fmincon with lower bound fails, even though solution is at initial point

Question

I'm trying to minimize a non-linear objective function (my actual function is much more complicated than that, but I found that even this simple function illustrates the point), where I know that minimum is obtained at the initial point x0:

fun = @(x) x(1)^2+x(2)^2;
x0 = [0 0];
lb1 = [0 0];
lb2 = [-1 -1];

[xc1 fvalc1] = fmincon(fun, x0, [],[],[],[], lb1, [Inf Inf])

Which outputs:

>> xc1 = 1.0e-03 * [0.6457    0.6457]
>> fvalc1 = 8.3378e-07

However, both using a different lower bound or using fminsearch instead work correctly:

[xc2 fvalc2] = fmincon(fun, x0, [],[],[],[], lb2, [Inf Inf])
>> xc2 = [0     0]
>> fvalc2 =  0

[xs fvals] = fminsearch(fun, x0)
>> xs =  [0     0]
>> fvals =   0

What goes wrong in the first fmincon call?

Wolfie Wolfie · Accepted Answer · 2018-03-05T11:50:14

We can diagnose this using the output output argument as specified in the docs

[xc1, fvalc1, ~, output] = fmincon(fun, x0, [],[],[],[], lb1, [Inf Inf])

The value output.stepsize is the final step size taken in the iterative solving process. In this case:

output.stepsize
>> ans = 6.586e-4

The estimated minima was at x = [6.457e-4, 6.457e-4] and the lower bounds you've permitted are [0 0], so the solver is not permitted to take another step! Another step would give x = [-1.29e-5, -1.29e-5] which is outside of the boundaries.

When you allow the lower bounds to be [-1, -1] the solver can over-shoot the minimum and approach it from all directions.

Moreover, we can use the options input to get even better insight!

options.Display = 'iter';
[xc1, fvalc1, ~, output] = fmincon(fun, x0, [],[],[],[], lb1, [Inf Inf], [], options);

Printed to the command window we see this:

Your initial point x0 is not between bounds lb and ub; FMINCON
shifted x0 to strictly satisfy the bounds.

                                            First-order      Norm of
 Iter F-count            f(x)  Feasibility   optimality         step
    0       3    1.960200e+00    0.000e+00    9.900e-01
    1       6    1.220345e-02    0.000e+00    8.437e-01    1.290e+00
    2       9    4.489374e-02    0.000e+00    4.489e-02    1.014e-01
    3      12    1.172900e-02    0.000e+00    1.173e-02    1.036e-01
    4      15    3.453565e-03    0.000e+00    3.454e-03    4.953e-02
    5      18    1.435780e-03    0.000e+00    1.436e-03    2.088e-02
    6      21    4.659097e-04    0.000e+00    4.659e-04    1.631e-02
    7      24    2.379407e-04    0.000e+00    2.379e-04    6.160e-03
    8      27    6.048934e-05    0.000e+00    6.049e-05    7.648e-03
    9      30    1.613884e-05    0.000e+00    1.614e-05    3.760e-03
   10      33    5.096660e-06    0.000e+00    5.097e-06    1.760e-03
   11      36    2.470360e-06    0.000e+00    2.470e-06    6.858e-04
   12      39    8.337765e-07    0.000e+00    8.338e-07    6.586e-04

So your x0 is invalid! This is why the solver doesn't return the result with 1 iteration and lower bounds of [0 0].

fminsearch also works for the same reason - you've not imposed a lower bound on which the solution sits.

fmincon with lower bound fails, even though solution is at initial point

1 Answers