hrrn.m

3个CPU调度算法的simulation. 分别是Short Job First,High Return Ration Next, Round Robin. 模拟得到R,U,S等重要指标。

function [newP,QReady]=HRRN(QReady,Qlen,GlobalCycle)
%It is SURE that there are processes in the queue now. now just to choose one
%The order is not important, so we can sort the queue first,then pick the
%first one. 
%
%NOTE: newP is not newp!!!!!!!!!!!!
%only for HRRN

newP=-1; % if newp is -1 when return, we can find some problem because we can not access TB(-1)!!

%calculate the return ratio=waitingtime/Tb, Note when there is empty
%element, we get 0/0=Inf, which Inf is a big number!!Not GOOD
% disp('------before sub----');
% QReady
for i=1:Qlen
  if QReady(i,1)~=0
     QReady(i,4)=GlobalCycle-QReady(i,3);  % get the waiting time
  end
end
% QReady

for i=1:Qlen
    if QReady(i,1)~=0
        QReady(i,4)=QReady(i,4)/QReady(i,2);   % return ratio=waitingtime/Tb
    end
end

QReady=sortrows(QReady,-4);            %by the return ratio,descending

% disp('------after sub----');
% QReady

%QReady(i,4)
%the result could be zero, so should differ for empty element 

[Y,Index]=max(QReady(:,4));    %find the first max return ratio
maxIndex=Index;
% Y
%--------------------------------------------
if Y==0                        %if 0 then there are two situation: empty element  or process no not waiting
%     fprintf('xxxxxxx\n');
   for i=1:Qlen                %find the first process with 0 waitingtime
       if QReady(i,1)>0
           newP=QReady(i,1);   %save the info
           maxIndex=i;
           break;
       end
       i=i+1;
   end
end
%----------------------------------------------
newP=QReady(Index,1);

QReady(:,4)=0;                  %restore to the 0 state

if newP<=0
    fprintf('Not good! newP=%d\n',newP);
    return;
end

%--------------------------------------------------
QReady(maxIndex,:)=0;  %move it out from the Queue by set 0
QReady=sortrows(QReady,-1);% move 0 out from the Queue
%QReady=sortrows(QReady,-3); %Resort. by descending order