No feasible solution in optimisation in linprog
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My code:
f= [89;82;82;29;76];
A1= [0 0 2 1 0
2 11 0 0 1
60 60 0 0 8
2 2 0 0 50
15 9 0 0 8
0 0 20 14 9
4 0 60 14 0];
A= [A1;-A1];
AEQ= [1 1 1 1 1];
b= [100;140;120;130;140;150;60; -10; -100;-12;-13;-80;-13;-45];
alpha = 19:0.1:22;
advertisement=zeros(length(alpha),1);
obj_values = [];
x_values = [];
for i = 1:length(alpha)
current_limit = alpha(i);
[x, obj, exitflag, output] = linprog(f, A, b, AEQ, current_limit, [],[]);
advertisement(i) = current_limit;
x_values = [x_values,x];
if exitflag == 1
obj_values = vertcat(obj_values, obj);
else
obj_values = vertcat(obj_values, NaN);
end
end
Hello, I'm not too sure where went wrong but it does not seems to produce any feasible solutions with regards to the problem that I have coded and there are no errors with the code. Would love if anyone could provide any advices.
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採用された回答
Matt J
2023 年 2 月 11 日
編集済み: Matt J
2023 年 2 月 11 日
Your problem is infeasible because of the input problem data you provided, not because of coding errors. In particular, in Market Group 3, it is clear that you must allocate less than 3 to TVL and TVP in order to stay below the Saturation Level of 120. However, if TVP abd TVL are <3, there is no way for Market Group 2 to reach its Minimum Exposures of 100.
2 件のコメント
Matt J
2023 年 2 月 11 日
If the optimization variables x(i) represent monetary expenditures, I'm assuming there should be non-negativity bounds on them as well, unless there are somehow ways to make negative expenditures in this scenario,
[x, obj, exitflag, output] = linprog(f, A, b, AEQ, current_limit, zeros(size(f)),[]);
その他の回答 (2 件)
Alan Stevens
2023 年 2 月 11 日
I don't know if this will fix your problem, but I think the first number in the second row of A1 should be 8 not 2.
John D'Errico
2023 年 2 月 11 日
編集済み: John D'Errico
2023 年 2 月 11 日
I posted a question recently to help explain some of the reasons when linprog fails. You should read my first answer, as it discusses infeasibility by linprog.
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