先来先服务调度和最短作业优先调度算法试验报告材料

1、实用文档实验概述：【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：向驱堤LLrs实用文档技至U法IM F nJ从刁 的J大X4 UtfTilt辛 了14=&迄出羽 亍*d !EUVjiaWFTd-r JW 各缨*iA 些堆 坦歧整行ITJ I 1MI E遂彼问腴：；：；2L 甲 2 并如时寻功地面HIxa 理巧若敏并讦*乎坷同酣咐冲试验总设计流

2、程图:实用文档进程等待时间=进程开始运行时间一进程提交时间（即进程处于就绪态时间）进程周转时间=进程结束时间一进程提交时间【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 屏幕显示：Please input job number, submit time and run time输入：19.00.228.50.538.01.049.10.1屏幕显示:What kind of algorithm do you want? Please input 1 to select FCFS, or 2 to

3、select SJF.输入：3 屏幕显示：You have input a wrong number, please input again.输入：1屏幕输出结果：submit run starting final wait turnaround38.01.08.09.00.01.028.50.59.09.50.51.0计成派计成派开始时间,结束时间等待时间,周转时间按照IO田排序_1FW报错继续输数计算各作业参数：弗雄蟀廊h据策憎间等待时间，周转时间并输出实用文档19.00.29.59.70.50.749.10.19.79.80.60.7屏幕显示：The average turnaround

4、 time is 0.85What kind of algorithm do you want? Please input 1 to select FCFS, or 2 to select SJF, or 0 to ex测试数据二：submitrun1 60.52 50.93 6.30.1FCFS和SJF算法结果一样：submitrun startingfinalwaitturnaround2 50.9 55.900.91 60.5 66.500.53 6.30.1 6.56.60.20.3The average turnaround time is 0.567测试数据三：submit run

5、150.224.20.335.10.345.20.1FCFS:submit runstartingfinalwait turnaround24.20.3 4.24.500.3150.2 55.200.235.10.3 5.25.50.10.445.20.1 5.55.60.30.4The average turnaround time is 0.325SJF:submit run starting final wait turnaround24.20.34.24.500.3150.255.200.245.20.15.25.300.135.10.35.35.60.20.5The average

6、turnaround time is 0.275源程序：#include stdio.h#include stdlib.h#include string.h#define M 50 struct sjf int jobnumber;实用文档float submittime;float runtime;float starttime;float finishtime;float waittime;float turnaroundtime;temp;static struct sjf stM;void input( struct sjf *p, int N)int i;printf( Please

7、 input the job number,submit time and run time:nFor exmple:1 8.5 2for (i =0;i N;i +)scanf( %d%f%f &pi . jobnumber, &pi . submittime, &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 %d ,pk . jobnumber);printf( nT

8、he processs information:n );printf(njobnumtsubmittruntstarttfinaltwaittturnaroundn);for (k =0;k N;k+)h+=pk . turnaroundtime;printf( %dt%-.1ft%-.1ft%-.1ft%-.1ft%-.1ft%-.1ftn,pk . jobnumber,pk . submittk . waittime,pk . turnaroundtime);g =h/ N;printf( nThe average turnaround time is %-.2fn,g);/*按提交时间从

9、小到大排序*/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;实用文档pi =pj;pj =temp;/*运行*/void deal( struct sjf *p, int N)(int k;for (k =0;k pk-1 . finishtime)(pk . starttime =pk . submittime;pk . finishtime =pk . submittime +pk . r

10、untime;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;实用文档void sort2( struct sjf *p, int N)(int next,m,n,k,i;float min;sort1(p,N);for (m=0

11、;mpm-1 . finishtime)(pm . finishtime =pm. submittime +pm. runtime;elsepm . finishtime =pm-1. finishtime +pm. runtime;for (n =m+1;n N;n+)(if (pn . submittime =pm. finishtime) /*判断内存中每次完成之后乂多少到达的进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 2兀一一JThe p

12、iocess，s information:obnurnubnitrunstartfinalwaitturnaround-8,01. 0a,og, o0 0L0Lc.0, 3Q. d9,09. S0.3L019.00. 29-59.70. 50.749.10. 1a. 79. S0.G0. 7The average traround tine s 0. 85ffhat kind of algorithm do you want? PLase input 1 to FCFS. or 2 cc selec t SJF or 0 toMH:2run order：3“142I：je proeesi7s

13、inf ornation:,obnuzsucnitrunstartfinalvaitturnaround-8.01. 0&09.00. 01.019.00 2g.o9.20 Q0.2y. 10. 19.30. 10.2-二8.50. 59,39. S0.81.3Ihe average turnaround time M 0* 67iVhat kind cf algorithm do vou want? PLease input 1 to select FCFS+or 2 to seleo测试二:Pleasel or exwle 111 9. 02 8 53 8.0the job nji

14、ntier, suLnit lirie and run time: & 5 2,0实用文档ncDlMXvYuVanXt inAitemp.ewe ithe total numiber of JDbs;3the job lumber, submit 1 ime and run tinted3. 5 2.03 6. 3 0. 1kind c-f algorithm do you want? Please inpuT 1 ro seLect FCFS. or 2 to selec t SJF or 0 ta ex t: r:in order：17一1-一3The proceus inform

15、ation：jobnumsuni trunstartfinal砒litturnaroundi-二5. 00. 95,05,9Q.O0.916. 00. 06.0&S0. 00.53630 I6.5& 60.20.3The aveiage t-irEarcund tine is 0. 57iV-iat kind al ori thm Ho 3PDUwant ? P .ease input 1 to select FCFSar 2 to select SJF or 0 to exit：2run order:2一1一3Izje pioces- - inf uznat ion；.obn

16、wnJUGElltrunstar!finalvaitturnaroundL5.00. 9.05. 90. 00. 91600 56.06.50 J0.536.30. L6.56.60.20.3The averturnaround t iw i 0. 57iVnat kind of algorithm do you wgi? FLcasc inpux 1 to scLcct FCFS* or 2 to sclec t SJF or 0 to exit;半：测试三:PledgeLlease inrut ForejiizpLe: 11 6 0.52 5 0. 9实用文档2 4, 2 0. 33 5.

17、 1 0. 34 5. 2 0- 1ffhot kind of nlsorihm do you wont? P ense input I toSFLCTFCFS, or 2 to selct SJF or 0 to exit：1run or der: 2一1 3一1I he prcfces i m infoymat ion :jobnun:Knitrunstartfinalvaitturnaround.A4.20 34.2150 00.315.00, 2o.Oo.20.00.235. 10. 35.25. 50. 10.1二二5.20 1三,a5, 60 sfl.4The average tu

18、rnaround time is 0.33Kiiit kind of al sat L t hm du you wantJ P lease input 1 to st Lee t FCS, ur 2 to selec t SJF ar 0 to exit:2 run order：2*1一一；一一 土The process* inforraaxian;zobnuzruntartfinalwaitturnaround二1. 20 34.250.00.35.00 2a,05,20 00.24o. 20, 10.23.3。00+1w5 10. 3二33.60. 20.5Ihe aveTage turn

19、around tine is 0. 27ri&t k;nd of alecrilrwi do you vajjt? Please input 1 to select FCF$. or 2 to selec t SJF nr ft to ?x:t:半,【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:实用文档ij 0；*inc2C 1Q4CYuVonbi nwwtcrnp cxcrm order: 3-21/4- The processs information：jobnun submit runstartfinalT?aitturnaround38. 01, 08. 09,

20、0Q 01.028.50.59. 09.50.51. 019.00.29. d9.70. o0,749一10,19 7璀o, e0.7The averagexurnarouncTime Is0, 8ohat kind of algorithm do you want? Please input 1 to select FCFS, or 2 t t SJF or 0 to exit:2 runorder;340-0The process1s information:j obnuiL sjbiiitrunstartfine!waitturnaround8.01.08. 09.00.01.09, 1

21、0. 19. 09. 1-0. 10.0(0.00,09. 10. 19, 19. 100.00.09. 19. 19. 19. 1The average turnaround time is 1. 80Khat kind of algorithm do you vant? Please input 1 tc select FCFS, or 2 tx SJF or 0 to exit:半：半：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;mN;m+)改为：int

22、next,m,n,k,i;float min;sort1(p,N);for (m=0;mpm-1 . finishtime)(pm . finishtime =pm. submittime +pm. runtime;elsepm . finishtime =pm-1. finishtime +pm. runtime;测试用例3的结果：错误1:Flense顷ILease inrut ForejiEule: 11 6 0, 52 5 0.9iVhat kind oft SJF or 0 torun order：2-M-3T5e procesi,information：jebnuraEubmi tr

23、unstartfinnlwaitturnaround5. 00. 95.05. 90.00.9I6.00. 55.96. 1T. 10.1r6,30 16,46,5在1fl. 2The average ffiiiit kind ofdl11lull t SJF or t) toeKLt：.tineci. . - .:i .Fltase inpui 1 to select FCFS. g 2 toturnaroundQr50实用文档elsepm . finishtime =pm-1. finishtime +pm. runtime;修改为：elseif (pm . submittime pm-1

24、 . finishtime)pm . finishtime =pm. submittime +pm. runtime;elsepm . 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 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：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:错误分析：同2一样,解决方法：将原来的：实用文档试验未考虑同一时间提交多个进程的情况，如:测试数据:

25、submitrun170.227.20.5370.1结果应该是:FCFS:submitrunstart finalwait turnaround170.27.07.50.00.5370.17.57.60.50.627.20.57.67.80.40.6The average turnaround time is 0.57.SJF:Jobnumsubmitrunstartingfinalwait turnaround370.17.07.10.00.1170.57.17.60.10.627.20.27.67.80.60.6The average turnaround time is 0.43.而程序

26、运行结果是：实用文档Please input the job number, t tine and run time;For exncle： 1&吕2, Q1 7 0. 5run order!1-*3?2The process* 5 information：-ebnumsubmitrunstartfinalwai tturnaround17.0O.o7.07. o0.00. 50. 637. 00. 17. 57. 60.52?, 20.27,67. 80,40 6The average turnaround 1imc is 0. 57kind of algorithm do you诉an匕!rit?i2The process s information;zobnmn submitrunat artfinal粗itt-jmaround17. 00.57.07. 50.005