Dynamic Optimization

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'$u$')

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'$u$')

(:toggle hide gekko2 button show="GEKKO (Python) Solution 2":)

1. this solution uses the m.integral function and produces the same answer

(:toggle hide gekko1 button show="GEKKO (Python) Solution 1":) (:div id=gekko1:)

m.options.IMODE = 6 m.solve()

1. 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)

1. 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)

1. Equations

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

1. Objective Function

m.Obj(m.integral(0.5*x1**2)*final)

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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)

1. 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)

1. Equations

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

1. Objective Function

m.Obj(x2*final)

m.options.IMODE = 6 m.solve()

1. 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:)

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

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

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

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.

$$x_1(0) = 1$$

$$x_1(0) = 1$$ $$x_2(0) = 0$$

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.

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

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

(: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

Luus Problem in MATLAB and Python

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References

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