$$ l = f_i(s) = a_{i0} + a_{i1} \cdot s + a_{i2} \cdot s^2 + a_{i3} \cdot s^3 + a_{i4} \cdot s^4 + a_{i5} \cdot s^5 (0 \leq s \leq d_{i}) $$
### 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:$$ \begin{vmatrix} -1 & -s_0 & -s_0^2 & -s_0^3 & -s_0^4&-s_0^5 \\ -1 & -s_1 & -s_1^2 & -s_1^3 & -s_1^4&-s_1^5 \\ ...&...-&...&...&...&... \\ -1 & -s_m & -s_m^2 & -s_m^3 & -s_m^4&-s_m^5 \\ \end{vmatrix} \cdot \begin{vmatrix} a_{i0} \\ a_{i1} \\ a_{i2} \\ a_{i3} \\ a_{i4} \\ a_{i5} \end{vmatrix} \geq -1 \cdot \begin{vmatrix} l_{ub,0}\\ l_{ub,1}\\ ...\\ l_{ub,m}\\ \end{vmatrix} $$