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Understanding Max & Min Capacity Problems: A Flow Network Approach

loren avatar
loren
9 diapositives · 241 Views · Jul 23, 2014

This guide explains max and min capacity problems in flow networks, detailing how to calculate maximum and minimum flows from source (S) to sink (T). It describes methods to ensure the minimum capacity conditions are met before maximizing flow and explores the concepts of excess flow and backflow within networks. By following systematic procedures, including labeling arcs and monitoring respective flows, we can effectively analyze and solve flow distribution in complex networks while adhering to capacity constraints.

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Understanding Max & Min Capacity Problems: A Flow Network Approach
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  1. Max & Min Capacity Problems Max flow = Add the upper capacities of edges along the cut directed from S to T but subtract lower capacities of edges directed from T to S. Min Flow = Add the lower capacities of edges along the cut directed from S to T but subtract upper capacities of edges directed from T to S. If negative then min flow is zero. It can be a good idea to first ensure that all the minimums are satisfied before trying to maximise the flow

  2. Max & Min Capacity Problems L1 = 10 + 11 = 21 L2 = 14 – 3 + 9 = 20 L3 = 12 – 3 + 9 = 18 (CT, DE must be saturated and EB must have at least the min) So the maximum flow is 18 in order that the minimum flow condition is met. The labelling procedure is used as before but if the flow in an arc is reduced then it must not be reduced below the minimum value.

  3. Excess and Backflow Network Excess flow = additional flow that may be possible Back flow = current flow

  4. Before 11 CT, DE must be saturated from the min cut line and EB have at least the min After Consider a route which satisfies the minimum capacity and determine whether this can be increased. Path SBCT min flow = 9 excess capacity 10 = min{10, 14, 12}. So use a backflow of 10 SB is then saturated.

  5. Before 0 0 6 11 0 11 0 9 5 After 0 11 Path SDEBCT : excess flow = 2 along CT However EB must have at least the min according to the min cut line and hence a flow of 3 must be used. CT only has an excess of 2 and so path SBCT must be reduced by 1 .

  6. Before After Now add in Path SDEBCT Path SDEBCT with EB saturated excess capacity 3 = min{11, 9, 6, 5, 3}. CT is saturated and EB has the min required.

  7. 0 0 11 0 Path SDET excess capacity 6 = min{8, 6, 11}. DE saturated .

  8. Redraw the diagram with the back flows as forward flows Flow In = 18 Flow Out = 18

  9. Now try and do it by satisfying the minimum capacity first and then maximising the flow starting from a) SBCT b) SDET