# A Mathematical Model for Efficient and Fair Resource Assignment in Multipath Transport
## 3 Mathematical Model
### Extension of the Objective Function for Fairness
#### 3.2.3 Network Flow Fair Allocation (NFF)
Based on these insights, the objective function is extended to Equation (15) in order to push the system towards multipath if flows are multipath capable:  
\[\text{NFF-objective-II:}\] \[\text{maximize}:\quad Alloc-\delta_{\text{all}}+M. \tag{15}\]  
\(M\) is the aggregated allocation of all multipath flows as defined in Equation (8). Though the objective function solves the above-mentioned problem for the scenario in Figure 4, adding \(M\) to the objective function might not be enough for ideal network flow fair allocation. Consider the network scenario in Figure 7. The network is composed of eight nodes \(\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e},\mathrm{f},\mathrm{m}\) and \(\mathrm{n}\). All link capacities are in Mbps. There are five flows \([\mathrm{e},\mathrm{a}],[\mathrm{a},\mathrm{d}],[\mathrm{b},\mathrm{f}],[ \mathrm{m},\mathrm{c}]\) and \([\mathrm{d},\mathrm{n}]\) inside the network. The ideal network flow fair solution would be flow \([\mathrm{e},\mathrm{a}]=10\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\), \([\mathrm{a},\mathrm{d}]=110\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\), \([\mathrm{b},\mathrm{f}]=10\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\), \([\mathrm{m},\mathrm{c}]=10\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\) and \([\mathrm{d},\mathrm{n}]=10\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\). Now, the value of the objective function NFF-objective-II can be calculated as follows:  
\[ \begin{array}{l}Alloc=10+110+10+10+10=150,\\ \delta_{all}=(110-10)+(110-10)+(110-10)+(110-10)=400,\\ M=110,\end{array} \]  
NFF-objective-II \(=150-400+110=-140\). The allocations of the flows are \([\mathrm{e},\mathrm{a}]=10\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\), \([\mathrm{a},\mathrm{d}]=10\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\), \([\mathrm{b},\mathrm{f}]=10\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\), \([\mathrm{m},\mathrm{c}]=10\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\) and \([\mathrm{d},\mathrm{n}]=10\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\) where multipath capable flow \([\mathrm{a},\mathrm{d}]\) does not utilize multipath then the value of the objective function NFF-objective-II is:  
\[ \begin{array}{l}Alloc=10+10+10+10=50,\\ \delta_{\text{all}}=0,\\ M=0,\end{array} \]  
NFF-objective-II \(=50\). This means the latter allocation where flow \([\mathrm{a},\mathrm{d}]\) does not utilize multipath is the optimum solution of the objective function NFF-objective-II for this scenario. If flow \([\mathrm{a},\mathrm{d}]\) was assigned to the \(100\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\) path via node \(\mathrm{c}\) in addition to the \(10\,\mathrm{M}\mathrm{b}\mathrm{p}\mathrm{s}\) path via node \(\mathrm{b}\), there would be a high allocation  
Figure 7: Example scenario for Network Flow Fair (NFF) with optimum allocation. Blue circles: nodes; black arrows/numbers: links with capacities; colored arrows/numbers: flows with capacity assignments. All numerical values in Mbps. difference between flow \([\mathrm{a},\mathrm{d}]\) and each of the other flows. The penalization of this high difference by the negative \(\delta_{\mathrm{all}}\) element in the objective function outweighs the multipath capacity gain \(M\). In order to tackle this problem, a multiplication factor \(|K|-1\) for the capacity gain is introduced, where \(|K|\) is the number of flows inside flow set \(K\) and \(M_{\mathrm{gain}}\) is the sum of the overall allocation gain due to multipath. The modified objective function is given in Equation (16):  
\[\text{\emph{NFF-objective-III:}}\] \[\text{\emph{maximize}}:\quad Alloc-\delta_{\mathrm{all}}+(|K|-1)M_{ \mathrm{gain}}. \tag{16}\]  
To formulate the equation sets for calculating \(M_{\mathrm{gain}}\), let \(max\_sf\_val^{st}\) be the maximum subflow allocation for the multipath flow \([s,t]\in K\). If the flow \([s,t]\in K\) is a single path flow, then \(max\_sf\_val^{st}=0\).