Studentized residuals (the linear case)

I have developed a linear and nonlinear regression solver for GAMS (see http://www.amsterdamoptimization.com/pdf/regression.pdf and http://www.amsterdamoptimization.com/pdf/nlregression.pdf). A user asked how to get studentized residuals from these. I probably should add these to the solver so they are written to the GDX file with statistics. But first lets investigate if we can do this at the GAMS level with the current information available.

The residualsare returned in the ls.gdx file for the linear regression solver. However to calculate the standard errors of the residuals we need the (diagonals of the) hat matrix:

The evaluation of this matrix needs a little bit of attention. It is known from calculating the estimates

that the evaluation of X’X is causing lots of numerical problems. Here is an example of this for the famous Longley data set: http://yetanothermathprogrammingconsultant.blogspot.com/2008/06/demonstration-of-numerical-issues-in-ls.html. However we also export the variance-covariance matrix. This matrix is defined by:

So a simple way to evaluate the Hat matrix could be:

Below is a complete GAMS example to calculate the (internally and externally) studentized residuals:

$ontext

 Regression example

 Cross-section data: weekly household expenditure on food and
 weekly household income from Griffiths, Hill and Judge,
 1993, Table 5.2, p. 182.


 Erwin Kalvelagen, october 2000

$offtext

option decimals=8;

* make sure elements are registered in this order:
set dummy /constant,income/;

set i /i1*i40/;

table data(i, *)
        expenditure income
i1        9.46       25.83
i2       10.56       34.31
i3       14.81       42.50
i4       21.71       46.75
i5       22.79       48.29
i6       18.19       48.77
i7       22.00       49.65
i8       18.12       51.94
i9       23.13       54.33
i10      19.00       54.87
i11      19.46       56.46
i12      17.83       58.83
i13      32.81       59.13
i14      22.13       60.73
i15      23.46       61.12
i16      16.81       63.10
i17      21.35       65.96
i18      14.87       66.40
i19      33.00       70.42
i20      25.19       70.48
i21      17.77       71.98
i22      22.44       72.00
i23      22.87       72.23
i24      26.52       72.23
i25      21.00       73.44
i26      37.52       74.25
i27      21.69       74.77
i28      27.40       76.33
i29      30.69       81.02
i30      19.56       81.85
i31      30.58       82.56
i32      41.12       83.33
i33      15.38       83.40
i34      17.87       91.81
i35      25.54       91.81
i36      39.00       92.96
i37      20.44       95.17
i38      30.10      101.40
i39      20.90      114.13
i40      48.71      115.46
;


variables
    constant     'estimate constant term coefficient'
    income       'estimate income coefficient'
    sse          'sum of squared errors'
;

equations
    fit(i)     'the linear model'
    obj        'objective'

;

obj..     sse =n= 0;
fit(i)..  data(i,'expenditure') =e= constant + income*data(i,'income');

option lp=ls;
model ols1 /obj,fit/;
solve ols1 minimizing sse using lp;

display constant.l, income.l, sse.l;


*
* load covariance matrix and sigma from ls.gdx
*
set k /constant,income/;
parameters
   covar(k,k) 'variance-covariance matrix'
   resid(i)   'residuals';
;
scalars
   sigma      'variance'
   rss        'residual sum of squares'
   df         'degrees of freedom'
;
execute_load 'ls.gdx',covar,sigma,resid,rss,df;
display covar,resid,sigma,rss,df;

*
* calculate W= inv(X'X) = COVAR/sigma^2
*
parameter W(k,k) "inv(X'X)";
alias (k,k2);
W(k,k2) = covar(k,k2)/sqr(sigma);
display W;

*
* form X
*
parameter X(i,k);
x(i,'constant') = 1;
x(i,'income') = data(i,'income');
display X;

*
* form H = X W X' (only diagonal)
*
parameter H(i);
h(i) =  sum(k2, sum(k, x(i,k)*w(k,k2)) * x(i,k2));
display h;

*
* (internally) studentized residuals
* this corresponds to stdres in ls.diag in R
*
parameter stud_res1(i) '(internally) studentized residuals';
stud_res1(i) = resid(i)/[sigma*sqrt(1-h(i))];
display stud_res1;

*
* (externally) studentized residuals
* this follows the Wikipedia definition
*
parameter stud_res2(i) '(externally) studentized residuals';
stud_res2(i) = resid(i)/[sqrt[(rss-sqr(resid(i)))/(df-1)]*sqrt(1-h(i))];
display stud_res2;

*
* (externally) studentized residuals
* this follows the R implementation of
* studres in ls.diag
*
parameter stud_res3(i) '(externally) studentized residuals R version';
stud_res3(i) = resid(i)/[sqrt[(rss-sqr(resid(i))/(1-h(i)))/(df-1)]*sqrt(1-h(i))];
display stud_res3;

The way I read it the R implementation differs a little bit from the Wikipedia definition. In the code above the calculation of stud_res2 will do what Wikipedia suggests and stud_res3 will mimic what R returns as studres in ls.diag. The difference is a factor 1/(1-h(i)) when taking out the i-th residual. Looking at equation 8.1.17 of Draper and Smith, Applied Regression Analysis, it looks like R is correct and Wikipedia (or rather how I read it) is not. So the version stud_res3 would be the correct one.

For non-linear regression I assume we just need to replace X by J (the Jacobian evaluated at the minimum residual sum of squares).

VSTO+Solver Foundation

Microsoft Solver Foundations provides a simple plug-in too to develop optimization applications with Excel as front-end. For more complex models it may be needed to use the API’s. Basically there are two alternatives: develop a .NET DLL and call this from VBA (an example is demonstrated here) or create a VSTO application. VSTO is a framework to develop .NET based tools for use in an Office environment. Actually the Excel plug-in is an example of an application level VSTO tool. Here I describe a document level VSTO application.

A delayed column generation algorithm for the cutting stock problem is a good example of a modeling tool that can not be coded in OML. Here is a simple example in GAMS. It is not very difficult to code this in the solver API of MSF.

The Excel demo application has two sheets. One with the input and the progress log and one with the results.

The algorithm is using the duals of the master problem to form the sub-problems. Access to duals is only available in the solver level API (for the current version of MSF). This means a very low-level way to describe the model. In this case the models are simple and well-structured so the pain is limited. For more complex models this approach may become burdensome.

In order to get duals we needed to create the following options:

The reason for setting the presolve level to zero is a bug in the solver. This is fixed in the next version of MSF.

Note that the algorithm used is different from the example My Documents\Microsoft Solver Foundation\Samples\Solvers\MixedIntegerProgramming\CuttingStock from the Solver Foundation distribution. That algorithm will generate all patterns in advance. In our algorithm we generate new patterns as we go.

VSTO applications are quite fun to develop. Excel integrates nicely in Visual Studio and debugging is easy.

A few issues I encountered:

• Sometimes there is still an Excel.exe process after terminating Visual Studio. Kill it using the process manager.
• C# does not really do COM very nicely. That means lots of type casts and no proper handling of optional parameters (need to use a lot of “missing” for them in calls). I believe this is much better in the next version of C#/Visual Studio.
• Reading the documentation on deployment and security gave me a headache. This is not for the faint-hearted.

GAMS: compile time vs execution time

if(ElasticityChg_switch = 1,
$CALL =GDXXRW.EXE IncomeElasticityAdjMatrix.xls …
$GDXIN ElastAdjFactor.gdx
$LOAD ElastAdjFactor=ElastAdjFactor
)

This code works differently than expected. The $ statements are carried out during compilation time while the if statement is executed during execution time. I.e. this actually is executed in the following order:

$CALL =GDXXRW.EXE IncomeElasticityAdjMatrix.xls …
$GDXIN ElastAdjFactor.gdx
$LOAD ElastAdjFactor=ElastAdjFactor

if(ElasticityChg_switch = 1,
)

This difference between compile time and execution time is not very well understood by most users. Unfortunately, the use of many of the newer gams constructs such as $call, $load requires a good understanding of these concepts. Originally GAMS had only a very few constructs that required such knowledge, so users could build models without being bothered by the difference between compilation and execution. This is no longer the case: nowadays a good GAMS modeler better understands this difference.

PS. We also should use $loaddc instead of $load (see http://yetanothermathprogrammingconsultant.blogspot.com/2009/08/gams-loaddc-vs-load.html).

Job Shop Scheduling

Although standard MIP formulations do not work very well on this problem, the current crop of MIP solvers can solve problems that used to be very difficult to solve. As example consider ft10 with 100 operations (10 machines, 10 jobs). This problem was unsolved for 26 years after its introduction in 1963. Now this can be solved in about 5 minutes to optimality using solvers like Gurobi and Cplex. (Some tabu search based methods have been able to find the optimal solution in seconds).

I wanted to solve this from Excel, but as Excel has no native GANTT charts, I created some VBA code so I could at least inspect the results.

Modeling question

I'm a new GAMS user. I am trying to model and solve a problem using
binary variables.

In the problem I have two sets:

set
t time periods /t1*t7/
l location /L1*L5/ ;

Currently, I do not have an objective function and I am just trying to
satisfy the constraints of my problem. I have a simplifed case of my
problem working.

Simplified case equations and constraints:
- I have a binary variable x with 35 entries
        binary variable x(t,l)

- "Equation1" Solves for the supply at time period for each location
(L1, L2, L3, L4, and L5). The supply depends on the binary variable x
(t,l). For example:
        x(t1,L1) = 1, then the supply value would INCREASE
        x(t1,L1) = 0, then the supply value would DECREASE

- "Equation2" The sum of all the binary variables x(t,l) over a time
period must equal 1 (only one location can have increasing supply
during a given time period). Example:
        equation2(t)    ..      sum(l,x(t,l)) =e= 1 ;

- "Equation3" Provides the upper limit of the supply value that can be
reached at each location.

- "Equation4" Provides the lower limit of the supply value that can be
reached at each location.

This simplified case solves correctly (satisfies all constraints).

Current problem:
My problem is that currently (in the simplified case), the binary
variable can move to any location from one time period to the next.
So, if the supply gets too low at location 5 (L5), the binary variable
can jump from any location to L5 (example, from L1 during time period
t4 to L5 during time period t5).

What I need to tell GAMS is that the binary variable can only move 1
spot from time period to time period. Example:

At t1, x(t1,'L1') = 1 (so the binary variable is at L1).
The supply value is low at L5.
But the binary variable can only move in the following manner:
t1 -> x(t1,'L1') = 1
t2 -> x(t2,'L2') = 1
t3 -> x(t3,'L3') = 1
t4 -> x(t4,'L4') = 1
t5 -> x(t5,'L5') = 1 (finally arrives at L5)

Also, the binary variable can't move from L1 to L5 (imagine the
locations are in a line, unable to wrap around).

Can anyone provide an idea (or similar example problem implementation,
if possible) on how to model this in GAMS? I tried the following:

m1(t)$((ord(t) lt card(t)) and (x(t,'b1') eq 1))        ..      x(t+1,'L1') + x(t
+1,'L2') =e= 1 ;
m2(t)$(ord(t) lt card(t) and x(t,'b2') = 1)     ..      x(t+1,'L2') + x(t
+1,'L1') + x(t+1,'L3') =e= 1 ;
m3(t)$(ord(t) lt card(t) and x(t,'b3') = 1)     ..      x(t+1,'L3') + x(t
+1,'L2') + x(t+1,'L4') =e= 1 ;
m4(t)$(ord(t) lt card(t) and x(t,'b4') = 1)     ..      x(t+1,'L4') + x(t
+1,'L3') + x(t+1,'L5') =e= 1 ;
m5(t)$(ord(t) lt card(t) and x(t,'b5') = 1)     ..      x(t+1,'L5') + x(t
+1,'L4') =e= 1 ;

But I got the following error from GLPK (using mip):
**** The following MIP errors were detected in model first:
****  57 equation m1 .. VAR operands relational or boolean
****  57 equation m2 .. VAR operands relational or boolean
****  57 equation m3 .. VAR operands relational or boolean
****  57 equation m4 .. VAR operands relational or boolean
****  57 equation m5 .. VAR operands relational or boolean

I don't understand these error messages...

In summary:
I would appreciate any help (examples, books, etc.) on how to
constrain the movement on the binary variable to the position that is
either 1) the same (L2 to L2 is legal) or 2) adjacent (L2 can only
move to L1 or L3, if not staying at L2). I am stuck and getting
frustrated. Thank you!

You can not have a variable in a dollar condition. That would make the model nonlinear: you are writing essentially an if-then expression.

If I read the description correctly you could do something like:

positive variable pos(t) 'position at t';
defpos(t)..   pos(t) =e= sum(L, ord(L)*x(t,L));
bound1pos(t-1)..  pos(t) =L= pos(t-1) +1;
bound2pos(t-1)..  pos(t) =G= pos(t-1) -1;

GAMS/Mosek error message

From a log file:

MOSEK Warning 700    : 8 zero elements are specified in sparse input data.
MOSEK Warning 700    : 8 zero elements are specified in sparse input data.
MOSEK Warning 700    : 8 zero elements are specified in sparse input data.
MOSEK Warning 700    : 8 zero elements are specified in sparse input data.
MOSEK Warning 700    : 8 zero elements are specified in sparse input data.
MOSEK Warning 700    : 8 zero elements are specified in sparse input data.
MOSEK Warning 700    : 8 zero elements are specified in sparse input data.
MOSEK Warning 700    : 8 zero elements are specified in sparse input data.
MOSEK Warning 700    : 8 zero elements are specified in sparse input data.
MOSEK Warning 700    : 8 zero elements are specified in sparse input data.

First: issuing this message one time is more than enough.

I suspect this is related to a very small value in the GAMS model, below some a[i,j] tolerance. As GAMS will not export zero coefficients, a strict reading of these messages would mean there is a bug somewhere. So lets hope these error messages are somewhat poorly formulated and indeed this a matter of small values being present in the model.

Formulating correct and succinct error messages is very important (but often underestimated by software developers, myself included). They provide feedback to user when it really matters: when something is wrong and the user is most likely confused. 

Sign problem

I'm trying to follow your implemantation of Benders Decomposition for SP with GAMS (http://www.amsterdamoptimization.com/pdf/stochbenders.pdf) but after looking deeply I can not discover the reason of a "-" sign. Sorry this basic question!

I look to Eq.7 (pag. 3), and it says theta > .... (no minus sign)

And during your implementation in GAMS you program:
cutconst(iter) = cutconst(iter) - prob(s)*sum(j,(-selmax.m(j))*demand(j,s));
cutcoeff(iter,j) = cutcoeff(iter,j) - prob(s)*(-selling.m(j));

I can not discover why we should have: "-" prob(s)
I know this the right programming, but I do not understand why we need this sign.

The sign of duals is not always well defined. In this case we needed to add a minus sign to get the correct results as GAMS is using a different sign convention than the text. It would have been better if the text and the code were in sync here or at least devote a remark about this.

GAMS/COINCBC: missing log

In releases 23.1 and 23.2 the log of the branch & bound algorithm is disappeared. After reporting about generated cuts no more information is displayed.

Log of 23.0:

GAMS/CoinCbc 2.2 LP/MIP Solver written by J. Forrest Problem statistics: 127 columns and 126 rows.                     108 variables have integrality restrictions. Calling CBC main solution routine... Coin Cbc and Clp Solver version 2.20.00, build Sep 20 2008 command line - GAMS/CBC -solve -quit Continuous objective value is 32.7059 - 0.00 seconds Optimal - objective value 32.7059 6 fixed, 0 tightened bounds, 0 strengthened rows, 0 substitutions processed model has 114 rows, 115 columns (98 integer) and 448 elements Objective coefficients multiple of 1 Cutoff increment increased from 1e-005 to 0.999 Pass   1: suminf.   10.30302 (63) obj. -29.2323 iterations 18 Pass   2: suminf.   10.30302 (63) obj. -29.2323 iterations 0 Pass   3: suminf.    8.39496 (51) obj. -26.535 iterations 43 Pass   4: suminf.    7.85121 (55) obj. -26.7163 iterations 4 Pass   5: suminf.    7.85121 (55) obj. -26.7163 iterations 0 Pass   6: suminf.    5.84706 (39) obj. -22 iterations 33 Pass   7: suminf.    5.84706 (39) obj. -22 iterations 0 Pass   8: suminf.    5.84706 (39) obj. -22 iterations 0 Pass   9: suminf.    5.03167 (39) obj. -19.6561 iterations 71 Pass  10: suminf.    5.03167 (39) obj. -19.6561 iterations 0 Pass  11: suminf.    5.03167 (39) obj. -19.6561 iterations 0 Pass  12: suminf.    3.05294 (26) obj. -15.2353 iterations 50 Pass  13: suminf.    3.05294 (26) obj. -15.2353 iterations 0 Pass  14: suminf.    3.05294 (26) obj. -15.2353 iterations 0 Pass  15: suminf.    3.63051 (27) obj. -15.1232 iterations 26 Pass  16: suminf.    3.63051 (27) obj. -15.1232 iterations 0 Pass  17: suminf.    3.63051 (27) obj. -15.1232 iterations 0 Pass  18: suminf.    2.75294 (21) obj. -13.4235 iterations 26 Pass  19: suminf.    2.74866 (20) obj. -13.4171 iterations 6 Pass  20: suminf.    2.74866 (20) obj. -13.4171 iterations 0 No solution found this major pass Before mini branch and bound, 7 integers at bound fixed and 0 continuous Full problem 114 rows 115 columns, reduced to 113 rows 108 columns - too large Mini branch and bound did not improve solution (0.11 seconds) Full problem 115 rows 115 columns, reduced to 115 rows 115 columns - too large After 0.11 seconds - Feasibility pump exiting - took 0.11 seconds 59 added rows had average density of 50.6441 At root node, 59 cuts changed objective from -29.8813 to -26.7458 in 100 passes Cut generator 0 (Probing) - 1834 row cuts, 0 column cuts (53 active)  in 0.775 seconds - new frequency is 1 Cut generator 1 (Gomory) - 5566 row cuts, 0 column cuts (6 active)  in 0.226 seconds - new frequency is 1 Cut generator 2 (Knapsack) - 14 row cuts, 0 column cuts (0 active)  in 0.007 seconds - new frequency is -100 Cut generator 3 (Clique) - 0 row cuts, 0 column cuts (0 active)  in 0.009 seconds - new frequency is -100 Cut generator 4 (MixedIntegerRounding2) - 643 row cuts, 0 column cuts (0 active)  in 0.020 seconds - new frequency is 1 Cut generator 5 (FlowCover) - 90 row cuts, 0 column cuts (0 active)  in 0.144 seconds - new frequency is -100 Cut generator 6 (TwoMirCuts) - 990 row cuts, 0 column cuts (0 active)  in 0.044 seconds - new frequency is -100 Optimal - objective value 26.7458 Optimal - objective value 26.7458 After 0 nodes, 1 on tree, 1e+050 best solution, best possible -26.7458 (2.22 seconds) Integer solution of -18 found after 11709 iterations and 84 nodes (4.46 seconds) Full problem 114 rows 115 columns, reduced to 98 rows 94 columns Integer solution of -21 found after 17199 iterations and 205 nodes (6.05 seconds) Full problem 114 rows 115 columns, reduced to 103 rows 99 columns - too large Search completed - best objective -21, took 25825 iterations and 367 nodes (7.34 seconds) Strong branching done 5428 times (117137 iterations), fathomed 39 nodes and fixed 233 variables Maximum depth 26, 1719 variables fixed on reduced cost Cuts at root node changed objective from 29.8813 to -26.7458 Probing was tried 660 times and created 7528 cuts of which 1317 were active after adding rounds of cuts (1.204 seconds) Gomory was tried 611 times and created 10689 cuts of which 626 were active after adding rounds of cuts (0.429 seconds) Knapsack was tried 100 times and created 14 cuts of which 0 were active after adding rounds of cuts (0.007 seconds) Clique was tried 100 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.009 seconds) MixedIntegerRounding2 was tried 651 times and created 2491 cuts of which 123 were active after adding rounds of cuts (0.090 seconds) FlowCover was tried 100 times and created 90 cuts of which 0 were active after adding rounds of cuts (0.144 seconds) TwoMirCuts was tried 100 times and created 990 cuts of which 0 were active after adding rounds of cuts (0.044 seconds) implication was tried 551 times and created 0 cuts of which 0 were active after adding rounds of cuts 5 bounds tightened after postprocessing Result - Finished objective 21 after 367 nodes and 25825 iterations - took 7.35 seconds (total time 7.39) Total time 7.39 Solved to optimality (within gap tolerances). MIP solution:                    21   (367 nodes, 7.395 seconds) Best possible:                   21 Absolute gap:                     0   (absolute tolerance optca: 0) Relative gap:                     0   (relative tolerance optcr: 0.1)

In gams 23.2 we see:

GAMS/CoinCbc 2.3 LP/MIP Solver written by J. Forrest Problem statistics: 127 columns and 126 rows.                     108 variables have integrality restrictions. Calling CBC main solution routine... Coin Cbc and Clp Solver version 2.3pre, build Apr 17 2009 command line - GAMS/CBC -solve -quit (default strategy 1) Continuous objective value is 32.7059 - 0.00 seconds 0 fixed, 0 tightened bounds, 91 strengthened rows, 0 substitutions 0 fixed, 0 tightened bounds, 4 strengthened rows, 0 substitutions 0 fixed, 0 tightened bounds, 4 strengthened rows, 0 substitutions 0 fixed, 0 tightened bounds, 4 strengthened rows, 0 substitutions 0 fixed, 0 tightened bounds, 4 strengthened rows, 0 substitutions 0 fixed, 0 tightened bounds, 4 strengthened rows, 0 substitutions 0 fixed, 0 tightened bounds, 4 strengthened rows, 0 substitutions 0 fixed, 0 tightened bounds, 4 strengthened rows, 0 substitutions 0 fixed, 0 tightened bounds, 4 strengthened rows, 0 substitutions processed model has 114 rows, 115 columns (98 integer) and 448 elements Objective coefficients multiple of 1 Cutoff increment increased from 1e-005 to 0.999 Pass   1: suminf.   11.18775 (60) obj. -28.9374 iterations 12 Pass   2: suminf.   11.18775 (60) obj. -28.9374 iterations 0 Pass   3: suminf.    7.03271 (40) obj. -21.3003 iterations 77 Pass   4: suminf.    6.88899 (42) obj. -21.037 iterations 4 Pass   5: suminf.    6.88899 (42) obj. -21.037 iterations 0 Pass   6: suminf.    4.86728 (32) obj. -17.0442 iterations 43 Pass   7: suminf.    4.86728 (32) obj. -17.0442 iterations 0 Pass   8: suminf.    4.86728 (32) obj. -17.0442 iterations 0 Pass   9: suminf.    3.82450 (18) obj. -14.6103 iterations 32 Pass  10: suminf.    3.82450 (18) obj. -14.6103 iterations 0 Pass  11: suminf.    3.82450 (18) obj. -14.6103 iterations 0 Pass  12: suminf.    4.04428 (12) obj. -9.88475 iterations 35 Pass  13: suminf.    2.36084 (11) obj. -9.87972 iterations 4 Pass  14: suminf.    2.36084 (11) obj. -9.87972 iterations 0 Pass  15: suminf.    5.41816 (25) obj. -14.5273 iterations 42 Pass  16: suminf.    4.90619 (25) obj. -14.6979 iterations 3 Pass  17: suminf.    4.90619 (25) obj. -14.6979 iterations 0 Pass  18: suminf.    5.63976 (21) obj. -11.9286 iterations 43 Pass  19: suminf.    5.63976 (21) obj. -11.9286 iterations 0 Pass  20: suminf.    5.63976 (21) obj. -11.9286 iterations 0 Pass  21: suminf.    5.60993 (28) obj. -13.8754 iterations 45 Pass  22: suminf.    5.60993 (28) obj. -13.8754 iterations 0 Pass  23: suminf.    5.60993 (28) obj. -13.8754 iterations 0 Pass  24: suminf.    4.23841 (23) obj. -14.553 iterations 48 Pass  25: suminf.    4.23841 (23) obj. -14.553 iterations 0 Pass  26: suminf.    4.23841 (23) obj. -14.553 iterations 0 Pass  27: suminf.    2.62284 (12) obj. -13 iterations 37 Pass  28: suminf.    2.62284 (12) obj. -13 iterations 0 Pass  29: suminf.    2.62284 (12) obj. -13 iterations 0 Pass  30: suminf.    4.52424 (26) obj. -13.3305 iterations 39 No solution found this major pass Before mini branch and bound, 1 integers at bound fixed and 0 continuous Full problem 114 rows 115 columns, reduced to 114 rows 114 columns - too large Mini branch and bound did not improve solution (0.10 seconds) Full problem 115 rows 115 columns, reduced to 115 rows 115 columns - too large After 0.11 seconds - Feasibility pump exiting - took 0.11 seconds 47 added rows had average density of 55.5957 At root node, 47 cuts changed objective from -29.7355 to -26.6652 in 100 passes Cut generator 0 (Probing) - 1402 row cuts, 0 column cuts (46 active)  in 0.755 seconds - new frequency is 1 Cut generator 1 (Gomory) - 5871 row cuts, 0 column cuts (1 active)  in 0.228 seconds - new frequency is 1 Cut generator 2 (Knapsack) - 0 row cuts, 0 column cuts (0 active)  in 0.013 seconds - new frequency is -100 Cut generator 3 (Clique) - 0 row cuts, 0 column cuts (0 active)  in 0.003 seconds - new frequency is -100 Cut generator 4 (MixedIntegerRounding2) - 389 row cuts, 0 column cuts (0 active)  in 0.020 seconds - new frequency is 1 Cut generator 5 (FlowCover) - 33 row cuts, 0 column cuts (0 active)  in 0.149 seconds - new frequency is -100 Cut generator 6 (TwoMirCuts) - 743 row cuts, 0 column cuts (0 active)  in 0.039 seconds - new frequency is -100 Solved to optimality (within gap tolerances). MIP solution:                    21   (98 nodes, 4.061 seconds) Best possible:                   21 Absolute gap:                     0   (absolute tolerance optca: 0) Relative gap:                     0   (relative tolerance optcr: 0.1)

With larger problems you have no clue how cbc is doing. The log is 100% of the “user interface” of a solver, so it is more important for a user than many developers think.

Scheduling question

>I thought it would be easy, and I could not find how to do it.
>Assume I have an ordered set of decisions x[100].
>I would like to express in OML these requirements:

> All these decisions should be equal to 1 (positive decisions), except for a subset where the
> decisions should be 0 (negative decisions).

>All the negative decisions should be adjacent in the ordered set of decisions.
>For example, all negative decisions could be from slots {13,14,15,16,17,18,19}.
>On the contrary, the set {5,6,7,13,14,15} could not be a valid set of negative
>decisions as they are not all adjacent (missing 8 to 12).

>Could such constraints be expressed in OML?

This problem is often seen in scheduling of machines where they can not be turned on more than a number of times. There are well-known standard formulations for this. E.g.

binary variable turnon[t];  
turnon[t] >= x[t]-x[t-1];
Sum[t, turnon[t]] <= 1;

The binary variable may be relaxed to a continuous variable between [0,1] to improve performance.

Postscript viewer

http://www.rampantlogic.com/psview/psview.html

After installing this tool, clicking on a postscript file will show it inside your pdf viewer. Works like a charm.