先來先服務調度和最短作業(yè)優(yōu)先調度算法實驗報告

1、實驗概述：【實驗目的及要求】理解并掌握處理機調度算法【實驗原理】基于先來先服務調度和最短作業(yè)優(yōu)先調度算法思想用C語言編程實現(xiàn)【實驗環(huán)境】（使用的軟件）Visual C+6.0實驗內容：本實驗模擬在單處理機情況下處理機調度，用C語言編程實現(xiàn)先來先服務和最短作業(yè)優(yōu)先調度算法?！緦嶒灧桨冈O計】FCFS流程圖：SJF流程圖：試驗總設計流程圖：輸入選擇數(shù)以選擇調度算法報錯繼續(xù)輸數(shù)先按提交時間排序按照提交時間排序1FCFS算法2SJF算法其他數(shù)進程等待時間進程開始運行時間進程提交時間(即進程處于就緒態(tài)時間) 輸入作業(yè)個數(shù)、作業(yè)號、提交時間、運行時間進程周轉時間進程結束時間進程提交時間 再按最短運行時間排序

2、計算各作業(yè)參數(shù)：開始時間，結束時間等待時間，周轉時間 并輸出 并輸出計算各作業(yè)參數(shù)：開始時間，結束時間等待時間，周轉時間 并輸出【實驗過程】（實驗步驟、記錄、數(shù)據(jù)、分析）測試用例1：屏幕顯示：Please input the total number of jobs輸入：4 <回車>屏幕顯示：Please input job number, submit time and run time 輸入：1 9.0 0.2<回車>2 8.5 0.5<回車>3 8.0 1.0<回車>4 9.1 0.1<回車>屏幕顯示：What kind of

3、algorithm do you want? Please input 1 to select FCFS, or 2 to select SJF.輸入：3 <回車>屏幕顯示：You have input a wrong number, please input again.輸入：1<回車>屏幕輸出結果： 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屏幕顯示：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.測試數(shù)據(jù)二： submit run 1 6 0.52 5 0.93 6.3 0.1FCFS和SJF算法結果一樣：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測試數(shù)據(jù)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); /*按提交時間從小到大排序

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; /*運行*/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)/*判斷內存中每

13、次完成之后又多少到達的進程*/ i+; min=pm+1.runtime; next=m+1; for(k=m+1;k<m+i;k+)/*找出到達后的進程中運行時間最小的進程*/ 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; 【結論】（結果）測試1：測試二：測試三：【小結】實驗中產(chǎn)生的錯誤及原因分析：測試用例1的結果：錯誤1：錯誤解決方式：主要是子函數(shù)sort2()中出的錯：i的作用域，程序修改：將原來：int next,m,n,k,i=0;float min;sort1(p,N);for(m=0;m<N;m+) 改為：int next,m,n,k,i;float min;sort1(p,N);for(m=0;m<N;m+) i=0;測試用例2的

16、結果：錯誤1：錯誤原因：未考慮到“提交時間（submittime）大于上個進程的結束時間”的情況：解決方法：將原來的：else pm.finishtime=pm-1.finishtime+pm.runtime; 修改為：else if(pm.submittime>pm-1.finishtime) pm.finishtime=pm.submittime+pm.runtime;else pm.finishtime=pm-1.finishtime+pm.runtime; 測試用例3的結果：錯誤1：錯誤分析：同2 一樣，未在子函數(shù)sort()中未考慮到“提交時間（submittime）大于上個進

17、程的結束時間”的情況：解決方法：將原來的：else pm.finishtime=pm-1.finishtime+pm.runtime; 修改為：else if(pm.submittime>pm-1.finishtime) pm.finishtime=pm.submittime+pm.runtime;else pm.finishtime=pm-1.finishtime+pm.runtime; 實驗的體會及收獲：通過這次試驗，我對處理機的調度算法特別是FCFS和SJF有了更深的理解，而且鍛煉了我的思維能力，使我能更全面地思考問題，以后還需要多做些這方面的練習。試驗不足之處：試驗未考慮同一時間

18、提交多個進程的情況，如：測試數(shù)據(jù)： submit run1 7 0.22 7.2 0.53 7 0.1結果應該是：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