Sunday, January 31, 2021

Solving a Multidimensional NLP via Line Search

Someone posted a nonconvex nonlinear optimization model on OR Stack Exchange and asked for advice about possible reformulations, piecewise linear approximations, using global optimizers, and other things. The model is as follows:\begin{alignat}{1} \max\,\, & q_{1}+q_{2}\\ \mathrm{s.t.} & \sum_{i=1}^{n}p_{i}x_{i}=\sum_{t=0}^{T}\frac{F_{t}}{(1+q_{1})^{t}} &(1)\\ & \sum_{i=1}^{n}p_{i}x_{i}=\sum_{t=0}^{T}\sum_{i=1}^{n}\frac{b_{t,i}x_{i}}{(1+q_{2})^{t}} &(2)\\ & \sum_{i=1}^{n}p_{i}x_{i}=\beta\sum_{t=0}^{T}\frac{F_{t}}{(1+q_{1}+q_{2})^{t}} &(3)\\ & q_{1,}q_{2}\ge0\\ & x\in\left[0,1\right]^{n} \end{alignat} All symbols other than $q_1$, $q_2$ and $x$ are model parameters (or indexes). The author originally had $x$ as binary variables, apparently believing that would facilitate linearization of products, but also expressed interest in the case where $x$ is continuous. I'm going to propose a possible "low-tech" solution procedure for the continuous case. The binary case is probably a bit too tough for me. The author supplied sample data for all parameters except $\beta$, with dimensions $n=3$ and $T=4$ but expressed a desire to solve the model for $n=10,000$ and $T=1,200$ (gulp).

Note that the left-hand sides (LHSes) of the three constraints are identical. Let $h()$ be the function on the right-hand side (RHS) of constraint (1), so that the RHS of (1) is $h(q_1)$. $h()$ is a monotonically decreasing function. The RHS of (3) is $\beta h(q_1 + q_2)$. Since the left sides are equal, we have $$\beta h(q_1 + q_2) = h(q_1) \quad (4)$$which tells us several things. First, $q_2 \ge 0 \implies h(q_1+q_2) \le h(q_1)$, so if $\beta<1$ it is impossible to satisfy (4). Second, if $\beta =1$, (4) implies that $q_2 = 0$, which simplifies the problem a bit. Lastly, let's assume $\beta > 1$. For fixed $q_1$ the LHS of (4) is monotonically decreasing in $q_2$, with the LHS greater than the RHS when $q_2 = 0$ and with $$\lim_{q_2\rightarrow \infty} \beta h(q_1+q_2) = \beta F_0.$$ If $\beta F_0 > h(q_1)$, there is no $q_2$ that can balance equation (4), and so the value of $q_1$ is infeasible in the model. If $\beta F_0 < h(q_1)$, then there is exactly one value of $q_2$ for which (4) holds, and we can find it via line search.

Next, suppose that we have a candidate value for $q_1$ and have found the unique corresponding value of $q_2$ by solving (4). We just need to find a vector $x\in [0,1]^n$ that satisfies (1) and (2). Equation (3) will automatically hold if (1) does, given (4). We can find $x$ by solving a linear program that minimizes 0 subject to (1), (2) and the bounds for $x$.

Thus, we have basically turned the problem into a line search on $q_1$. Let's set an arbitrary upper limit of $Q$ for $q_1$ and $q_2$, so that our initial "interval of uncertainty" for $q_1$ is $[0, Q]$. It's entirely possible that neither 0 nor $Q$ is a feasible value for $q_1$, so our first task is to search upward from $0$ until we find a feasible value (call it $Q_\ell$) for $q_1$, then downward from $Q$ until we find a feasible value (call it $Q_h$) for $q_1$. After that, we cross our fingers and hope that all $q_1 \in [Q_\ell,Q_h]$ are feasible. I think this is true, although I do not have a proof. (I'm much less confident that it is true if we require $x$ to be binary.) Since $q_2$ is a function of $q_1$ and the objective function does not contain $x$, we can search $[Q_\ell,Q_h]$ for a local optimum (for instance, by golden section search) and hope that the objective function is unimodal as a function of $q_1$, so that the local optimum is a global optimum. (Again, I do not have proof, although I would not be surprised if it were true.)

I put this to the test with an R script, using the author's example data. The linear programs were solved using CPLEX, with the model expressed via the OMPR package for R and using ROI as the intermediary between OMPR and CPLEX. I first concocted an arbitrary feasible solution and used it to compute $\beta$, so that I would be sure that the problem was feasible with my choice of $\beta$. Using $\beta = 1.01866$ and 100 (arbitrarily chosen) as the initial upper bounds for $q_1$ and $q_2$, my code produced an "optimal" solution of $q_1= 5.450373$, $q_2 = 0.4664311$, $x = (1, 0.1334608, 0)$ with objective value $5.916804$. There is a bit of rounding error involved: the common LHS of (1)-(3) evaluated to 126.6189, while the three RHSes were 126.6186, 126.6188, and 126.6186. (In my graduate student days, our characterization of this would be "good enough for government work".) Excluding loading libraries, the entire process took under three seconds on my desktop computer.

You can access my R code from my web site. It is in the form of an R notebook (with the code embedded), so even if you are not fluent in R, you can at least read the "prose" portions and see some of the nagging details involved.


Thursday, January 28, 2021

A Monotonic Assignment Problem

 A question posted on Stack Overflow can be translated to an assignment problem with a few "quirks". First, the number of sources ($m$) is less than the number of sinks ($n$), so while every source is assigned to exactly one sink, not every sink is assigned to a source. Second, there are vectors $a\in\mathbb{R}^m$ and $b\in\mathbb{R}^n$ containing weights for each source and sink, and the cost of assigning source $i$ to sink $j$ is $a_i \times b_j$. Finally, there is a monotonicity constraint. If source $i$ is assigned to sink $j$, then source $i+1$ can only be assigned to one of the sinks $j+1,\dots,n$.

Fellow blogger Erwin Kalvelagen posted a MIP model for the problem and explored some approaches to solving it. A key takeaway is that for a randomly generated problem instance with $m=100$ and $n=1,000$, CPLEX needed about half an hour to get a provably optimal solution. After seeing Erwin's post, I did some coding to cook up a network (shortest path) solution in Java. Several people proposed similar and in some cases essentially the same model in comments on Erwin's post. Today, while I was stuck on a Zoom committee call and fighting with various drawing programs to get a legible diagram, Erwin produced a follow-up post showing the network solution (including the diagram I was struggling to produce ... so I'll refer readers to Erwin's post and forget about drawing it here).

The network is a layered digraph (nodes organized in layers, directed arcs from nodes in one layer to nodes in the next layer). It includes two dummy nodes (a start node in layer 0 and a finish node in layer $m+1$). All nodes in layer $i\in \lbrace 1,\dots,m \rbrace$ represent possible sink assignments for source $i$. The cost of an arc entering a node representing sink $j$ in layer $i$ is $a_i \times b_j$, regardless of the source of the arc. All nodes in layer $m$ connect to the finish node via an arc with cost 0. The objective value of any valid assignment is the sum of the arc costs in the path from start to finish corresponding to that assignment, and the optimal solution corresponds to the shortest path from start to finish.

The monotonicity restriction is enforced simply by omitting arcs from any node in layer $i$ to a lower-index node in layer $i+1$. It also allows us to eliminate some nodes. In the first layer after the start node (where we assign source 1), the available sinks are $1,\dots,n-m+1$. The reason sinks $n-m+2,\dots,n$ are unavailable is that assigning source 1 to one of them and enforcing monotonicity would cause us to run out of sinks before we had made an assignment for every source. Similarly, nodes in layer $i>1$ begin with sink $i$ (because the first $i-1$ sinks must have been assigned or skipped in earlier layers) and end with sink $n-m+i$ (to allow enough sinks to cover the remaining $m-i$ nodes).

For the dimensions $m=100$, $n=1000$, the network has 90,102 nodes and 40,230,551 arcs. That may sound like a lot, but my Java code solves it in under four seconds, including the time spent setting up the network. I used the excellent (open-source) algs4 library, and specifically the AcyclicSP class for solving the shortest path problem. Erwin reports even faster solution time for his network model (0.9 seconds, coded in Python), albeit on different hardware. At any rate, he needed about half an hour to solve the MIP model, so the main takeaway is that for this problem the network model is much faster.

If anyone is interested, my Java code is available for download from my Git repository. The main branch contains just the network model, and the only dependency is the algs4 library. There is also a branch named "CPLEX" which contains the MIP model, in case you either want to compare speeds or just confirm that the network model is getting correct results. If you grab that branch, you will also need to have CPLEX installed.

Friday, January 22, 2021

Rainbow Parentheses in RStudio

I use the open-source edition of the RStudio IDE for any R coding I do, and I'm a big fan of it. The latest version (1.4.1103) introduced a new feature that was previously only available in alpha and beta versions: rainbow parentheses. I'd never heard the term before, but the meaning turns out to be remarkably simple. When turned on, if you enter an expression with nested parentheses, brackets or braces, RStudio automatically color codes the parentheses (brackets, braces) to make it easier to see matching pairs. This is in addition to the existing feature that highlights the matching delimiter when you put the cursor after a delimiter.

I was geeked to try this out, but when I first installed the latest version and turned it on, I did not see any changes. Eventually I figured out that it was color coding the delimiters, but the differences were too subtle for me to see. (This was with the default Textmate theme for the IDE.) So I hacked a new theme which makes the colors easier to see. I'll go through the steps here.

First, let me point to some documentation. In a blog post, the folks at RStudio explain how to turn rainbow parentheses on, either globally or just for specific files, and near the end tell which CSS classes need to be tweaked to customize the colors (.ace_paren_color_0 to .ace_paren_color_6). A separate document discusses how to create custom themes.

Theme selection in RStudio is done via Tools > Global Options... > Appearance > Editor theme. Since I use the default (Textmate) theme, my first step was to track down that file and make a copy with a new name. On my Linux Mint system, the file is /usr/lib/rstudio/resources/themes/textmate.rstheme. On Windows (and Macs?) the built-in themes will be lurking somewhere else. The customization document linked above alluded to a ~/.R/rstudio/themes directory on Linux and Macs, but that directory did not exist for me. (I created it, and parked my hacked theme file there.) Put the copied file someplace under a new name. I'm not sure whether the file name is significant to RStudio, but better safe than sorry.

Open the copy you made of your preferred theme file in a text editor. The first two lines are comments that define the theme name (as it will appear in the list of editor themes in RStudio) and whether it is a dark theme or not. Change the name to something that won't conflict with existing themes. In my case, the first line was

/* rs-theme-name: Textmate (default) */
which I changed to
/* rs-theme-name: Paul */
(not very clever, but it got the job done).

Now add code at the bottom to define colors for the seven "ace_paren_color" styles. Here's what I used:

.ace_paren_color_0 {
  color: #000000 
  /* black */

.ace_paren_color_1 {
  color: #ff00ff
  /* magenta */

.ace_paren_color_2 {
  color: #ffff00
  /* yellow */

.ace_paren_color_3 {
  color: #0080ff
  /* light blue */

.ace_paren_color_4 {
  color: #FF0000
  /* red */

.ace_paren_color_5 {
  color: #004f39
  /* Spartan green */

.ace_paren_color_6 {
  color: #0000ff
  /* dark blue */

Once you have a candidate style to test, go to the editor themes settings and use the Add... button to add it. I had to fight through a lot of complaints from RStudio, and I needed to restart RStudio to get the new theme to appear. In the same dialog, I selected it, and then put it to the test by typing some nonsense with lots of nested parentheses in a file.

There are two things to watch out for if you try to remove a theme (using the Remove button in that dialog). First, you cannot remove the currently selected theme, so you will need to select a different theme and click Apply, then go back and select the theme to remove. Second, if you remove a theme, RStudio will delete the theme file. So be sure you have a backup copy if you think you might want to use it again (or edit it).

One good thing: once you have added your theme, you can edit your theme without having to remove it and then add it back. After saving any changes, you just have to switch to some other theme and then switch back to your theme to see the impact of the changes in your documents.