$title LGO Interface Example (FCT,SEQ=265)

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This example is taken from the LGO Interface Guide. It is a
5-variable, 3-constraint test problem with the global solution
at x* = 0 f(x*) = 0.

Janos Pinter, LGO - Users Guide, Pinter Consulting Services, Halifax,
Canada, 2003.

Keywords: nonlinear programming, discontinuous derivatives, mathematics
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Scalar scaleaux / 2 /;

Variable obj, aux1, aux1a, aux2, aux2a, aux, fct, x1, x2, x3, x4, x5;

Equation defobj, deffct, defaux, defaux1, defaux1a, defaux2, defaux2a, con1, con2, con3;

defobj..   obj   =e= fct + scaleaux*aux;

deffct..   fct   =e= sqr(x1) + sqr(x2) + sqr(x3) + sqr(x4) + sqr(x5);

defaux..   aux   =e= aux1a + aux2a;

defaux1..  aux1  =e= sqr(sqr(x1) - x2) + sqr(x3) + 2*sqr(x4) + sqr(x5 - x2);

defaux1a.. aux1a =e= abs(sin(4*mod(aux1,pi)));

defaux2..  aux2  =e= sqr(x1 + x2 - x3 + x4 - x5) + 2*sqr(-x1 + x2 + x3 - x4 + x5);

defaux2a.. aux2a =e= abs(sin(3*mod(aux2,pi)));

con1..     x1 + 3*sqr(x2) + sqr(x3) - 2*sqr(x4) + sqr(x5) =e= 0;

con2..     x1 + 4*x2 - x3 + x4 - 3*x5 =e= 0;

con3..     sqr(x1) - sqr(x3) + 2*sqr(x2) - sqr(x4) - sqr(x5) =e= 0;

Model m / all /;

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x1.lo = -10; x1.l = 2; x1.up = 5;
x2.lo = -10; x2.l = 2; x2.up = 5;
x3.lo = -10; x3.l = 2; x3.up = 5;
x4.lo = -10; x4.l = 2; x4.up = 5;
x5.lo = -10; x5.l = 2; x5.up = 5;
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solve m using dnlp min obj;

Parameter report 'diff from global solution';
report('x1') = round(0 - x1.l,6);
report('x2') = round(0 - x2.l,6);
report('x3') = round(0 - x3.l,6);
report('x4') = round(0 - x4.l,6);
report('x5') = round(0 - x5.l,6);

display report;