Optimisation Problems Hsc

[Catherin Bergan]

For each combinatorial optimization problem, there is a corresponding decision problem that asks whether there is a feasible solution for some particular measure m0. For example, if there is a graph G which contains vertices u and v, an optimization problem might be "find a path from u to v that uses the fewest edges". This problem might have an answer of, say, 4. A corresponding decision problem would be "is there a path from u to v that uses 10 or fewer edges?" This problem can be answered with a simple 'yes' or 'no'.

In the field of approximation algorithms, algorithms are designed to find near-optimal solutions to hard problems. The usual decision version is then an inadequate definition of the problem since it only specifies acceptable solutions. Even though we could introduce suitable decision problems, the problem is more naturally characterized as an optimization problem.[2]

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).

I am slightly confused by some terminology I have encountered regarding the complexity of optimization problems. In an algorithms class, I had the large parsimony problem described as NP-complete. However, I am not exactly sure what the term NP-complete means in the context of an optimization problem. Does this just mean that the corresponding decision problem is NP-complete? And does that mean that the optimization problem may in fact be harder (perhaps outside of NP)?

In particular, I am concerned about the fact that while an NP-complete decision problem is polynomial time verifiable, a solution to a corresponding optimization problem does not appear to be polynomial time verifiable. Does that mean that the problem is not really in NP, or is polynomial time verifiability only a characteristic of NP decision problems?

You have to be careful when carrying over the concepts from decision problems. It can be done and a precise notion of NP-completeness for optimization problems can be given. Look at this answer. It is of course different from the NP-completeness for decision problems, but it is based on the sames ideas (reductions).

If you want to verify that a solution is not just feasible, but also optimal, I would say that this is as hard as solving the original optimization problem because, in order to refute a given feasible and possibly optimal solution as non-optimal, you have to give a better solution, which might require you to find the true optimal solution.

For the relation between decision and optimization (search) problems, take a peek at Bellare and Goldwasser's The Complexity of Decision versus Search, SIAM Journal of Computing 23(1), feb 1994. In a nutshell: If the decision problem is NP-complete, the search problem is hard too, and can be solved calling the decision problem's solver a polynomial number of times.

You can easily define NP-hard optimisation problems: We call a decision problem D NP-hard if any decision problem in NP can be reduced to D in polynomial time. And we can use the same definition for any problem P: P is NP-hard if any decision problem in NP can be reduced to P in polynomial time.

The reason most optimization problems can be classed as P, NP, NP-complete, etc., is the Kuhn-Tucker conditions. I'll talk in terms of linear-programming problems, but the KTC apply in many other optimization problems. For each optimization problem there is a dual. If the goal in the original problem is to maximize some function, then the dual (usually) has a function to be minimized.* Feasible, but non-optimal solutions to the original problem will be infeasible/invalid for the dual problem, and vice-versa. If, and only if, a solution is feasible for the primary and dual, it is an optimal solution for both. (Technically, may be one of a large number of optimal solutions which give the same result.)

So finding an optimal solution of an optimization problem is equivalent to finding a valid solution for the primary and dual. You may use optimization algorithms to find that solution, but the overall process is an existence proof.

So far, I have been able to align my two optimization problems vertically. Searching other questions/answers here, I tried using the multicol package but the end result is really bad (see below for the code). I cannot figure out a nice alignment.

The screenshot you posted suggests that you want to top-align the two optimization problem statements. The multicol package may not be the best-suited tool for this typesetting objective. I would like to suggest that you use two side-by-side top-aligned minipage environments, each of width 0.45\textwidth. A separate comment: since there really is just one alignment column in the optimization problem statements, align should do just as well as alignat.

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There are two types of problems in this world. Optimisation problems and Unbounded problems and most creative people are only good at solving one of those types of problems. More on how this impacts Twitter and Elon, but first some definitions.

An optimisation problem is the one most people think of. It's the type of problem where you ultimately want the measurement that reflects success to trend to either zero (this never happens) or 1 (this happens all the time)

Solving these types of problems requires creative thinking on how to get the thing being measured to 0 or 1, it might not feel like creative thinking as we've been conditioned to think that creativity is only what we observe in the other type of problem, what I call unbounded problems.

An unbounded problem is where the solution can't be reached by trying to optimise the existing system. It happens when you reach outside the existing mode of thinking or system entirely. It requires leaps of imagination, iteration in strange or unfamiliar directions, an intuition regarding the human condition and quite often for the zeitgeist to be aligned .. what we often call an idea whose time has come.

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