# Why CLRS example on residual networks does not follows its formula?

I am learning algorithms to solve Maximum Flow problem by reading the CLRS book and confused by the following figure: That is:

A flow in a residual network provides a roadmap for adding flow to the original flow network. If $$f$$ is a flow in $$G$$ and $$f'$$ is a flow in the corresponding residual network $$G_f$$, we define $$f \uparrow f'$$, the augmentation of flow $$f$$ by $$f'$$, to be a function from $$V \times V$$ to $$R$$, defined by

$$(f \uparrow f')(u, v) = \begin{cases} f(u,v) + f'(u, v) - f'(v, u) & > \text{if (u,v) \in E} \\ 0 & \text{otherwise} \end{cases}$$

How the flow network in (c), for example $$(s, v_2)$$ got the flow 12 ? If we follow the formula, it must have a flow 5: $$8 + 5 - 8 = 5$$

That's not what the formula gives you. As the caption says, the capacity of the augmenting path in the residual network in (b) is $$4$$. Therefore we send 4 units of flow along the augmenting path from $$s$$ to $$t$$, namely, the path $$s \to v_2 \to v_3 \to t$$. In particular, $$f(s,v_2)=8$$, $$f'(s,v_2)=4$$, and $$f'(v_2,s)=0$$, so the updated flow is $$8+4-0=12$$.
The residual network $$G_f$$ with augmenting path $$p$$ shaded; its residual capacity is $$c_f(p)=c_f(v_2,v_3)=4$$.
Since the capacity of path $$p$$ is 4 (not 5), we find a flow $$f'$$ in the residual network $$G_f$$ that is defined by $$f'(s,v_2)=f'(v_2,v_3)=f'(v_3,t)=4$$. So for the network flow $$f\uparrow f'$$ in (c), we have $$(f\uparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$