proc optmodel options

[dali74]

Hi for all!

I´m migrating proc nlp to proc optmodel code. I can show  Gradient Objective Function with PROC NLP  but i didnt found it in proc OPTMODEL.

Anyone could help? 

thanks

[RobPratt]

Dual variables (or Lagrange multipliers) are available from the .DUAL suffix.

See the Dual Values and Suffixes sections of the documentation for details.

[dali74]

Yes i saw it but it´s not same than gradient function.

I reply my code with proc nlp in proc optmodel.

[RobPratt]

Can you please share code and data for both versions (PROC NLP and PROC OPTMODEL)?

[dali74]

Yes Rob, you can see it :

Proc NLP

proc nlp ;
array par1[2] 5.5 4.1   ; 
array par2[2] 3.1 2.1; 
array z[2] z1-z2;
min y;
decvar z1-z2;
do n=1 to 2;
y= y+( par1[n]*par2[n]*z[n]**3);
end;
run;

And using proc optmodel now.

PROC OPTMODEL;
set I = 1..2;
number n=2;
number parm1 {I} =[5.545455 4.47841 ];
number parm2 {I} =[3.121211 2.04111];
var z{I};
MIN y = SUM{k IN I} (z[k] * (parm1[k] + (z[k] / parm2[k])));
SOLVE with nlp;
PRINT parm1 parm2 z.sol z.dual;
QUIT;

[RobPratt]

The optimization problem you specified in PROC OPTMODEL is different than the one that uses PROC NLP.  Also, the PROC NLP problem is unbounded, which you can see by letting z[n] approach negative infinity.

[dali74]

sorry it was my fault. I´m trying with multiple models

You will see with this code.

proc nlp ;
array par1[2] 5.545455 4.47841 ; 
array par2[2] 3.121211 2.04111; 
array z[2] z1-z2;
min y;
decvar z1-z2;
do n=1 to 2;
y= y+(z[n]*(par1[n]+z[n]/par2[n]));
end;
run;

PROC OPTMODEL;
set I = 1..2;
number n=2;
number parm1 {I} =[5.545455 4.47841 ];
number parm2 {I} =[3.121211 2.04111];
var z{I};
MIN y = SUM{k IN I} (z[k] * (parm1[k] + (z[k] / parm2[k])));
SOLVE with nlp;
PRINT parm1 parm2 z.sol z.dual;
QUIT;

[RobPratt]

I don't see any discrepancy between the results.  The optimal solutions and objective values agree, and the gradients are essentially 0.

SAS Output

PROC NLP: Nonlinear Minimization

Optimization Results
Parameter Estimates
N	Parameter	Estimate	Gradient Objective Function
1	z1	-8.654268	4.440892E-16
2	z2	-4.570464	-4.44089E-16

Value of Objective Function = -34.2301309

SAS Output

The OPTMODEL Procedure

Solution Summary
Solver	NLP
Algorithm	Interior Point
Objective Function	y
Solution Status	Optimal
Objective Value	-34.2301309
Optimality Error	2.391962E-15
Infeasibility	0
Iterations	4
Presolve Time	0.00
Solution Time	0.00

[1]	parm1	parm2	z.SOL	z.DUAL
1	5.5455	3.1212	-8.6543	1.1546E-14
2	4.4784	2.0411	-4.5705	-2.3093E-14

[dali74]

and how can i obtain Max Abs Gradient Element in proc optmodel?

[RobPratt]

For your problem:

print (max {k in I} abs(z[k].dual));

More generically:

print (max {j in 1.._NVAR_} abs(_VAR_[j].dual));

[dali74]

I think gradient of proc nlp and lagrangian of proc optmodel are similar but not the same.

[RobPratt]

Are you looking for a symbolic formula for each partial derivative?  Or do you just want to compute the gradient at a specific point?

[dali74]

i was looking for the second option (using the same start points that proc nlp) but if you can specify first option too it will be really help!

thanks

[RobPratt]

This blog post discusses symbolic derivatives with SAS.

And here's an undocumented way to compute the gradient at a given point (in this case, the initial solution from PROC NLP):

PROC OPTMODEL;
   set I = 1..2;
   number parm1 {I} =[5.545455 4.47841];
   number parm2 {I} =[3.121211 2.04111];
   var z{I};
   z[1] = 0.478582;
   z[2] = 0.451506; 
   MIN y = SUM{k IN I} (z[k] * (parm1[k] + (z[k] / parm2[k])));
   SOLVE with none;
   PRINT parm1 parm2 z.sol z.dual;
QUIT;

SAS Output

The SAS System

The OPTMODEL Procedure

Solution Summary
Solver	None
Objective Function	y
Solution Status	Optimal
Objective Value	4.8492418235
Iterations	0
Presolve Time	0.00
Solution Time	0.00

[1]	parm1	parm2	z.SOL	z.DUAL
1	5.5455	3.1212	0.47858	5.8521
2	4.4784	2.0411	0.45151	4.9208

You can see that the values match the PROC NLP initial solution, shown here:

SAS Output

The SAS System

PROC NLP: Nonlinear Minimization

Optimization Start
Parameter Estimates
N	Parameter	Estimate	Gradient Objective Function
1	z1	0.478582	5.852119
2	z2	0.451506	4.920822

Value of Objective Function = 4.8492430842

[dali74]

Hi Rob!

thanks for your quick answer. I´m triying to reproduce your results but i dont get same solution:

1º Using proc NLP

 proc nlp ;
array par1[2] 5.545455 4.47841 ; 
array par2[2] 3.121211 2.04111; 
array z[2] z1-z2;
min y;
decvar z1-z2;
do n=1 to 2;
y= y+(z[n]*(par1[n]+(z[n]/par2[n])));
end;
run;

Then i use proc optmodel with -8.654268 and -4.570464

PROC OPTMODEL;
   set I = 1..2;
   number parm1 {I} =[5.545455 4.47841];
   number parm2 {I} =[3.121211 2.04111];
   var z{I};
   z[1] = -8.654268	;
   z[2] = -4.570464;
   MIN y = SUM{k IN I} (z[k] * (parm1[k] + (z[k] / parm2[k])));
   SOLVE with none;
   PRINT parm1 parm2 z.sol z.dual;
QUIT;

As you can see it not the same

what do i do wrong?