Solving SVM without QP solver I've consulted three textbooks on support vector machines, and it is unclear to me how to solve such a problem without a QP solver (which is something my professor has asked us to to). Can anyone help me see the workings through a toy example?
Given toy data:
$$X = \begin{bmatrix} 0 & 0\\ 2 & 3\end{bmatrix} \qquad\qquad y = \begin{bmatrix} -1\\ 1\end{bmatrix}$$
My understanding proceeds as follows:
$$\min_w w^Tw \ \ \ \ \ s.t. \ \ y_i(w^Tx_i + b) \ge 1$$
Express the previous constrained problem as an unconstrained problem:
$$L(w,b,\alpha) = \frac{1}{2}w^Tw - \sum_i\alpha_i[y_i(w^Tx_i + b)-1]$$
Dual problem is equivalent to but easier than the primal problem:
$$\min_w  \max_{\alpha \ge 0} \frac{1}{2}w^Tw - \sum_{i}\alpha_i[y_i(w^Tx_i+b)-1] = \max_{\alpha \ge 0} \min_w \frac{1}{2}w^Tw - \sum_{i}\alpha_i[y_i(w^Tx_i+b)-1]$$
Start by minimizing $w$ in terms of $\alpha$:
$$\frac{\delta L}{\delta w} = 0 = w - \sum_i\alpha_iy_ix_i \implies w = \sum_i\alpha_iy_ix_i$$
Plug this value for $w$ into the Lagrangian from earlier:
$$L(w,b,\alpha) = \frac{1}{2}(\sum_i\alpha_iy_ix_i)(\sum_i\alpha_iy_ix_i) - \sum_ja_j[y_j[(\sum_i\alpha_iy_ix_i)x_j + b]-1]$$
Simplify:
$$= \frac{1}{2}\sum_{i,j}\alpha_i\alpha_jy_iy_jx_ix_j - \sum_{i,j}a_ia_jy_iy_jx_ix_j + b\sum_ia_iy_i+\sum_i\alpha_i$$
The third term equals zero because:
$$\frac{\delta L}{\delta b} = 0 = -\sum_i\alpha_iy_i$$
Ergo:
$$L(w,b,\alpha) = \sum_i\alpha_i - \frac{1}{2}\sum_{i,j}\alpha_i\alpha_jy_iy_jx_ix_j$$
Plug in $X$ and $Y$:
$$= a_i + a_2 - \frac{1}{2}[ (-1)(-1)a_1^2(0) + (-1)(1)a_1a_2(0) + (-1)(1)a_2a_1(0) + (1)(1)a_2^2(13) ]$$
$$= a_1 + a_2 - \frac{13}{2}a_2^2$$
I imagined that my goal now was to solve for values of $a_i$ that maximize the foregoing equation. I thought I would take partial derivatives with respect to $a_i$ and set the derivatives to zero, but that can't be right because I would end up with $\frac{\delta L}{\delta a_1} = 0 = 1$
 A: We have the quadratic program (QP) in $\mathrm w \in \mathbb R^2$ and $b \in \mathbb R$
$$\begin{array}{ll} \text{minimize} & w_1^2 + w_2^2\\ \text{subject to} & - (0 w_1 + 0 w_2 - b) \geq 1\\ & + (2 w_1 + 3 w_2 - b) \geq 1\end{array}$$
which can be written as follows
$$\begin{array}{ll} \text{minimize} & w_1^2 + w_2^2\\ \text{subject to} & b \geq 1\\ & 2 w_1 + 3 w_2 \geq b + 1\end{array}$$
If $b \geq 1$, then $b + 1 \geq 2$. Thus, 
$$\begin{array}{ll} \text{minimize} & w_1^2 + w_2^2\\ \text{subject to} & b \geq 1\\ & 2 w_1 + 3 w_2 \geq 2\end{array}$$
Decoupling $\mathrm w$ and $b$, we solve the QP in $\mathrm w$
$$\begin{array}{ll} \text{minimize} & w_1^2 + w_2^2\\ \text{subject to} & 2 w_1 + 3 w_2 \geq 2\end{array}$$
and then choose a $b \geq 1$. Rewriting the inequality constraint as an equality constraint, we then have the least-norm problem 
$$\begin{array}{ll} \text{minimize} & w_1^2 + w_2^2\\ \text{subject to} & 2 w_1 + 3 w_2 = 2\end{array}$$
whose solution is 
$$\bar{\mathrm w} := \dfrac{2}{13} \begin{bmatrix} 2\\ 3\end{bmatrix}$$
