I'm having a trouble understanding example problem 13.1 on page 371 of the 2nd edition of Nocedal & Wright:
\begin{aligned} \begin{equation} \min_x -4x_1 - 2x_2 \text{ s.t } \\ x_1 + x_2 + x_3 = 5 \\ 2x_1 + \tfrac {x_2}{2} + x_4 = 8 \\ x \geq 0 \end{equation} \end{aligned}
but ran into a calculation discrepancy described below that I thought might be preventing me from seeing how the update works in action:
NW assume that the initial basic basis is given by elements $\{3,4\}$ and the non-basic basis is given by elements $\{1,2\}$.
In NW's notation, they compute $$ s_N = \begin{bmatrix} s_1 \\ s_2 \end{bmatrix} = c_N - N^T \lambda = \begin{bmatrix} -3 \\ -2 \end{bmatrix} $$
However, I get $\lambda = (B^T)^{-1}c_B = I(0,0)^T$ so that $$ s_N = c_N - N^T \lambda = \begin{bmatrix} -4 \\ -2 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 1 & 1/2 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} -4 \\ -2 \end{bmatrix}. $$
I checked the errata but didn't see anything listed, so I'm confused about what I'm missing.
You are right, that does not make any sense. Another error in this example is that they claim that $\lambda=[-5/3; -2/3]$ found at the start of the third iteration is optimal, but it clearly is not since strong duality does not hold: $c^Tx = -4\cdot 11/3 - 2\cdot 4/3 = -52/3$, while $b^T\lambda = 5\cdot -5/3 + 8\cdot -2/3 = -41/3$. They report the correct optimal $x$, but $\lambda$ should be $[-4/3; -4/3]$.
If you replace $-4$ in the objective function with $-3$, the entire example makes sense.
The errata have not been updated for a while. You should contact the authors though, maybe they still update them.