Optimization Number Problems

[Griselda Humbarger]

Whenthe amount of complex data that I want to recover becomes smaller, the program will remind me that the calculated data is sparse and can not be processed later. How do I deal with this error to eliminate?

Regarding item 1, what is GEZZ(U) ? As for speed, how much of the run time is prior to seeing the first solver output vs. while solver output is occurring? If you are not using Mosek and have access to it, try using it, because it is more reliable and numerically robust, and likely faster, especially on difficult problems, than SeDuMi or SDPT3. Try every SDP solver you have available. If you show the CVX and solver output, perhaps some of the more expert forum posters could determine something from seeing it. I think forum readers will need to know more about the numerical values in GEZZ(U) to determine whether there might be poor scaling, condition number, or otherwise which could be contributing to solution difficulty, and therefore to long solver run time. That said, CVX is not designed for speed, and in particular it is not designed to be fast in transforming model inputs for submission to the solver; rather, it is designed to save human model building (entry) time.

You need to be more specific regarding item 2. Can you show the smallest possible reproducible problem, complete with all input data, which exhibits the error, as well as showing all CVX and solver output, including the error messages?

In this section we are going to look at optimization problems. In optimization problems we are looking for the largest value or the smallest value that a function can take. We saw how to solve one kind of optimization problem in the Absolute Extrema section where we found the largest and smallest value that a function would take on an interval.

In this problem we want to maximize the area of a field and we know that will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. The two equations for these are,

Setting this equal to zero and solving gives a lone critical point of \(y = 125\). Plugging this into the area gives an area of \(A\left( 125 \right) = 31250\,\mboxf\mboxt^2\). So according to the method from Absolute Extrema section this must be the largest possible area, since the area at either endpoint is zero.

Next, the vast majority of the examples worked over the course of the next section will only have a single critical point. Problems with more than one critical point are often difficult to know which critical point(s) give the optimal value. There are a couple of examples in the next two sections with more than one critical point including one in the next section mentioned above in which none of the methods discussed above easily work. In that example you can see some of the ideas you might need to do in order to find the optimal value.

We want to minimize the cost of the materials subject to the constraint that the volume must be 50ft3. Note as well that the cost for each side is just the area of that side times the appropriate cost.

Now we need the critical point(s) for the cost function. First, notice that \(w = 0\) is not a critical point. Clearly the derivative does not exist at \(w = 0\) but then neither does the function and remember that values of \(w\) will only be critical points if the function also exists at that point. Note that there is also a physical reason to avoid \(w = 0\). We are constructing a box and it would make no sense to have a zero width of the box.

Note as well here that provided \(w > 0\), which from a physical standpoint we know must be true for the width of the box, then the volume function will be concave down and so if we get a single critical point then we know that it will have to be the value that gives the absolute maximum.

In this case we can exclude the negative critical point since we are dealing with a length of a box and we know that these must be positive. Do not however get into the habit of just excluding any negative critical point. There are problems where negative critical points are perfectly valid possible solutions.

Now, as noted above we got a single critical point, 1.2910, and so this must be the value that gives the maximum volume and since the maximum volume is all that was asked for in the problem statement the answer is then : \[V\left( 1.2910 \right) = 2.1517\,\mboxm^3\].

Also, as seen in the last example we used two different methods of verifying that we did get the optimal value. Do not get too locked into one method of doing this verification that you forget about the other methods.

As an interesting side problem and extension to the above example you might want to show that for a given volume, \(L\), the minimum material will be used if \(h = 2r\) regardless of the volume of the can.

So, knowing that whatever \(h\) is it must be in the range \(0 \le h \le 5\) we can see that the second critical point is outside this range and so the only critical point that we need to worry about is 1.9183.

Finally, since the volume is defined and continuous on \(0 \le h \le 5\) all we need to do is plug in the critical points and endpoints into the volume to determine which gives the largest volume. Here are those function evaluations.

This problem is a little different from the previous problems. Both the constraint and the function we are going to optimize are areas. The constraint is that the overall area of the poster must be 200 in2 while we want to optimize the printed area (i.e. the area of the poster with the margins taken out).

Mathematical optimization (alternatively spelled optimisation) or mathematical programming is the selection of a best element, with regard to some criteria, from some set of available alternatives.[1][2] It is generally divided into two subfields: discrete optimization and continuous optimization. Optimization problems arise in all quantitative disciplines from computer science and engineering[3] to operations research and economics, and the development of solution methods has been of interest in mathematics for centuries.[4]

In the more general approach, an optimization problem consists of maximizing or minimizing a real function by systematically choosing input values from within an allowed set and computing the value of the function. The generalization of optimization theory and techniques to other formulations constitutes a large area of applied mathematics.

Problems formulated using this technique in the fields of physics may refer to the technique as energy minimization,[5] speaking of the value of the function f as representing the energy of the system being modeled. In machine learning, it is always necessary to continuously evaluate the quality of a data model by using a cost function where a minimum implies a set of possibly optimal parameters with an optimal (lowest) error.

Typically, A is some subset of the Euclidean space R n \displaystyle \mathbb R ^n , often specified by a set of constraints, equalities or inequalities that the members of A have to satisfy. The domain A of f is called the search space or the choice set, while the elements of A are called candidate solutions or feasible solutions.

While a local minimum is at least as good as any nearby elements, a global minimum is at least as good as every feasible element.Generally, unless the objective function is convex in a minimization problem, there may be several local minima.In a convex problem, if there is a local minimum that is interior (not on the edge of the set of feasible elements), it is also the global minimum, but a nonconvex problem may have more than one local minimum not all of which need be global minima.

asks for the maximum value of the objective function 2x, where x may be any real number. In this case, there is no such maximum as the objective function is unbounded, so the answer is "infinity" or "undefined".

The term "linear programming" for certain optimization cases was due to George B. Dantzig, although much of the theory had been introduced by Leonid Kantorovich in 1939. (Programming in this context does not refer to computer programming, but comes from the use of program by the United States military to refer to proposed training and logistics schedules, which were the problems Dantzig studied at that time.) Dantzig published the Simplex algorithm in 1947, and also John von Neumann and other researches worked on the theoretical aspects of linear programming (like the theory of duality) around the same time.[7]

Adding more than one objective to an optimization problem adds complexity. For example, to optimize a structural design, one would desire a design that is both light and rigid. When two objectives conflict, a trade-off must be created. There may be one lightest design, one stiffest design, and an infinite number of designs that are some compromise of weight and rigidity. The set of trade-off designs that improve upon one criterion at the expense of another is known as the Pareto set. The curve created plotting weight against stiffness of the best designs is known as the Pareto frontier.

A design is judged to be "Pareto optimal" (equivalently, "Pareto efficient" or in the Pareto set) if it is not dominated by any other design: If it is worse than another design in some respects and no better in any respect, then it is dominated and is not Pareto optimal.

The choice among "Pareto optimal" solutions to determine the "favorite solution" is delegated to the decision maker. In other words, defining the problem as multi-objective optimization signals that some information is missing: desirable objectives are given but combinations of them are not rated relative to each other. In some cases, the missing information can be derived by interactive sessions with the decision maker.

Optimization problems are often multi-modal; that is, they possess multiple good solutions. They could all be globally good (same cost function value) or there could be a mix of globally good and locally good solutions. Obtaining all (or at least some of) the multiple solutions is the goal of a multi-modal optimizer.

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