如何使用 Gekko 求解可重復使用的再入飛行器的最優控制

Question

我正在尋找最佳控制（aoa 和坡度角），以最大化使用 Gekko 的穿梭型再入飛行器的橫向范圍。 下面是我目前的代碼，我得到一個帶有“退出：超過最大迭代次數”的“未找到解決方案”。 模擬假設點質量具有非旋轉地球框架。 EOMS 是 6 個耦合的非線性 ODE。 我嘗試使用不同的求解器，實現/刪除 state 和控制約束，增加最大迭代次數等。我對我在 Gekko 中問題的設置和實現沒有信心，並希望對我接下來可以嘗試的問題提供一些反饋。 我嘗試遵循APMonitor 的示例 11. Optimal Control with Integral Objective 、 Inverted Pendulum Optimal Control和示例 13. Optimal Control: Minimize Final Time中的設置和布局。 我正在尋找的解決方案如下。

很感謝任何形式的幫助！

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

pi = math.pi

########################
######FRONT MATTER######
########################
m = GEKKO() # initialize GEKKO
nt = 2501 #simulation time is 2500 seconds
tfin = 2500
m.time = np.linspace(0,tfin,nt) #time array

#==================#
#PARAMS
#==================#
Re  = m.Param(value = 6371203.92)           # radius of the earth, m
S   = m.Param(value = 249.9091776)          # vehicle surface area, m^2
cl0 = m.Param(value = -0.2070)              # coeff lift param 1
cl1 = m.Param(value = 1.6756)               # coeff lift param 2
cd0 = m.Param(value = 0.0785)               # coeff drag param 1
cd1 = m.Param(value = -0.3529)              # coeff drag param 2
cd2 = m.Param(value = 2.0400)               # coeff drag param 3
H   = m.Param(value = 7254.24)              # density scale height, m
rho0= m.Param(value = 1.225570827014494)    # sea level atmospheric density, kg/m^3
mu  = m.Param(value = 3.986031954093051e14) #earth gravitational param, m^3/s^2
mass= m.Param(value = 92079.2525560557)     #vehicle mass, kg

#===============================#
#BOUNDARY CONDITIONS
#===============================#
t0     = 0
alt0   = 79248
rad0   = alt0+Re
altf   = 24384
radf   = altf+Re
lon0   = 0
lat0   = 0
speed0 = +7802.88
speedf = +762
fpa0   = -1*pi/180
fpaf   = -5*pi/180
azi0   = +90*pi/180
azif   = -90*pi/180

#===============================#
#LIMITS ON VARIABLES
#===============================#
tfMin = 0;              tfMax = 3000;
radMin = Re;            radMax = rad0;
lonMin = -pi;           lonMax = -lonMin;
latMin = -70*pi/180;    latMax = -latMin;
speedMin = 10;          speedMax = 45000;
fpaMin = -80*pi/180;    fpaMax =  80*pi/180;
aziMin = -180*pi/180;   aziMax =  180*pi/180;
aoaMin = -90*pi/180;    aoaMax = -aoaMin;
bankMin = -90*pi/180;   bankMax =   1*pi/180;

#===============================#
#VARIABLES
#===============================#

#state variables and bounds
rad = m.Var(value=rad0, lb=radMin, ub=radMax)       # radius, m
lon = m.Var(value=lon0, lb=lonMin, ub=lonMax)       # longitude, rad
lat = m.Var(value=lat0, lb=latMin, ub=latMax)       # latitude, rad
vel = m.Var(value=speed0, lb=speedMin, ub=speedMax) # velocity, m/sec
fpa = m.Var(value=fpa0, lb=fpaMin, ub=fpaMax)       # flight path angle, rad
azi = m.Var(value=azi0, lb=aziMin, ub=aziMax)       # azimuth angle, rad

#control variables
aoa     = m.MV(value=-20, lb=aoaMin, ub=aoaMax)    # angle of attack, rad
aoa.STATUS = 1
aoa.DCOST = 1e-2
bank    = m.MV(value=0, lb=bankMin, ub=bankMax)    # bank angle, rad
bank.STATUS = 1
bank.DCOST = 1e-2

#===============================#
#INTERMEDIATE VARIABLES
#===============================#
altitude = m.Intermediate(rad - Re)
CD       = m.Intermediate(cd0+cd1*aoa+cd2*aoa**2)
rho      = m.Intermediate(rho0*m.exp(-altitude/H))
CL       = m.Intermediate(cl0+cl1*aoa)
q        = m.Intermediate(0.5*rho*vel**2)
D        = m.Intermediate(q*S*CD/mass)
L        = m.Intermediate(q*S*CL/mass)
gravity  = m.Intermediate(mu/rad**2)

#===============================#
#EOMS
#===============================#

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

m.Equation(rad.dt() == vel*m.sin(fpa))
m.Equation((rad*m.cos(lat))*lon.dt() == vel*m.cos(fpa)*m.sin(azi))
m.Equation(rad*lat.dt() == vel*m.cos(fpa)*m.cos(azi))
m.Equation(vel.dt() == -D-gravity*m.sin(fpa))
m.Equation(vel*fpa.dt() == (L*m.cos(bank)-m.cos(fpa)*(gravity-vel**2/rad)))
m.Equation(vel*azi.dt() == (L*m.sin(bank)/m.cos(fpa)+vel**2*m.cos(fpa)*m.sin(azi)*m.tan(lat)/rad))

#===============================#
#OPTIMIZATION SOLVER
#===============================#
m.Obj(-lat*final)


# m.options.SOLVER = 3
# m.options.IMODE = 6
# m.solve(disp=True)

m.options.MAX_ITER = 500
m.options.IMODE = 6
# m.options.NODES = 3
# m.options.MV_TYPE = 1
m.options.SOLVER = 3
# m.open_folder()
m.solve()

高度、速度、迎角和坡度角的解決方案應該是這樣的： Altitude Velocity AngleofAttack Bank

Answer 1

我通過減少最終時間（成功解決方案最大 = 0.04）並重新排列方程以避免可能的被零除來獲得成功的解決方案：

m.Equation(rad.dt() == vel*m.sin(fpa))
m.Equation((rad*m.cos(lat))*lon.dt() == vel*m.cos(fpa)*m.sin(azi))
m.Equation(rad*lat.dt() == vel*m.cos(fpa)*m.cos(azi))
m.Equation(vel.dt() == -D-gravity*m.sin(fpa))
m.Equation(vel*fpa.dt() == (L*m.cos(bank)-m.cos(fpa)*(gravity-vel**2/rad)))
m.Equation(m.cos(fpa)*rad*vel*azi.dt() == \
           (L*m.sin(bank)*rad+vel**2*(m.cos(fpa))**2*m.sin(azi)*m.tan(lat)))

從一小段時間開始並觀察狀態有助於對 model 進行故障排除。

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

pi = math.pi

########################
######FRONT MATTER######
########################
m = GEKKO() # initialize GEKKO
nt = 101 # 2501 #simulation time is 2500 seconds
tfin = 0.04
m.time = np.linspace(0,tfin,nt) #time array

#==================#
#PARAMS
#==================#
Re  = m.Param(value = 6371203.92)           # radius of the earth, m
S   = m.Param(value = 249.9091776)          # vehicle surface area, m^2
cl0 = m.Param(value = -0.2070)              # coeff lift param 1
cl1 = m.Param(value = 1.6756)               # coeff lift param 2
cd0 = m.Param(value = 0.0785)               # coeff drag param 1
cd1 = m.Param(value = -0.3529)              # coeff drag param 2
cd2 = m.Param(value = 2.0400)               # coeff drag param 3
H   = m.Param(value = 7254.24)              # density scale height, m
rho0= m.Param(value = 1.225570827014494)    # sea level atmospheric density, kg/m^3
mu  = m.Param(value = 3.986031954093051e14) #earth gravitational param, m^3/s^2
mass= m.Param(value = 92079.2525560557)     #vehicle mass, kg

#===============================#
#BOUNDARY CONDITIONS
#===============================#
t0     = 0
alt0   = 79248
rad0   = alt0+Re
altf   = 24384
radf   = altf+Re
lon0   = 0
lat0   = 0
speed0 = +7802.88
speedf = +762
fpa0   = -1*pi/180
fpaf   = -5*pi/180
azi0   = +90*pi/180
azif   = -90*pi/180

#===============================#
#LIMITS ON VARIABLES
#===============================#
tfMin = 0;              tfMax = 3000;
radMin = Re;            radMax = rad0;
lonMin = -pi;           lonMax = -lonMin;
latMin = -70*pi/180;    latMax = -latMin;
speedMin = 10;          speedMax = 45000;
fpaMin = -80*pi/180;    fpaMax =  80*pi/180;
aziMin = -180*pi/180;   aziMax =  180*pi/180;
aoaMin = -90*pi/180;    aoaMax = -aoaMin;
bankMin = -90*pi/180;   bankMax =   1*pi/180;

#===============================#
#VARIABLES
#===============================#

#state variables and bounds
rad = m.Var(value=rad0, lb=radMin, ub=radMax)       # radius, m
lon = m.Var(value=lon0, lb=lonMin, ub=lonMax)       # longitude, rad
lat = m.Var(value=lat0, lb=latMin, ub=latMax)       # latitude, rad
vel = m.Var(value=speed0, lb=speedMin, ub=speedMax) # velocity, m/sec
fpa = m.Var(value=fpa0, lb=fpaMin, ub=fpaMax)       # flight path angle, rad
azi = m.Var(value=azi0, lb=aziMin, ub=aziMax)       # azimuth angle, rad

#control variables
aoa     = m.MV(value=-20, lb=aoaMin, ub=aoaMax)    # angle of attack, rad
aoa.STATUS = 1
bank    = m.MV(value=0, lb=bankMin, ub=bankMax)    # bank angle, rad
bank.STATUS = 1

#===============================#
#INTERMEDIATE VARIABLES
#===============================#
altitude = rad - Re
CD       = cd0+cd1*aoa+cd2*aoa**2
rho      = rho0*m.exp(-altitude/H)
CL       = cl0+cl1*aoa
q        = 0.5*rho*vel**2
D        = q*S*CD/mass
L        = q*S*CL/mass
gravity  = mu/rad**2

#===============================#
#EOMS
#===============================#

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

m.Equation(rad.dt() == vel*m.sin(fpa))
m.Equation((rad*m.cos(lat))*lon.dt() == vel*m.cos(fpa)*m.sin(azi))
m.Equation(rad*lat.dt() == vel*m.cos(fpa)*m.cos(azi))
m.Equation(vel.dt() == -D-gravity*m.sin(fpa))
m.Equation(vel*fpa.dt() == (L*m.cos(bank)-m.cos(fpa)*(gravity-vel**2/rad)))
m.Equation(m.cos(fpa)*rad*vel*azi.dt() == \
           (L*m.sin(bank)*rad+vel**2*(m.cos(fpa))**2*m.sin(azi)*m.tan(lat)))

#===============================#
#OPTIMIZATION SOLVER
#===============================#
m.Maximize(lat*final)
m.options.SOLVER = 3
m.options.IMODE = 6
m.solve(disp=True)

plt.subplot(4,2,1)
plt.plot(m.time,rad.value,label='rad')
plt.legend()
plt.subplot(4,2,2)
plt.plot(m.time,lon.value,label='lon')
plt.legend()
plt.subplot(4,2,3)
plt.plot(m.time,lat.value,label='lat')
plt.legend()
plt.subplot(4,2,4)
plt.plot(m.time,vel.value,label='vel')
plt.legend()
plt.subplot(4,2,5)
plt.plot(m.time,fpa.value,label='fpa')
plt.legend()
plt.subplot(4,2,6)
plt.plot(m.time,azi.value,label='azi')
plt.legend()
plt.subplot(4,2,7)
plt.plot(m.time,aoa.value,label='aoa')
plt.xlabel('Time')
plt.legend()
plt.subplot(4,2,8)
plt.plot(m.time,bank.value,label='bank')
plt.xlabel('Time')
plt.legend()
plt.show()

一個建議是關閉自由度以驗證具有較小時間范圍的解決方案 ( STATUS=0 )。 可能還需要額外的約束來將三角函數保持在 -2 pi 到 2 pi 區域。 還有關於初始化具有挑戰性的問題的其他信息：

• Safdarnejad, SM, Hedengren, JD, Lewis, N.R., Haseltine, E., 動態系統優化的初始化策略, 計算機與化學工程, 2015, Vol. 78，第 39-50 頁，DOI：10.1016/j.compchemeng.2015.04.016。

Answer 2

成功.. 感謝您的反饋和指點@John Hedengren。 在結合了我在這里學到的知識並恢復了狀態的界限之后，我能夠成功地使用 Gekko 產生與以前的解決方案一致的答案。 對 Gekko 和這里的潛力感到非常興奮！