### tp067v3.map

fmcTitle ("tp067v3"):
# Source version 1
# FMC's first native input language.
# The Hock & Schittkowski test problem #67.
# This is a free, more intuitive formulation of #67,
# without discontinuities, as it would probably be defined
# by a modeler of our days.
# The solution is equal to the one obtained with tp067v1,
# i.e., it is sensible but not exact in the sense of #67.
y2 := tp067v3x (fmc_ident_tcb, 2, 0, x1, x2, x3):
y3 := tp067v3x (fmc_ident_tcb, 3, 0, x1, x2, x3):
y4 := tp067v3x (fmc_ident_tcb, 4, 0, x1, x2, x3):
y5 := tp067v3x (fmc_ident_tcb, 5, 0, x1, x2, x3):
y6 := tp067v3x (fmc_ident_tcb, 6, 0, x1, x2, x3):
y7 := tp067v3x (fmc_ident_tcb, 7, 0, x1, x2, x3):
y8 := tp067v3x (fmc_ident_tcb, 8, 0, x1, x2, x3):
fmcFunctionDiffHint ([ tp067v3x, 0, 0, 0,
tp067v3x ( fmcFunctionArg1, fmcFunctionArg2, 1,
fmcFunctionArg4, fmcFunctionArg5, fmcFunctionArg6 ),
tp067v3x ( fmcFunctionArg1, fmcFunctionArg2, 2,
fmcFunctionArg4, fmcFunctionArg5, fmcFunctionArg6 ),
tp067v3x ( fmcFunctionArg1, fmcFunctionArg2, 3,
fmcFunctionArg4, fmcFunctionArg5, fmcFunctionArg6 ) ]):
fmcExternalCodePath ("../../doc/RevisedHockSchittkowski/src/tp067v3x.c"):
fmcMinimum (-(0.063*y2*y5 - 5.04*x1 - 3.36*y3 - 0.035*x2 - 10*x3)):
fmcInequality (i1, y2 - 0):
fmcInequality (i2, y3 - 0):
fmcInequality (i3, y4 - 85):
fmcInequality (i4, y5 - 90):
fmcInequality (i5, y6 - 3):
fmcInequality (i6, y7 - 1/100):
fmcInequality (i7, y8 - 145):
fmcInequality (i8, 5000 - y2):
fmcInequality (i9, 2000 - y3):
fmcInequality (i10, 93 - y4):
fmcInequality (i11, 95 - y5):
fmcInequality (i12, 12 - y6):
fmcInequality (i13, 4 - y7):
fmcInequality (i14, 162 - y8):
fmcStrongLowerBound (x1, 10^(-5)):
fmcStrongLowerBound (x2, 10^(-5)):
fmcStrongLowerBound (x3, 10^(-5)):
fmcStrongUpperBound (x1, 2000):
fmcStrongUpperBound (x2, 16000):
fmcStrongUpperBound (x3, 120):
fmcInitialValue (x1, 1745):
fmcInitialValue (x2, 12000):
fmcInitialValue (x3, 110):
# best known objective = -1162.02698005969
# begin of best known solution
# x[1] = x1 = 1728.371443241086
# x[2] = x2 = 16000
# x[3] = x3 = 98.13617652300942
# end of best known solution

Stephan K.H. Seidl