## Saturday, October 23, 2010

### Change to Math Rendering

I just got done switching the handling of mathematical notation in the blog from LaTeXMathML to MathJax. I think it may be a bit easier for me to use as an author, and the fonts seem to render faster in Firefox. The main reason for the switch, though, is that statistics Google compiled on my site visitors showed 69% of viewers using Windows but only 25% using Internet Explorer. I'd like to interpret that as a sign that readers of this blog have above average intelligence (or at least common sense), but it occurred to me that the fact that IE users needed a plug-in to see the mathematical notation in some posts might have been a deterrent. Not wanting to scare anyone away, I switched to MathJax, which I believe will work on a copy of IE that lacks the plug-in (and thus cannot display MathML natively) by switching to web-based fonts automatically. I'm writing this in Firefox (on a Mint laptop that lacks a copy of IE), so testing will have to wait until my next foray in Windows.

## Friday, October 22, 2010

### Backing Up My Blog

I just spent an hour or so searching for options to print Blogger entries (printing just the blog entry itself, not all the widgets and other clutter) and/or back up the content. Ideally, I'd hoped to add a "print this" widget, but apparently none exists, and the best solution I found (which involved adding custom code to the template) didn't work for me because I couldn't figure out the right IDs to use for the bits I wanted to make vanish. There's still the browser print feature, but that gets you all sorts of things you probably don't want.

On the backup issue, I had better luck. There are various options for (essentially) spidering the site and storing pages, including the ScrapBook extension for Firefox (which I already use for other purposes). In my case, I found a simpler option. The key is that I already have source documents for any images I use (because I have to create them myself). For lengthier posts, particularly those with lots of math, I sometimes compose the post in LyX and transfer it to Blogger, which means the LyX document serves as a backup. So my main need is to back up just the text itself (with or without all the extra clutter), primarily for the posts that I create directly in Blogger's editor (such as this one).

Enter the DownThemAll! extension for Firefox. Once you've installed it, the process is pretty simple:

If there's a better way to do this (and particularly if there's a simply way to print just what's in the actual post), I'd love to hear it.

On the backup issue, I had better luck. There are various options for (essentially) spidering the site and storing pages, including the ScrapBook extension for Firefox (which I already use for other purposes). In my case, I found a simpler option. The key is that I already have source documents for any images I use (because I have to create them myself). For lengthier posts, particularly those with lots of math, I sometimes compose the post in LyX and transfer it to Blogger, which means the LyX document serves as a backup. So my main need is to back up just the text itself (with or without all the extra clutter), primarily for the posts that I create directly in Blogger's editor (such as this one).

Enter the DownThemAll! extension for Firefox. Once you've installed it, the process is pretty simple:

- In the Blogger dashboard, select Posting > Edit Posts and list all your posts (or all your recent posts if you've backed up the older ones). There's a limit of something like 300 posts on any one page, so if you're very prolific, you'll need to iterate a few times. (Then again, if you're that prolific, you obviously have no shortage of idle time.)
- While staring at the list of your posts (each post title being a link to view the corresponding entry), click Tools > DownThemAll! Tools > DownThemAll! in Firefox to open the a dialog in which you can select the files to download. Highlight the ones that say "View" in the description field. A quick filter on "*.html" selects them all without (so far at least) grabbing anything else.
- Select where to save them, futz with other settings if you wish, then click Start! and watch DownThemAll! suck down all your postings.

If there's a better way to do this (and particularly if there's a simply way to print just what's in the actual post), I'd love to hear it.

## Tuesday, October 19, 2010

### Piecewise-linear Functions in Math Programs

How to handle a piecewise-linear function of one variable in an optimization
model depends to a large extent on whether it represents an economy
of scale, a diseconomy of scale, or neither. The concepts apply generally,
but in this post I'll stick to cost functions (where smaller is better,
unless you are the federal government) for examples; application to
things like productivity measures (where more is better) requires
some obvious and straightforward adjustments. Also, I'll stick to
continuous functions today.

Let's start with a diseconomy of scale. On the cost side, this is something that grows progressively more expensive at the margin as the argument increases. Wage-based labor is a good example (where, say, time-and-a-half overtime kicks in beyond some point, and double-time beyond some other point). A general representation for such a function is \[f(x)=a_{i}+b_{i}x\mbox{ for }x\in[d_{i-1},d_{i}],\: i\in\{1,\ldots,N\}\] where $d_{0}=-\infty$ and $d_{N}=+\infty$ are allowed and where the following conditions must be met: \begin{eqnarray*}d_{i-1} & < & d_{i}\:\forall i\in\{1,\ldots,N\}\\b_{i-1} & < & b_{i}\:\forall i\in\{2,\ldots,N\}\\a_{i-1}+b_{i-1}d_{i-1} & = & a_{i}+b_{i}d_{i-1}\:\forall i\in\{2,\ldots,N\}.\end{eqnarray*}The first condition just enforces consistent indexing of the intervals, the second condition enforces the diseconomy of scale, and the last condition enforces continuity. The diseconomy of scale ensures that $f$ is a convex function, as illustrated in the following graph.

Letting $f_{i}(x)=a_{i}+b_{i}x$, we can see from the graph that $f$ is just the upper envelope of the set of functions $\{f_{1},\ldots,f_{N}\}$ (easily verified algebraically). To incorporate $f(x)$ in a mathematical program, we introduce a new variable $z$ to act as a surrogate for $f(x)$, and add the constraints\[z\ge a_{i}+b_{i}x\;\forall i\in\{1,\ldots,N\}\] to the model. Anywhere we would use $f(x)$, we use $z$ instead. Typically $f(x)$ is either directly minimized in the objective or indirectly minimized (due to its influence on other objective terms),

and so the optimal solution will contain the minimum possible value of $z$ consistent with the added constraints. In other words, one of the constraints defining $z$ will be binding. On rare occasions this will not hold (for instance, cost is relevant only in satisfying a budget, and the budget limit is quite liberal). In such circumstances, the optimal solution may have $z>f(x)$, but that will not cause an incorrect choice of $x$, and the value of $z$ can be manually adjusted to $f(x)$ after the fact.

Now let's consider an economy of scale, such as might occur when $x$ is a quantity of materials purchases, $f(x)$ is the total purchase cost, and the vendor offers discounts for larger purchases. The marginal rate of change of $f$ is now nonincreasing as a function of $x$, and as a result $f$ is concave, as in the next graph.

Now $f$ is the lower envelope of $\{f_{1},\ldots,f_{N}\}$. We define the segments as before, save with the middle condition reversed to $b_{i-1}>b_{i}\:\forall i\in\{2,\dots,N\}$. We cannot simply add constraints of the form \[z\le a_{i}+b_{i}x\] if we are (directly or indirectly) minimizing $z$, since $z=-\infty$ would be optimal (or, if we asserted a finite lower bound for $z$, the lower bound would automatically be optimal). In the presence of an economy of scale, we are forced to add binary variables to the formulation, either directly or indirectly. The easiest way to do this is to use a type 2 specially ordered set (SOS2) constraint, provided that the solver (and any modeling language/system) understand SOS2, and provided that $d_{0}$ and $d_{N}$ are finite. Introduce new (continuous) variables $\lambda_{0},\ldots,\lambda_{N}\ge0$ along with $z$, and introduce the following constraints:\begin{eqnarray*}\lambda_{0}+\cdots+\lambda_{N} & = & 1 \\ \lambda_{0}d_{0}+\cdots+\lambda_{N}d_{N} & = & x\\ \lambda_{0}f(d_{0})+\cdots+\lambda_{N}f(d_{N}) & = & z.\end{eqnarray*} Also communicate to the solver that $\{\lambda_{0},\ldots,\lambda_{N}\}$ is an SOS2 set. The SOS2 constraint says that at most two of the $\lambda$ variables can be nonzero, and that if two are nonzero they must be consecutive. Suppose that $\lambda_{i}$ and $\lambda_{i+1}$ are positive and all others are zero. Then the new constraints express $x$ as a convex combination of $d_{i}$ and $d_{i+1}$ (with weights $\lambda_{i}$ and $\lambda_{i+1}=1-\lambda_{i}$), and express $z=f(x)$ as a convex combination of $f(d_{i})$ and $f(d_{i+1}$) using the same weights.

If SOS2 constraints are not an option, then binary variables have to be added explicitly. Let $\delta_{i}\in\{0,1\}$ designate whether $x$ belongs to the interval $[d_{i-1},d_{i}]$ ($\delta_{i}=1$) or not ($\delta_{i}=0$). Add the following constraints:\begin{eqnarray*}\delta_{1}+\cdots+\delta_{N} & = & 1\\ x & \le & d_{i}+M(1-\delta_{i})\:\forall i\in\{1,\ldots,N\}\\ x & \ge & d_{i-1}-M(1-\delta_{i})\:\forall i\in\{1,\ldots,N\}\\ z & \ge & a_{i}+b_{i}x-M(1-\delta_{i})\:\forall i\in\{1,\ldots,N\}\end{eqnarray*} where $M$ is a suitably large constant. (This formulation will again require that $d_{0}$ and $d_{N}$ be finite.) The first constraint says that $x$ must belong to exactly one interval. If $\delta_{j}=1$, the second and third constraints force $d_{j-1}\le x\le d_{j}$ and are vacuous for all $i\neq j$ (assuming $M$ is correctly chosen). The final constraint forces $z\ge f_{j}(x)$ and is again vacuous for all $i\neq j$. Since $z$ represents something to be minimized, as before we expect $z=f_{j}(x)$ in almost all cases, and can manually adjust it in the few instances were the optimal solution contains padding.

Let's start with a diseconomy of scale. On the cost side, this is something that grows progressively more expensive at the margin as the argument increases. Wage-based labor is a good example (where, say, time-and-a-half overtime kicks in beyond some point, and double-time beyond some other point). A general representation for such a function is \[f(x)=a_{i}+b_{i}x\mbox{ for }x\in[d_{i-1},d_{i}],\: i\in\{1,\ldots,N\}\] where $d_{0}=-\infty$ and $d_{N}=+\infty$ are allowed and where the following conditions must be met: \begin{eqnarray*}d_{i-1} & < & d_{i}\:\forall i\in\{1,\ldots,N\}\\b_{i-1} & < & b_{i}\:\forall i\in\{2,\ldots,N\}\\a_{i-1}+b_{i-1}d_{i-1} & = & a_{i}+b_{i}d_{i-1}\:\forall i\in\{2,\ldots,N\}.\end{eqnarray*}The first condition just enforces consistent indexing of the intervals, the second condition enforces the diseconomy of scale, and the last condition enforces continuity. The diseconomy of scale ensures that $f$ is a convex function, as illustrated in the following graph.

Letting $f_{i}(x)=a_{i}+b_{i}x$, we can see from the graph that $f$ is just the upper envelope of the set of functions $\{f_{1},\ldots,f_{N}\}$ (easily verified algebraically). To incorporate $f(x)$ in a mathematical program, we introduce a new variable $z$ to act as a surrogate for $f(x)$, and add the constraints\[z\ge a_{i}+b_{i}x\;\forall i\in\{1,\ldots,N\}\] to the model. Anywhere we would use $f(x)$, we use $z$ instead. Typically $f(x)$ is either directly minimized in the objective or indirectly minimized (due to its influence on other objective terms),

and so the optimal solution will contain the minimum possible value of $z$ consistent with the added constraints. In other words, one of the constraints defining $z$ will be binding. On rare occasions this will not hold (for instance, cost is relevant only in satisfying a budget, and the budget limit is quite liberal). In such circumstances, the optimal solution may have $z>f(x)$, but that will not cause an incorrect choice of $x$, and the value of $z$ can be manually adjusted to $f(x)$ after the fact.

Now let's consider an economy of scale, such as might occur when $x$ is a quantity of materials purchases, $f(x)$ is the total purchase cost, and the vendor offers discounts for larger purchases. The marginal rate of change of $f$ is now nonincreasing as a function of $x$, and as a result $f$ is concave, as in the next graph.

Now $f$ is the lower envelope of $\{f_{1},\ldots,f_{N}\}$. We define the segments as before, save with the middle condition reversed to $b_{i-1}>b_{i}\:\forall i\in\{2,\dots,N\}$. We cannot simply add constraints of the form \[z\le a_{i}+b_{i}x\] if we are (directly or indirectly) minimizing $z$, since $z=-\infty$ would be optimal (or, if we asserted a finite lower bound for $z$, the lower bound would automatically be optimal). In the presence of an economy of scale, we are forced to add binary variables to the formulation, either directly or indirectly. The easiest way to do this is to use a type 2 specially ordered set (SOS2) constraint, provided that the solver (and any modeling language/system) understand SOS2, and provided that $d_{0}$ and $d_{N}$ are finite. Introduce new (continuous) variables $\lambda_{0},\ldots,\lambda_{N}\ge0$ along with $z$, and introduce the following constraints:\begin{eqnarray*}\lambda_{0}+\cdots+\lambda_{N} & = & 1 \\ \lambda_{0}d_{0}+\cdots+\lambda_{N}d_{N} & = & x\\ \lambda_{0}f(d_{0})+\cdots+\lambda_{N}f(d_{N}) & = & z.\end{eqnarray*} Also communicate to the solver that $\{\lambda_{0},\ldots,\lambda_{N}\}$ is an SOS2 set. The SOS2 constraint says that at most two of the $\lambda$ variables can be nonzero, and that if two are nonzero they must be consecutive. Suppose that $\lambda_{i}$ and $\lambda_{i+1}$ are positive and all others are zero. Then the new constraints express $x$ as a convex combination of $d_{i}$ and $d_{i+1}$ (with weights $\lambda_{i}$ and $\lambda_{i+1}=1-\lambda_{i}$), and express $z=f(x)$ as a convex combination of $f(d_{i})$ and $f(d_{i+1}$) using the same weights.

If SOS2 constraints are not an option, then binary variables have to be added explicitly. Let $\delta_{i}\in\{0,1\}$ designate whether $x$ belongs to the interval $[d_{i-1},d_{i}]$ ($\delta_{i}=1$) or not ($\delta_{i}=0$). Add the following constraints:\begin{eqnarray*}\delta_{1}+\cdots+\delta_{N} & = & 1\\ x & \le & d_{i}+M(1-\delta_{i})\:\forall i\in\{1,\ldots,N\}\\ x & \ge & d_{i-1}-M(1-\delta_{i})\:\forall i\in\{1,\ldots,N\}\\ z & \ge & a_{i}+b_{i}x-M(1-\delta_{i})\:\forall i\in\{1,\ldots,N\}\end{eqnarray*} where $M$ is a suitably large constant. (This formulation will again require that $d_{0}$ and $d_{N}$ be finite.) The first constraint says that $x$ must belong to exactly one interval. If $\delta_{j}=1$, the second and third constraints force $d_{j-1}\le x\le d_{j}$ and are vacuous for all $i\neq j$ (assuming $M$ is correctly chosen). The final constraint forces $z\ge f_{j}(x)$ and is again vacuous for all $i\neq j$. Since $z$ represents something to be minimized, as before we expect $z=f_{j}(x)$ in almost all cases, and can manually adjust it in the few instances were the optimal solution contains padding.

## Saturday, October 9, 2010

### Binary Variables and Quadratic Terms

Generally speaking, if $x$ and $y$ are variables in a mathematical program, and you want to deal with their product $z=x y$, you will end up having a quadratic problem (which may create problems with convexity if the quadratic terms appear in constraints). There is a special case, though, when at least one of $x$ and $y$ is binary, and when the other variable is bounded. Under those assumptions, the product can be linearized.

Let $x$ be binary and let $L\le y \le U$ where $L$ and $U$ are bounds known in advance. Introduce a new variable $z$. (Programming note: Regardless of whether $y$ is real-valued, integer-valued or binary, $z$ can be treated as real-valued.) Add the following four constraints: \begin{eqnarray*} z & \le & Ux \\ z & \ge & Lx \\ z & \le & y - L(1-x) \\ z & \ge & y - U(1-x). \end{eqnarray*}

Consider first the case $x=0$, which means the product $z = x y$ should be $0$. The first pair of inequalities says $0 \le z \le 0$, forcing $z=0$. The second pair of inequalities says $y-U \le z \le y - L$, and $z=0$ satisfies those inequalities.

Now consider the case $x=1$, so that the product should be $z = y$. The first pair of inequalities becomes $L \le z \le U$, which is satisfied by $z = y$. The second pair says $y \le z \le y$, forcing $z= y$ as desired.

Let $x$ be binary and let $L\le y \le U$ where $L$ and $U$ are bounds known in advance. Introduce a new variable $z$. (Programming note: Regardless of whether $y$ is real-valued, integer-valued or binary, $z$ can be treated as real-valued.) Add the following four constraints: \begin{eqnarray*} z & \le & Ux \\ z & \ge & Lx \\ z & \le & y - L(1-x) \\ z & \ge & y - U(1-x). \end{eqnarray*}

Consider first the case $x=0$, which means the product $z = x y$ should be $0$. The first pair of inequalities says $0 \le z \le 0$, forcing $z=0$. The second pair of inequalities says $y-U \le z \le y - L$, and $z=0$ satisfies those inequalities.

Now consider the case $x=1$, so that the product should be $z = y$. The first pair of inequalities becomes $L \le z \le U$, which is satisfied by $z = y$. The second pair says $y \le z \le y$, forcing $z= y$ as desired.

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