Hanging chain problem (GEKKO)

This page contains a solution of the energy minimization of the Hanging chain problem in GEKKO Python format. The GEKKO package is available with pip install gekko. The Python code uses orthogonal collocation and a simultaneous optimization method. The end point constraints are imposed as soft constraints (objective terms).

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
 
n = 51
Lp = 4
b = 3
a = 1
m = GEKKO()
m.time = np.linspace(0, 1, n)
 
x1 = m.Var(value=a, lb=0, ub=10)
x2 = m.Var(value=0, lb=0, ub=10)
x3 = m.Var(value=0, lb=0, ub=10)
 
u = m.MV(value=0, lb=-10, ub=20) # integer=True,
 
tmp = np.zeros(n)
tmp[-1] = 1
 
final = m.Param(value=tmp)
 
m.Equation(x1.dt() == u)
m.Equation(x2.dt() == x1 * (1+u**2)**0.5)
m.Equation(x3.dt() == (1+u**2)**0.5)
# m.Equation((x1-b)*final == 0)
# m.Equation((x3-Lp)*final == 0)
 
m.Obj(1000*(x1-b)**2*final)
m.Obj(1000*(x3-Lp)**2*final)
 
#m.fix(x1, n-1, b)
#m.fix(x3, n-1, Lp)
 
m.options.SOLVER = 3
m.options.NODES = 3
m.options.IMODE = 6
m.Obj(x2*final)
 
# initialize
m.options.TIME_SHIFT = 0
u.STATUS = 0
m.solve()
 
# solve
u.STATUS = 1
u.DCOST = 1e-3
m.solve()
 
plt.subplot(2,1,1)
plt.plot(m.time,x1.value,'r--',label=r'$x_1$')
plt.plot(m.time,x2.value,'g-',label=r'$x_2$')
plt.plot(m.time,x3.value,'k:',label=r'$x_3$')
plt.ylabel('x')
plt.legend()
plt.xlim([0,1])
plt.subplot(2,1,2)
plt.plot(2,1,2)
plt.plot(m.time,u.value,'k-',label=r'$u$')
plt.ylabel('u')
plt.xlim([0,1])
plt.show()

Solver Results

An MINLP solution is calculated with IPOPT with an objective function value of 5.13266.