Cost from table Topic is solved

[Alexanre.ito]

Hi I'm with a problem, I don't know how to declare an equation to calculate the cost.
I have to find a diameter, this diameter have a cost
exemple:

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SET 
 d /d1*d7 /;
 nn /n1.n2, n1.n3, n3.n1, n3.n2/;
 
PARAMETERS
         dc(d) diam. comercial (m)
               / d1  0.100
                 d2  0.150
                 d3  0.200
                 d4  0.250
                 d5  0.300
                 d6  0.350
                 d7  0.400 /
         pd(d) cost of each diam.
               / d1  1.000
                 d2  3.000
                 d3  9.000
                 d4  11.00
                 d5  13.00
                 d6  19.00
                 d7  21.00 /

I tried this equations.
Variables: b (binary), X (find the diameter), costtub (find the cost), L (length), cost (OF)

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*Finde one diameter of PARAMETERS 
 Equation1(nn).. X(nn)=e=sum(d,dc(d)*b(d,nn));
 
 Equation2(nn).. costtub(nn)=e=L(nn)*sum(d,pd(d)*b(d,nn));
 
 OF..      cost =e= sum(nn ,costtub(nn));
 

Equation 1 is not working well, at least it brings a result, however Equation 2 returns zero.
I have equations and inequations to define the diameters, and my Objective function is the cost, that I want to minimize the cost.
My OF is right? Should I minimize the sum of COSTTUB and the X (Diameter)?
Some one have any idea why my equation is not working? and how I should sove this problem?

Some of the results

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X(n1,n2)
                (.LO, .L, .UP = 0.1, 0.1, 0.4)
        1       Equation1(n1,n2)
       (0)      Equation2(n1,n2)
       
---- VAR X  diam.                (m)

         LOWER     LEVEL     UPPER    MARGINAL

n1.n2     0.100     0.133     0.400      .         
n1.n3     0.100     0.225     0.400      .         
n3.n1     0.100     0.275     0.400      .         
n3.n2     0.100     0.112     0.400      .        

---- VAR ,costtub           ($)

         LOWER     LEVEL     UPPER    MARGINAL

n1.n2     -INF       .        +INF       .         
n1.n3     -INF       .        +INF       .         
n3.n1     -INF       .        +INF       .         
n3.n2     -INF       .        +INF       . 

,

The LOG

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 Node  Act. Lev.  Objective  IInf  Best Int.   Best Bound   Gap  (3 secs)
    200     6   4       0.0000   9            -       0.0000         -
    201     5   5   infeasible   -            -       0.0000         -
    202     4   5   infeasible   -            -       0.0000         -
    203     3   5   infeasible   -            -       0.0000         -
    204     2   5   infeasible   -            -       0.0000         -
    205     3   4       0.0000  12            -       0.0000         -
    206     4   4       0.0000   9            -       0.0000         -
    207     3   5   infeasible   -            -       0.0000         -
    208     2   5   infeasible   -            -       0.0000         -
    209     1   5   infeasible   -            -       0.0000         -
    210     0   5   infeasible   -            -            -         -
 Integer infeasible
 Statistics:
    Iterations    :              3848
    NLP Seconds   :          2.863281
    B&B nodes     :               210
    MIP solution  :                 -
    Best possible :                 -
    Absolute gap  :                 -     optca :  0.000000
    Relative gap  :                 -     optcr :  0.100000
    Model Status  :                10
    Solver Status :                 1

 NLP Solver Statistics
    Total Number of NLP solves  :        212
    Total Number of NLP failures:          0
    Details:         conopt
      # execs           212
      # failures          0
 Terminating.
***
*** Status: Normal completion

I may be wrong, but I have the intuition that equation 2 is not being able to define the cost of the diameter
Thanks

[GFA]

Don't think I understand your problem correctly (would be easier if you can post your full GAMS code), but it seems like if you minimize the cost (or costtub(nn)) L(nn) can be 0. If that is the case than costtub is 0 too.
Maybe try to write down your mathematical model first.

GFA