fix_final works for x_f=[0,0,0,0,0,0] but for absolutely no other final state – 'Solution Not Found'

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英文:

fix_final works for x_f=[0,0,0,0,0,0] but for absolutely no other final state - 'Solution Not Found'

问题

'x_f'在一个单独的文件中定义，但这里是GEKKO实现的代码：

  1. m = GEKKO(remote=solverParams["remote"])
  2. m.time = timeSeq
  3. x = [m.Var(value=x_0[i], fixed_initial=True) for i in range(len(x_0))]
  4. u = [m.Var(value=u_0[i], fixed_initial=False) for i in range(len(u_0))]
  5. w = m.Param(value=w)
  7. for i in range(len(x)):
  8.     m.fix_final(x[i], val=x_f[i])
  9.     if stateBounds[i]["lower"] != "-Inf":
  10.         x[i].lower = stateBounds[i]["lower"]
  11.     if stateBounds[i]["upper"] != "+Inf":
  12.         x[i].upper = stateBounds[i]["upper"]
  13. for i in range(len(u)):
  14.     if forceBounds[i]["lower"] != "-Inf":
  15.         u[i].lower = forceBounds[i]["lower"]
  16.     if forceBounds[i]["upper"] != "+Inf":
  17.         u[i].upper = forceBounds[i]["upper"]
  19. eqs = [x[i].dt() == np.matmul(A[i], x) + np.matmul(np.atleast_1d(B[i]), u)
  20.        for i in range(len(x))]
  21. eqs = m.Equations(eqs)
  23. startTime = time.time()
  24. m.Minimize(np.add(np.matmul(np.subtract(x, x_f), np.atleast_1d(np.matmul(np.atleast_1d(Q), np.subtract(x, x_f)))),
  25.                   np.matmul(u, np.atleast_1d(np.matmul(np.atleast_1d(R), u))))
  26.            * w)
  27. m.options.IMODE = 6
  28. m.options.SOLVER = 3
  29. m.solve(disp=solverParams["disp"])
  30. stopTime = time.time()
  32. x_p = [x[i].value for i in range(len(x))]
  33. u_p = [u[i].value for i in range(len(u))]
  34. x_p = np.transpose(x_p)
  35. u_p = np.transpose(u_p)

我的测试案例是：

  1. x_0 = np.array([50.00, -25.00, 80.00, 0.00, 0.00, 0.00])

如果我将x_f设置为精确的零向量，它会给我一个解决方案，但如果我将其设置为，比如：

  1. x_f = np.array([0.04, 0.00, 0.00, 0.00, 0.00, 0.00])

它会说它遇到了局部不可行的点。我怀疑这不是实际的不可行情况，因为这已经发生了对于多个x_0和x_f的情况。

在这里，Q矩阵是6x6的零矩阵，R是3x3的10*I矩阵，系统是线性时不变的。

英文:

'x_f' is defined in a separate file but here's the code for the GEKKO implementation:

  1. m = GEKKO(remote = solverParams["remote"])
  2. m.time = timeSeq
  3. x = [m.Var(value = x_0[i], fixed_initial =  True) for i in range(len(x_0))]
  4. u = [m.Var(value = u_0[i], fixed_initial = False) for i in range(len(u_0))]
  5. w = m.Param(value = w)
  7. for i in range(len(x)):
  8.   m.fix_final(x[i], val = x_f[i])
  9.   if stateBounds[i]["lower"] != "-Inf":
  10.     x[i].lower = stateBounds[i]["lower"]
  11.   if stateBounds[i]["upper"] != "+Inf":
  12.     x[i].upper = stateBounds[i]["upper"]
  13. for i in range(len(u)):
  14.   if forceBounds[i]["lower"] != "-Inf":
  15.     u[i].lower = forceBounds[i]["lower"]
  16.   if forceBounds[i]["upper"] != "+Inf":
  17.     u[i].upper = forceBounds[i]["upper"]
  18.   
  19. eqs = [x[i].dt() == np.matmul(A[i], x) + np.matmul(np.atleast_1d(B[i]), u)
  20.         for i in range(len(x))]
  21. eqs = m.Equations(eqs)
  23. startTime = time.time()
  24. m.Minimize(np.add(np.matmul(np.subtract(x, x_f), np.atleast_1d(np.matmul(np.atleast_1d(Q), np.subtract(x, x_f)))),
  25.                   np.matmul(            u,       np.atleast_1d(np.matmul(np.atleast_1d(R),             u))))
  26.           *w)
  27. m.options.IMODE  = 6
  28. m.options.SOLVER = 3
  29. m.solve(disp = solverParams["disp"])
  30. stopTime = time.time()
  32. x_p = [x[i].value for i in range(len(x))]
  33. u_p = [u[i].value for i in range(len(u))]
  34. x_p = np.transpose(x_p)
  35. u_p = np.transpose(u_p)

My test case is for:

  1. x_0 = np.array([50.00, -25.00, 80.00, 0.00, 0.00, 0.00])

If I set x_f to be the exact zero vector, it gives me a solution but if I set it to, say,

  1. x_f = np.array([ 0.04,   0.00,  0.00, 0.00, 0.00, 0.00])

it says that it hits a point of local infeasibility. I doubt that this is an actual scenario of infeasibility since this has happened for several cases of x_0 and x_f.

Here, the Q matrix is 0_{6x6} and R is 10*I_{3x3} and the system is linear time-invariant.

答案1

得分: 1

修复最终状态也会将最终状态的导数修复为零。这是Gekko的已知问题，已在其他地方报告过。

不要修复最终值，尝试使用终端方程，例如：

  1. p = np.zeros_like(m.time); p[-1]=1
  2. final = m.Param(p)
  3. m.Equation(final*(x[i]-x_f[i])==0)

如果这不起作用，请尝试将软终端约束包括在目标函数中。这通常对求解器来说是最容易的，不会导致潜在的不可行解决方案。

  1. m.Minimize(final*(x[i]-x_f[i])**2)

这是一个使用软终端约束来达到最终目标的双倒立摆示例。

fix_final works for x_f=[0,0,0,0,0,0] but for absolutely no other final state – 'Solution Not Found'

英文:

Fixing the final state also fixes the derivative of the final state to zero. This is a known issue with Gekko that has been reported elsewhere.

Instead of fixing the final value, try a terminal equation such as:

  1. p = np.zeros_like(m.time); p[-1]=1
  2. final = m.Param(p)
  3. m.Equation(final*(x[i]-x_f[i])==0)

If this doesn't work, try a soft terminal constraint that is included in the objective function. This is typically the easiest for the solver and doesn't result in a potential infeasible solution.

  1. m.Minimize(final*(x[i]-x_f[i])**2)

Here is an example of a double inverted pendulum that uses the soft terminal constraint to reach a final target.

fix_final works for x_f=[0,0,0,0,0,0] but for absolutely no other final state – 'Solution Not Found'

huangapple
  • 本文由 发表于 2023年4月20日 06:03:35
  • 转载请务必保留本文链接：https://go.coder-hub.com/76059134.html
  • gekko
  • optimization
  • python
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