Question

我已经在GAMS中使用非线性互补建模对运输分配问题进行了编码。该代码非常适合单个OD（原始目标）和多O Single-D情况。但是，当我切换到具有多个OD对的Grid网络时，出现了问题。

我尝试修复某些变量，但仍然要求修复更多变量。

SET     N  NODES /1*9/;

ALIAS (I,N), (J,N), (K,N), (L,N);

SET     DEST(J) IDENTIFICATION OF DESTINATION NODES,
        ACTIVE(I,J,K) IDENTIFIES THE SET OF ACTIVE ARCS,
        A(N,N) ARCS;


SET PARAM /A, B, K/

TABLE ARC_COST(I,J,PARAM)  Arc cost data
           A        B          K
1.5        4        0.6        28
5.2        2        0.3        33
2.3        5        0.75       22
3.9        5        0.75       24
1.6        2        0.3        29
6.7        4        0.6        33
7.8        3        0.45       27
8.9        5        0.75       26
5.4        3        0.45       22
4.8        4        0.6        22
6.4        3        0.45       25
4.3        5        0.75       25;


Parameter COEF_A(I,J), COEF_B(I,J), COEF_K(I,J);
COEF_A(I,J) = ARC_COST(I,J,"A");
COEF_B(I,J) = ARC_COST(I,J,"B");
COEF_K(I,J) = ARC_COST(I,J,"K");

TABLE DMD(I,J)  Trip matrix
         1        2        3        4        5        6        7        8        9
1        0        0        0        0        0        0        0        0        40
2        0        0        0        0        0        0        0        0        0
3        0        0        0        0        0        0        0        0        0
4        0        0        0        0        0        0        0        0        0
5        0        0        20       0        0        0        0        0        0
6        0        0        0        0        0        0        0        25       0
7        0        0        0        0        0        0        0        0        0
8        0        0        0        0        0        0        0        0        0
9        0        0        0        0        0        0        0        0        0;



*       Identify arcs using flow cost parameter:
A(I,J) = YES$COEF_A(I,J);
*       Identify destination nodes using the trip table:
DEST(J) = YES$SUM(I, DMD(I,J));



ACTIVE(A,K) = YES$DEST(K);
ACTIVE(I,J,I) = NO;
ACTIVE(I,I,J) = NO;


Display  ACTIVE, COEF_A, COEF_B, COEF_K ;

VARIABLES        T(I,J)          TIME TO GET FROM NODE I TO NODE J,
                 X(I,J,K)        FLOW TO K ALONG ARC I-J,
                 F(I,J)          AGGREGATE FLOW ON ARC I-J;


EQUATION        RATIONAL(I,J,K)         COST MINIMIZATION
                BALANCE(I,J)            MATERIAL BALANCE
                FDEF(I,J)               AGGREGATE FLOW DEFINITION;

*       The time to reach node K from node I is no greater than
*       the time required to travel from node I to node J and then
*       from node J to node K.

RATIONAL(I,J,K)$ACTIVE(I,J,K).. COEF_A(I,J) + COEF_B(I,J) * POWER(F(I,J)/COEF_K(I,J),4) + T(J,K)=G= T(I,K);

*       The flow into a node equals demand plus flow out:


BALANCE(I,K)$T.UP(I,K)..SUM(A(I,J)$ACTIVE(A,K),X(A,K)) =G=SUM(A(J,I)$ACTIVE(A,K),X(A,K)) + DMD(I,K);




*       Flow on a given arc constitutes flows to all destinations K:

FDEF(A)..F(A) =E= SUM(K$ACTIVE(A,K), X(A,K));

*       Here is the MCP model:

MODEL TRAFFIC /RATIONAL.X, BALANCE.T, FDEF.F/;

*       Initial levels for arc flows are needed so that we can
*       properly evaluate the nonlinear functions:

F.L(A) = COEF_K(A);
X.L(A,K) = 0.0;
T.L(I,J)   = COEF_A(I,J)$A(I,J) + SMIN(K$A(I,K),COEF_A(I,K))$(NOT A(I,J));

*       Lower bounds are zero for flows, positive for times:

X.LO(A,K) = 0.0;
T.LO(I,J) = 0.0;


*       Fixing values causes corresponding equilibrium conditions
*       to be dropped:


T.FX(I,I) = 0;
T.FX(I,J)$(NOT A(I,J)) = 0;
F.FX(I,J)$(NOT A(I,J)) = 0;


option mcp=miles;
SOLVE TRAFFIC USING MCP;

Answer 1

此模型有点混乱。但是这里有些编辑至少可以为您提供解决方案。

消息非常好：

 **** MCP pair BALANCE.T has empty equation but associated variable is NOT fixed
      T(1,5)

这是因为当DEST（K）不存在时BALANCE（I，K）为空。解决方法并不困难：

代替

T.FX(I,J)$(NOT A(I,J)) = 0;

写

T.FX(I,K)$(not DEST(K)) = 0;

T还有另一个问题：某些T.LO（I，J）.LO和T.L（I，J）值是INF：

T.L(I,J)   = COEF_A(I,J)$A(I,J) + SMIN(K$A(I,K),COEF_A(I,K))$(NOT A(I,J));
display T.L;

显示：

----     98 VARIABLE T.L  TIME TO GET FROM NODE I TO NODE J

            1           2           3           4           5           6           7           8           9

1       2.000       2.000       2.000       2.000       4.000       2.000       2.000       2.000       2.000
2       5.000       5.000       5.000       5.000       5.000       5.000       5.000       5.000       5.000
3       5.000       5.000       5.000       5.000       5.000       5.000       5.000       5.000       5.000
4       4.000       4.000       5.000       4.000       4.000       4.000       4.000       4.000       4.000
5       2.000       2.000       2.000       3.000       2.000       2.000       2.000       2.000       2.000
6       3.000       3.000       3.000       3.000       3.000       3.000       4.000       3.000       3.000
7       3.000       3.000       3.000       3.000       3.000       3.000       3.000       3.000       3.000
8       5.000       5.000       5.000       5.000       5.000       5.000       5.000       5.000       5.000
9        +INF        +INF        +INF        +INF        +INF        +INF        +INF        +INF        +INF

（课程：始终，请检查并再次检查每个步骤的结果是否符合您的预期。）将模型传递给求解器时，不允许使用这些+ INF值（正确地如此）。我们可以按如下方法使用创可贴：

T.L(i,j)$(T.L(i,j)=INF) = 1e6;
T.LO(i,j)$(T.LO(i,j)=INF) = 1e6;

（我在solve语句之前添加了它）。更好的是：将分配固定为T.L（I，J）。提示：空集上的SMIN是多少？您可能需要考虑一下如何以更结构化的方式来解决这个问题。

如何在GAMS中使用非线性互补程序对多OD对网格网络建模

1 个答案: