GAMS中索引为“ i”和“ i + 1”的求和

Question

我不知道如何在GAMS中编写此求和，因为它的索引的差为一。

我的意思是我有索引 i 和 i + 1 。对于“ i = 1，.. 54;

i，j是集合，a（i），b（i），c是参数，x，y是变量

sum(j, a(i)*x(i,j))-b(i)*y(i)+ c * (sum(j, a(i+1)*x(i+1,j))-b(i+1)*y(i)) <= 136

换句话说，我想拥有

 sum(j, a(1)*x(1,j))-b(1)*y(1)+ c * (sum(j, a(2)*x(2,j))-b(2)*y(2) )<= 136

 sum(j, a(3)*x(3,j))-b(3)*y(3)+ c * (sum(j, a(4)*x(4,j))-b(4)*y(4) )<= 136
 .
 .
 .


 sum(j, a(53)*x(53,j))-b(53)*y(53)+ c * (sum(j, a(54)*x(54,j))-b(54)*y(54) )<= 136

Answer 1

那呢？

eq1(i)$mod(ord(i),2).. sum(j, a(i)*x(i,j))-b(i)*y(i)+ c * (sum(j, a(i+1)*x(i+1,j))-b(i+1)*y(i)) =L= 136;