Asked by Roberto
on 28 Mar 2013

I have a network with 10 nodes, each of which knows exactly the distance from other nodes. Suppose node 1 starts to transmit. All nodes at a distance greater than d from another node can transmit at the same time. Observing the adjacent array of node 1, we can see NODE(1).adj = [0 1 1 1 0 1 0 1 0 0]. It means that node 2, 3, 4, 6, 8 are at a distance greater than d from node 1 (if the distance is greater than d the value is 1). We put these nodes in another array called CANDIDATES, so CANDIDATES = [2 3 4 6 8]. But node 2 and node 6 cannot transmit at the same time because their mutual distance is lower than d, as we can see by analyzing NODE(2).adj. My idea is: the first element of CANDIDATES transmit, in this case node 2, then the second element, node 3, controls its distance from the active node, node 2: if the distance between node 2 and node 3 is greater than d then node 3 can transmit; after that, the third element, node 4, controls its distance from all active nodes, node 2 and node 3: if the distance from both nodes is greater than d, node 4 can transmit at the same time. In the test I've made only node 1, 2 and 3 can transmit simultaneously. how can I make this comparison? here is the code I've written so far

%%%%%%%%%%%%%%%%%%%% PARAMETERS %%%%%%%%%%%%%%%%%%%%%%%%%%%%

%field dimensions xm = 50; ym = 50;

%numeber of nodes n = 10;

%treshold distance d = 30;

%%%%%%%%%%%%%%%%%%%% END PARAMETERS %%%%%%%%%%%%%%%%%%%%%%%

%create a network for i=1:1:n %node position NODE(i).xd = rand(1,1)*xm; NODE(i).yd = rand(1,1)*ym;

%adjacent array NODE(i).adj = zeros(1,n); end

for i=1:1:n %node state: 1 tx/rx, 0 dozen NODE(i).state(1,1) = 0; %packet type: 1, 2, 3 NODE(i).pkt(1,1) = 0; %destination NODE(i).pkt(2,1) = 0; end

%adjacency matrix: array which relates only to the node i; matrix built by rows for i=1:1:n for j=1:1:n if(i ~= j) distance = sqrt((NODE(i).xd - NODE(j).xd)^2+(NODE(i).yd - NODE(j).yd)^2); NODE(i).adj(1,j) = distance; end end end

%pick a random destination (random number between o and n) dst = floor((n+1)*rand(1));

%a random node begin transmissions temp = ceil(n*rand);

NODE(temp).state(1,1) = 1; NODE(temp).pkt(1,1) = 1; NODE(temp).pkt(2,1) = dst;

for i=1:1:n if(NODE(temp).adj(1,i) > d) provvisorio(1,i) = i; else provvisorio(1,i) = 0; end end

j = 1; for i=1:1:n if(provvisorio(1,i) ~= 0) candidate(1,j) = provvisorio(1,i); j = j+1; end end

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Answer by Andrei Bobrov
on 28 Mar 2013

Edited by Andrei Bobrov
on 28 Mar 2013

Accepted answer

n = 10; xm = 40; ym = 60; ij = [randi(xm,n,1), randi(ym,n,1)];

d = bsxfun(@minus,ij,permute(ij,[3,2,1])); dist1 = squeeze(bsxfun(@hypot,d(:,1,:),d(:,2,:))); dist1(1:n+2:end) = inf; out = find(all(dist1 > 30));

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