source: trunk/sample/optimization/sample_minlp2.mso @ 1007

#*
   Structural Optimization of Process Flowsheets (example from GAMS)

   The goal is the profitable production of chemical C.
   which can be produced from chemical B where B may be
   the raw material that can be purchased from the
   external market or an intermediate that is produced
   from raw material A. There are two alternative paths
   of producing B from A. A mixed-integer nonlinear
   formulation is presented to solve the optimal
   production and capacity expansion problem.


Kocis, G R, and Grossmann, I E, Relaxation Strategy for the Structural
Optimization of Process Flow Sheets. Independent Engineering Chemical
Research 26, 9 (1987), 1869-1880.

Morari, M, and Grossmann, I E, Eds, Chemical Engineering Optimization
Models with GAMS. Computer Aids for Chemical Engineering Corporation,
1991.

   Process flowsheet


         A2    +-----+  B2      BP
        +----->|  2  |----->+    |
   A    |      +-----+      |    |  B1    +-----+    C1
   ---->|                   +----+------->|  1  |-------->
        |      +-----+      |             +-----+
        +----->|  3  |----->+
         A3    +-----+  B3


*#

using "types";
binary as Integer (Lower=0, Upper=1);

Optimization procsel
        VARIABLES
# Positive Variables
    a2 as positive;   # consumption of chemical a in process 2
    a3 as positive;   # consumption of chemical a in process 3
    b2 as positive;   # production capacity of chemical b in process 2
    b3 as positive;   # production capacity of chemical b in process 3
    bp as positive;   # amount of chemical b purchased in external market
    b1 as positive;   # consumption of chemical b in process 1
    c1 as positive;   # production capacity of chemical c in process 1 ;

# Binary Variables
    y1 as binary;   # denotes potential existence of process 1
    y2 as binary;   # denotes potential existence of process 2
    y3 as binary;   # denotes potential existence of process 3  ;

# Variable
    pr as Real;   # total profit in million $ per year ;

        MAXIMIZE
        pr;
        
        EQUATIONS

#*
* the original constraint for inout2 is b2 = log(1+a2)
* but this has been convexified to the form used below.
* the same is true for inout3. so b2 and b3 are the
* output variables from units 2 and 3 respectively
*#

 c1 = 0.9*b1;
 exp(b2) - 1 = a2;
 exp(b3/1.2) - 1 = a3;

 b1 = b2 + b3 + bp;

 c1 <= 2*y1;
 b2 <= 4*y2;
 b3 <= 5*y3;

 pr = 11*c1                             # sales revenue
        - 3.5*y1 - y2 - 1.5*y3  # fixed investment cost
        - b2 - 1.2*b3           # operating cost
        - 1.8*(a2+a3) - 7*bp;   # purchases

# demand constraint on chemical c based on market requirements

  c1 <= 1;

        OPTIONS
        Dynamic = false;
        NLPSolveNLA = true;
        
        NLPSolver(File = "minlp_emso", derivative_test = "second-order",
        bonmin_algorithm = "B-BB",
        print_level = 5);

end

# same as above but without convexification
Optimization procsel1
        VARIABLES
# Positive Variables
    x1 as positive;   # consumption of chemical a in process 2
    x2 as positive;   # consumption of chemical a in process 3
    x3 as positive;   # production capacity of chemical c in process 1 ;

# Binary Variables
    y1 as binary;   # denotes potential existence of process 1
    y2 as binary;   # denotes potential existence of process 2
    y3 as binary;   # denotes potential existence of process 3  ;

# Variable
    pr as Real;   # total profit in million $ per year ;

        MAXIMIZE
        pr;

        EQUATIONS
        -y1+0.9*ln(1+x1)+1.08*ln(1+x2)+0.9*x3 <= 0;
        -10*y2+ln(1+x1) <= 0;
        -10*y3+1.2*ln(1+x2) <= 0;
        y2+y3-1 <= 0;
        pr = 2.9*x3+8.9*ln(1+x1)+10.44*ln(1+x2)-1.8*(x1+x2)-3.5*y1-y2-1.5*y3;

# redundant Lower bound

        x1 >= 0;

        x2 >= 0;

        x3 >= 0;

# Initial guess
        GUESS
        x1 = 0;
        x2 = 0;
        x3 = 1;
        y1 = 0;
        y2 = 1;
        y3 = 0;

        OPTIONS
        Dynamic = false;
        NLPSolveNLA = true;
        
        NLPSolver(File = "minlp_emso", derivative_test = "second-order",
        bonmin_algorithm = "B-BB",
        print_level = 5);

end