### tp067v1.apm

```Model tp067v1
! Source version 1

! This is the poor man's free formulation of #67, intuitive,
! without discontinuities. The poor man's one because the sense
! of #67 is to prove whether a solving software has procedure
! handling capabilities. Since up to now, procedures cannot
! directly be expressed in APM, this formulation has to be thought
! of as a workaround. This is further a free formulation because
! the obtained solution is sensible but not exact.
! Both the auxiliaries below, x and x, are initialized in such
! a manner that the initial point belongs to the feasible domain.
! Recall, the initial point of the original #67 is also feasible.

Parameters
ivx4 = 3048.289708675017
ivx5 =   89.19762025827802
End Parameters

Variables
x =  1745, >= 10^(-5), <=  2000
x = 12000, >= 10^(-5), <= 16000
x =   110, >= 10^(-5), <=   120
x =  ivx4
x =  ivx5
obj
End Variables

Intermediates
y2  = x
y3  = 1.22*y2 - x
y6  = (x + y3)/x
y2c = 0.01*x*(112 + 13.167*y6 - 0.6667*y6^2)
y4  = x
y5  = 86.35 + 1.098*y6 - 0.038*y6^2 + 0.325*(y4 - 89)
y8  = 3*y5 - 133
y7  = 35.82 - 0.222*y8
y4c = 98000*x/(y2*y7 + 1000*x)
c[ 1] =  y2c -  y2
c[ 2] =  y4c -  y4
c[ 3] =   y2 -   0
c[ 4] =   y3 -   0
c[ 5] =   y4 -  85
c[ 6] =   y5 -  90
c[ 7] =   y6 -   3
c[ 8] =   y7 -   1/100
c[ 9] =   y8 - 145
c = 5000 -  y2
c = 2000 -  y3
c =   93 -  y4
c =   95 -  y5
c =   12 -  y6
c =    4 -  y7
c =  162 -  y8
mf = -(0.063*y2*y5 - 5.04*x - 3.36*y3 - 0.035*x - 10*x)
End Intermediates

Equations
c[1: 2]  = 0
c[3:16] >= 0

obj = mf

! best known objective = -1162.02698005969
! begin of best known solution
! x =  1728.371443241086
! x = 16000
! x =    98.13617652300942
! x =  3056.042166591054
! x =    90.61853974698703
! end of best known solution
End Equations
End Model
```

Stephan K.H. Seidl