Subsections

• Formulation
• Performance

9 Particle Steering

Minimize the time taken for a particle, acted upon by a thrust of constant magnitude, to achieve a given altitude and terminal velocity.

Formulation

The equations of motion are

$$\ddot{y}_{1} = a \cos (u), \qquad \ddot{y}_{2} = a\sin (u),$$ (6)

where $( y_{1} , y_{2} )$ is the position of the particle, $u$ is the control angle with

$$\vert u(t) \vert \leq \frac{\pi }{2} ,$$

and $a$ is the constant magnitude of thrust. The particle is initially at rest so that

$$y_{1}(0) = y_{2}(0) = \dot{y}_{1}(0) = \dot{y}_{2}(0) = 0 .$$

The problem is to minimize the travel time $t_f$ so that the particle achieves a given height $y_2(t_f)$ and terminal velocity $( \dot y_1 (t_f) , \dot y_2 (t_f) )$.

This is a classical (see Bryson and Ho [7, pages 59-62]) problem in dynamic optimization. We use for the magnitude of thrust and the boundary conditions [4]

$$y_{2}(t_{f}) = 5, \qquad \dot{y}_{1}(t_{f}) = 45, \qquad \dot{y}_{2}(t_{f}) = 0 .$$

Discretization is done using a uniform time step and the trapezoidal rule for the integration of the system over intervals. Data for this problem is shown in Table 9.1.

Table 9.1: Particle steering problem data

Variables
Constraints
Bounds
Linear equality constraints	0
Linear inequality constraints	0
Nonlinear equality constraints
Nonlinear inequality constraints	0
Nonzeros in	0
Nonzeros in

Performance

Table 9.2: Performance on particle steering problem

Solver	$n_h=50$	$n_h=100$	$n_h=200$	$n_h=400$
LANCELOT	30.83 s	124.62 s	412.89 s	2997.88 s
$f$	5.54672e-01	5.54594e-01	5.54588e-01	5.54552e-01$\dagger$
$c$ violation	2.69440e-06	1.94910e-06	6.12610e-06	8.56120e-06$\dagger$
iterations	366	416	410	575
LOQO	923.3 s	$\ddagger$	$\ddagger$	$\ddagger$
$f$	5.54668e-01	$\ddagger$	$\ddagger$	$\ddagger$
$c$ violation	4.1e-10	$\ddagger$	$\ddagger$	$\ddagger$
iterations	20163	$\ddagger$	$\ddagger$	$\ddagger$
MINOS	1.58 s	5.62 s	27.86 s	143.09 s
$f$	5.54668e-01	5.54595e-01	5.54577e-01	5.54572e-01
$c$ violation	2.5e-13	1.1e-08	2.4e-13	7.9e-10
iterations	10	15	29	60
SNOPT	3.25 s	13.43 s	53.81 s	147.37 s
$f$	5.54668e-01	5.54595e-01	5.54577e-01	5.54573e-01
$c$ violation	5.7e-08	1.9e-09	6.8e-09	5.4e-07
iterations	36	43	42	25
$\dagger$ Errors or warnings. $\ddagger$ Timed out.

Results for the AMPL implementation are given in Table 9.2. The initial values for $y_2$ and $y_3 = \dot y_ 1$ are chosen as the functions

$$y_1(t) = 5 \left ( \frac {t}{t_f} \right ) , \qquad y_3(t) = 45 \left ( \frac {t}{t_f}\right ) .$$

Initial values for $y_2$, $y_4 = \dot y_2,$ and $u$ are set to zero. The initial value for the final time is . Plots of the height $y_2$ and control $u$ as a function of the horizontal position $y_1$ are in Figure 9.1.

Only LANCELOT returns an error here, for $n_h = 400$, of step got too small. Even so, it comes near to the optimal solution value.

Figure 9.1: Height and control as a function of position for the particle steering problem