GAMS — The Solver Manuals - Available Software

136 CONOPT
EQUATION E1, E2, E3, E4;
E1 .. LOG(X1) + X2 =E= 1.6;
E2 .. 5 * X2 =E= 3;
E3 .. OBJ =E= SQR(X1) + 2 * SQR(X2) + 3 * SQR(X3);
E4 .. X1 + X2 =E= 3.318;
X1.LO = 0.1;
MODEL DEMO / ALL /; SOLVE DEMO USING NLP MINIMIZING OBJ;
that is derived from the first model by the addition of equation E4. This model produces the following output
** An equation is inconsistent with other equations in the
pre-triangular part of the model.
Residual=
2.81828458E-04
Tolerance (RTNWTR)= 6.34931126E-07
The pre-triangular feasibility tolerance may be relaxed with
a line:
SET RTNWTR X.XX
in the CONOPT control program.
E4: Inconsistency in pre-triangular part of model.
The solution order of the critical equations and
variables is:
E2 is solved with respect to
X2. Solution value = 6.0000000000E-01
E1 is solved with respect to
X1. Solution value = 2.7182818285E+00
All variables in equation E4 are now fixed
and the equation is infeasible. Residual = 2.8182845830E-04
First E2 is solved with respect to X2, then E1 is solved with respect to X1 as indicated by the last part of the
output. At this point all variables that appear in equation E4, namely X1 and X2, are fixed, but the equation
is not feasible. E4 is therefore inconsistent with E1 and E2 as indicated by the first part of the output. In this
case the inconsistency is fairly small, 2.8E-04, so it could be a tolerance problem. CONOPT will always report
the tolerance that was used, rtnwtr - the triangular Newton tolerance, and if the infeasibility is small it will also
tell how the tolerance can be relaxed. Section 5 in the main text on ”The CONOPT Options File” gives further
details on how to change tolerances, and a complete list of options is given in Appendix B.
You can turn the identification and solution of pre-triangular variables and equations off by adding the line
”lspret = f” in the CONOPT control program. This can be useful in some special cases where the point
defined by the pre-triangular equations gives a function evaluation error in the remaining equations. The following
example shows this:
VARIABLE X1, X2, X3, X4, OBJ;
EQUATION E1, E2, E3, E4;
E1 .. LOG(1+X1) + X2 =E= 0;
E2 .. 5 * X2 =E= -3;