1、精選優質文檔-傾情為你奉上學號E 專業計算機科學與技術 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】最低松弛度優先調度算法【實驗目的】1. 學習最低松弛度算法思想。2. 用代碼實現改算法，解決最低松弛度優先調度問題。【實驗原理】最低松弛度優先（LLF）算法是根據任務緊急（或松弛）的程度，來確定任務的優先級。任務的緊急程度愈高，為該任務所賦予的優先級就愈高，使之優先執行。在實現該算法時要求系統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，被優先調度。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束還需多少時間，如果任務已經運行了一部分，則： 任務松弛度=任務的處理時間-任務已經運行的時間 當前時間【實驗內容】數據結構：本實驗進程采用結構體來存儲線信息，采用結構體數組來存儲進程序列。進程結構定義如下：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;/已經服務時間int songchi;/松弛度int label;/等于0未被處理，等于1已經被處理完畢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;/已經服務時間int songchi;/松弛度int label;/等于0未被處理，等于1已經被處理完畢pcb;void pcb_A(pcb * A,int n)/A進程創建函數 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進程創建函數 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"<<