How can I set a starting point in a situation where I have multiple variables?

I am trying to use the fmincon function in Optimtool.

My objective function is objecfun (w, d) and my constraint function is nonlconstr (w, d, g, x).

The number of variable w is 30(w11, w12, w13, w14, w15, w21, ... , w65), the number of d is 30(d11, d12, d13, d14, d15, d21, ... , d65), g is 6(g1, g2, g3, g4, g5, g6), and x is 3(x1, x2, x3).

I entered a value within the constraint range of each variable at the starting point, but an error of 'x0: matched dimension of the concatenated matrix' occurred.

How do I set the starting point in optimtool?

I look forward to your reply. Thank you.

[Walter Roberson]

When you say that the variable w is a total of 30, do you mean that it is a vector of 30 elements? And d, g, and x are constant additional inputs, rather than variables that are to be optimized?

[TaeWan Kim]

The number of variable w is 30(w11, w12, w13, w14, w15, w21, ... , w65), the number of d is 30(d11, d12, d13, d14, d15, d21, ... , d65), g is 6(g1, g2, g3, g4, g5, g6), and x is 3(x1, x2, x3).

I attached a photograph together. please check.

I look forward to your reply. Thank you.

[Walter Roberson]

Your objfun requires two parameters, but you specify @objecfun as the objective function. optimtool is only going to pass a single parameter to the objective function. You need to specify @(w) objecfun(w,d) ... though I have concerns about whether it will be able to find d inside the tool.

[TaeWan Kim]

I originally had 4 variables (w, d, g, x) in the objective function, but I changed all the variables to x to use the fmincon solver (all 69 variables are represented by x1 to x69) , The optimum value was obtained. But I have three questions.

1) Is it ok to enter linear constraints in c and ceq, where you enter nonlinear constraints? (It has already been confirmed that fmincon solver is activated.)

2) Should I set a starting point within the constraints of all variables?

3) If the optimization result is out of the constraint range, will this result be obtained only if I set the objective function and constraint range incorrectly?

I look forward to your reply. Thank you.

Best regards.

[Walter Roberson]

1) There is no linear constraints c and ceq, at least under that name. The typical names used are

• A -- something x vars matrix used for linear inequalities. Use one row for each linear inequality.
• b -- something x 1 matrix used for linear inequalities. A*x' <= b . Use one row for each linear inequality
• Aeq -- something x vars matrix used for linear equalities. Use one row for each linear equality
• beq -- something x 1 matrix used for linear equalities. Aeq*x' = beq . Use one row for each linear equality/
• lb -- 1 x vars matrix of lower bounds
• ub -- 1 x vars matrix of upper bounds
• nonlcon -- function handle to nonlinear constraints function. Must return two outputs, usually called c and ceq . c <= 0 is used for nonlinear inequalities. ceq = 0 is used for nonlinear equalities

lb and ub are the most efficient constraints. Most routines enforce them at all times

A and b are the second most efficient constraints. Sometimes routines temporarily permit them to be violated, especially during initial gradient estimation.

Aeq and beq are more difficult to work with. When they are enforced, they are enforced within a tolerance, and some of the minimizers might need to let them be violated for a while to find a region where they hold.

nonlinear constraints are the most difficult to work with. It is common for the minimizers to permit nonlinear constraints to be violated deliberately in hopes of finding regions that the constraints hold in. If you are counting on the nonlinear constraints being enforced in order to make one of the calculations in the objective function valid (e.g. avoiding square root of a negative number) then you will probably run into problems.

The preferred order is lb/ub, A/b, Aeq/beq, and finally worst nonlcon . Express any given constraint in the first of these that can handle it. All of them can be used together. Starting point and A, b are the only ones that must be passed in (if you have no constraints at all then use fminunc)

2) If it is feasible to use a starting point that satisfies all constraints, then that is a good idea. But it is common for it to only be feasible to satisfy lb/ub and A/b and perhaps Aeq/Beq, and it is common it to not be known what point satisfies the nonlinear constraints as well. When you have nonlinear constraints, a lot of the time can get spent hunting for a point that satisfies all of the constraints.

3) There is a constraint tolerance and especially considering round-off it can be inefficient to tighten the constraint tolerance too much.

Especially if you have nonlinear constraints, ut can be difficult for the function to find an area that satisfies all of the constraints. This does not mean that the constraints are set incorrectly, but the hunting might not go well; and while chasing a direction that seems to have the potential to find a location that satisfies all constraints, the search might get stuck in a local minima. A different starting point can make a real difference. Sometimes a lot of different starting points are needed just to find a single feasible spot.

[TaeWan Kim]

Your answer has been a great help. Thank you.

Thanks to your answers, I was able to solve some of the problems.

Last time I defined lb and ub for each variable, and linear equality constraints and linear inequality constraints for nonlinear constraint functions.(file name: nonconstr)

But this time, lb and ub are entered in the lb space and the ub space, which are prepared in the Optimization App. As a result, the variables were within the constraints I wanted.

To acquire right result, I realized that I need to enter the linear equation constraint and the linear inequality constraint into the others space(A, b, Aeq, beq).

I refer to the examples provided by mathwork to find out how to enter linear equation constraints and linear inequality constraints. (https://kr.mathworks.com/help/optim/ug/optimization-tool-with-the-fmincon-solver.html#d117e13816)

However, this example only shows two variables, only one linear inequality constraint, and there is no explanation for even linear equality constraints.

For my problem, there are variables x1 to x38. My function is declared only x like this. → function f = objecfun5(x). Therefore, I inputed 38 values ​​in each row.

To represent -x (5) + x (33) <= 3 and -x(6)+x(33)<=6, I entered the following values.

A: [0 0 0 0 –1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0; 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0]

b: [3 6]

However, only the command 'Aineq: Error: Input characters not valid for Matlab statements or expressions' is printed.

So, I want to know how to enter linear equation constraints and linear inequality constraints in the Optimization App in situations where there are many variables and constraints.

I look forward to your reply. Thank you.

Best regards.

Taewan Kim

[Walter Roberson]

In your A, your first - (minus) character is not a - character and is instead U+2013, en-dash .

[TaeWan Kim]

Thank you for your reply.

As a result of checking, it was really - problem.

I have another question. I want only one of several variables to be positive and all others to be zero. Can I add this constraint now?

I look forward to your reply. Thank you.

Best regards.

Taewan Kim

[Walter Roberson]

Set the upper and lower bound for the variable to 0 in order to force the variable value to be 0. It is, however, better to substitute constants into the equations and use the modified versions of those.