Lagrangian with inequalities

I have a toy question on SVM , where i have to find the weight $w$ by solving the Lagrangian multiplier method by hand . I know Lagrangain with equalities only . Here I have to deal with inequalities and I dont know how to proceed .

We have to minimize $$Q(\alpha)=\alpha_1+\alpha_2+\alpha_3+\alpha_4-\frac{1}{2}(\alpha_2^2-4\alpha_2\alpha_3+4\alpha_3^2+4\alpha_4^2)$$

subjected to $$\alpha_1+\alpha_2-\alpha_3-\alpha_4 = 0$$ and $$\alpha_i\ge0,\ \ \ \ \ i=1,2,3,4$$ .

• for personal background : i am from electrical background (junior), i had a course on calculus in my freshman year , there i studied lagrangian with equalities only . – abkds Feb 28 '14 at 0:58
• Can you check if your function needs to be maximized? – suresh Feb 28 '14 at 14:56

I believe this problem is unbounded. Let $\alpha_2=\alpha_3=0$. Then, your objective function reduces to:

$$\alpha_1+\alpha_4−2\alpha_4^2$$

$$\alpha_1-\alpha_4 = 0.$$
Note, since $\alpha_2=\alpha_3=0$ we also have $\alpha_2\geq 0$ and $\alpha_3\geq 0$. Now, set $\alpha_1=\alpha_4$ and let both go to infinity. We remain feasible $\alpha_1-\alpha_4=0$ and $\alpha_1\geq 0$ and $\alpha_4\geq 0$. However, the objective function goes to $-\infty$ since $-2\alpha_4^2$ grows to $-\infty$ much faster than $\alpha_1+\alpha_4$ grows to $\infty$.
Eliminate $\alpha_4$ using the constraint in the form of an equality but add a new inequality $\alpha_1+\alpha_2-\alpha_3\geq 0$ that replaces $\alpha_4\geq0$. Call the function $Q(\alpha)$ after the elimination of $\alpha_4$, $P(\alpha)$. The problem now reduces to minimizing $P(\alpha)$ in a region $R$ in the positive $(\alpha_1,\alpha_2,\alpha_3)$ octant bounded by the plane $\alpha_1+\alpha_2-\alpha_3=0$.
The next steps are as follows: (i) Solve for $dP=0$ and see if there are any solutions in the region $R$. (ii) Compare the value for any such solution that you obtain with values on the boundaries of $R$ and pick the minimum one. The boundaries of $R$ are the $(\alpha_i=0)\cap R$ for $i=1,2,3$ and $(\alpha_1+\alpha_2-\alpha_3=0)\cap R$.