I am solving a large scale optimization problem using Python GEKKO. Some variables of the model have to not change during the optimization process. So I was wondering what is the difference between using the `m.fix`

GEKKO method on the variables (`m.Var`

) that I want to be constant, or replacing the variables with constants declared with `m.Const`

.

I use IMODE = 3 (default) as I do not do anything that requires simulation or estimation, derivatives, time, etc.

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## Answer

When possible, it is best to use `int`

or `float`

numbers in the model for large-scale problems. It is also possible to use `m.Const()`

or `m.Var()`

and fix the value. The `m.Var()`

option is least efficient because it builds the 1st and 2nd derivative information with operator overloading. If you do need to include simple equations, such as used with `c4`

definition, then use `m.options.REDUCE=1`

to do a pre-solve to identify and eliminate variables that associated with simple pre-solvable constraints. Setting `m.options.REDUCE=2`

scans the model twice and it is possible to do additional scans.

from gekko import GEKKO m = GEKKO() c1 = 2.0 c2 = m.Const(3.0) c3 = m.Var(); m.fix(c3,4.0) c4 = m.Var(); m.Equation(c4==5) v = m.Var() m.Minimize((v-c1-c2-c3-c4)**2) m.solve(disp=False) print(c1,c2.value,c3.value[0],c4.value[0],v.value[0])

Results:

2.0 3.0 4.0 5.0 14.0

**Constant definition for improved model performance**

- Best:
`c1`

– Python number - 2nd Best:
`c2`

– Gekko constant - 3rd Best:
`c3`

– Gekko variable, fixed - Worst:
`c4`

– Gekko variable, equation

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