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gekko - MATLAB 中的局部不可行性

转载 作者:行者123 更新时间:2023-12-05 07:07:23 25 4
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我需要有关动态优化问题的帮助,该问题涉及具有此最优控制问题的 UAV 消耗能量优化。

enter image description here

enter image description here

我的代码是这样的

教育:

Parameters
tf

#Velocidad de rotores rad/s
#Las condiciones iniciales permiten igualar la acción de la gravedad
#Se tomo 4000rad/s como la velocidad maxima de los rotores
w1 = 912.32, >=0, <=3000
w2 = 912.32, >=0, <=3000
w3 = 912.32, >=0, <=3000
w4 = 912.32, >=0, <=3000
t1 = 0, >=0
t2 = 0, >=0
t3 = 0, >=0
t4 = 0, >=0

Constants
!----------------COEFICIENTES DEL MODELO-----------------!
#Gravedad
g = 9.81 !m/s^2
pi = 3.14159265359

#Motor Coefficients
J = 4.1904e-5 !kg*m^2
kt = 0.0104e-3 !N*m/A
kv = 96.342 !rad/s/volt
Dv = 0.2e-3 !N*m*s/rad
R = 0.2 !Ohms

#Battery parameters
Q = 1.55 !Ah
Rint = 0.02 !Ohms
E0 = 1.24 !volt
K = 2.92e-3 !volt
A = 0.156
B =2.35

#Quadrotor parameters
l = 0.175 !m
m = 1.3 !kg
Ix = 0.081 !kg*m^2
Iy = 0.081 !kg*m^2
Iz = 0.142 !kg*m^2
kb = 3.8305e-6 !N/rad/s
ktau = 2.2518e-8 !(N*m)/rad/s

#Parametrizacion del polinomio
a1 = -1.72e-5
a2 = 1.95e-5
a3 = -6.98e-6
a4 = 4.09e-7
b1 = 0.014
b2 = -0.0157
b3 = 5.656e-3
b4 = -3.908e-4
c1 = -0.8796
c2 = 0.3385
c3 = 0.2890
c4 = 0.1626

Variables
!------------------CONDICONES INICIALES------------------!
x = 0
xp = 0
y = 0
yp = 0
z = 0
zp = 0
pitch = 0, >=-pi/2, <=pi/2 !theta - restricciones
pitchp = 0
roll = 0, >=-pi/2, <=pi/2 !phi - restricciones
rollp = 0
yaw = 0 !psi
yawp = 0%, >=-200/180, <=200/180

#Función objetivo
of = 0 !condición inicial de la función objetivo
Intermediates

#Motor 1
aw1 = a1*w1^2 + b1*w1 + c1
bw1 = a2*w1^2 + b2*w1 + c2
cw1 = a3*w1^2 + b3*w1 + c3
dw1 = a4*w1^2 + b4*w1 + c4
#Motor 2
aw2 = a1*w2^2 + b1*w2 + c1
bw2 = a2*w2^2 + b2*w2 + c2
cw2 = a3*w2^2 + b3*w2 + c3
dw2 = a4*w2^2 + b4*w2 + c4
#Motor 3
aw3 = a1*w3^2 + b1*w3 + c1
bw3 = a2*w3^2 + b2*w3 + c2
cw3 = a3*w3^2 + b3*w3 + c3
dw3 = a4*w3^2 + b4*w3 + c4
#Motor 4
aw4 = a1*w4^2 + b1*w4 + c1
bw4 = a2*w4^2 + b2*w4 + c2
cw4 = a3*w4^2 + b3*w4 + c3
dw4 = a4*w4^2 + b4*w4 + c4
#frj(wj(t),Tj(t))
fr1=aw1*t1^3 + bw1*t1^2 + cw1*t1 + dw1
fr2=aw2*t2^3 + bw2*t2^2 + cw2*t2 + dw2
fr3=aw3*t3^3 + bw3*t3^2 + cw3*t3 + dw3
fr4=aw4*t4^3 + bw4*t4^2 + cw4*t4 + dw4
!---------------------CONTROL INPUTS---------------------!
T = kb * (w1^2 + w2^2 + w3^2 + w4^2)
u1 = kb * (w2^2 - w4^2)
u2 = kb * (w3^2 - w1^2)
u3 = ktau * (w1^2 - w2^2 + w3^2 - w4^2)
wline = w1 - w2 + w3 - w4
!-------------------ENERGIA POR ROTOR--------------------!
Ec1 = ((J*$w1 + ktau*w1^2 + Dv*w1)/fr1)*w1
Ec2 = ((J*$w2 + ktau*w2^2 + Dv*w2)/fr2)*w2
Ec3 = ((J*$w3 + ktau*w3^2 + Dv*w3)/fr3)*w3
Ec4 = ((J*$w4 + ktau*w4^2 + Dv*w4)/fr4)*w4
Ectotal = Ec1 + Ec2 + Ec3 + Ec4
Equations
!---------------MINIMIZAR FUNCIÓN OBJETIVO---------------!
minimize tf * of
!-----------------RELACION DE VARIABLES------------------!
xp = $x
yp = $y
zp = $z
pitchp = $pitch
rollp = $roll
yawp = $yaw
!-----------------CONDICONES DE FRONTERA-----------------!
#Condiciones finales del modelo
tf * x = 4
tf * y = 5
tf * z = 6
tf * xp = 0
tf * yp = 0
tf * zp = 0
tf * roll = 0
tf * pitch = 0
tf * yaw = 0
!-----------------TORQUE DE LOS MOTORES------------------!
t1 = J*$w1 + ktau*w1^2 + Dv*w1
t2 = J*$w2 + ktau*w2^2 + Dv*w2
t3 = J*$w3 + ktau*w3^2 + Dv*w3
t4 = J*$w4 + ktau*w4^2 + Dv*w4
!------------------------SUJETO A------------------------!
#Modelo aerodinámico del UAV
m*$xp = (cos(roll)*sin(pitch)*cos(yaw) + sin(roll)*sin(yaw))*T
m*$yp = (cos(roll)*sin(pitch)*sin(yaw) - sin(roll)*cos(yaw))*T
m*$zp = (cos(roll)*cos(pitch))*T-m*g
Ix*$rollp = ((Iy - Iz)*pitchp*yawp + J*pitchp*wline + l*u1)
Iy*$pitchp = ((Iz - Ix)*rollp*yawp - J*rollp*wline + l*u2)
Iz*$yawp = ((Ix - Iy)*rollp*pitchp + u3)
!--------------------FUNCIÓN OBJETIVO--------------------!
$of = Ectotal

MATLAB:

clear all; close all; clc

server = 'http://127.0.0.1';
app = 'traj_optima';

addpath('C:/Program Files/MATLAB/apm_matlab_v0.7.2/apm')
apm(server,app,'clear all');
apm_load(server,app,'ecuaciones_mod.apm');
csv_load(server,app,'tiempo2.csv');

apm_option(server,app,'apm.max_iter',200);
apm_option(server,app,'nlc.nodes',3);
apm_option(server,app,'apm.rtol',1);
apm_option(server,app,'apm.otol',1);
apm_option(server,app,'nlc.solver',3);
apm_option(server,app,'nlc.imode',6);
apm_option(server,app,'nlc.mv_type',1);


costo=1e-5;%1e-5
%VARIABLES CONTROLADAS
%Velocidades angulares
apm_info(server,app,'MV','w1');
apm_option(server,app,'w1.status',1);
apm_info(server,app,'MV','w2');
apm_option(server,app,'w2.status',1);
apm_info(server,app,'MV','w3');
apm_option(server,app,'w3.status',1);
apm_info(server,app,'MV','w4');
apm_option(server,app,'w4.status',1);

% Torques
apm_info(server,app,'MV','t1');
apm_option(server,app,'t1.status',1);
apm_info(server,app,'MV','t2');
apm_option(server,app,'t2.status',1);
apm_info(server,app,'MV','t3');
apm_option(server,app,'t3.status',1);
apm_info(server,app,'MV','t4');
apm_option(server,app,'t4.status',1);

%Salida
output = apm(server,app,'solve');
disp(output)
y = apm_sol(server,app);
z = y.x;

tiempo2.csv

time,tf
0,0
0.001,0
0.2,0
0.4,0
0.6,0
0.8,0
1,0
1.2,0
1.4,0
1.6,0
1.8,0
2,0
2.2,0
2.4,0
2.6,0
2.8,0
3,0
3.2,0
3.4,0
3.6,0
3.8,0
4,0
4.2,0
4.4,0
4.6,0
4.8,0
5,0
5.2,0
5.4,0
5.6,0
5.8,0
6,0
6.2,0
6.4,0
6.6,0
6.8,0
7,0
7.2,0
7.4,0
7.6,0
7.8,0
8,0
8.2,0
8.4,0
8.6,0
8.8,0
9,0
9.2,0
9.4,0
9.6,0
9.8,0
10,1

最后得到的答案是:

enter image description here

我需要帮助解决这个局部不可行性问题。

最佳答案

不可行解是由终端约束引起的:

  tf * z = 4
tf * z = 5
tf * z = 6

tf=0时,约束被评估为0=40=50=6 并且求解器报告求解器无法满足这些要求。相反,您可以将约束设置为:

  tf * (x-4) = 0
tf * (y-5) = 0
tf * (z-6) = 0

这样,约束在 tf=0 和最后一次 tf=1 时有效。一种可能更好的方法是使用 f=1000 将终端约束转换为客观术语,例如:

  minimize f*tf*((x-4)^2 + (y-5)^2 + (z-6)^2)
minimize f*tf*(xp^2 + yp^2 + zp^2)
minimize f*tf*(roll^2 + pitch^2 + yaw^2)

这样,如果优化器无法达到终端约束,则优化器不会报告不可行的解决方案 discussed in the pendulum problem .我对您的模型和脚本进行了一些其他修改以实现成功的解决方案。这是一个总结:

  • 将终端约束转换为目标函数(软约束)
  • 参数t1-t4应该是变量
  • 通过制作 w1-w4 变量和 w1p-w4p 变量修复了自由度问题。 w1-w4为微分态。
  • 在 -10 和 10 之间为 w1p-w4p 添加约束以帮助求解器收敛
  • 添加了初始化步骤以在优化之前模拟模型。本文中有更多关于初始化策略的详细信息:Safdarnejad, S.M., Hedengren, J.D., Lewis, N.R., Haseltine, E., Initialization Strategies for Optimization of Dynamic Systems , 计算机与化学工程, 2015, Vol. 78, pp. 39-50, DOI: 10.1016/j.compchemeng.2015.04.016

型号

Parameters
tf

w1p = 0 > -10 < 10
w2p = 0 > -10 < 10
w3p = 0 > -10 < 10
w4p = 0 > -10 < 10


Constants
!----------------COEFICIENTES DEL MODELO-----------------!
#Gravedad
g = 9.81 !m/s^2
pi = 3.14159265359

#Motor Coefficients
J = 4.1904e-5 !kg*m^2
kt = 0.0104e-3 !N*m/A
kv = 96.342 !rad/s/volt
Dv = 0.2e-3 !N*m*s/rad
R = 0.2 !Ohms

#Battery parameters
Q = 1.55 !Ah
Rint = 0.02 !Ohms
E0 = 1.24 !volt
K = 2.92e-3 !volt
A = 0.156
B =2.35

#Quadrotor parameters
l = 0.175 !m
m = 1.3 !kg
Ix = 0.081 !kg*m^2
Iy = 0.081 !kg*m^2
Iz = 0.142 !kg*m^2
kb = 3.8305e-6 !N/rad/s
ktau = 2.2518e-8 !(N*m)/rad/s

#Parametrizacion del polinomio
a1 = -1.72e-5
a2 = 1.95e-5
a3 = -6.98e-6
a4 = 4.09e-7
b1 = 0.014
b2 = -0.0157
b3 = 5.656e-3
b4 = -3.908e-4
c1 = -0.8796
c2 = 0.3385
c3 = 0.2890
c4 = 0.1626

Variables
!------------------CONDICONES INICIALES------------------!
x = 0
xp = 0
y = 0
yp = 0
z = 0
zp = 0
pitch = 0, >=-pi/2, <=pi/2 !theta - restricciones
pitchp = 0
roll = 0, >=-pi/2, <=pi/2 !phi - restricciones
rollp = 0
yaw = 0 !psi
yawp = 0 %, >=-200/180, <=200/180

#Velocidad de rotores rad/s
#Las condiciones iniciales permiten igualar la acción de la gravedad
#Se tomo 4000rad/s como la velocidad maxima de los rotores
w1 = 912.32, >=0, <=3000
w2 = 912.32, >=0, <=3000
w3 = 912.32, >=0, <=3000
w4 = 912.32, >=0, <=3000

t1 = 0, >=0
t2 = 0, >=0
t3 = 0, >=0
t4 = 0, >=0

#Función objetivo
of = 0 !condición inicial de la función objetivo
Intermediates

#Motor 1
aw1 = a1*w1^2 + b1*w1 + c1
bw1 = a2*w1^2 + b2*w1 + c2
cw1 = a3*w1^2 + b3*w1 + c3
dw1 = a4*w1^2 + b4*w1 + c4
#Motor 2
aw2 = a1*w2^2 + b1*w2 + c1
bw2 = a2*w2^2 + b2*w2 + c2
cw2 = a3*w2^2 + b3*w2 + c3
dw2 = a4*w2^2 + b4*w2 + c4
#Motor 3
aw3 = a1*w3^2 + b1*w3 + c1
bw3 = a2*w3^2 + b2*w3 + c2
cw3 = a3*w3^2 + b3*w3 + c3
dw3 = a4*w3^2 + b4*w3 + c4
#Motor 4
aw4 = a1*w4^2 + b1*w4 + c1
bw4 = a2*w4^2 + b2*w4 + c2
cw4 = a3*w4^2 + b3*w4 + c3
dw4 = a4*w4^2 + b4*w4 + c4
#frj(wj(t),Tj(t))
fr1=aw1*t1^3 + bw1*t1^2 + cw1*t1 + dw1
fr2=aw2*t2^3 + bw2*t2^2 + cw2*t2 + dw2
fr3=aw3*t3^3 + bw3*t3^2 + cw3*t3 + dw3
fr4=aw4*t4^3 + bw4*t4^2 + cw4*t4 + dw4
!---------------------CONTROL INPUTS---------------------!
T = kb * (w1^2 + w2^2 + w3^2 + w4^2)
u1 = kb * (w2^2 - w4^2)
u2 = kb * (w3^2 - w1^2)
u3 = ktau * (w1^2 - w2^2 + w3^2 - w4^2)
wline = w1 - w2 + w3 - w4
!-------------------ENERGIA POR ROTOR--------------------!
Ec1 = ((J*$w1 + ktau*w1^2 + Dv*w1)/fr1)*w1
Ec2 = ((J*$w2 + ktau*w2^2 + Dv*w2)/fr2)*w2
Ec3 = ((J*$w3 + ktau*w3^2 + Dv*w3)/fr3)*w3
Ec4 = ((J*$w4 + ktau*w4^2 + Dv*w4)/fr4)*w4
Ectotal = Ec1 + Ec2 + Ec3 + Ec4

! scaling factor for terminal constraint
f = 1000

Equations
!---------------MINIMIZAR FUNCIÓN OBJETIVO---------------!
minimize tf * of
!-----------------RELACION DE VARIABLES------------------!
xp = $x
yp = $y
zp = $z
pitchp = $pitch
rollp = $roll
yawp = $yaw

w1p = $w1
w2p = $w2
w3p = $w3
w4p = $w4
!-----------------CONDICONES DE FRONTERA-----------------!
#Condiciones finales del modelo
#tf * (x-4) = 0
#tf * (y-5) = 0
#tf * (z-6) = 0
#tf * xp = 0
#tf * yp = 0
#tf * zp = 0
#tf * roll = 0
#tf * pitch = 0
#tf * yaw = 0

minimize f*tf*((x-4)^2 + (y-5)^2 + (z-6)^2)
minimize f*tf*(xp^2 + yp^2 + zp^2)
minimize f*tf*(roll^2 + pitch^2 + yaw^2)
!-----------------TORQUE DE LOS MOTORES------------------!
t1 = J*w1p + ktau*w1^2 + Dv*w1
t2 = J*w2p + ktau*w2^2 + Dv*w2
t3 = J*w3p + ktau*w3^2 + Dv*w3
t4 = J*w4p + ktau*w4^2 + Dv*w4
!------------------------SUJETO A------------------------!
#Modelo aerodinámico del UAV
m*$xp = (cos(roll)*sin(pitch)*cos(yaw) + sin(roll)*sin(yaw))*T
m*$yp = (cos(roll)*sin(pitch)*sin(yaw) - sin(roll)*cos(yaw))*T
m*$zp = (cos(roll)*cos(pitch))*T-m*g
Ix*$rollp = ((Iy - Iz)*pitchp*yawp + J*pitchp*wline + l*u1)
Iy*$pitchp = ((Iz - Ix)*rollp*yawp - J*rollp*wline + l*u2)
Iz*$yawp = ((Ix - Iy)*rollp*pitchp + u3)
!--------------------FUNCIÓN OBJETIVO--------------------!
$of = Ectotal

MATLAB 脚本

clear all; close all; clc

server = 'http://byu.apmonitor.com';
app = 'traj_optima';

addpath('apm')
apm(server,app,'clear all');
apm_load(server,app,'ecuaciones_mod.apm');
csv_load(server,app,'tiempo2.csv');

apm_option(server,app,'apm.max_iter',1000);
apm_option(server,app,'apm.nodes',3);
apm_option(server,app,'apm.rtol',1e-6);
apm_option(server,app,'apm.otol',1e-6);
apm_option(server,app,'apm.solver',3);
apm_option(server,app,'apm.imode',6);
apm_option(server,app,'apm.mv_type',1);


costo=1e-5;%1e-5
%VARIABLES CONTROLADAS
%Velocidades angulares
apm_info(server,app,'MV','w1p');
apm_option(server,app,'w1p.status',1);
apm_info(server,app,'MV','w2p');
apm_option(server,app,'w2p.status',1);
apm_info(server,app,'MV','w3p');
apm_option(server,app,'w3p.status',1);
apm_info(server,app,'MV','w4p');
apm_option(server,app,'w4p.status',1);

%Salida
disp('')
disp('------------- Initialize ----------------')
apm_option(server,app,'apm.coldstart',1);
output = apm(server,app,'solve');
disp(output)

disp('')
disp('-------------- Optimize -----------------')
apm_option(server,app,'apm.time_shift',0);
apm_option(server,app,'apm.coldstart',0);
output = apm(server,app,'solve');
disp(output)
y = apm_sol(server,app);
z = y.x;

这提供了一个成功的解决方案,但不满足终端约束。求解器优化了 w1p-w4p 的使用以最小化目标,但没有解决方案可以达到终端约束。

 The solution was found.

The final value of the objective function is 50477.4537378181

---------------------------------------------------
Solver : IPOPT (v3.12)
Solution time : 3.06940000000759 sec
Objective : 50477.4537378181
Successful solution
---------------------------------------------------

作为下一步,我建议您增加时间点的数量或 allow the final time to change以满足终端约束。您可能还想考虑切换到 Python Gekko使用与 APM MATLAB 相同的底层引擎。在这种情况下,建模语言与 Python 完全集成。

关于gekko - MATLAB 中的局部不可行性,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/62160913/

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