Does intlinprog find a local minimum or global minimum?
显示 更早的评论
I have a problem that can mathematically be described as a MILP. However, it is essential that I find the global optimum (if it exists) and not just a feasible point. So I understand that the relevant matlab function is intlinprog. According to Matlab documentation (https://www.mathworks.com/help/optim/ug/local-vs-global-optima.html), I know that linprog guarantees global optimum and not just a local one. My question is does intlinprog guarantee the same or not?
采纳的回答
In ideal math, yes, however real world computers can't do ideal math, so you can be significantly off from the global solution depending on how you set your tolerances.. This is illustrated in the example below. The ideal global minimum is y=1, but intlinprog finds y=0, because it is within the default numerical tolerances.
x=optimvar('x',1,'type','integer','LowerBound',0);
y=optimvar('y',1,'type','integer','LowerBound',0);
prob=optimproblem('Objective',y, 'Constraints', y>=x+1e-4);
sol=solve(prob)
5 个评论
Not that I know of. The global minimum itself is an uncertain thing that can vary discontinuously due to small uncertainties in the problem data. Consider the two solutions below. If the difference between b1 and b2 is meant to be significant, then the solver has correctly given us different solutions. However, if the difference between b1 and b2 is just the result of floating point calculation errors, and they are really supposed to be the same, then one of the solutions is wrong. We can't know which one, because we don't know which b is the correct one.
x=optimvar('x',1,'type','integer','LowerBound',0);
y=optimvar('y',1,'type','integer','LowerBound',0);
b1=1e-7;
b2=0;
prob1=optimproblem('Objective',y,'Constraints', y>=x+b1);
prob2=optimproblem('Objective',y, 'Constraints', y>=x+b2);
opts=optimoptions('intlinprog','ConstraintTolerance',1e-8,'IntegerTolerance',1e-6);
solve(prob1,'options',opts)
solve(prob2,'options',opts)
John D'Errico
2021-11-8
Since intlinprog still lives in a double precision world, you can never set tolerances that will insure there will never be a problem of the type @Matt J pointed out.
I don't think so, because let's apply your proposal to my earlier example, adding an upper bound of 2 on both x and y.
x=optimvar('x',1,'type','integer','LowerBound',0,'UpperBound',2);
y=optimvar('y',1,'type','integer','LowerBound',0,'UpperBound',2);
prob=optimproblem('Objective',y, 'Constraints', y>=x+1e-8);
In this case, there are no non-integer variables, so your method reduces to simply evaluating all 9 combinations 0<=(x,y)<=2.
But how will you decide whether the combination (x,y)=(0,0) is supposed to be feasible or not, bearing in mind that the 1e-8 might just be floating point noise? The decision you make will change the solution and its optimum value by 1.
更多回答（2 个）
Chunru
2021-11-8
1 个投票
MILP is NP-hard problem. All solvers require some heuristic rules and the global optimum can not guarenteed for larger problems.
John D'Errico
2021-11-8
1 个投票
Another issue is that it is easy to formulate a problem with multiple solutions, all equally good. intlinprog should find one of them, but any solution is as good as another. These will typically lie along a constraint boundary, or parallel to one. So is that a global solution or a local one? It depends on how you choose to define what a local solution means to you.
类别
在 帮助中心 和 File Exchange 中查找有关 Linear Programming and Mixed-Integer Linear Programming 的更多信息
Translated by 
