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I need to find the global maximum of the function \begin{align} f\left(x\right) & = p_1 \max\left(\sum a_{1i} x_{1i}, \sum b_{1i} x_{1i}\right) - \sum c_{1i} x_{1i} \\ &+\ldots \\ &+ p_n \max\left(\sum a_{ni} x_{ni}, \sum b_{ni} x_{ni}\right) - \sum c_{ni} x_{ni} \end{align} where all the coefficients $a_{ji}, b_{ji}$ and $c_{ji}$ are nonnegative and each variable $x_{ji}$ lives in $[0,1]$.
The number of terms $n$ is between 500 and 100,000.
I have some flexibility in choosing this $n$.
The index $i$ runs over 30 terms, so each summation in $f$ contains 30 terms.

What are my options to solve this problem?

I think $f$ is a convex function, so the maximization is an integer programming problem. But, at the same time, it has a linear structure, so I wonder if it is better to view as a linear programming problem.

I'm considering a branch-and-bound algorithm, but I don't see what will be good lower an upper bounding functions.

A gradient descent may work, but I need to be sure that it founds the global maximum.

Edit
The variable $x_{ji}$ are not all independents.
Some variables are repeated. For example, \begin{align} &x_{11}=x_{21}=x_{31}=x_{41} \\ &x_{51}=x_{61}=x_{71}=x_{81} \\ &\ldots \end{align} and \begin{align} &x_{21}=x_{22} \\ &x_{23}=x_{24} \\ &\ldots \end{align}

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    $\begingroup$ What is your flexibility in choosing n? Do you get to pick k and any l subterms p_i..., or the first k, or is there some other constraint? Are the 30n variables I see independent from each other? What have you tried for k=1 or 2? $\endgroup$ Feb 8, 2014 at 19:56
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    $\begingroup$ Each variable lives in $[0, 1]$ or in $\{0, 1\}?$ If the former, in what way is this an integer programming problem? $\endgroup$
    – Igor Rivin
    Feb 8, 2014 at 20:38
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    $\begingroup$ I am not sure what you mean still. The convex function $f(x) = -x^2$ on the domain $[-1, 1]$ does not appear to achieve its maximum on an endpoint (which is, I assume, what you mean by "vertex"). Integer programming is far harder than general convex programming. $\endgroup$
    – Igor Rivin
    Feb 8, 2014 at 20:49
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    $\begingroup$ Igor, your example is not convex linear. Unless I am reading things wrong, the poster has to decide on an n, produce 90n coefficients, solve n relatively straightforward optima on n independent domains, and then sum the results and decide whether to choose a different n and repeat the process or not. Even though the poster believes his F and his approximations f are convex, we don't know enough of the heuristics or the process to give him/her reassurance, much less guarantee, that the optimal value will be approached this way. $\endgroup$ Feb 8, 2014 at 21:50
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    $\begingroup$ Ah, I see, so your program is not a convex program, but it does have "convex epigraphs", which should make it more tractable than some... $\endgroup$
    – Igor Rivin
    Feb 9, 2014 at 2:59

2 Answers 2

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This is not an answer, and my expertise in tackling these problems is very small. My commentary above seemed to help, however; Nicolas has a refined version of the problem in any event. I have another opinion to render which may inspire someone else to post a solution.

The version of the problem looks like optimization of a real valued function on a (piece of a) high dimensional manifold, with two features that might be used in developing a quick method for finding a global optimum: the first is that function is composition of an order preserving operator (sum from 1 to n) of other functions of "low dimension", and the second is that these other functions are relatively simple in terms of multivariable optimization; were it not for the relations between the labeled variables, you could tell the computer to solve all the low dimensional problems, tot up the results on an adding machine, and call it a day.

However (at this writing), the relations between the slices (my temporary name for the projections of the manifold into compact subsets of various lower dimensional spaces) also seem to be order preserving, and suggest promise of finding quick optima. The suggestion below is to solve the problem on a slice, then on a related but initially disjoint slice, then laminate the two slices together and adjust the result.

From the example posted in the question, I take the subterm of f that involves variables x_1i and work on that domain of those variables and call that slice S1; I do something similar with 1 replaced by 2 and get slice S2. Lets solve on the two slices and call the optimal inputs xb1 and xb2, assuming no dependencies between the variables for S1 and the variables for S2.

Time to laminate. To simplify matters let's take a simple relation of x_11=x_21. (If one had x_11=-x_21, that would not be an order preserving relation in my view; one might use a different coordinate system or some other artifice to "get everything headed in the same direction", but I will shy away from such difficulties for now.) If the first coordinates of xb1 and xb2 agree, then an appropriate stitching of the solutions should yield a solution xb12 to the sum of the two terms on S12, a lamination of slices S1 and S2 (verification needed). Otherwise, one needs to combine xb1 and xb2 in some way to produce xb12.

I propose for each slice Sj to compute the differential cost d_jc between the solutions already computed and a new optimum under the additional constraint x_j1=c. I won't embarrass myself by working out the algebra and getting it wrong: instead I will (hopefully not) embarrass myself by asserting d_jc is affine in c, or linear in abs(c - xbj_j1). Now we have a one dimensional linear minimization which answer c is applied to the two restricted problems on the slices, producing new values for xb1 and xb2 which agree on the variables x_j1, which can be stitched together to form xb12.

Now the programme is to slice the (domain of the) manifold into Si, solve the simple problems on these slices, and then laminate them back together, hopefully in a computationally cheap way. This is likely highly unoriginal as idea. I hope it proves to be more motivational than unoriginal.

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  • $\begingroup$ This is smart, especially the stiching by minimizing the cost. I hesitate on the 'best' implementation of your solution, because, when $n$ grows, the slice $S_1$ grows too: $x_{11}=x_{12}=\ldots=x_{1,20}$. Anyway, your approach shows that I can do a progressive branch-and-bound to solve the problem. First, I prune with all variables independent, then I prune with the smallest slice, then the next smallest slice, and so on. Thank you. $\endgroup$
    – user24451
    Feb 9, 2014 at 15:13
  • $\begingroup$ Also, when the feasibility set is small enough, I do the stiching by minimization. $\endgroup$
    – user24451
    Feb 9, 2014 at 15:15
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You are trying to maximize a convex function over a convex set (in your case a slice of a high-dimensional cube. This has been studied a lot, and you can see this stackexchange discussion for references (though there are new ones almost daily, since this is of crucial importance in compressed sensing, distance matrix reconstruction, etc).

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