findoptt.m

这是一个超宽带的源码

%
% function optt = findoptt(L,C,pmax,rP,a,w,maxw)
%
% Arguments:
% L - links in the network
% C - coordinates of nodes
% pmax - max. transmission power
% rP - see rndtopo2d
% a - fading coefficient (a<-2)
%
% Output:
% optt(i) - optimal blocking distance around destination of L(i), 
%           i.e. radius of the circle in which there should be no active nodes
%           when L(i) is active.
%
%
% Description:
% We try to estimate the optimal blocking distance for the source of each link. 
% For every node we start with a candidate distance t. We then estimate a number of slots N needed
% for the network as a number of link sources in that circle. We also estimate the interference
% generated by the rest of the network as I = 2*pi* lambda * int_t^infty x^-alpha, that is
% I = 2*pi*lambda * s^{-alpha+1} / (alpha-1). We then try different possible t until we maximize
% N / (1+I) which is our mimic of the rate objective function.
%
% ADDENDUM: New and improved heuristics. Assume t. Count the number of link sources in the blocking 
% region (it-1). Count the interference of the nodes outside (I). Divide I by (it-1) since only
% every (it-1) node is going to send at a time. Add white noise (1) to I. Minimize (it-1)*I since
% rate is inversly proportional to this.
%
%

function optt = findoptt(L,C,pmax,rP,a,w,maxw,rat)

if nargin < 8
	rat = 0;
end
if nargin < 7
	maxw = 1;
end
if nargin < 6
	w = ones(size(L,1),1);
end


eps = 1e-5;

nl = size(L,1);
nn = size(C,1);

optt = zeros(1,nl);

for i=1:nl
    an = [1:L(i,1)-1,L(i,1)+1:nn];
	dd = dist(an, L(i,2)+zeros(size(an)), C);
    [dd,indd] = sort(dd);
	rPi = rP(indd);


    it = 2;
    mmin = Inf;
    ch = 1;

    while ch & it <= nn-1
        ch = 0;
        t = dd(it);

        
        % calculate interference of the other nodes
        % this is not possible in a real network!
        
        I = 0;
        for j=it+1:nn-1
            I = I + rPi(j)*dd(j)^a;
        end

        I = I / (it-1);
        
        I = 1 + pmax * I;
        
        
        % veoma los estimator na bazi point processa (al zapravo iz moje glave)
        %I = 2*pi/nn;
        %I = I * t^(a+1) / (-a-1);

        
        if mmin > (it-1)*I
            mmin = (it-1)*I;
            ch = 1;
        end
    
        it = it+1;
    end

    if it == nn
        optt(i) = dd(nn-1) + eps;
    else
        optt(i) = dd(it-2) + eps;
    end
        
	optt(i) = optt(i) * (1 + rat * w(i)/maxw);
    
end