GEKKO how to solve a optimal lunar soft landing problem?

Question

I tried to solve a 2D optimal lunar soft landing problem by GEKKO. I assume that the moon is not rotating. The spacecraft is supposed to land softly on the surface of the moon, i.e. the final vertical velocity v and the final horizontal velocity u should be zero, the final height r should be the radius of the moon.

The problem can be illustrated as follows:

The state and control variables as well as equations are listed as follows (control variables are thrust force F and attitude angle φ):

I constructed the fuel optimal problem as follows (control variable φ is written as f):

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
import time
import os

r0=1753000
v0=1675

time_start = time.time()

model = GEKKO()
nt = 501
model.time = np.linspace(0,1,nt)

# optimize final time
tf = model.FV(value=1.0,lb=0.1,ub=1000.0)
tf.STATUS = 1

# controls
F = model.MV(value=0,lb=0,ub=2000)
F.STATUS = 1
f = model.MV(value=-0.5*np.pi,lb=-0.5*np.pi,ub=0.5*np.pi)
f.STATUS = 1

# state variables
r = model.Var(value=r0,lb=1738000) # height
s = model.Var(value=0) # tru anomaly
v = model.Var(value=0) # vertical velocity
u = model.Var(value=v0) # horizional velocity
m = model.Var(value=600,lb=0,ub=600) # mass

# constants
mu = 4.90275*10**(12) # lunar gravitational constant
Isp = 300*9.8 # specific impulse

# Equations
model.Equation( r.dt() == tf * v )
model.Equation( s.dt() == tf * u/r )
model.Equation( v.dt() == tf * (F/m*model.cos(f)-mu/(r**2)+u**2/r) )
model.Equation( u.dt() == tf * (F/m*model.sin(f)-u*v/r) )
model.Equation( m.dt() == tf * (-F/Isp) )

# terminal constraints
_finalMask = np.zeros(nt)
_finalMask[-1] = 1.0
finalMask = model.Param(value=_finalMask)
model.Equation(v*finalMask>=0)
model.Equation(v*finalMask<=0.5)
model.fix_final(r,val=1738000)
model.fix_final(u,val=0)

model.Obj(-m*finalMask) # Objective function to be minimized

model.options.IMODE = 6
model.solver_options = ['max_iter 5000']
model.solve() # solve

time_end=time.time()
print('Calculation Time: ',time_end-time_start)

# scaled time
print('Landing Time: ' + str(tf.value[0]))
tm = np.linspace(0,tf.value[0],nt)
finaltime = tm[-1]

# PLOT
fig = plt.figure(1)
ax1 = fig.add_subplot(2, 3, 1)
ax2 = fig.add_subplot(2, 3, 2)
ax3 = fig.add_subplot(2, 3, 3)
ax4 = fig.add_subplot(2, 3, 4)
ax5 = fig.add_subplot(2, 3, 5)
ax6 = fig.add_subplot(2, 3, 6)
ax1.plot(tm,r.value,'k-',label=r'$r$')
ax1.set_xlabel('Time')
ax1.set_ylabel('r')
ax1.set_xlim(0,finaltime)
ax2.plot(tm,v.value,'b-',label=r'$v$')
ax2.set_xlabel('Time')
ax2.set_ylabel('v')
ax2.set_xlim(0,finaltime)
ax3.plot(tm,u.value,'g-',label=r'$w$')
ax3.set_xlabel('Time')
ax3.set_ylabel('u')
ax3.set_xlim(0,finaltime)
ax4.plot(tm,m.value,'y-',label=r'$m$')
ax4.set_xlabel('Time')
ax4.set_ylabel('m')
ax4.set_xlim(0,finaltime)
ax5.plot(tm,f.value,'c-',label=r'$f$')
ax5.set_xlabel('Time')
ax5.set_ylabel('f')
ax5.set_xlim(0,finaltime)
ax6.plot(tm,F.value,'r-',label=r'$F$')
ax6.set_xlabel('Time')
ax6.set_ylabel('F')
ax6.set_xlim(0,finaltime)
plt.tight_layout()
plt.show()

It can found a solution very similar to other's research.

but there is a sharp turn in f. It is unacceptable because the angle φ should be continuously changed .

Also, I tried to scale the function

scale = 1e-6
model.Equation( r.dt() == tf * v )
model.Equation( r*s.dt()*scale == tf * u*scale )
model.Equation( m*(r**2)*v.dt()*scale**2 == tf * ((r**2)*F*model.cos(f)-mu*m+(u**2)*r*m)*(scale**2) )
model.Equation( m*r*u.dt()*scale == tf * (F*r*model.sin(f)-u*v*m)*scale )
model.Equation( Isp*m.dt() == tf * (-F) )

but failed with

Solution Not Found
EXIT: Converged to a point of local infeasibility. Problem may be infeasible.

In order to get a smooth f, I changed the second control variable to angular acceleration a, and the state equation became

the code became:

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
import time
import os

r0=1753000
v0=1675

time_start = time.time()

model = GEKKO() # remote=False
nt = 501
model.time = np.linspace(0,1,nt)

tf = model.FV(value=1.0,lb=0.1,ub=1000.0)
tf.STATUS = 1

# controls
F = model.MV(value=0,lb=0,ub=2000)
F.STATUS = 1
a = model.MV(value=0,lb=-0.5*np.pi/180,ub=0.5*np.pi/180)
a.STATUS = 1

# state variables
r = model.Var(value=r0,lb=1738000) # height
s = model.Var(value=0) # tru anomaly
v = model.Var(value=0) # vertical velocity
u = model.Var(value=v0) # horizional velocity
f = model.Var(value=-0.5*np.pi) # angle
w = model.Var(value=0, lb=-10*np.pi/180,ub=10*np.pi/180) # angular velocity
m = model.Var(value=600,lb=0,ub=600) # mass

# constants
mu = 4.90275*10**(12) # lunar gravitational constant
Isp = 300*9.8 # specific impulse

# Equations
model.Equation( r.dt() == tf * v )
model.Equation( s.dt() == tf * u/r )
model.Equation( v.dt() == tf * (F/m*model.cos(f)-mu/(r**2)+u**2/r) )
model.Equation( u.dt() == tf * (F/m*model.sin(f)-u*v/r) )
model.Equation( f.dt() == tf * (w - u/r) ) # ---newly added
model.Equation( w.dt() == tf * a ) # ---newly added
model.Equation( m.dt() == tf * (-F/Isp) )

# terminal constraints
_finalMask = np.zeros(nt)
_finalMask[-1] = 1.0
finalMask = model.Param(value=_finalMask)
model.Equation(v*finalMask>=0)
model.Equation(v*finalMask<=0.5)
model.Equation(f*finalMask>=-5*np.pi/180) # ***newly added
model.Equation(f*finalMask<=5*np.pi/180) # ***newly added
model.fix_final(r,val=1738000)
model.fix_final(u,val=0)

model.Obj(-m*finalMask) # Objective function to be minimized

model.options.IMODE = 6
model.solver_options = ['max_iter 99999']
model.solve() # solve

time_end=time.time()
print('Calculation Time: ',time_end-time_start)

# scaled time
print('Landing Time: ' + str(tf.value[0]))
tm = np.linspace(0,tf.value[0],nt)
finaltime = tm[-1]

# PLOT
fig = plt.figure(1)
ax1 = fig.add_subplot(2, 4, 1)
ax2 = fig.add_subplot(2, 4, 2)
ax3 = fig.add_subplot(2, 4, 3)
ax4 = fig.add_subplot(2, 4, 4)
ax5 = fig.add_subplot(2, 4, 5)
ax6 = fig.add_subplot(2, 4, 6)
ax7 = fig.add_subplot(2, 4, 7)
ax8 = fig.add_subplot(2, 4, 8)
ax1.plot(tm,r.value,'k-',label=r'$r$')
ax1.set_xlabel('Time')
ax1.set_ylabel('r')
ax1.set_xlim(0,finaltime)
ax2.plot(tm,v.value,'b-',label=r'$v$')
ax2.set_xlabel('Time')
ax2.set_ylabel('v')
ax2.set_xlim(0,finaltime)
ax3.plot(tm,u.value,'g-',label=r'$u$')
ax3.set_xlabel('Time')
ax3.set_ylabel('u')
ax3.set_xlim(0,finaltime)
ax4.plot(tm,m.value,'y-',label=r'$m$')
ax4.set_xlabel('Time')
ax4.set_ylabel('m')
ax4.set_xlim(0,finaltime)
ax5.plot(tm,F.value,'r-',label=r'$F$')
ax5.set_xlabel('Time')
ax5.set_ylabel('F')
ax5.set_xlim(0,finaltime)
ax6.plot(tm,f.value,'k-',label=r'$f$')
ax6.set_xlabel('Time')
ax6.set_ylabel('f')
ax6.set_xlim(0,finaltime)
ax7.plot(tm,w.value,'b-',label=r'$w$')
ax7.set_xlabel('Time')
ax7.set_ylabel('w')
ax7.set_xlim(0,finaltime)
ax8.plot(tm,a.value,'r-',label=r'$a$')
ax8.set_xlabel('Time')
ax8.set_ylabel('a')
ax8.set_xlim(0,finaltime)
plt.tight_layout()
plt.show()

however, I've ran the code for many hours but didn't get a solution or result. More specifically, it stops at certain iterations:

I tried to use soft terminal constraints but also failed to get any result:

model.Equation(r*finalMask>=1738000)
model.Equation(r*finalMask<=1738001)
model.Equation(v*finalMask>=0)
model.Equation(v*finalMask<=0.5)
model.Equation(u*finalMask>=-0.5)
model.Equation(u*finalMask<=0.5)
model.Equation(f*finalMask>=-5*np.pi/180)
model.Equation(f*finalMask<=5*np.pi/180)

Here is a project report from a similar project: https://apmonitor.com/do/uploads/Main/Report_rocket_landing2019.pdf and the source code: https://github.com/ericman314/dynamic-optimization/tree/master/rocket — John Hedengren, Jan 28 '21 at 15:27
@JohnHedengren Thank you! It helps a lot. I will take a few days to work with it. — abmin, Jan 29 '21 at 02:31
@JohnHedengren I have updated my question as well as my codes, could you please have a look at the question again? I will appreciate it if you could give any further suggestions. — abmin, Jan 29 '21 at 07:23

score 0 · Answer 1 · answered Feb 06 '21 at 19:05

I couldn't get your code to run because w wasn't defined but maybe that was changed to u. Also, there are many places where you needed a semi-colon to separate the statements on the same line. I tried to get a version that would run but haven't been successful yet. Could you verify that the posted code runs to show the problem that you described or else post your updated code that does run successfully?

I modified the equations to be consistent with your problem statement and avoid potential issues that may occur with divide by zero. I also included a scaling factor because some of the equations have r**2 that is a very large number.

# equations
scale = 1e-6
model.Equation( r.dt() == tf * v )
model.Equation( r * s.dt()*scale == tf * u*scale )
model.Equation( m * (r**2) * v.dt() * scale**2 \
                == tf * ((r**2)*F*model.cos(f)-mu+(u**2)*(r**3)) *(scale**2))
model.Equation( m * r * u.dt()*scale == tf * (-u*v*m+F*model.cos(f))*scale )
model.Equation( Isp * m.dt() == -tf * F )

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt

model = GEKKO() # initialize gekko
nt = 501
model.time = np.linspace(0,1,nt)

# optimize final time
tf = model.FV(value=100.0,lb=1.0,ub=1000.0)
tf.STATUS = 1

# controls
F = model.MV(value=0,lb=0,ub=1500)
F.STATUS = 1
f = model.MV(value=0,lb=-0.5*np.pi,ub=0.5*np.pi)
f.STATUS = 1

# state variables
# height = 1753 km
r = model.Var(value=1753000,lb=1738000) 
# vertical velocity = 0 m/s
v = model.Var(value=0)
# azimuth = 0 rad
s = model.Var(value=0)
# azimuth angular velocity rad/s
u = model.Var(value=9.65e-4)
# mass = 600 kg
m = model.Var(value=600,lb=0,ub=600) 

# constants
mu = 4.90275e12 # lunar gravitational constant
Isp = 300*9.8 # specific impulse

# equations
scale = 1e-6
model.Equation( r.dt() == tf * v )
model.Equation( r * s.dt()*scale == tf * u*scale )
model.Equation( m * (r**2) * v.dt() * scale**2 \
                == tf * ((r**2)*F*model.cos(f)-mu+(u**2)*(r**3)) *(scale**2))
model.Equation( m * r * u.dt()*scale == tf * (-u*v*m+F*model.cos(f))*scale )
model.Equation( Isp * m.dt() == -tf * F )

# terminal conditions
_finalMask = np.zeros(nt)
_finalMask[-1] = 1.0
finalMask = model.Param(value=_finalMask)

# Terminal Constraint 1
if False:
    model.Equation((r-1738000)*finalMask==0)
    model.Equation(v*finalMask==0)
    model.Equation(u*finalMask==0)

# Terminal Constraint 2
if False:
    model.Equation((r-1738000)*finalMask<=0)
    model.Equation(v*finalMask<=0)
    model.Equation(u*finalMask<=0)

if True:
    # terminal constraints
    _finalMask = np.zeros(nt)
    _finalMask[-1] = 1.0
    finalMask = model.Param(value=_finalMask)
    model.Equation(v*finalMask>=0)
    model.Equation(v*finalMask<=0.2)
    model.fix_final(r,val=1738000)
    model.fix_final(u,val=0)

model.Minimize(tf)
model.options.IMODE = 6
model.solve()

print('Final Time: ' + str(tf.value[0]))
tm = np.linspace(0,tf.value[0],nt)

# PLOT
fig = plt.figure(1)
ax1 = fig.add_subplot(2, 3, 1)
ax2 = fig.add_subplot(2, 3, 2)
ax3 = fig.add_subplot(2, 3, 3)
ax4 = fig.add_subplot(2, 3, 4)
ax5 = fig.add_subplot(2, 3, 5)
ax6 = fig.add_subplot(2, 3, 6)
ax1.plot(tm,r.value,'k-',label=r'$r$')
ax1.set_xlabel('Time'); ax1.set_ylabel('r')
ax2.plot(tm,v.value,'b-',label=r'$v$')
ax2.set_xlabel('Time'); ax2.set_ylabel('v')
ax3.plot(tm,s.value,'g-',label=r'$s$')
ax3.set_xlabel('Time'); ax3.set_ylabel('s')
ax4.plot(tm,f.value,'y-',label=r'$f$')
ax4.set_xlabel('Time'); ax4.set_ylabel('f')
ax5.plot(tm,F.value,'c-',label=r'$F$')
ax5.set_xlabel('Time'); ax5.set_ylabel('F')
ax6.plot(tm,m.value,'r-',label=r'$m$')
ax6.set_xlabel('Time'); ax6.set_ylabel('m')
plt.tight_layout()
plt.show()

Numerical solutions are solved more easily if all variable are around the value of 1 and equation residuals start at about that same level. Gekko automatically scales variables based on the initial condition.

As you observed, it can be easier to find a solution that is within a small range versus exactly reaching an exact final condition. One other thing that helps is to include an objective (soft constraint) and equality constraint (hard constraint). This combination typically works better for these types of terminal constraint problems with adjustable final time.

You may also want to initialize with a specified final time and then solve it again. You can turn off degrees of freedom such as:

# solve first time without adjustable tf
tf.STATUS = 0
model.solve()

# solve again with tf adjustable
tf.STATUS = 1
# don't advance the initial condition
model.options.TIME_SHIFT=0
model.solve()

Another option is to solve with F.UPPER=1500 and then attempt F.UPPER=2000 for another solve. The Jennings optimal control problem is another example that may help.

Thank you for your explanation. I'm sorry that my previous code cannot run. I've updated my question as well as my codes. There are two versions, both of them can run. The first version takes `phi (f)` as control variable and the solution shows a sharp turn of it. I constructed a second version and run a couple of hours but failed to get any solution. Please refer to the updated question for more information. — abmin, Feb 11 '21 at 07:12

GEKKO how to solve a optimal lunar soft landing problem?

1 Answers1