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

figure 26.4

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: $8 + 5 - 8 = 5$


2 Answers 2


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 $8+4-0=12$.


It is explained in part (b) of the caption of Figure 26.4.

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 $$ (f\uparrow f')(v_2, v_3)=f(v_2,v_3)+f'(v_2,v_3) = 8+4=12.$$


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