*----------------------------------------------------------- * Date created: February 19, 1996 * * Floudas & Pardalos, * A collection of test problems for constrained global * optimization. * Springer Verlag - Berlin, 1990. * Haverly Pooling Problem - Case III. * Test problem 3, pp.61 * * Dr. Neculai Andrei * Research Institute for Informatics, * 8-10, Averescu Avenue, Bucharest 1, Romania * E-mail: nandrei@u3.ici.ro * web: www.ici.ro/camo *----------------------------------------------------------- * subroutine ini(n,m,mb,me,sb,x,icgc,ipgc,icge,ipge, 1 nszc,nsze,nb) * double precision x(n) integer sb(mb) integer icgc(nszc),ipgc(n+1) integer icge(nsze),ipge(n+1) * * Information about the problem. * write(1,10) 10 format(5x,'Example Floudas & Pardalos') write(1,11) 11 format(5x,'Test Problem 3, pp.61, Case III') * * Dimension of the problem: * n=9 m=2 me=4 mb=11 nszc=8 nsze=15 * * Initial point: * x(1)=2.d0 x(2)=40.d0 x(3)=1.d0 x(4)=1.d0 x(5)=1.d0 x(6)=1.d0 x(7)=1.d0 x(8)=1.d0 x(9)=1.d0 * * Index vector for simple bounds: cb >= 0. * sb(1)=1 sb(2)=-1 sb(3)=2 sb(4)=-2 sb(5)=3 sb(6)=4 sb(7)=5 sb(8)=6 sb(9)=7 sb(10)=8 sb(11)=9 * * Rows indices of the nonzeros of the Jacobian of the Inequalities * icgc(1)=1 icgc(2)=2 icgc(3)=1 icgc(4)=2 icgc(5)=1 icgc(6)=2 icgc(7)=1 icgc(8)=2 * * Starting address of columns of the Jacobian of the Inequalities * ipgc(1)=1 ipgc(2)=2 ipgc(3)=0 ipgc(4)=0 ipgc(5)=3 ipgc(6)=4 ipgc(7)=5 ipgc(8)=6 ipgc(9)=7 ipgc(10)=9 * * Rows indices of the nonzeros of the Jacobian of the Equalities * icge(1)=2 icge(2)=3 icge(3)=1 icge(4)=4 icge(5)=1 icge(6)=4 icge(7)=2 icge(8)=3 icge(9)=1 icge(10)=2 icge(11)=4 icge(12)=1 icge(13)=3 icge(14)=4 icge(15)=4 * * Starting address of columns of the Jacobian of the Equalities * ipge(1)=1 ipge(2)=2 ipge(3)=3 ipge(4)=5 ipge(5)=7 ipge(6)=8 ipge(7)=9 ipge(8)=12 ipge(9)=15 ipge(10)=16 * return end * *-------------------------------------------------------------- * Date created: February 19, 1996 * Example from Floudas & Pardalos. 6.2. Case III. * Haverly Pooling Problem *-------------------------------------------------------------- * subroutine prob(n,m,mb,me,sb,x,objf,gobj,c,gc,cb,e,ge, 1 nszc,nsze,nb) * * Calculate problem function at iterate x. * double precision x(n),objf,gobj(n),c(m),gc(nszc) double precision cb(mb),e(me),ge(nsze) * * Objective function and its gradient: * * objf=-9.d0*x(1)-15.d0*x(2)+6.d0*x(3)+13.d0*x(4)+ 1 10.d0*(x(5)+x(6)) * gobj(1)=-9.d0 gobj(2)=-15.d0 gobj(3)=6.d0 gobj(4)=13.d0 gobj(5)=10.d0 gobj(6)=10.d0 gobj(7)=0.d0 gobj(8)=0.d0 gobj(9)=0.d0 * * Bounds on variables: * cb(1)=x(1) cb(2)=100.d0-x(1) cb(3)=x(2) cb(4)=200-x(2) cb(5)=x(3) cb(6)=x(4) cb(7)=x(5) cb(8)=x(6) cb(9)=x(7) cb(10)=x(8) cb(11)=x(9) * * Constraints. (Inequalities): * c(1)=-x(7)*x(9)-2.d0*x(5)+2.5d0*x(1) c(2)=-x(8)*x(9)-2.d0*x(6)+1.5*x(2) * * Jacobian of the inequalities constraints: * gc(1)=2.5d0 gc(2)=1.5d0 gc(3)=-2.d0 gc(4)=-2.d0 gc(5)=-x(9) gc(6)=-x(9) gc(7)=-x(7) gc(8)=-x(8) * * Constraints. (Equalities): * e(1)=-x(3)-x(4)+x(7)+x(8) e(2)=x(1)-x(5)-x(7) e(3)=x(2)-x(6)-x(8) e(4)=-3.d0*x(3)-x(4)+x(7)*x(9)+x(8)*x(9) * * Jacobian of the equality conbstraints: * ge(1)=1.d0 ge(2)=1.d0 ge(3)=-1.d0 ge(4)=-3.d0 ge(5)=-1.d0 ge(6)=-1.d0 ge(7)=-1.d0 ge(8)=-1.d0 ge(9)=1.d0 ge(10)=-1.d0 ge(11)=x(9) ge(12)=1.d0 ge(13)=-1.d0 ge(14)=x(9) ge(15)=x(7)+x(8) * return end *------------------------------------------------FP62C.for