The issue is that the problem is probably infeasible.  The form of the problem is to minimize a sum-of-squares objective sum {i in OBS} (y[i] – a[i])^2 subject to equality constraints y[i] = a[i].  If these constraints were feasible, the optimal objective value would be 0.  Even running multistart in distributed mode on a grid with more than 100 workers, I was not able to find a feasible solution.  The constraints are not ignored but are difficult (probably impossible) to satisfy.  Both the iteration log and the Solution Summary report an Infeasibility of 0.5.  If I omit the constraints, the smallest objective value I get is 33.119976397, and the resulting variable values are not all 0. ... View more

As indicated in the log, you were reading only one observation:   NOTE: There were 1 observations read from the data set WORK.EQUATION. If you add a PRINT statement after the READ DATA, you can see that only the first observation was read:   print x y; x y 1 14   Instead, you need an index set for the observations: set OBS; number x {OBS}, y {OBS}; read data equation into OBS=[_N_] x y; print x y; [1] x y 1 1 14 2 94 851 3 52 473 4 70 635 5 57 518 6 28 257 7 42 383 8 47 428 9 39 356 10 7 68 11 28 257 12 27 248 13 21 194 14 9 86 15 84 761 16 31 284 17 72 653 18 39 356 19 35 320   Similarly, the objective function needs to sum over the observations:   min sq_diff = sum{i in OBS}(y[i]-(((((x[i]*w[1]-b[1])*w[3])+((x[i]*w[2]-b[1])*w[4])-b[2])*w[7])+((((x[i]*w[1]-b[1])*w[5])+((x[i]*w[2]-b[1])*w[6])-b[2])*w[8])-b[3]))**2; From your solution, it looks like you wanted w and b to be >= 0.  Here's a compact way to declare that:     var w {1..8} >= 0, b {1..3} >= 0; By default, decision variables have infinite lower and upper bounds.  If you do not specify >= 0, it turns out the solver actually finds an optimal objective value of 0 for this problem.     I'm not sure what you intended with EXPAND / IIS because you are not using the IIS solver option.  But EXPAND by itself will show you the full problem as a sanity check.   The problem appears to be nonconvex, so I recommend using the MULTISTART (or MS) option to try to avoid getting stuck in a local minimum.   Putting it all together: proc optmodel; var w {1..8} >= 0, b {1..3} >= 0; set OBS; number x {OBS}, y {OBS}; read data equation into OBS=[_N_] x y; print x y; min sq_diff = sum{i in OBS}(y[i]-(((((x[i]*w[1]-b[1])*w[3])+((x[i]*w[2]-b[1])*w[4])-b[2])*w[7])+((((x[i]*w[1]-b[1])*w[5])+((x[i]*w[2]-b[1])*w[6])-b[2])*w[8])-b[3]))**2; expand; solve with nlp / ms; print w b; quit; The SAS System The OPTMODEL Procedure Problem Summary Objective Sense Minimization Objective Function sq_diff Objective Type Nonlinear     Number of Variables 11 Bounded Above 0 Bounded Below 11 Bounded Below and Above 0 Free 0 Fixed 0     Number of Constraints 0     The SAS System   The OPTMODEL Procedure Solution Summary Solver Multistart NLP Algorithm Interior Point Direct Objective Function sq_diff Solution Status Optimal Objective Value 126.64605979     Number of Starts 100 Number of Sample Points 1600 Number of Distinct Optima 30 Random Seed Used 1679670 Optimality Error 3.1111693E-8 Infeasibility 0     Presolve Time 0.00 Solution Time 0.55 [1] w b 1 43.23150674 -.0000000021 2 34.96859624 0.0000017686 3 102.26490433 -.0000000082 4 80.22402775   5 180.63897324   6 97.11066371   7 0.00055746   8 0.00045163     The resulting solution is different than what you expected, but the optimal objective value matches, so this looks like a case of multiple optimal solutions. ... View more

If I understand correctly, the following changes should do what you want:   num input_slack {GARAGES, INPUTS}; num output_slack {GARAGES, OUTPUTS}; do k = GARAGES; solve; efficiency_number[k] = 1 / Inefficiency.sol; for {j in GARAGES} weight_sol[k,j] = (if Weight[j].sol > 1e-6 then Weight[j].sol else .); for {i in INPUTS} input_slack[k,i] = round(Input_con[i].ub - Input_con[i].body, 1e-6); for {i in OUTPUTS} output_slack[k,i] = round(Output_con[i].body - Output_con[i].lb, 1e-6); end; set EFFICIENT_GARAGES = {j in GARAGES: efficiency_number[j] >= 1 - 1e-6}; set INEFFICIENT_GARAGES = GARAGES diff EFFICIENT_GARAGES; create data slack_data from [garage]=INEFFICIENT_GARAGES garage_name efficiency_number {i in INPUTS} <col(i)=input_slack[garage,i]> {i in OUTPUTS} <col(i)=output_slack[garage,i]>;       ... View more

Note that GEODIST defaults to kilometers.  If you want miles, include 'M' as a fifth argument.   When I run with distance threshold 10000, I get one facility per customer, as expected from the assign_def constraints.   When I run with distance threshold 800, I get infeasible, as you did.  These notes in the log indicate several customers that are not within 800 miles of any facility: NOTE: The constraint 'assign_def[478]' is empty and infeasible. NOTE: The constraint 'assign_def[486]' is empty and infeasible. NOTE: The constraint 'assign_def[488]' is empty and infeasible. NOTE: The constraint 'assign_def[489]' is empty and infeasible. NOTE: The constraint 'assign_def[490]' is empty and infeasible. NOTE: The constraint 'assign_def[491]' is empty and infeasible. NOTE: The constraint 'assign_def[492]' is empty and infeasible. NOTE: The constraint 'assign_def[902]' is empty and infeasible.  The minimum feasible threshold turns out to be 926.21776127 miles, the distance from customer 492 to facility 6. ... View more

Infeasibility is essentially 0 here, so all constraints are satisfied within tolerance.  Iteration Limit Reached means that the solution might not be locally optimal.  Multisource Weber is a difficult problem, so you might consider solving it approximately by first solving a p-median problem (MILP) with the customers as the candidate locations and then solving a single-source Weber problem for each resulting cluster (p independent NLP problems). ... View more