intlinprog is returning noninteger values
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I have my code where DDD is a 35x60 matrix I'm trying to get integer values for w(26:60) but its giving me
w= -1.0000
0
-1.0000
1.0000
0
-0.6667
-0.2222
-1.0000
-0.6667
-1.0000
-0.5556
-1.0000
-0.6667
-1.0000
-1.0000
-0.3333
-1.0000
0.7778
0
-1.0000
-1.0000
-0.7778
-0.8890
-0.6667
-0.3333
-1.0000
-1.0000
-1.0000
-2.0000
-1.0000
-1.0000
0
-1.0000
-1.7778
-1.0000
-2.0000
-1.6667
0
1.0000
0
0
-1.0000
-1.0000
-2.0000
-2.0000
-2.0000
-2.0000
-1.0000
-2.0000
-3.0000
-1.6667
-2.0000
-2.0000
-1.2222
-2.0000
-2.0000
-3.0000
-2.0000
-2.3333
-15.0000
For my Code:
intcon= zeros(35,1);
for i= 1 : 35
intcon(i,1)=25+i;
end
D = [ DDD ; eye(25), zeros(25,35); -eye(25), zeros(25,35) ] ;
f= [zeros(25,1) ; g ; -1 ] ;
b=[zeros(34,1); [1-0.0001 ] ; ones(25,1); ones(25,1) ] ;
w = intlinprog(-f,intcon,D,b) ;
4 个评论
What do the exitflag and other output arguments say?
[x,fval,exitflag,output] = intlinprog(...)
In any case, intlinprog will have an easier time of things if you use lb,ub to express the upper and lower bounds on your first 25 unknowns,
b=[zeros(34,1); 0.9999];
f= [zeros(25,1) ; g ; -1 ] ;
lb(1:25,1)=-1; lb(26:60)=-inf;
ub(1:25,1)=+1; ub(26:60)=+inf;
[w,fval,exitflag,output] = intlinprog(-f,26:60,DDD,b,[],[],lb,ub) ;
Taner Cokyasar
2016-7-14
Writing this just for future users: The intcon code you are using seems to be wrong. Intcon should be (1,35) for your problem. So all numbers should in the first row. The following code could solve this problem. As Matt mentioned, ub and lb are also necessary for particular bounds on variables.
intcon = 26:1:60;
Brendan Hamm
2016-7-15
@Taner For future users, it should be known that the intcon being a row or column vector does not matter. Internally, intlinprog uses:
intcon = intcon(:)
Maybe at some point in the past this was different, I am not sure, but this would mean future users using past versions :).
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