Quick start - Gams

Notes
When a statement like cc(j)=bc(j)+10; is executed this is done for all elements in j so if j
had 100,000 elements this would define values for each and every one.
These assignments can be the sole entry of a data item or may redefine items.
If an item is redefined then it has the new value from then on and does not retain the
original data.
The example cc("j1")=1; shows how one addresses a single specific element not the
whole set, namely one puts the entry in quotes (single or double). This is further
discussed in the Sets chapter.
Calculations do not have to cover all set element cases of the parameters involved
(through partial set references as discussed in the Sets chapter). Set elements that are not
computed over retain their original values if defined or a zero if never defined by entry or
previous calculation.
A lot more on calculations appears in the Calculating chapter.
Algebraic nature of variable and equation specifications
When one moves to algebraic modeling the variable and equation declarations can have an added
element of set dependency as illustrated in our examples and reproduced below
POSITIVE VARIABLES x(j);
VARIABLES PROFIT ;
EQUATIONS OBJective ,
constraint(i) ;
POSITIVE VARIABLES P(commodities)
Qd(commodities)
Qs(commodities) ;
EQUATIONS PDemand(commodities)
PSupply(commodities)
Equilibrium(commodities) ;
Such definitions indicate that these variables and equations are potentially defined for every
element of the defining set (also called the domain) thus x could exist for each and every element
in j. However the actual definition of variables does not occur until the .. equation specifications
are evaluated as discussed next. More on set dependent variable and equation definitions
appears in the Variables, Equations, Models and Solves chapter.
Algebra and model .. specifications
The equations and variables in a model are defined by the evaluation of the .. equation
specifications. The .. equations for our examples are
OBJective.. PROFIT=E= SUM(J,c(J)*x(J)) ;
constraint(i).. SUM(J,a(i,J) *x(J)) =L= b(i);
Courtesy of B.A. McCarl, October 2002 31