QP-Spline-Path Optimizer¶
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Quadratic programming + Spline interpolation
1. Objective function¶
1.1 Get path length¶
Path is defined in station-lateral coordination system. The s range from vehicle’s current position to default planning path length.
1.2 Get spline segments¶
Split the path into n segments. each segment trajectory is defined by a polynomial.
1.3 Define function for each spline segment¶
Each segment i has accumulated distance \(d_i\) along reference line. The trajectory for the segment is defined as a polynomial of degree five by default.
$$
l = f_i(s)
= a_{i0} + a_{i1} \cdot s + a_{i2} \cdot s^2 + a_{i3} \cdot s^3 + a_{i4} \cdot s^4 + a_{i5} \cdot s^5 (0 \leq s \leq d_{i})
$$
1.4 Define objective function of optimization for each segment¶
$$
cost = \sum_{i=1}^{n} \Big( w_1 \cdot \int\limits_{0}^{d_i} (f_i')^2(s) ds + w_2 \cdot \int\limits_{0}^{d_i} (f_i'')^2(s) ds + w_3 \cdot \int\limits_{0}^{d_i} (f_i^{\prime\prime\prime})^2(s) ds \Big)
$$
1.5 Convert the cost function to QP formulation¶
QP formulation:
$$
\begin{aligned}
minimize & \frac{1}{2} \cdot x^T \cdot H \cdot x + f^T \cdot x \\
s.t. \qquad & LB \leq x \leq UB \\
& A_{eq}x = b_{eq} \\
& Ax \geq b
\end{aligned}
$$
Below is the example for converting the cost function into the QP formulation.
$$
f_i(s) ＝
\begin{vmatrix} 1 & s & s^2 & s^3 & s^4 & s^5 \end{vmatrix}
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix}
$$
And
$$
f_i'(s) =
\begin{vmatrix} 0 & 1 & 2s & 3s^2 & 4s^3 & 5s^4 \end{vmatrix}
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix}
$$
And
$$
f_i'(s)^2 =
\begin{vmatrix} a_{i0} & a_{i1} & a_{i2} & a_{i3} & a_{i4} & a_{i5} \end{vmatrix}
\cdot
\begin{vmatrix} 0 \\ 1 \\ 2s \\ 3s^2 \\ 4s^3 \\ 5s^4 \end{vmatrix}
\cdot
\begin{vmatrix} 0 & 1 & 2s & 3s^2 & 4s^3 & 5s^4 \end{vmatrix}
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix}
$$
then we have,
$$
\int\limits_{0}^{d_i} f_i'(s)^2 ds ＝
\int\limits_{0}^{d_i}
\begin{vmatrix} a_{i0} & a_{i1} & a_{i2} & a_{i3} & a_{i4} & a_{i5} \end{vmatrix}
\cdot
\begin{vmatrix} 0 \\ 1 \\ 2s \\ 3s^2 \\ 4s^3 \\ 5s^4 \end{vmatrix}
\cdot
\begin{vmatrix} 0 & 1 & 2s & 3s^2 & 4s^3 & 5s^4 \end{vmatrix}
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} ds
$$
extract the const outside the integration, we have,
$$
\int\limits_{0}^{d_i} f'(s)^2 ds ＝
\begin{vmatrix} a_{i0} & a_{i1} & a_{i2} & a_{i3} & a_{i4} & a_{i5} \end{vmatrix}
\cdot
\int\limits_{0}^{d_i}
\begin{vmatrix} 0 \\ 1 \\ 2s \\ 3s^2 \\ 4s^3 \\ 5s^4 \end{vmatrix}
\cdot
\begin{vmatrix} 0 & 1 & 2s & 3s^2 & 4s^3 & 5s^4 \end{vmatrix} ds
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix}
$$
$$
＝\begin{vmatrix} a_{i0} & a_{i1} & a_{i2} & a_{i3} & a_{i4} & a_{i5} \end{vmatrix}
\cdot \int\limits_{0}^{d_i}
\begin{vmatrix}
0 & 0 &0&0&0&0\\
0 & 1 & 2s & 3s^2 & 4s^3 & 5s^4\\
0 & 2s & 4s^2 & 6s^3 & 8s^4 & 10s^5\\
0 & 3s^2 & 6s^3 & 9s^4 & 12s^5&15s^6 \\
0 & 4s^3 & 8s^4 &12s^5 &16s^6&20s^7 \\
0 & 5s^4 & 10s^5 & 15s^6 & 20s^7 & 25s^8
\end{vmatrix} ds
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix}
$$
Finally, we have
$$
\int\limits_{0}^{d_i}
f'_i(s)^2 ds =\begin{vmatrix} a_{i0} & a_{i1} & a_{i2} & a_{i3} & a_{i4} & a_{i5} \end{vmatrix}
\cdot \begin{vmatrix}
0 & 0 & 0 & 0 &0&0\\
0 & d_i & d_i^2 & d_i^3 & d_i^4&d_i^5\\
0& d_i^2 & \frac{4}{3}d_i^3& \frac{6}{4}d_i^4 & \frac{8}{5}d_i^5&\frac{10}{6}d_i^6\\
0& d_i^3 & \frac{6}{4}d_i^4 & \frac{9}{5}d_i^5 & \frac{12}{6}d_i^6&\frac{15}{7}d_i^7\\
0& d_i^4 & \frac{8}{5}d_i^5 & \frac{12}{6}d_i^6 & \frac{16}{7}d_i^7&\frac{20}{8}d_i^8\\
0& d_i^5 & \frac{10}{6}d_i^6 & \frac{15}{7}d_i^7 & \frac{20}{8}d_i^8&\frac{25}{9}d_i^9
\end{vmatrix}
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix}
$$
Please notice that we got a 6 x 6 matrix to represent the derivative cost of 5th order spline.
Similar deduction can also be used to calculate the cost of second and third order derivatives.
2 Constraints¶
2.1 The init point constraints¶
Assume that the first point is (\(s_0\), \(l_0\)), (\(s_0\), \(l'_0\)) and (\(s_0\), \(l''_0\)), where \(l_0\) , \(l'_0\) and \(l''_0\) is the lateral offset and its first and second derivatives on the init point of planned path, and are calculated from \(f_i(s)\), \(f'_i(s)\), and \(f_i(s)''\).
Convert those constraints into QP equality constraints, using:
$$
A_{eq}x = b_{eq}
$$
Below are the steps of conversion.
$$
f_i(s_0) =
\begin{vmatrix} 1 & s_0 & s_0^2 & s_0^3 & s_0^4&s_0^5 \end{vmatrix}
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5}\end{vmatrix} = l_0
$$
And
$$
f'_i(s_0) =
\begin{vmatrix} 0& 1 & 2s_0 & 3s_0^2 & 4s_0^3 &5 s_0^4 \end{vmatrix}
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} = l'_0
$$
And
$$
f''_i(s_0) =
\begin{vmatrix} 0&0& 2 & 3\times2s_0 & 4\times3s_0^2 & 5\times4s_0^3 \end{vmatrix}
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} = l''_0
$$
where i is the index of the segment that contains the \(s_0\).
2.2 The end point constraints¶
Similar to the init point, the end point \((s_e, l_e)\) is known and should produce the same constraint as described in the init point calculations.
Combine the init point and end point, and show the equality constraint as:
$$
\begin{vmatrix}
1 & s_0 & s_0^2 & s_0^3 & s_0^4&s_0^5 \\
0&1 & 2s_0 & 3s_0^2 & 4s_0^3 & 5s_0^4 \\
0& 0&2 & 3\times2s_0 & 4\times3s_0^2 & 5\times4s_0^3 \\
1 & s_e & s_e^2 & s_e^3 & s_e^4&s_e^5 \\
0&1 & 2s_e & 3s_e^2 & 4s_e^3 & 5s_e^4 \\
0& 0&2 & 3\times2s_e & 4\times3s_e^2 & 5\times4s_e^3
\end{vmatrix}
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix}
=
\begin{vmatrix}
l_0\\
l'_0\\
l''_0\\
l_e\\
l'_e\\
l''_e\\
\end{vmatrix}
$$
2.3 Joint smoothness constraints¶
This constraint is designed to smooth the spline joint. Assume two segments \(seg_k\) and \(seg_{k+1}\) are connected, and the accumulated s of segment \(seg_k\) is \(s_k\). Calculate the constraint equation as:
$$
f_k(s_k) = f_{k+1} (s_0)
$$
Below are the steps of the calculation.
$$
\begin{vmatrix}
1 & s_k & s_k^2 & s_k^3 & s_k^4&s_k^5 \\
\end{vmatrix}
\cdot
\begin{vmatrix}
a_{k0} \\ a_{k1} \\ a_{k2} \\ a_{k3} \\ a_{k4} \\ a_{k5}
\end{vmatrix}
=
\begin{vmatrix}
1 & s_{0} & s_{0}^2 & s_{0}^3 & s_{0}^4&s_{0}^5 \\
\end{vmatrix}
\cdot
\begin{vmatrix}
a_{k+1,0} \\ a_{k+1,1} \\ a_{k+1,2} \\ a_{k+1,3} \\ a_{k+1,4} \\ a_{k+1,5}
\end{vmatrix}
$$
Then
$$
\begin{vmatrix}
1 & s_k & s_k^2 & s_k^3 & s_k^4&s_k^5 & -1 & -s_{0} & -s_{0}^2 & -s_{0}^3 & -s_{0}^4&-s_{0}^5\\
\end{vmatrix}
\cdot
\begin{vmatrix}
a_{k0} \\ a_{k1} \\ a_{k2} \\ a_{k3} \\ a_{k4} \\ a_{k5} \\ a_{k+1,0} \\ a_{k+1,1} \\ a_{k+1,2} \\ a_{k+1,3} \\ a_{k+1,4} \\ a_{k+1,5}
\end{vmatrix}
= 0
$$
Use \(s_0\) = 0 in the equation.
Similarly calculate the equality constraints for:
$$
f'_k(s_k) = f'_{k+1} (s_0)
\\
f''_k(s_k) = f''_{k+1} (s_0)
\\
f'''_k(s_k) = f'''_{k+1} (s_0)
$$
2.4 Sampled points for boundary constraint¶
Evenly sample m points along the path, and check the obstacle boundary at those points. Convert the constraint into QP inequality constraints, using:
$$
Ax \geq b
$$
First find the lower boundary \(l_{lb,j}\) at those points \((s_j, l_j)\) and \(j\in[0, m]\) based on the road width and surrounding obstacles. Calculate the inequality constraints as:
$$
\begin{vmatrix}
1 & s_0 & s_0^2 & s_0^3 & s_0^4&s_0^5 \\
1 & s_1 & s_1^2 & s_1^3 & s_1^4&s_1^5 \\
...&...&...&...&...&... \\
1 & s_m & s_m^2 & s_m^3 & s_m^4&s_m^5 \\
\end{vmatrix} \cdot \begin{vmatrix}a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix}
\geq
\begin{vmatrix}
l_{lb,0}\\
l_{lb,1}\\
...\\
l_{lb,m}\\
\end{vmatrix}
$$
Similarly, for the upper boundary \(l_{ub,j}\), calculate the inequality constraints as:
$$
\begin{vmatrix}
-1 & -s_0 & -s_0^2 & -s_0^3 & -s_0^4&-s_0^5 \\
-1 & -s_1 & -s_1^2 & -s_1^3 & -s_1^4&-s_1^5 \\
...&...-&...&...&...&... \\
-1 & -s_m & -s_m^2 & -s_m^3 & -s_m^4&-s_m^5 \\
\end{vmatrix}
\cdot
\begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix}
\geq
-1 \cdot
\begin{vmatrix}
l_{ub,0}\\
l_{ub,1}\\
...\\
l_{ub,m}\\
\end{vmatrix}
$$