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.
... View more
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
... View more
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.
... View more
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
... View more
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
... View more
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.
... View more
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.
... View more
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?
... View more
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
... View more
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;
... View more
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
... View more
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;
... View more
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.
... View more
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.
... View more
I suspect that you want to enforce the following proposition: if V = 1 and R = 1 and A = 1 and B = 1 and C = 1 then RV = 1 The corresponding linear constraint would be: V + R + A + B + C - 4 <= RV But you currently have (Con New): V + R + A + B + C + RV >= 5
... View more
