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Yet Another Math Programming Consultant

 3 years ago
source link: https://yetanothermathprogrammingconsultant.blogspot.com/2021/03/a-difficult-multiline-regression-data.html
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A difficult multi-line regression data set

 In [1], I discussed some models for a small 2 line regression problem. Here I want to dive into a more challenging problem.

Data generation process

I started with the following five lines: f1(x)=200−3xf2(x)=150−xf3(x)=10f4(x)=−50+2xf5(x)=−120f1(x)=200−3xf2(x)=150−xf3(x)=10f4(x)=−50+2xf5(x)=−120 To generate points, I used xi∼U(0,100)ki∼U{1,5}ϵi∼N(0,15)yi=aki+bki⋅xi+ϵixi∼U(0,100)ki∼U{1,5}ϵi∼N(0,15)yi=aki+bki⋅xi+ϵi
This corresponds to the following picture:
Now, we drop the lines and just keep the points. Together with K=5K=5 (the number of lines), this is all the data we give to the model:
Looking at this picture it seems a very difficult task to recover our 5 lines.
The data set is thus:
----     32 PARAMETER x  data

i1    5.141,    i2    0.601,    i3   40.123,    i4   51.988,    i5   62.888,    i6   22.575,    i7   39.612
i8   27.601,    i9   15.237,    i10  93.632,    i11  42.266,    i12  13.466,    i13  38.606,    i14  37.463
i15  26.848,    i16  94.837,    i17  18.894,    i18  29.751,    i19   7.455,    i20  40.135,    i21  10.169
i22  38.389,    i23  32.409,    i24  19.213,    i25  11.237,    i26  59.656,    i27  51.145,    i28   4.507
i29  78.310,    i30  94.575,    i31  59.646,    i32  60.734,    i33  36.251,    i34  59.407,    i35  67.985
i36  50.659,    i37  15.925,    i38  65.689,    i39  52.388,    i40  12.440,    i41  98.672,    i42  22.812
i43  67.565,    i44  77.678,    i45  93.245,    i46  20.124,    i47  29.714,    i48  19.723,    i49  24.635
i50  64.648,    i51  73.497,    i52   8.544,    i53  15.035,    i54  43.419,    i55  18.694,    i56  69.269
i57  76.297,    i58  15.481,    i59  38.938,    i60  69.543,    i61  84.581,    i62  61.272,    i63  97.597
i64   2.689,    i65  18.745,    i66   8.712,    i67  54.040,    i68  12.686,    i69  73.400,    i70  11.323
i71  48.835,    i72  79.560,    i73  49.205,    i74  53.356,    i75   1.062,    i76  54.387,    i77  45.113
i78  97.533,    i79  18.385,    i80  16.353,    i81   2.463,    i82  17.782,    i83   6.132,    i84   1.664
i85  83.565,    i86  60.166,    i87   2.702,    i88  19.609,    i89  95.071,    i90  33.554,    i91  59.426
i92  25.919,    i93  64.063,    i94  15.525,    i95  46.002,    i96  39.334,    i97  80.546,    i98  54.099
i99  39.072,    i100 55.782


----     32 PARAMETER y  data

i1    192.922,    i2   -122.916,    i3     21.992,    i4     70.273,    i5     91.422,    i6    109.528
i7    129.230,    i8   -140.463,    i9    180.013,    i10    24.614,    i11  -108.546,    i12    10.211
i13  -148.679,    i14    24.443,    i15   122.773,    i16   164.505,    i17   164.359,    i18   132.833
i19   -19.868,    i20    15.210,    i21   155.782,    i22   127.528,    i23   122.108,    i24   149.691
i25    -0.143,    i26  -120.337,    i27    94.868,    i28   -39.217,    i29    96.683,    i30    46.506
i31    35.514,    i32    22.589,    i33    97.811,    i34  -113.523,    i35    88.707,    i36    94.006
i37     9.232,    i38    73.173,    i39    26.600,    i40     0.429,    i41     3.943,    i42   119.583
i43   101.666,    i44   -38.179,    i45    39.630,    i46   137.980,    i47     4.808,    i48    18.422
i49   -10.888,    i50    83.361,    i51    65.726,    i52   -39.174,    i53     0.179,    i54   126.818
i55   139.561,    i56   -18.520,    i57   105.170,    i58     8.206,    i59    95.491,    i60    87.668
i61   -64.973,    i62    12.739,    i63   -13.237,    i64   -70.810,    i65   136.727,    i66   178.416
i67     7.676,    i68   -51.541,    i69    78.491,    i70   143.387,    i71   115.522,    i72    66.261
i73    52.631,    i74  -100.755,    i75   151.703,    i76    47.462,    i77   127.759,    i78   -65.191
i79    -6.051,    i80   137.610,    i81   130.662,    i82   128.182,    i83   148.801,    i84   -23.712
i85   -64.504,    i86    27.111,    i87   150.200,    i88   171.403,    i89    72.107,    i90  -105.643
i91  -112.767,    i92   112.218,    i93   102.355,    i94     5.007,    i95    86.645,    i96  -127.749
i97   -30.270,    i98    79.430,    i99    78.474,    i100   -7.158
The questions we want to address are:
  • Can we use an optimization model to find the best fit for five different lines?
  • Do the regression lines we find this way, correspond to the original functions we used in our data generation process?
The model has to make two decisions:
  1. Assign points to lines.
  2. Fitting: find the coefficients (constant term and slope) of each line.
I expected this to be not too difficult. This turned out to be wrong. 
We have 5×100=5005×100=500 binary variables to model the assignment.  This does seem small for a good MIP solver, but remember 25002500 is a big number [3]:
This is a number with 151 decimal digits.
Note: a better measure of size would be 51005100 which is a bit less: 7888609052210118054117285652827862296732064351090230047702789306640625 (70 decimal digits).

Solution process

To make things a bit easier, compared to a least squares objective, I used a linear MIP model based on LAD regression [2]:
Linear LAD Modelmin∑izi−zi≤ri≤zi∀iabsolute valueri=yi−α0,k−α1,k⋅xi+si,k∀i,kfit (possibly relaxed)−M⋅(1−δi,k)≤si,k≤M⋅(1−δi,k)∀i,kbig-M constraint∑kδi,k=1∀ipick one regression lineα0,k≤α0,k+1∀kexcept lastorder by constant termδi,k∈{0,1}min∑izi−zi≤ri≤zi∀iabsolute valueri=yi−α0,k−α1,k⋅xi+si,k∀i,kfit (possibly relaxed)−M⋅(1−δi,k)≤si,k≤M⋅(1−δi,k)∀i,kbig-M constraint∑kδi,k=1∀ipick one regression lineα0,k≤α0,k+1∀kexcept lastorder by constant termδi,k∈{0,1}
The big-M constraints implement the implication δi,k=1⇒si,k=0δi,k=1⇒si,k=0i.e. if δi,k=1δi,k=1 then point ii is used in fitting line kk. This model could not be solved to proven optimality. The best bound started out very bad (at zero) and moved very slowly.  In the end, I used a heuristic to find a good solution and then used the MIP model to improve upon this using MIPSTART.
The heuristic looked like:
  1. If time or iteration limit reached: stop
  2. Generate a random assignment. Evaluate the objective using an L1norm regression procedure (for each of the five lines).
  3. For each point ii, see if it improves the objective if we assign it to a different line.
  4. If there is no longer an improvement possible, go to step 1. 
This only looks at one point at a time, so it will not deliver optimal solutions. Hence we restart with a fresh random solution if we are unable to further improve the solution. In step 3 I use initially a cheap test: keep the lines constant and only check residuals without rerunning regressions. A more expensive test is to compare the residuals after rerunning regressions. This more expensive test is used once the simple test does not produce reassignments anymore.
The results were:
  • Heuristic: obj=957.1482
  • MIP improvement: obj=946.787
The picture below is from this solution.
This picture is very much like the picture we started with! This is not bad at all.
The final LAD estimates are:
----    473 PARAMETER est  estimates

               a0          a1

line1    -123.012       0.160
line2     -66.229       2.246
line3      -0.078       0.041
line4     155.979      -1.123
line5     211.432      -3.213
We can compare this against the coefficients we used to generate the data. 

Sidenote: checking the heuristic

Remember, our heuristic would try to move single points to another line. When it finishes it delivers a solution where such a move is no longer profitable. We can check if the heuristic is correct, by solving a slightly different version of the MIP model: restrict the number of changes in the assignment variable δi,kδi,k to one. If this model stops without an improvement, we have some confidence in the solution our heuristic produces. 
The modeling involved is as follows: di,k≥δi,k−δ0i,kdi,k≥δ0i,k−δi,k∑i,kdi,k≤1di,k∈[0,1]di,k≥δi,k−δi,k0di,k≥δi,k0−δi,k∑i,kdi,k≤1di,k∈[0,1] Here δ0i,kδi,k0 is the solution found by the heuristic. The variable di,k=1di,k=1 if we change an assignment from the heuristic solution.

GA heuristic

I also tried the GA algorithm in R. This heuristic only allows real-valued or binary variables. I used a real-valued vector of length 5 with bounds [1,5.9999][1,5.9999]. The fitness function used this algorithm:
  1. Use the floor function to convert the reals to integers between 1 and 5.
  2. Perform an L1 regression on each line.
  3. The objective is minus the sum of all absolute residuals.
fitness <- function(rx) {
  ix <- floor(rx)
  # rx are n=100 real numbers in [1,K+0.999]
  # ix are n=100 integers in [1,K]

  obj <- 0 
  for (k in 1:K) {
     cnt <- length(ix[ix==k]);
     if (cnt>2) {
        r <- l1fit(x[ix==k],y[ix==k])
        obj <- obj + sum(abs(r$residuals))
     }
  }
  return(-obj);
}
The simplicity of this is very attractive. Also, it is easy to change the fitness function to work with a least-squares fit.
Note that sometimes the GA algorithm generated an assignment with less than two points to a line. In that case, we don't use regression (that would fail) but conclude the residuals are zero for that line. With two points assigned to a line, we also can prevent doing a regression.
The best objective was: 1092.948. This is not as good as the solution we found before.
As a final test, I did provide our best solution (obj=946.787), and let the GA algorithm run:
No improvement was found.

Conclusions

  • This problem was much more challenging than I expected. 
  • The MIP model based on LAD regression has real troubles, but in the end, it was a good tool to improve the heuristic solution. Sometimes you have to be happy with just good solutions. Looking at the picture, the final solution is matching the original lines quite well.
  • I'll keep thinking about this a bit to see if I can come up with a better way to do this.
  • Is there a way to find proven optimal solutions for this problem? Maybe even for the least-squares problem? I suspect the combinatorics are just difficult here.

References

Appendix

The best solution I found:
----    216 PARAMETER sol4b  line number, best solution found

i1   5,    i2   1,    i3   2,    i4   2,    i5   4,    i6   4,    i7   4,    i8   1,    i9   5,    i10  3,    i11  1
i12  3,    i13  1,    i14  2,    i15  5,    i16  2,    i17  5,    i18  4,    i19  3,    i20  2,    i21  4,    i22  4
i23  4,    i24  5,    i25  3,    i26  1,    i27  4,    i28  2,    i29  2,    i30  4,    i31  5,    i32  5,    i33  5
i34  1,    i35  2,    i36  4,    i37  3,    i38  2,    i39  5,    i40  3,    i41  3,    i42  4,    i43  2,    i44  5
i45  4,    i46  4,    i47  3,    i48  3,    i49  2,    i50  4,    i51  4,    i52  2,    i53  3,    i54  4,    i55  4
i56  5,    i57  2,    i58  3,    i59  5,    i60  2,    i61  5,    i62  5,    i63  3,    i64  2,    i65  4,    i66  5
i67  3,    i68  2,    i69  4,    i70  4,    i71  4,    i72  4,    i73  5,    i74  1,    i75  4,    i76  2,    i77  4
i78  5,    i79  3,    i80  4,    i81  4,    i82  4,    i83  4,    i84  3,    i85  5,    i86  5,    i87  4,    i88  5
i89  4,    i90  1,    i91  1,    i92  4,    i93  4,    i94  3,    i95  4,    i96  1,    i97  5,    i98  4,    i99  5
i100 3


----    216 PARAMETER a  estimated coefficients, best solution found

       const.term       slope      sum|r|

line1    -123.012       0.160     105.780
line2     -66.229       2.246     154.042
line3      -0.078       0.041     156.192
line4     155.979      -1.123     324.395
line5     211.432      -3.213     206.378
total                             946.787

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