Figure 3: Freshwater flowrate vs flowrate of SR1 (a) Conflicting SR1 and SR2 (b) Non-conflicting SR1 and SR2
The source allocation pattern becomes slightly more complicated if more than two contaminants are involved. As established in the two-contaminant problem, the interception point where all impurity constraints are active represents the minimum freshwater point. This is similar to any problem with ’k’ contaminants [27].
Based on the geometric property of straight lines, if there exists a common intersection point among all ’k’ straight lines, then no other intersection points exist. The converse of the property is also true, i.e. If the intersection point for lines of contaminant pairs A-B, lines B-C/A-C are the same, then there exists a common intersection point [27]. This property is important as it determines whether a point exists where all impurity constraints are active. If such a point exists, then it represents the minimum freshwater point (if all the sources have enough flowrate). The common point is the minimum freshwater point, only if there is at least one conflicting sequence of the two sources. If there are no sources which have a conflicting sequence, the common point is not the minimum point.
However, not every source allocation problem has a common intersection point. To achieve optimum recycle with minimum fresh resource, the concentration or load limits for all contaminants for all sinks should be reached (if possible), or at least several limits are reached- see Chin et al. [27]. This complicates the problem as the source arrangement is not based on the concentration order any more. It becomes tedious if ones have to determine the source allocation amount manually. Section 2.4 is devoted to the explanation of the proposed heuristics to determine the number of sources to be allocated.

Network design: source allocation through the graphical plot

An efficient graphical targeting or design method for single contaminant material recycle/reuse problem is the cumulative load vs cumulative flowrate diagram [9]. This strategy is equivalent to sequential, fulfilling each sink with internal resources, which are prioritised with contaminant concentrations. Each segment of the line represents each sink/source, with a horizontal length of the line represents the flowrate and vertical length represents the contaminant load. The figures are usually plotted by compositing each sink/source line based on the ascending order of the contaminants’ concentration.
In the case of single contaminant, the recycle strategy first starts with the cleanest (highest quality) sink with the use of the cleanest source. The source line is moved horizontally (pure fresh resource) until it touches the sink line, and the source is located below the sink line – see Figure 4. The overlapped flow rate represents how many sources can be used for the sink. The remaining flow of the sink can only be fulfilled by fresh resource. The remaining flowrate of the source is transferred to the next sink – Figure 4. For the third sink, it can be fully fulfilled by the mix of the remainder of the second source and third source, without the use of fresh resource – Figure 4. Adding together the fresh resource flow, Figure 4 shows the typical representation of the Composite Curves in Load vs Flow diagram. The total freshwater target is so (F1 + F2), which is the fresh resource requirement.
In fact, the water network design with source allocation can already be determined from Figures 4a-c. For example, Figure 4a shows only part of source 1 can be allocated to sink 1. Figure 4b shows the allocation of sources 1 and 2 to sink 2. The benefits of determining the freshwater target for each sink sequentially allows for that sink the source allocation to be determined simultaneously.
If a similar strategy is applied in the multi-contaminant problem, slight adjustment on the Source CC has to be done. For illustrative purpose, Figure 5 shows that for the first sink, the limit for contaminant A is reached. There is still room for contaminant B for the first sink as its limit is