A recent question on OR Stack Exchange has to do with getting an $L_1$ regression fit to some data. (I'm changing notation from the original post very slightly to avoid mixing sub- and super-scripts.) The author starts with $K$ observations $y_1, \dots, y_K$ of the dependent variable and seeks to find $x_{i,k} \ge 0$ ($i=1,\dots,N$, $k=1,\dots,K$) so as to minimize the $L_1$ error $$\sum_{k=1}^K \left|y_k - \sum_{i=1}^N \frac{e^{x_{i,k}}}{\sum_{j=1}^K e^{x_{i,j}}}\right|.$$ The author was looking for a way to linearize the objective function.
The solution I proposed there begins with a change of variables: $$z_{i,k}=\frac{e^{x_{i,k}}}{\sum_{j=1}^K e^{x_{i,j}}}.$$ The $z$ variables are nonnegative and must obey the constraint $$\sum_{k=1}^{K}z_{i, k}=1\quad\forall i=1,\dots,N.$$ With this change of variables, the objective becomes $$\sum_{k=1}^K \left|y_k - \sum_{i=1}^N z_{i,k} \right|.$$ Add nonnegative variables $w_k$ ($k=1,\dots, K$) and the constraints $$-w_k \le y_k - \sum_{i=1}^N z_{i,k} \le w_k \quad \forall k=1,\dots,K,$$ and the objective simplifies to minimizing $\sum_{k=1}^K w_k$, leaving us with an easy linear program to solve.
That leaves us with the problem of getting from the LP solution $z$ back to the original variables $x$. It turns out the transformation from $x$ to $z$ is invariant with respect to the addition of constant offsets. More precisely, for any constants $\lambda_i$ ($i=1,\dots,N$), if we set $$\hat{x}_{i,k}=x_{i,k} + \lambda_i \quad \forall i,k$$ and perform the $x\rightarrow z$ transformation on $\hat{x}$, we get $$\hat{z}_{i,k}=\frac{e^{\lambda_{i}}e^{x_{i,k}}}{\sum_{j=1}^{K}e^{\lambda_{i}}e^{x_{i,j}}}=z_{i,k}\quad\forall i,k.$$ This allows us to convert from $z$ back to $x$ as follows. For each $i$, set $j_0=\textrm{argmin}_j z_{i,j}$ and note that $$\log\left(\frac{z_{i,k}}{z_{i,j_0}}\right) = x_{i,k} - x_{i, j_0}.$$ Given the invariance to constant offsets, we can set $x_{i, j_0} = 0$ and use the log equation to find $x_{i,k}$ for $k \neq j_0$.
Well, almost. I dealt one card off the bottom of the deck. There is nothing stopping the LP solution $z$ from containing zeros, which will automatically be the smallest elements since $z \ge 0$. That means the log equation involves dividing by zero, which has been known to cause black holes to erupt in awkward places. We can fix that with a slight fudge: in the LP model, change $z \ge 0$ to $z \ge \epsilon$ for some small positive $\epsilon$ and hope that the result is not far from optimal.
I tested this with an R notebook. In it, I generated values for $y$ uniformly over $[0, 1]$, fit $x$ using the approach described above, and also fit it using a genetic algorithm for comparison purposes. In my experiment (with dimensions $K=100$, $N=10$), the GA was able to match the LP solution if I gave it enough time. Interestingly, the GA solution was dense (all $x_{i,j} > 0$) while the LP solution was quite sparse (34 of 1,000 values of $x_{i,j}$ were nonzero). As shown in the notebook (which you can download here), the LP solution could be made dense by adding positive amounts $\lambda_i$ as described above, while maintaining the same objective value. I tried to make the GA solution sparse by subtracting $\lambda_i = \min_k x_{i,k}$ from the $i$-th row of $x$. It preserved nonnegativity of $x$ and maintained the same objective value, but reduce density only from 1 to 0.99.
