Monday, February 27, 2017

MSU Students Do Microfinance

Several years ago, I found out about, an online "microfinance" site where individuals can make small loans ($25 is the standard increment at Kiva) to entrepreneurs in low income settings. The entrepreneurs actually apply for larger amounts, which they typically receive from third-party "field partners". The entrepreneurs repay the loans with interest to the field partners, who in turn repay the principal (no interest) to the original donors. A modest deposit by a donor can be turned over repeatedly into loans, enjoying a version of what in my long-ago undergrad econ class was termed the "multiplier effect". I've made a fair number of loans over the years, nearly all of which have been fully repaid. It's a bit of fun as well as personally satisfying.

So I was geeked to learn from one of my colleagues at the Broad College of Business, Professor Paulette Stenzel, that we have a student organization on campus dedicated to microfinance. The Spartan Global Development Fund is a 501(c)(3) nonprofit organization, created and run by students, that makes microloans both through their own field partners and through Kiva. It's a registered student organization, meaning that any Michigan State University student can join. Students and nonstudents alike can donate through PayPal.

Next time I'm on the Kiva site, I plan to join their Kiva "team", and I encourage anyone interested in microfinance to check out their site.

Thursday, February 23, 2017

Another Absolute Value Question

Someone asked whether it is possible to eliminate the absolute value from the constraint$$Lx\le |y| \le Ux$$(where $L\ge 0$ and $U>0$ are constants, $x$ is a binary variable, and $y$ is a continuous variable). The answer is yes, but what it takes depends on whether $L=0$ or $L>0$.

The easy case is when $L=0$. There are two possibilities for the domain of $y$: either $y\in [0,0]$ (if $x=0$) or $y\in [-U,U]$ (if $x=1$). One binary variable is enough to capture two choices, so we don't need any new variables. We just need to rewrite the constraint as$$-Ux\le y \le Ux.$$If $L>0$, then there are three possibilities for the domain of $y$: $[-U, -L] \cup [0,0] \cup [L, U]$. That means we'll need at least two binary variables to cover all choices. Rather than change the interpretation of $x$ (which may be used elsewhere in the questioner's model), I'll introduce two new binary variables ($z_1$ for the left interval and $z_2$ for the right interval) and link them to $x$ via $x=z_1+z_2$. That leads to the following constraints:$$-Uz_1 \le y \le -Lz_1 + U(1-z_1)$$and$$Lz_2 - U(1-z_2) \le y \le Uz_2.$$ If $x=0$, both $z_1$ and $z_2$ must be 0, and the new constraints force $y=0$. If $x=1$, then either $z_1=1$ or $z_2=1$ (but not both). Left to the reader as an exercise: verify that $z_1=1\implies -U\le y \le -L$ while $z_2=1 \implies L\le y\le U$.