Main

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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.

## Integral Objective (Luus)

## Main.IntegralObjective History

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Added lines 86-89:

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(: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:)

(: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:)

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Changed line 9 from:

{$\min_u \frac{1}{2} \int_0^2 x_1(t) \, dt$}

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:

{$\frac{dx_2}{dt} = \frac{1}{2} x_1^2(t)$}

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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.

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.

Changed line 12 from:

{$x(0) = 1$}

to:

{$x_1(0) = 1$}

Changed lines 25-26 from:

{$x(0) = 1$}

to:

{$x_1(0) = 1$}

{$x_2(0) = 0$}

{$x_2(0) = 0$}

Added lines 29-30:

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.

Changed line 17 from:

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

to:

{$x_2 = \frac{1}{2} \int_0^2 x_1(t) \, dt$}

Changed line 24 from:

{$\frac{dx_2}{dt} = \frac{1}{2} x_1(t)

to:

{$\frac{dx_2}{dt} = \frac{1}{2} x_1(t)$}

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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: http://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. [[http://dx.doi.org/10.1016/j.compchemeng.2014.04.013|Article]]

(: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

----

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>

(:htmlend:)

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!!!! References

# Beal, L., GEKKO for Dynamic Optimization, 2018, URL: http://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. [[http://dx.doi.org/10.1016/j.compchemeng.2014.04.013|Article]]