how do i create a binary [0,1]?
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how can i write to declare that x will be variable for binary 0,1
if x(i,j)=1 ;
else (x(k,q)= 0;
end;
i want to put it in my coding below
clc;
clear;
%sum sum sum sum(fik*djq*xij*xkq)
%i,k= facilities
%j,q= location
%f(i,k)= flow between facilities i and k
%d(j,q)= distance between locations j and q
%xij = 1 if facility i is assigned to location j and if otherwise, xij = 0
% Flow matrix: flow assigning facility i (column) to facility k (row)
f = [0 5 7 9;
5 0 4 6;
7 4 0 3;
9 6 3 0];
%Distance matrix: distance assigning location j (column) to location q (row)
d = [0 6 8 9;
6 0 5 1;
8 5 0 2;
9 1 2 0];
z= 0;
nf= 4;
nd= 4;
for i=1:nf
for j=1:nf
for k=1:nd
for q=1:nd
z = min('z','f(i,k)*d(j,q)*x(i,j)*x(k,q)');
end
end
end
end
%Constraints
%The first set of constraints requires that each facility gets exactly one
%location, that is for each facility, the sum of the location values
%corresponding to that facility is exactly one
Constraints.constr1 = sum(x,2) == 1;
%The second set of constraints are inequalities. These constraints specify
%that each office has no more than one facility in it.
Constraints.constr2 = sum(x,1) == 1;
disp (z);
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回答(2 个)
Image Analyst
2018-5-16
Do you mean like this???
% x(i, j) = 1 if facility i is assigned to location j and if otherwise, xij = 0
% Flow matrix: flow assigning facility i (column) to facility k (row)
f = [0 5 7 9;
5 0 4 6;
7 4 0 3;
9 6 3 0];
x = zeros(size(f));
[rows, columns] = size(f)
for row = 1 : rows
for col = 1 : columns
facility_i = col;
facility_k = row;
if f(row, col) == facility_k
x(row, col) = 1;
end
end
end
x
Not exactly what i, j, and k mean. How are they related to row and column. If you say xij, then i=row and j=column, but then in your flow matrix explanation you say i is column, not row and you don't even mention j - you mention k which is row, but i was row. I'm confused (or you are).
semi legend
2018-11-11
编辑:semi legend
2018-11-11
imagine if you have 5 virables , and you want to set the last 2 either on of them 0 or 1
you could do it, but that depends in what you want for that problem "to be or not to be"
[Inf;Inf;Inf;1;1]; to be % it will force (4) and (5) to be binary by 1 or 0 or [Inf;Inf;Inf;0;0]; or not to be
inf ; means array of infinite ( integers)
hope that at least helps
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