I’m a newbie in gekko, and want to use it in my linear programming problems.
I have variable names, costs, minimum and maximum bounds in separate dictionaries (my_vars, Cost, Min and Max) with variable names as their keys, and the objective is minimizing total cost with determining the amount of variables satisfying the constraints.
I did as below;
LP = GEKKO(remote=False)
vars = LP.Array(LP.Var, (len(my_vars)))
i=0
for xi in vars:
xi.lower = Min[list(my_vars)[i]]
xi.upper = Max[list(my_vars)[i]]
i += 1
Here I’d like to use variable original names instead of xi, is there any way?
it continues as;
LP.Minimize(sum(float(Cost[list(my_vars)[i]])*vars[i] for i in range(len(my_vars))))
LP.Equation(sum(vars) == 100)
Also I have constraint’s left hand side (LHS) (coefficients of variables) and right hand side (RHS) numbers in two pandas data frame files, and like to construct equations using a for loop.
I don’t know how to do this?
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Answer
Here is one way to use your dictionary values to construct the problem:
from gekko import GEKKO
# stored as list
my_vars = ['x1','x2']
# stored as dictionaries
Cost = {'x1':100,'x2':125}
Min = {'x1':0,'x2':0}
Max = {'x1':70,'x2':40}
LP = GEKKO(remote=False)
va = LP.Array(LP.Var, (len(my_vars))) # array
vd = {} # dictionary
for i,xi in enumerate(my_vars):
vd[xi] = va[i]
vd[xi].lower = Min[xi]
vd[xi].upper = Max[xi]
# Cost function
LP.Minimize(LP.sum([Cost[xi]*vd[xi] for xi in my_vars]))
# Summation as an array
LP.Equation(LP.sum(va)==100)
# This also works as a dictionary
LP.Equation(LP.sum([vd[xi] for xi in my_vars])==100)
LP.solve(disp=True)
for xi in my_vars:
print(xi,vd[xi].value[0])
print ('Cost: ' + str(LP.options.OBJFCNVAL))
This produces a solution:
EXIT: Optimal Solution Found.
The solution was found.
The final value of the objective function is 10750.00174236579
---------------------------------------------------
Solver : IPOPT (v3.12)
Solution time : 0.012199999999999996 sec
Objective : 10750.00174236579
Successful solution
---------------------------------------------------
x1 69.999932174
x2 30.0000682
Cost: 10750.001742
Here are a few examples of efficient linear programming with Gekko by exploiting problem sparsity.