Your Major_cost data set has missing values, and these values are used in the first sum in Z. By the way, in the future, it is probably better to attach a file (see the "Use advanced editor" button) rather than copy and paste. Looking ahead to the inv_flow* constraints, I see that you have used j as both a constraint index and a summation index within the same constraint. This does not make sense and will cause errors.
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You can narrow down which coefficient is problematic by commenting out all but one sum, considering one sum at a time. Alternatively, please post your full code and data here.
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Also, as of SAS/OR 13.1, the network solver in PROC OPTMODEL provides access to the same algorithms from PROC OPTNET: SAS/OR(R) 13.2 User's Guide: Mathematical Programming
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Can you please attach the ret data set? Note that <> is the MAX operator, so your IMPVAR declaration is not doing what you want. Instead, use ne or ~= or ^= for not equals. Also, I suspect that replacing the IMPVAR with explicit variables and constraints will reduce the solve time.
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Glad to help. You might also be interested in this book of 29 examples that illustrate various features of PROC OPTMODEL: SAS/OR(R) 13.2 User's Guide: Mathematical Programming Examples
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If you want sigtsq to depend on decision variables, you should declare it as an implicit variable: impvar sigtsq {j in 1..n} = (if j = 1 then mean(of epsilonsq ) else omega + alpha*epsilonsq[j- 1] + beta*sigtsq[j-1]); See the IMPVAR statement in the PROC OPTMODEL documentation: SAS/OR(R) 13.2 User's Guide: Mathematical Programming
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A good search phrase is "cutting stock problem." Here is a SAS/OR documentation example that addresses the classical one-dimensional version: http://support.sas.com/documentation/cdl/en/ormpug/67517/HTML/default/viewer.htm#ormpug_optmodel_examples05.htm Yours is two-dimensional and more complicated, but there is a large body of literature on both heuristics and exact algorithms.
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Alternatively, you can keep the string indices if you define PAIRS as: set PAIRS = {i in ISET, iprime in ISET: i < iprime}; where ISET is the common, string-valued index set for i and iprime.
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Looks like the code got mangled in the cut and paste. I get other errors like this one: ERROR: DO expression not given value. Can you please try attaching the code as a file instead?
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It is discussed in the "More on Index Sets" section of the PROC OPTMODEL documentation: SAS/OR(R) 13.2 User's Guide: Mathematical Programming See also this "Manpower Planning" example that shows three equivalent forms: SAS/OR(R) 13.2 User's Guide: Mathematical Programming Examples
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Yes, you can do this using PROC OPTMODEL. If you have declared W as: var W {1..i}; You can express the constraint that W sums to zero as follows: con SumToZero: sum {j in 1..i} W = 0; To make p negative, include a slightly negative upper bound in the variable declaration: var p <= -1e-6;
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That expression is the minimum of the empty set and is a constant that is equal to the largest double-precision floating-point number (8-bytes) that is representable on your computer. Later versions of the documentation (starting in 13.1) use instead the simpler but equivalent constant('BIG'): SAS/OR(R) 13.1 User's Guide: Mathematical Programming Legacy Procedures
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You can diagnose infeasibility by using the IIS= option in the SOLVE WITH LP statement: SAS/OR(R) 13.2 User's Guide: Mathematical Programming You can then examine the .STATUS suffix by using PRINT: SAS/OR(R) 13.2 User's Guide: Mathematical Programming or EXPAND / IIS: SAS/OR(R) 13.2 User's Guide: Mathematical Programming or CREATE DATA: create data var_status_data from name=_VAR_.name status=_VAR_.status; create data con_status_data from name=_CON_.name status=_CON_.status;
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Yes, they are equivalent, and neither one is more efficient than the other. In both cases, the LP solver sees a single vector of objective coefficients and optimizes the x and y parts simultaneously. The variable names do not influence the solver, which internally considers all problems as having one set of variables indexed from 0 to n - 1.
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Then I think you want V + R + A + B - 3 <= RV instead of your current Con New. If that change doesn't correct the issue, you might try fixing variables to a known feasible solution, call the solver, and see which constraints are flagged as infeasible.
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