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