## 3 Transport System Modelling and Optimization
### Nash Equilibrium  
In game theory, a Nash equilibrium refers to a non-cooperative game with two or more participants, assuming that each participant knows the strategy of the other participants. Every static game has at least one Nash equilibrium [59].  
The Nash equilibrium assumes that each transport system user independently makes a decision regarding the choice of the transport route, focusing on minimizing travel time or transport costs. This corresponds to a road traffic situation, in which a vehicle driver, having current information on the volume of traffic on the individual link of the transport network (\(i\), \(j\)) and the predicted travel time \(t^{p,ab}\big{(}x^{p,ab}\big{)}\) selects the route of the route of movement \(p\). Road situation can be obtained on an ongoing basis (on-line) from navigation applications [60,61] or intelligent systems [62].  
Traffic flows occurring at origin node \(a\in A\) are assumed to be displaced to destination node \(b\in B\), as described by Equation (8):  
\[x^{(a,b)}=\sum_{p\in P^{ab}}x^{p,ab},\ \forall(a,b)\in E. \tag{8}\]  
The sum of traffic flows entering node \(j\in W\) is equal to the sum of the traffic flows leaving that node, according to the relation (9):  
\[\sum_{i\in W}x_{ij}=\sum_{k\in W}x_{jk},\ \forall j\in W. \tag{9}\]  
The aggregate flow on the link (\(i\), \(j\)) is equal to the sum of the flows in all O-D pairs (\(a\), \(b\))\(\in E\) on all routs \(p\in P^{ab}\) containing link (\(i\), \(j\)), according to Equation (10) [11]:  
\[x_{ij}=\sum_{(a,b)\in E}\sum_{p\in P^{ab}}a_{ij}^{p,ab}\delta^{p,ab}=\sum_{(a, b)\in E}\sum_{p\in P^{ab}}x^{p,ab},\ \forall(i,j)\in L, \tag{10}\]  
Under the non-negativity condition of the traffic flow (11) [11]:  
\[x_{ij}\geq 0,\ \forall(i,j)\in L. \tag{11}\]  
The Nash equilibrium (user equilibrium) can also be interpreted as a potential stable point in a dynamic adjustment process in which individuals (road users) adapt their behaviour to that of road users, looking for strategies that give them better results [63].  
The system is in a state of equilibrium, if the average transport times for all possible travel routes in relation to individual movement O-D pairs are equal. Assuming that \(t^{p,ab}\) is the average movement time of the transport unit within the O-D pair (\(a\), \(b\)) on route \(p\), the Nash equilibrium equation takes the following form [27,50,64]:  
\[min\sum_{(i,j)\in L}\int_{0}^{x_{ij}}t_{ij}(x)\ dx, \tag{12}\]  
\[\forall(a,b)\in E,\ \forall p\in P^{ab}_{x>0}t^{p,ab}\big{(}x^{p,ab}\big{)}= \alpha^{ab}\wedge\alpha^{ab}\to min, \tag{13}\]  
where: \(E\)--set of movement pairs O-D within a transport system, \(P^{ab}\)--set of roads for movement pairs O-D (\(a\), \(b\)), \(x^{p,ab}\)--traffic flow volume on road \(p\) within a pair O-D (\(a\), \(b\)), \(t^{p,ab}\)--travel time of a traffic flow unit by route \(p\) within a pair O-D (\(a\), \(b\)), \(\alpha^{ab}\)--minimum travel time within a pair O-D (\(a\), \(b\)).  
If Nash equilibrium is achieved by the transport system, then any change in the path of movement by any of the traffic participants cannot be individually beneficial to them. In this situation, each traffic participant, guided by individual profit (minimum travel time), remains at his own choice.