操作系統(tǒng)實驗九

1、精選優(yōu)質(zhì)文檔-傾情為你奉上學號E 專業(yè)計算機科學與技術(shù) 姓名施飛宇實驗日期2018/12/20 教師簽字 成績實驗報告【實驗名稱】最低松弛度優(yōu)先調(diào)度算法【實驗目的】1. 學習最低松弛度算法思想。2. 用代碼實現(xiàn)改算法，解決最低松弛度優(yōu)先調(diào)度問題?！緦嶒炘怼孔畹退沙诙葍?yōu)先（LLF）算法是根據(jù)任務緊急（或松弛）的程度，來確定任務的優(yōu)先級。任務的緊急程度愈高，為該任務所賦予的優(yōu)先級就愈高，使之優(yōu)先執(zhí)行。在實現(xiàn)該算法時要求系統(tǒng)中有一個按松弛度排序的實時任務就緒隊列，松弛度最低的任務排在隊列最前面，被優(yōu)先調(diào)度。松弛度的計算方法如下： 任務的松弛度=必須完成的時間-其本身的運行時間-當前時間&

2、#160;其中其本身運行的時間指任務運行結(jié)束還需多少時間，如果任務已經(jīng)運行了一部分，則： 任務松弛度=任務的處理時間-任務已經(jīng)運行的時間 當前時間【實驗內(nèi)容】數(shù)據(jù)結(jié)構(gòu)：本實驗進程采用結(jié)構(gòu)體來存儲線信息，采用結(jié)構(gòu)體數(shù)組來存儲進程序列。進程結(jié)構(gòu)定義如下：typedef structint start;/進程申請服務開始時間int over;/進程截止時間int survey1;/需要服務總時間int survey2;/已經(jīng)服務時間int songchi;/松弛度int label;/等于0未被處理，等于1已經(jīng)被處理完畢pcb;程序流程圖：程序代碼：%2018/12/20 篤行南樓 A20

3、2 E 施飛宇#include "iostream"#include <stdio.h>using namespace std;typedef structint start;/進程申請服務開始時間int over;/進程截止時間int survey1;/需要服務總時間int survey2;/已經(jīng)服務時間int songchi;/松弛度int label;/等于0未被處理，等于1已經(jīng)被處理完畢pcb;void pcb_A(pcb * A,int n)/A進程創(chuàng)建函數(shù) int i; for(i=0;i<n+1;i+) Ai.start=i*20; Ai.s

4、urvey2=0; Ai.label=0; Ai.songchi=10; Ai.survey1=10; Ai.over=Ai.start+20; if(i!=n) cout<<"A"<<i+1<<"到達時間"<<Ai.start<<" 截止時間"<<Ai.over<<endl; void pcb_B(pcb * B,int n)/B進程創(chuàng)建函數(shù) int i; for(i=0;i<n+1;i+) Bi.start=i*50; Bi.survey2=

5、0; Bi.label=0; Bi.songchi=25; Bi.survey1=25; Bi.over=Bi.start+50; if(i!=n) cout<<"B"<<i+1<<"到達時間"<<Bi.start<<" 截止時間"<<Bi.over<<endl; int test(pcb *X,int n,int time) int i,j; for(i=0;i<n;i+) if(Xi.label=0&&Xi.start<

6、=time) Xi.songchi=Xi.over-(Xi.survey1-Xi.survey2)-time; return i; return -1;void run_AB(pcb *A,int n1,pcb *B,int n2) int time; int ai,bi,ka,kb; int max_time; int AB=1; int label=0; if(n1*20>=n2*50) max_time=n1*20; else max_time=n2*50; cout<<"時間 "<<" A"<<"

7、;截止時間 "<<" 服務時間 "<<" 松弛度 "<<"B 截止時間 "<<" 服務時間 "<<" 松弛度 "<<endl; cout<<"0"<<" A"<<"1"<<" "<<A0.over<<" "<<A0.survey2&l

8、t;<" "<<A0.songchi<<" B"<<"1"<<" "<<B0.over<<" "<<B0.survey2<<" "<<B0.songchi<<endl; for(time=0;time<=max_time;time=time + 1) ka = test(A,n1,time); if(ka = -1) ai=ai+1; else

9、ai=ka; / cout<<"A"<<ai+1<<" "<<Aai.songchi<<endl; kb=test(B,n2,time); if(kb=-1) bi=bi+1; else bi=kb; / cout<<"B"<<bi+1<<" "<<Bbi.songchi<<endl; if(Bbi.songchi=0|ka=-1|label=1) AB=0;label=0; if(Aai.son

10、gchi=0|kb=-1|label=2) AB=1;label=0; if(AB=1&&ai<n1) if(Aai.survey1=(+Aai.survey2) cout<<time+1<<" A"<<ai+1<<" "<<Aai.over<<" "<<Aai.survey2<<" "<<Aai.songchi/5*5<<" B"<<bi+1

11、<<" "<<Bbi.over<<" "<<Bbi.survey2<<" "<<Bbi.songchi/5*5<<endl; cout<<time+1<<" A"<<ai+1<<"進程結(jié)束"<<endl; Aai.label=1; label=1; else if(time+1)%5=0) cout<<time+1<<"

12、 A"<<ai+1<<" "<<Aai.over<<" "<<Aai.survey2<<" "<<Aai.songchi/5*5<<" B"<<bi+1<<" "<<Bbi.over<<" "<<Bbi.survey2<<" "<<Bbi.songchi/5*5<

13、<endl; if(AB=0&&bi<n2) if(Bbi.survey1=(+Bbi.survey2) cout<<time+1<<" A"<<ai+1<<" "<<Aai.over<<" "<<Aai.survey2<<" "<<Aai.songchi/5*5<<" B"<<bi+1<<" "<&

14、lt;Bbi.over<<" "<<Bbi.survey2<<" "<<Bbi.songchi/5*5<<endl; cout<<time+1<<" B"<<bi+1<<"進程結(jié)束"<<endl; Bbi.label=1; label=2; else if(time+1)%5=0) cout<<time+1<<" A"<<ai+1<&l

15、t;" "<<Aai.over<<" "<<Aai.survey2<<" "<<Aai.songchi/5*5<<" B"<<bi+1<<" "<<Bbi.over<<" "<<Bbi.survey2<<" "<<Bbi.songchi/5*5<<endl; if(kb=-1) bi=bi-1; if(ka=-1) ai=ai-1; int main() int n1,n2; pcb A10,B10; cout<<"輸入A進程個數(shù)"<<