\$title LGO Test t1000_10 (T1000,SEQ=266) \$onText This test problem is due to J Pinter. A Drud has pointed out that solvers using derivatives will have a hard time with this model. There is a primal solution but the dual does not exist and the global solution does not satisfy the Kuhn-Tucker conditions. sqr(x) = 0 implies x = 0, but the marginal on this constraint is usually infinite and the solution depends wildly on any finite tolerance. This model can be reformulated as shown by Drud. His reformulation is given below. The global solutions is x(i) = 0. Janos Pinter, LGO - Users Guide, Pinter Consulting Services, Halifax, Canada, 2003. needs include file t1000d.inc Keywords: nonlinear programming, mathematics \$offText Set i / 1*1000 / i10(i) / 2*10 /; Variable objf, sum_sq, sum_noise, x(i); Equation objdef, defsq, defnoise, con0, con1, con2, con3, con4, con5, con6, con7, con8, con9; Parameter scal, tol; scal = 100; tol = 0.001; objdef.. objf =e= sum_sq + scal*sum_noise; defsq.. sum_sq =e= sum(i, sqr(x(i))); defnoise.. sum_noise =e= sum(i, sqr(sin(sqr(x(i))))); con0.. sqr(sum(i10(i), (x(i)-x(i-1))*(x(i+1)-x(i-1)))) =l= tol; con1.. sqr(x('111')+x('122')-x('133')-x('144')-x('155')+x('166')-x('177')+x('188')) =l= tol; con2.. sqr(x('199')-x('203')*x('215')-x('227')*x('239')-x('242')*x('254')+x('266')) =l= tol; con3.. sqr(x('279')-x('283')*x('295')-x('7')*x('23')-x('42')*x('54')+x('66')) =l= tol; con4.. sqr(x('311')+x('322')-x('333')-x('344')-x('355')+x('366')-x('377')+x('388')) =e= 0; con5.. sqr(x('599')-x('603')*x('615')-x('627')*x('539')-x('542')*x('654')+x('666')) =l= tol; con6.. sqr(x('679')-x('783')*x('795')-x('7')*x('23')-x('742')*x('754')+x('666')) =l= tol; con7.. sqr(x('811')+x('322')-x('833')-x('344')-x('855')+x('366')-x('387')+x('888')) =e= 0; con8.. sqr(x('899')-x('903')*x('915')-x('627')*x('939')-x('942')*x('654')+x('966')) =e= 0; con9.. sqr(x('679')-x('783')*x('795')-x('7')*x('23')-x('742')*x('754')+x('666')) =e= 0; Model t1000 / all /; \$include t1000d.inc x.lo(i) = xinit(i,'lower'); x.up(i) = xinit(i,'upper'); x.l (i) = xinit(i,'level'); option limCol = 0, limRow = 0, solPrint = off; solve t1000 min objf using nlp; Parameter report(i,*) 'summary report showing non optimal variables'; report(i,'t1000')\$(abs(x.l(i)) > 1e-5) = x.l(i); * The following is an alternative formulation that produces * a proper dual at the optimal point (suggested by A Drud). Equation con0a, con1a, con2a, con3a, con4a, con5a, con6a, con7a, con8a, con9a, con0b, con1b, con2b, con3b, con5b, con6b; con0a.. (sum(i10(i), (x(i)-x(i-1))*(x(i+1)-x(i-1)))) =l= sqrt(tol); con0b.. (sum(i10(i), (x(i)-x(i-1))*(x(i+1)-x(i-1)))) =g= -sqrt(tol); con1a.. (x('111')+x('122')-x('133')-x('144')-x('155')+x('166')-x('177')+x('188')) =l= sqrt(tol); con1b.. (x('111')+x('122')-x('133')-x('144')-x('155')+x('166')-x('177')+x('188')) =g= -sqrt(tol); con2a.. (x('199')-x('203')*x('215')-x('227')*x('239')-x('242')*x('254')+x('266')) =l= sqrt(tol); con2b.. (x('199')-x('203')*x('215')-x('227')*x('239')-x('242')*x('254')+x('266')) =g= -sqrt(tol); con3a.. (x('279')-x('283')*x('295')-x('7')*x('23')-x('42')*x('54')+x('66')) =l= sqrt(tol); con3b.. (x('279')-x('283')*x('295')-x('7')*x('23')-x('42')*x('54')+x('66')) =g= -sqrt(tol); con4a.. (x('311')+x('322')-x('333')-x('344')-x('355')+x('366')-x('377')+x('388')) =e= 0; con5a.. (x('599')-x('603')*x('615')-x('627')*x('539')-x('542')*x('654')+x('666')) =l= sqrt(tol); con5b.. (x('599')-x('603')*x('615')-x('627')*x('539')-x('542')*x('654')+x('666')) =g= -sqrt(tol); con6a.. (x('679')-x('783')*x('795')-x('7')*x('23')-x('742')*x('754')+x('666')) =l= sqrt(tol); con6b.. (x('679')-x('783')*x('795')-x('7')*x('23')-x('742')*x('754')+x('666')) =g= -sqrt(tol); con7a.. (x('811')+x('322')-x('833')-x('344')-x('855')+x('366')-x('387')+x('888')) =e= 0; con8a.. (x('899')-x('903')*x('915')-x('627')*x('939')-x('942')*x('654')+x('966')) =e= 0; con9a.. (x('679')-x('783')*x('795')-x('7')*x('23')-x('742')*x('754')+x('666')) =e= 0; Model t1000a / objdef, defsq, defnoise con0a, con1a, con2a, con3a, con4a, con5a, con6a, con7a, con8a, con9a con0b, con1b, con2b, con3b, con5b, con6b /; x.l(i) = xinit(i,'level'); solve t1000a min objf using nlp; report(i,'t1000a')\$(abs(x.l(i)) > 1e-5) = x.l(i); display report;