5 The diagram shows a flow through a network of directed arcs. The amount that can flow in each arc is limited by its upper capacity, and the lower capacity of each arc is 0 . The labelled arrows by the arcs show how much more (excess capacity) and how much less (potential backflow) could flow in each arc, in litres per second, and the objective is to find the maximum flow from \(S\) to \(T\).
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1. How many litres per second are currently flowing along the route SACHT?
2. How many litres per second are currently flowing from \(S\) to \(T\) ?
3. What is the maximum by which the flow along the route SBGIT can be increased? Use the copy of the network in your answer book to show what happens to the labels on the arrows when the flow along this route is augmented by this amount.
4. Find the maximum amount that can flow through the network. Explain, by using a cut, how you know that your flow is a maximum. The network had vertices \(S , A , B , C , D , E , F , G , H , I\) and \(T\). The single vertex \(E\) is replaced by the pair of vertices \(E _ { 1 }\) and \(E _ { 2 }\), together with the \(\operatorname { arc } E _ { 1 } E _ { 2 }\).
5. What does the \(\operatorname { arc }\) joining \(E _ { 1 }\) and \(E _ { 2 }\) tell you about the flow at \(E\) ?
6. Complete the diagram in your answer book to show the flow resulting after part (iii) on a directed network with vertices \(S , A , B , C , D , E , F , G , H , I\) and \(T\).