Why if we don't use residual graph in Ford-Fulkerson to finding maximum flow give no any approximation of optimal solution?
With such a modification our algorithm do find a augmenting path and increase flow,continue until we have no augmenting path,I read many papers but I can't prove that without residual graph we have no any approximation of optimal solution.As the result suppose $F^*$ be optimal maximum flow,i want prove that this strategy give us $\alpha F^*$ such that $\alpha$ be a number in $0<\alpha\leq 1$.Any hint will be constructive.
The issue is making the "wrong" choices, in those cases the residual graph corrects them.
To understand the issue, consider the digraph with vertices $s,t,a,b$ and arcs $\overrightarrow{sa}, \overrightarrow{sb}, \overrightarrow{ab}, \overrightarrow{ta}, \overrightarrow{tb}$ of capacity $1$ each.
This graph has an obvious flow of value $2$ namely $1$ on the route $s \to a \to t$ and $1$ on the route $s \to b \to t$.
But, and here is the issue, what if you start by setting a flow of $1$ on the route $s \to a \to b \to t$. Try to increase it, and you will see that, without using the residual graph (or undoing some of the choices) you cannot.
This can be easily extrended the following way: add vertices $c,t'$ and arcs $\overrightarrow{sc}, \overrightarrow{tc}, \overrightarrow{ct'}$ of capacity $1$ and $\overrightarrow{tt'}$ of capacity $2$. Then, this netweork has a flow of 3, but the flow $s \to a \to b \to t \to c \to t'$ of value $1$ cannot be increased. And the idea of this step can be extended inductively, to create a network which admits a flow of value $n \in \mathbb N$, but which also has a flow of value 1 which cannot be increased (unless we use the residual graph/undo some choices).