Question About The Conditional Statement ('m.if3') In The GEKKO

I would like to add some conditional statements in the GEKKO code. I added the below statement with 'm.if3'. However, it has returned syntax error that I didn't have without the c

Solution 1:

You need just one correction (remove the <0) to use the if3 function in Gekko.

R1_1 = m.if3(R1, 0, R1)

The m.if3 function uses a condition that switches what is used (argument 2 or 3) based on whether the condition is less than zero or greater than or equal to zero. Here is the result of help (m.if3) with some additional explanation:

if3(condition, x1, x2) method of gekko.gekko.GEKKO instance

IF conditional with a binary switch variable.
The traditional method for IF statements is not continuously
differentiable and can cause a gradient-based optimizer to fail
to converge.

Usage: y = m.if3(condition,x1,x2)
Inputs:
   condition: GEKKO variable, parameter, or expression
   x1 and x2: GEKKO variable, parameter, or expression
Output: GEKKO variable y = x1 when condition<0
                       y = x2 when condition>=0

One other thing to consider is that the if3 function uses binary variables and you'll need to use a Mixed Integer solver to find an integer solution. You can either remove the line that switches the solver to IPOPT (if3 switches to the APOPT solver by default) or you can switch the option to APOPT manually.

m.options.SOLVER = 1

Some of the constants were missing from your script. I added some dummy variables just to make it run.

from gekko import GEKKO
import numpy as np

m = GEKKO()

help(m.if3)

tstep = 1.0
cs0=[1,1,1,1]
r0 = 1.0
cFeMax = 1.0
kd = [1,1,1]
Keq = [1,1,1]
Dif = [1,1,1,1]
cg0 = [1,1,1]
nus = np.array([[1,1,1],[1,1,1],[1,1,1],[1,1,1]])

m.time = np.array([0,tstep])

cH = m.CV(value=cs0[0])
cM = m.CV(value=cs0[1])
cW = m.CV(value=cs0[2])
cF = m.CV(value=cs0[3])

R1_1 = m.Var()

r3 = m.Intermediate(r0*(1-cF/cFeMax)**(1/3))
r2 = m.Intermediate(r0*((2*cH + 3*cM)/cFeMax)**(1/3))
r1 = m.Intermediate(r0*(2*cH/cFeMax)**(1/3))
x = m.Intermediate(r1/r0)
y = m.Intermediate(r2/r0)
z = m.Intermediate(r3/r0)
A1 = m.Intermediate(1/x**2/(kd[0]*(1+1/Keq[0])))
A2 = m.Intermediate(1/y**2/(kd[1]*(1+1/Keq[1])))
A3 = m.Intermediate(1/z**2/(kd[2]*(1+1/Keq[2])))
B1 = m.Intermediate((y-x)/x/y*r0/Dif[1])
B2 = m.Intermediate((z-y)/y/z*r0/Dif[2])
B3 = m.Intermediate((1-z)/z*r0/Dif[3])
F = 0
W = m.Intermediate((A1+B1)*(A3*(A2+B2+B3+F)+(A2+B2)*(B3+F))+A2*(A3*(B2+B3+F))+B2*(B3+F))
ceq1 = m.Intermediate((cg0[0]+cg0[1])/(1+Keq[0]))
ceq2 = m.Intermediate((cg0[0]+cg0[1])/(1+Keq[1]))
ceq3 = m.Intermediate((cg0[0]+cg0[1])/(1+Keq[2]))

R1 = m.Intermediate(3/r0/W*((A3*(A2+B2+B3+F)+(A2+B2)*(B3+F))*(cg0[0]-ceq1) \
                     -(A3*(B2+B3+F)+B2*(B3+F))*(cg0[0]-ceq2) \
                     -A2*(B3+F)*(cg0[0]-ceq3)))
R2 = m.Intermediate(3/r0/W*(-(B2*(A3+B3+F)+A3*(B3+F))*(cg0[0]-ceq1) \
                     +((A1+B1+B2)*(A3+B3+F)+A3*(B3+F))*(cg0[0]-ceq2) \
                     -(A1+B1)*(B3+F)*(cg0[0]-ceq3)))
R3 = m.Intermediate(3/r0/W*(-A2*(B3+F)*(cg0[0]-ceq1) \
                    -(A1+B1)*(B3+F)*(cg0[0]-ceq2) \
                    +((A1+B1)*(A2+B2+B3+F)+A2*(B2+B1+F))*(cg0[0]-ceq3)))

R1_1 = m.if3(R1, 0, R1)

m.Equation(cH.dt() == nus[0].dot([R1_1, R2, R3]))
m.Equation(cM.dt() == nus[1].dot([R1_1, R2, R3]))
m.Equation(cW.dt() == nus[2].dot([R1_1, R2, R3]))
m.Equation(cF.dt() == nus[3].dot([R1_1, R2, R3]))

m.options.IMODE = 4
m.options.SOLVER = 1
m.options.nodes = 2

m.solve()