printf("Pleaseinputthejobnumber,submittimeandruntime:\nForexmple:18.52.0\n"); for(i=0;i<N;i++) { scanf("%d%f%f",&p[i].jobnumber,&p[i].submittime,&p[i].runtime);}}voidprint(structsjf*p,intN){ intk; floath,g; printf("runorder:"); printf("%d",p[0].jobnumber);for(k=1;k<N;k++)printf("-->%d",p[k].jobnumber);printf("\nTheprocess'sinformation:\n");printf("\njobnum\tsubmit\trun\tstart\tfinal\twait\tturnaround\n");for(k=0;k<N;k++){ h+=p[k].turnaroundtime;printf("%d\t%-.1f\t%-.1f\t%-.1f\t%-.1f\t%-.1f\t%-.1f\t\n",p[k].jobnumber,p[k].submittime,p[k].runtime,p[k].starttime,p[k].finishtime,p[k].waittime,p[k].turnaroundtime);}g=h/N;printf("\nTheaverageturnaroundtimeis%-.2f\n",g);}/*按提交时间从小到大排序*/voidsort1(structsjf*p,intN){inti,j;for(i=0;i<N;i++)for(j=0;j<=i;j++)if(p[i].submittime<p[j].submittime){temp=p[i];p[i]=p[j];p[j]=temp;}}/*运行*/voiddeal(structsjf*p,intN){ intk; for(k=0;k<N;k++) { if(k==0) { p[k].starttime=p[k].submittime; p[k].finishtime=p[k].submittime+p[k].runtime; } else { if(p[k].submittime>p[k-1].finishtime) { p[k].starttime=p[k].submittime; p[k].finishtime=p[k].submittime+p[k].runtime; } else { p[k].starttime=p[k-1].finishtime; p[k].finishtime=p[k-1].finishtime+p[k].runtime; } } } for(k=0;k<N;k++) { p[k].turnaroundtime=p[k].finishtime-p[k].submittime; p[k].waittime=p[k].starttime-p[k].submittime;}}voidsort2(structsjf*p,intN){ intnext,m,n,k,i; floatmin; sort1(p,N); for(m=0;m<N;m++) { i=0; if(m==0) p[m].finishtime=p[m].submittime+p[m].runtime; else { if(p[m].submittime>p[m-1].finishtime) { p[m].finishtime=p[m].submittime+p[m].runtime; } else p[m].finishtime=p[m-1].finishtime+p[m].runtime; } for(n=m+1;n<N;n++) { if(p[n].submittime<=p[m].finishtime)/*判断内存中每次完成之后又多少到达的进程*/ i++; } min=p[m+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(p[k+1].runtime<min) { min=p[k+1].runtime; next=k+1; } } temp=p[m+1]; p[m+1]=p[next]; p[next]=temp; }deal(p,N);print(p,N); }voidmain(){ intN,i; printf("Pleaseinputthetotalnumberofjobs:"); scanf("%d",&N); input(st,N);loop2: printf("Whatkindofalgorithmdoyouwant?Pleaseinput1toselectFCFS,or2toselectSJFor0toexit:");loop: scanf("%d",&i); if(i==0) exit(1); elseif(i==1) { sort1(st,N); deal(st,N); print(st,N); gotoloop2; } elseif(i==2){ sort2(st,N); gotoloop2;}else{ printf("Youhaveinputawrongnumber,pleaseinputagain:"); gotoloop;}}【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：intnext,m,n,k,i=0; floatmin; sort1(p,N); for(m=0;m<N;m++) {改为：intnext,m,n,k,i; floatmin; sort1(p,N); for(m=0;m<N;m++) { i=0;测试用例2的结果：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 { p[m].finishtime=p[m-1].finishtime+p[m].runtime; }修改为：else { if(p[m].submittime>p[m-1].finishtime) { p[m].finishtime=p[m].submittime+p[m].runtime; } else p[m].finishtime=p[m-1].finishtime+p[m].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）大于上个进程的结束时间”的情况：解决方法：将原来的： else p[m].finishtime=p[m-1].finishtime+p[m].runtime;修改为：else { if(p[m].submittime>p[m-1].finishtime) { p[m].finishtime=p[m].submittime+p[m].runtime; } else p[m].finishtime=p[m-1].finishtime+p[m].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提交多个进程的情况，如：测试数据：submitrun170.227.20.5370.1结果应该是：FCFS：submitrunstartfinalwaitturnaround170.27.07.50.00.5370.17.57.60.50.627.20.57.67.80.40.6Theaverageturnaroundtimeis0.57.SJF:Jobnumsubmitrunstartingfinalwaitturnaround370.17.07.10.00.1170.57.17.60.10.62