# data file for space-25-r.mod
# AMPL coding: S. Leyffer, University of Dundee, April 2000.

# ... modelling discrete sizes with SOS1 variables
param Nsize := 30;		# ... No. of different sizes

# ... definition of set lengths
param Ndof     := 18;
param Nmembers := 25;

# ... group members of same areas
set group1 :=   2  3  4  5 ;
set group2 :=   6  7  8  9 ;
set group3 :=  10 11 ;
set group4 :=  12 13 ;
set group5 :=  14 15 16 17 ;
set group6 :=  18 19 20 21 ;
set group7 :=  22 23 24 25 ;

param     F :=
    1     1
    2    10
    3   -10
    5    10
    6   -10
    7     0.5
    16    0.6 ;

param truss:
         dx          dy          dz :=
 1     75.0000      0.          0.
 2     75.0000     37.5000   -100.0000
 3    -75.0000     37.5000   -100.0000
 4     75.0000    -37.5000   -100.0000
 5    -75.0000    -37.5000   -100.0000
 6      0.         37.5000   -100.0000
 7      0.        -37.5000   -100.0000
 8      0.         37.5000   -100.0000
 9      0.        -37.5000   -100.0000
 10     0.        -75.0000      0.
 11     0.        -75.0000      0.
 12    75.0000      0.          0.
 13   -75.0000      0.          0.
 14   -62.5000   -137.5000   -100.0000
 15   -62.5000    137.5000   -100.0000
 16    62.5000   -137.5000   -100.0000
 17    62.5000    137.5000   -100.0000
 18  -137.5000     62.5000   -100.0000
 19   137.5000     62.5000   -100.0000
 20  -137.5000    -62.5000   -100.0000
 21   137.5000    -62.5000   -100.0000
 22   -62.5000    -62.5000   -100.0000
 23   -62.5000     62.5000   -100.0000
 24    62.5000     62.5000   -100.0000
 25    62.5000    -62.5000   -100.0000;

param lv:
        1     2     3     4     5     6  :=

 1      1     2     3     4     5     6
 2      1     2     3    10    11    12
 3      4     5     6     7     8     9
 4      1     2     3    13    14    15
 5      4     5     6    16    17    18
 6      4     5     6    10    11    12
 7      4     5     6    13    14    15
 8      1     2     3     7     8     9
 9      1     2     3    16    17    18
 10     7     8     9    16    17    18
 11    10    11    12    13    14    15
 12     7     8     9    10    11    12
 13    13    14    15    16    17    18
 14     7     8     9     0     0     0
 15    16    17    18     0     0     0
 16    10    11    12     0     0     0
 17    13    14    15     0     0     0
 18    10    11    12     0     0     0
 19     7     8     9     0     0     0
 20    13    14    15     0     0     0
 21    16    17    18     0     0     0
 22    16    17    18     0     0     0
 23     7     8     9     0     0     0
 24    10    11    12     0     0     0
 25    13    14    15     0     0     0 ;

# ... discrete sizes are set here
let{i in 1..26} size[i] := i/10.0;
let size[27] := 2.8;	let size[28] := 3.0;
let size[29] := 3.2;	let size[30] := 3.4;
## display size;

# ... lower/upper bounds on u
let ul[1] := -0.35; 	let uu[1] := 0.35;
let ul[2] := -0.35;	let uu[2] := 0.35;
let ul[4] := -0.35; 	let uu[4] := 0.35;
let ul[5] := -0.35; 	let uu[5] := 0.35;

# ... initialize N
let{m in members} N[m,1] :=  1.0;
let{m in members} N[m,2] := -1.0;

# ... initialize mC
let{m in members} mC[m,1] := -ll[m];
let{m in members} mC[m,2] := -mm[m];
let{m in members} mC[m,3] := -nn[m];
let{m in members} mC[m,4] :=  ll[m];
let{m in members} mC[m,5] :=  mm[m];
let{m in members} mC[m,6] :=  nn[m];

# ... generate C
for{m in members, md in mdof} 
   if lv[m,md] > 0 then {let C[m,lv[m,md]] := mC[m,md]};
## display C;

## solution from Tabu Search paper (infeasible ?)
## fix A[1]  := 0.1;
## fix A[2]  := 0.6;
## fix A[6]  := 3.4;
## fix A[10] := 0.1;
## fix A[12] := 1.8;
## fix A[14] := 0.9;
## fix A[18] := 0.4;
## fix A[22] := 3.4;

## solution from NLP solve by SL
## fix A[1]  :=   0.1;
## fix A[2]  :=   0.4;
## fix A[6]  :=   3.4;
## fix A[10] :=   0.1;
## fix A[12] :=   2.2;
## fix A[14] :=   1.0;
## fix A[18] :=   0.4;
## fix A[22] :=   3.4;