No, the names do not have to be the same. If they are different, say org1 and dep1 in the data set, then you can use org=org1 dep=dep1 in the READ DATA statement.
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I don't see anything wrong with s[1] = ... = s[8] = 0. That is feasible to your bounds and constraints. You can use the following CON statement to declare your constraints: con c {i in w diff {card(w)}: mod(i,8) ne 0}: s = s[i+1]; Alternatively, you can omit the constraints and declare only three s variables in the first place: var s{1..ceil(card(w)/8)} >= 0 <= .8 init 1; impvar ads{i in w} = org * s[ceil(i/8)];
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In program 2, you have declared ads as a numeric parameter, so the resulting objective is constant. Instead, you should declare ads as an implicit variable (and omit the FOR loop) as follows: impvar ads {i in w} = org * s;
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Thanks for providing the larger data set. I found that your OPTMODEL code runs roughly 8 times as fast in 9.3 as compared to 9.2. Although the NLP solvers have improved significantly in 9.3, this problem has only 3 variables, and you will typically not see much gain from the solvers when there is a small number of variables and constraints. Instead, the 8-fold speedup here is due to OPTMODEL problem generation improvements.
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I'm glad this helped. There is currently no feature to obtain multiple solutions from a MILP solver call, but we are considering such a feature for a future release.
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Attached are two files. In both, I have rewritten the first READ DATA statement more simply as follows: READ DATA &TABLE_NAME_VALUES(OBS=&OBS_COUNT) into {j in WSET} <values[i,j]=col('v'||j)>; I have also rewritten the objective function using implicit variables, as follows: impvar r {i in VSET} = (sum {j in WSET} values[i,j] * weights ) - margins; impvar avg = (sum {k in VSET} r ) / NOBS; min min_std = sum {i in VSET} (r - avg)**2; The following numeric parameter then calculates the standard deviation from the optimal objective value: num std = sqrt(min_std.sol / NOBS); The first file (tom022012.sas) includes the usual linearization sketched earlier, but the resulting MILP is big and the solver runs out of memory on my machine. So I added the compact linearization as well, and it solves in a couple of minutes. For your input data, the optimal solution still turns out to be trivial, putting all weight on j = 2. The second file (tom022012_heur.sas) contains a heuristic approach that is much faster and scalable and might suit your needs better. The idea is to first solve the QP without discrete variables, and then solve a small MILP to find the closest (in absolute value) grid point to that optimal solution. The resulting solution is not guaranteed to be optimal to the original problem, but because your objective function is convex and you have such a fine discretization (2% increments), it will be very close to optimal. I hope this helps.
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Attached is a sketch of the idea. The optimal solution turns out to be trivial because I don't know what your other constraints are. If this linearization makes your problem too big, an alternative "compact" linearization sometimes performs better. I can show you that, too, but please try this idea first.
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IML does have a QP solver, but it does not allow discrete variables. The QP solver in OPTMODEL also does not allow discrete variables, but your problem can be linearized by introducing binary variables. And then you can use the MILP solver in OPTMODEL. If you can give more details, I would be glad to show you how to do this. Rob Pratt
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Here are the seven solutions returned by the MILP solver: 12321, 33124, 34225 12544, 34225, 44521 12321, 44521, 55225 34225, 44521, 52441 12544, 34225, 52441 12321, 12544, 55225 12321, 52441, 55225 I interpreted the last statement in the puzzle as a rule to help narrow down these seven candidates to one solution. Solutions 2 through 7 all have the property that digit 3 appears just once. But solution 1 is the only one where digit 5 appears just once. Maybe I'm reading too much into the last statement, and the puzzle is just to find all seven solutions.
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Attached is a way to find all solutions by using the MILP solver in SAS/OR. There are seven solutions if you ignore the last clue: "If you knew which digit I have used just once you could deduce my three squares with certainty." But six of the solutions use digit 3 exactly once, and only one solution uses digit 5 exactly once. So that is the solution we want, namely 12321, 33124, 34225.
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