Infeasible start Newton's method I am implementing infeasible start Newton's method from the information in the slides (slide 11 of the link) posted here. It requires us to calculate primal and dual Newton steps, denoted by, $\Delta x_{nt}$ and $\Delta v_{nt}$ from following expressions:
$$
\begin{align}
\begin{bmatrix}
\nabla^2 f(x) & A^T \\
A & 0
\end{bmatrix}
\begin{bmatrix}
\Delta x_{nt} \\
\Delta v_{nt}
\end{bmatrix}
= -\begin{bmatrix}
\nabla f(x) + A^Tv \\
Ax-b
\end{bmatrix}
\end{align}
$$ 
Now we have to calculate $\Delta x_{nt}$ and $\Delta v_{nt}$. $\Delta x_{nt}$ ca be calculated as $A\Delta x_{nt}=Ax-b$ but how do I calculate $\Delta v_{nt}$ since it involves $v$. I do not know from where do I get $v$. Can anybody help me in this regard?
EDIT: In this method, I have to calculate $||r(x,v)||$. Now $r(x,v)$ contains two parts $r_{primal}$ and $r_{dual}$. The problem is how do I calculate $||r(x,v)||$ if both are of different sizes. Do I add two norms or just take the dual norm into consideration? You can see the method here on slide 11.

This is a single linear system. You're solving one large linear system to find $\Delta v$ and $\Delta x$ simultaneously. You can simply start at $v=0$, then update $v=v+\alpha\Delta v$ at each iteration. $v$ is the Lagrange multiplier for the equality constraints in the underlying optimization problem.
You can also solve for $v+\Delta v$ simultaneously. After all, you can rewrite the linear system this way, too:
$$\begin{bmatrix} \nabla^2 f(x) & A^T \\ A & 0 \end{bmatrix} \begin{bmatrix} \Delta x \\ v + \Delta v \end{bmatrix} = - \begin{bmatrix} \nabla f(x) \\ A x - b \end{bmatrix}$$
A final point: you may not be able to take a full step $x+\Delta x$, due to implicit constraints on $x$ buried in $f(x)$. Instead, you will need to choose a step size $\alpha<1$ and update: $x\rightarrow x+\alpha \Delta x$. The slides should point out how best to compute the step size.
