Integral Objective (Luus)

Main.IntegralObjective History

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March 27, 2023, at 05:46 PM by 10.35.117.248 -
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Attach:download.png [[Attach:dynamic_optimization_luus.zip|Luus Problem in MATLAB and Python]]

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Attach:download.png [[Attach:dynamic_optimization_luus.zip|Luus Problem in MATLAB and Python]] and [[https://youtu.be/WIe31dTaW6g|Video]]
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November 17, 2021, at 12:52 AM by 10.35.117.248 -
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plt.plot(m.time,x1.value,'k-',LineWidth=2,label=r'$x_1$')
plt.plot(m.time,x2.value,'b-',LineWidth=2,label=r'$x_2$')
plt.plot(m.time,u.value,'r--',LineWidth=2,label=r'$u$')
to:
plt.plot(m.time,x1.value,'k-',lw=2,label=r'$x_1$')
plt.plot(m.time,x2.value,'b-',lw=2,label=r'$x_2$')
plt.plot(m.time,u.value,'r--',lw=2,label=r'$u$')
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plt.plot(m.time,x1.value,'k-',LineWidth=2,label=r'$x_1$')
plt.plot(m.time,x2.value,'b-',LineWidth=2,label=r'$x_2$')
plt.plot(m.time,u.value,'r--',LineWidth=2,label=r'$u$')
to:
plt.plot(m.time,x1.value,'k-',lw=2,label=r'$x_1$')
plt.plot(m.time,x2.value,'b-',lw=2,label=r'$x_2$')
plt.plot(m.time,u.value,'r--',lw=2,label=r'$u$')
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December 23, 2019, at 08:43 PM by 136.36.211.159 -
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(:toggle hide gekko2 button show="GEKKO (Python) Solution 2: m.integral":)
to:
(:toggle hide gekko2 button show="GEKKO (Python) Solution 2":)
Added line 90:
# this solution uses the m.integral function and produces the same answer
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December 23, 2019, at 08:42 PM by 136.36.211.159 -
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(:toggle hide gekko button show="GEKKO (Python) Solution":)
(:div id=gekko:)
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(:toggle hide gekko1 button show="GEKKO (Python) Solution 1":)
(:div id=gekko1:)
Added lines 71-112:

m.options.IMODE = 6
m.solve()
# m.solve(remote=False)  # for local solve

plt.figure(1)
plt.plot(m.time,x1.value,'k-',LineWidth=2,label=r'$x_1$')
plt.plot(m.time,x2.value,'b-',LineWidth=2,label=r'$x_2$')
plt.plot(m.time,u.value,'r--',LineWidth=2,label=r'$u$')
plt.legend(loc='best')
plt.xlabel('Time')
plt.ylabel('Value')
plt.show()
(:sourceend:)
(:divend:)

(:toggle hide gekko2 button show="GEKKO (Python) Solution 2: m.integral":)
(:div id=gekko2:)
(:source lang=python:)
import numpy as np
import matplotlib.pyplot as plt
from gekko import GEKKO

m = GEKKO()

nt = 101
m.time = np.linspace(0,2,nt)

# Variables
x1 = m.Var(value=1)
u = m.Var(value=0,lb=-1,ub=1)

p = np.zeros(nt)
p[-1] = 1.0
final = m.Param(value=p)

# Equations
m.Equation(x1.dt()==u)
x2 = m.Intermediate(m.integral(0.5*x1**2))

# Objective Function
m.Obj(m.integral(0.5*x1**2)*final)
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February 13, 2018, at 02:45 PM by 45.56.3.173 -
Added lines 86-89:

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<iframe width="560" height="315" src="https://www.youtube.com/embed/23cTqkkJbIU" frameborder="0" allow="autoplay; encrypted-media" allowfullscreen></iframe>
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February 13, 2018, at 02:11 PM by 45.56.3.173 -
Changed lines 44-85 from:
to:
(:toggle hide gekko button show="GEKKO (Python) Solution":)
(:div id=gekko:)
(:source lang=python:)
import numpy as np
import matplotlib.pyplot as plt
from gekko import GEKKO

m = GEKKO()

nt = 101
m.time = np.linspace(0,2,nt)

# Variables
x1 = m.Var(value=1)
x2 = m.Var(value=0)
u = m.Var(value=0,lb=-1,ub=1)

p = np.zeros(nt)
p[-1] = 1.0
final = m.Param(value=p)

# Equations
m.Equation(x1.dt()==u)
m.Equation(x2.dt()==0.5*x1**2)

# Objective Function
m.Obj(x2*final)

m.options.IMODE = 6
m.solve()
# m.solve(remote=False)  # for local solve

plt.figure(1)
plt.plot(m.time,x1.value,'k-',LineWidth=2,label=r'$x_1$')
plt.plot(m.time,x2.value,'b-',LineWidth=2,label=r'$x_2$')
plt.plot(m.time,u.value,'r--',LineWidth=2,label=r'$u$')
plt.legend(loc='best')
plt.xlabel('Time')
plt.ylabel('Value')
plt.show()
(:sourceend:)
(:divend:)
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February 13, 2018, at 02:09 PM by 45.56.3.173 -
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{$\min_u \frac{1}{2} \int_0^2 x_1(t) \, dt$}
to:
{$\min_u \frac{1}{2} \int_0^2 x_1^2(t) \, dt$}
Changed lines 17-18 from:
{$x_2 = \frac{1}{2} \int_0^2 x_1(t) \, dt$}
to:
{$x_2 = \frac{1}{2} \int_0^2 x_1^2(t) \, dt$}
Changed line 24 from:
{$\frac{dx_2}{dt} = \frac{1}{2} x_1(t)$}
to:
{$\frac{dx_2}{dt} = \frac{1}{2} x_1^2(t)$}
Changed line 30 from:
The initial condition of the integral starts at zero and becomes the integral in the time range of 0 to 2 at the final point for the optimal control problem.
to:
The initial condition of the integral starts at zero and becomes the integral in the time range of 0 to 2. The value that is minimized is at the final point in the time horizon of the optimal control problem.
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February 13, 2018, at 01:59 PM by 45.56.3.173 -
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{$x(0) = 1$}
to:
{$x_1(0) = 1$}
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{$x(0) = 1$}
to:
{$x_1(0) = 1$}
{$x_2(0) = 0$}
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The initial condition of the integral starts at zero and becomes the integral in the time range of 0 to 2 at the final point for the optimal control problem.
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February 13, 2018, at 01:57 PM by 45.56.3.173 -
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{$x_2 = \frac{1}{2} \int_0^2 x(t) \, dt$}
to:
{$x_2 = \frac{1}{2} \int_0^2 x_1(t) \, dt$}
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February 13, 2018, at 01:57 PM by 45.56.3.173 -
Changed line 24 from:
{$\frac{dx_2}{dt} = \frac{1}{2} x_1(t)
to:
{$\frac{dx_2}{dt} = \frac{1}{2} x_1(t)$}
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February 13, 2018, at 01:56 PM by 45.56.3.173 -
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(:title Integral Objective (Luus):)
(:keywords nonlinear control, Luus, optimal control, dynamic optimization, engineering optimization, MATLAB, Python, GEKKO, optimization suite:)
(:description Luus optimal control problem solved with MATLAB, Python, and GEKKO.:)

'''Objective:''' Set up and solve the Luus optimal control benchmark problem. Create a program to optimize and display the results. ''Estimated Time: 30 minutes''

The Luus optimal control problem has an integral objective function. Integrals are a natural expression of many systems where the accumulation of a quantity is maximized or minimized.

{$\min_u \frac{1}{2} \int_0^2 x_1(t) \, dt$}
{$\mathrm{subject \; to}$}
{$\frac{dx_1}{dt}=u$}
{$x(0) = 1$}
{$-1 \le u(t) \le 1$}

Integral expressions are reformulated in differential and algebraic form by defining a new variable.

{$x_2 = \frac{1}{2} \int_0^2 x(t) \, dt$}

The new variable and integral are differentiated and included as an additional equation. The problem then becomes a minimization of the new variable {`x_2`} at the final time.

{$\min_u x_2\left(t_f\right)$}
{$\mathrm{subject \; to}$}
{$\frac{dx_1}{dt}=u$}
{$\frac{dx_2}{dt} = \frac{1}{2} x_1(t)
{$x(0) = 1$}
{$t_f = 2$}
{$-1 \le u(t) \le 1$}

!!!! Solutions

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Attach:download.png [[Attach:dynamic_optimization_luus.zip|Luus Problem in MATLAB and Python]]

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<iframe width="560" height="315" src="https://www.youtube.com/embed/WIe31dTaW6g" frameborder="0" allowfullscreen></iframe>
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!!!! References

# Beal, L., GEKKO for Dynamic Optimization, 2018, URL: https://gekko.readthedocs.io/en/latest/

# Hedengren, J. D. and Asgharzadeh Shishavan, R., Powell, K.M., and Edgar, T.F., Nonlinear Modeling, Estimation and Predictive Control in APMonitor, Computers and Chemical Engineering, Volume 70, pg. 133–148, 2014. [[https://dx.doi.org/10.1016/j.compchemeng.2014.04.013|Article]]
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