Here is the least absolute deviation problem under concerned: $ \underset{\textbf{w}}{\arg\min} L(w)=\sum_{i=1}^{n}|y_{i}-\textbf{w}^T\textbf{x}|$. I know it can be rearranged as LP problem in following way:

$\min \sum_{i=1}^{n}u_{i}$

$u_i \geq \textbf{x}^T\textbf{w}- y_{i} \; i = 1,\ldots,n$

$u_i \geq -\left(\textbf{x}^T\textbf{w}-y_{i}\right) \; i = 1,\ldots,n$

But I have no idea to solve it step by step, as I am a newbie to LP. Do you have any idea? Thanks in advance!


Here is the latest stage I have reached at this problem. I am trying to solve the problem following this note:

Step 1: Formulating it to a standard form

$\min Z=\sum_{i=1}^{n}u_{i}$

$ \textbf{x}^T\textbf{w} -u_i+s_1=y_{i} \; i = 1,\ldots,n$

$ \textbf{x}^T\textbf{w} +u_i+s_2=-y_{i} \; i = 1,\ldots,n$

subject to $s_1 \ge 0; s_2\ge 0; u_i \ge 0 \ i=1,...,n$

Step 2: Construct a initial tableau

           |      |    0      |    1   |  0  |   0   |   0    
 basic var | coef |  $p_0$    |  $u_i$ |  W  | $s_1$ | $s_2$ 
      $s_1$| 0    |  $y_i$    |   -1   |  x  |   1   |   0
      $s_2 | 0    |  $-y_i$   |    1   |  x  |   0   |   1
      z    |      |    0      |    -1  |  0  |   0   |   0

Step 3: Choose basic variables

$u_i$ is chosen as input base variable. Here comes a problem. When choosing the output base variable, it is obvious $y_i/-1=-y_i/1=-y_i$. According to the note, if $y_i\ge0$, the problem has unbounded solution.

I am totally lost here. I wonder if there is anything wrong and how should I continue the following steps.

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    $\begingroup$ Pragmatically, you use a linear program solver instead of writing your own. I reccomend gurobi. $\endgroup$ – Matthew Drury Feb 7 '17 at 15:25
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    $\begingroup$ @MatthewDrury Thanks for you reply. But I want to know exactly how LP works in this problem, instead of just taking the answer. $\endgroup$ – southdoor Feb 7 '17 at 15:26
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    $\begingroup$ Do you know or did you Google 'simplex method'? $\endgroup$ – user83346 Feb 7 '17 at 15:32
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    $\begingroup$ Linear program is just a formulation of your problem in terms of maximizing (or minimizing) of linear goal function subject to some linear constraints. It doesn't "solve" itself. There are a bunch of algorithms that solve these specially formulated programs, one of most commonly used is Simplex $\endgroup$ – Łukasz Grad Feb 7 '17 at 15:33
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    $\begingroup$ @fcop Yes, indeed I have read some notes of simplex method. But I have no idea how to generate it to this problem. As the examples in those notes are very simple and specific. I can not find one begin with general problems. I have already spent two nights in this problem, but still being confused. Sorry. $\endgroup$ – southdoor Feb 7 '17 at 15:34

You want an example for solving least absolute deviation by linear programming. I will show you an simple implementation in R. Quantile regression is a generalization of least absolute deviation, which is the case of the quantile 0.5, so I will show a solution for quantile regression. Then you can check the results with the R quantreg package:

rq_LP  <-  function(x, Y, r=0.5, intercept=TRUE) {
    if (intercept) X  <-  cbind(1, x) else X <-  cbind(x)
    N   <-  length(Y)
    n  <-  nrow(X)
    stopifnot(n == N)
    p  <-  ncol(X)
    c  <-  c(rep(r, n), rep(1-r, n), rep(0, 2*p))  # cost coefficient vector
    A  <- cbind(diag(n), -diag(n), X, -X)
    res  <-  lp("min", c, A, "=", Y, compute.sens=1)
### Desempaquetar los coefs:
    sol <- res$solution
    coef1  <-  sol[(2*n+1):(2*n+2*p)]
    coef <- numeric(length=p)
    for (i in seq(along=coef)) {
         coef[i] <- (if(coef1[i]<=0)-1 else +1) *  max(coef1[i], coef1[i+p])

Then we use it in a simple example:

Y  <- starsCYG[, 2]
x  <- starsCYG[, 1]
rq_LP(x, Y)
[1]  8.1492045 -0.6931818

then you yourself can do the check with quantreg.

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    $\begingroup$ +1 I am a big fan of doing things manually and differently then compare! $\endgroup$ – Haitao Du Mar 27 '17 at 16:47
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    $\begingroup$ For a post with a little more explanation see quantile regression $\endgroup$ – Stop Closing Questions Fast Dec 29 '18 at 22:05

Linear Programming can be generalized with convex optimization, where in addition to simplex, many more reliable algorithms are available.

I would suggest you to check The Convex Optimization Book and the CVX toolbox they provided. Where you can easily formulate least absolute deviation with regularization.



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    $\begingroup$ Thanks for your answer. But when I try to search the term "simplex method" in the book, I can't find any. And the CVX toolbox is just a tool for take input as the LP problem and run the algorithm. But what I really want is how the algorithm works in this problem. Neither the final result, nor how to formulate the problem. But the step to get the result. thanks $\endgroup$ – southdoor Feb 7 '17 at 17:03

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