## MODEL TRANSPORT. MINLP

Problems with modeling
Jonás
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Joined: 1 year ago

### MODEL TRANSPORT. MINLP

Hello, I am starting to work with GAMS and I am with an exercise that must be wrong in modeling because the objective value to be minimized is 0 and hardly gives values, I expose the exercise and the document lst.

Code: Select all

``````OPTION OPTCR=0.00001;
SETS
I products / PR1, PR2 /
J trucks / CAM1*CAM5 /
K markets / MER1, MER2 /;
PARAMETERS
A(I) offer
/ PR1 50
PR2 50 /
B(J) transport
/ CAM1 25
CAM2 25
CAM3 50
CAM4 50
CAM5 100 / ;

TABLE D(K,I) min demand
PR1  PR2
MER1   25   25
MER2   25   25 ;

SCALAR L transport cost proportional to the weight. L = proportionality constant  /1/;

PARAMETERS
C(J) fixed transportation cost
/ CAM1 20
CAM2 20
CAM3 40
CAM4 40
CAM5 80 / ;

VARIABLES
X (I, J, K) quantities transported of product I to market K in truck J
Y(J,K) total quantities of product I to market K in truck J
ALFA (J, K) binary variable by which a truck can only go to a single market
BETA (J) binary variable that marks with 1 the use of a truck and with zero the non-use
Z cost of transportation ;
POSITIVE VARIABLE X,Y ;
BINARY VARIABLE ALFA,BETA ;

EQUATIONS
COSTE objetive function
OFERTA(I) offer of the product I

DEMANDA(K,I) demand of the merkat K
TRANSPORTE(J) transport of the truck  J
DEFYJK(J,K) matching equation Y(J,K) and X(I,J,K)
CAMER(J,K) equation of binary variable ALFA
USOCAM(J); equation of binary variable BETA

COSTE .. Z =E= SUM((J,K), Y(J,K)*BETA(J)*C(J)*L);
OFERTA(I) .. SUM((J,K), X(I,J,K)) =L= A(I) ;
DEMANDA(K,I) .. SUM(J, X(I,J,K) ) =G= D(K,I) ;
TRANSPORTE(J) .. SUM((I,K) ,X(I,J,K)) =L= B(J);
DEFYJK(J,K) .. Y(J,K) =E= SUM(I,X(I,J,K));
CAMER(J,K) .. Y(J,K) =E= SUM(I,X(I,J,K)*ALFA(J,K));
USOCAM(J) ..  BETA(J) =E= SUM(K,ALFA(J,K));
MODEL DICIEMBRE2 /ALL/ ;
SOLVE DICIEMBRE2 USING MINLP MINIMIZING Z ;
DISPLAY X.L, Z.L, Y.L, ALFA.L, BETA.L;

The lst. document tell me this........

MODEL   DICIEMBRE2          OBJECTIVE  Z
TYPE    MINLP               DIRECTION  MINIMIZE
SOLVER  DICOPT              FROM LINE  80

**** SOLVER STATUS     1 Normal Completion
**** MODEL STATUS      5 Locally Infeasible
**** OBJECTIVE VALUE                0.0000
And this.......
** Warning **  The number of nonlinear derivatives equal to zero
in the initial point is large (= 27 percent).
A better initial point will probably help the
optimization.
Pre-triangular equations:   0
Post-triangular equations:  1
** Infeasible solution. There are no superbasic variables.
** Warning **  The number of nonlinear derivatives equal to zero
in the final point is large (= 22 percent).
Better bounds on the variables may help the
optimization.
** Warning **  The variance of the derivatives in the final
point is large (= 4.1 ). A better scaling or
better bounds on the variables will probably
help the optimization.``````
Obviusly, something is wrong. If someone can help me I will be very grateful.
Thanks