How to convert interval from to scalar. The problem is:

k = -0.5:.05:-0.1; (interval)
prob.Objective = k * f1 * x(1) + f2 * x(2);

Result:

Objective must be a scalar OptimizationExpression or a struct containing a scalar OptimizationExpression.

Answer 1

[Walter Roberson]

You are trying to do multi-objective optimization, which is not supported by surrogate optimization or Problem Based Optimization

At that point in the code, whatever f1 and f2 are, they are effectively constants for the purpose of considering the values of k * f1 * x(1) + f2 * x(2) .

If f1*x(1) is positive, then the minimum of the combination k * f1 * x(1) + f2 * x(2) would always be at the point where k is most negative, and there is no point examining larger k.

If f1*x(1) is negative, then the minimum of the combination k * f1 * x(1) + f2 * x(2) would always be at the point where k is least negative, and there is no point examining smaller k.

Therefore, while there might be a point in evaluating at both k = -0.5 and k = -0.1, there is no point at evaluating the other members of the interval. And there is not even a point examining both of those k values:

temp = f1 * x(1);
if temp < 0
   out = -0.1 * f1 * x(1) + f2 * x(2);
else
   out = -0.5 * f1 * x(1) + f2 * x(2);
end
prob.Objective = out;

[Indulis Straume]

but if k * f1 * x(1) + f2 * x(2) need a maximum?

The value of K must then be the least negative?

[Walter Roberson]

If f1*x(1) is positive, then the maximum of the combination k * f1 * x(1) + f2 * x(2) would always be at the point where k is most positive, and there is no point examining smaller k.

If f1*x(1) is negative, then the maximum of the combination k * f1 * x(1) + f2 * x(2) would always be at the point where k is least positive, and there is no point examining larger k.

Therefore, while there might be a point in evaluating at both k = -0.5 and k = -0.1, there is no point at evaluating the other members of the interval. And there is not even a point examining both of those k values:

temp = f1 * x(1);
if temp > 0
    out = -0.1 * f1 * x(1) + f2 * x(2);
else
    out = -0.5 * f1 * x(1) + f2 * x(2);
end
prob.Objective = out;

Answer 2

[Chunru]

Do you mean a loop over a number of interals?

k = -0.5:.05:-0.1;  %(interval)
for i=1:length(k)
    prob.Objective = k(i) * f1 * x(1) + f2 * x(2);
end

[Walter Roberson]

No, the user is trying to create an objective for Problem Based Optimization. Your code is overwriting the objective each time, and you cannot use

k = -0.5:.05:-0.1;  %(interval)
for i=1:length(k)
    prob.Objective(i) = k(i) * f1 * x(1) + f2 * x(2);
end

because problem-based optimization only permits a single scalar value for the Objective