About Equilibrium - Simplified bottleneck model

Model assumptions and scenario brief

   Travelers choose their own departure time to minimize their travel cost during the peak travel period , From the place of departure to the destination .

1. It is assumed that there are a fixed number of travelers during the peak travel period （ Such as morning rush hour ） Want to drive from the same starting point to the same destination （ Users in the same residential area have the travel demand to the same work area ）.
2. Assume that during peak hours , The road capacity of the travel road is insufficient to meet the travel needs of travelers , That is, there must be a period of road congestion .
3. Suppose there is only one road between the starting point and the destination .
4. Suppose all travelers drive at the same speed , All travelers have the same time value , Early arrivals to the work area 、 Late value （ It is assumed here that the perceived utility of travelers at home is higher than that at work , There will be a penalty for being late ）.
5. The number of vehicles per unit time on the travel road is fixed , That is, the unit time service rate of the travel road is a fixed value （ It is not considered that there are travel faults on the travel road , Such as rear end collision , Bad weather, etc ）.
   Travel scenario As shown below ：
   
   Early arrival and late arrival explain
   Schematic diagram of early travelers ： Take the first traveler for example
   

Schematic diagram of late travelers ： Take the last traveler as an example

Symbol definition

For the convenience of the later model description , Here we first define the symbols needed in the article ：

$N$： Total number of trips during the study period
$\alpha$： The traveler's unit time value （ Unit such as ： element / Hours ）
$\beta$ ： Unit early arrival cost
$\gamma$ ： Unit late cost
$t$ ： The departure time of the traveler
$T(t)$： Travelers in $t$ Total travel time of departure and arrival at destination at any time
$D(t)$ ：$t$ The queue length on the road at any time （ Cumulative number of vehicles ）
$s$ ： The number of vehicles per unit on the road （ The service rate of the travel road ）
$r(t)$：$t$ The departure rate of travelers at any time
$c(t)$： $t$ The total travel cost of travelers departing at any time
$t_{qb}$： The moment when queues begin to form on the road
$t_{qe}$： The moment when the queue dissipates on the road
$t^{\ast }$： When you expect to arrive at your destination
$\tilde{t}$： The departure time that can arrive at the expected time

Travelers in $t$ Time total travel time from departure to destination $T(t)$ Can be expressed as
$$T(t)=T_0+T^v(t),(1)$$ among ,$T_0$ Is a fixed constant , It indicates the travel time of the traveler from the starting place to the destination when there is no vehicle queue . When the traffic capacity of the travel road is insufficient to meet the travel needs of travelers ( namely $r(t)>s$) when , Cumulative vehicle queues will be generated on the travel road .$T^v(t)$ by $t$ The time for travelers to queue up on the road ,$T^v(t)$ It can be mathematically expressed as
$$T^v(t)=\frac{D(t)}{s},(2)$$ On the road $t$ The number of travel vehicles accumulated at any time shall be as of $t$ The number of vehicles entering the road at any time minus the total number of vehicles leaving , namely
$$D(t)={\int }_{t_{qb}}^{t}r(u)du-s(t-t_{qb}),(3)$$ The cumulative number of vehicles on the unit trip road is
$$\dot{D}(t)=r(t)-s.(4)$$ The total travel cost of travelers is the sum of the travel time cost of travelers and the utility cost of travelers arriving early or late , The mathematical representation is ：
$$c(t)=\alpha T(t)+\beta max\{0,t^{\ast }-t-T(t)\}+\gamma max\{0,t+T(t)-t^{\ast }\}.(5)$$ Empirical experience shows that $\gamma >\alpha >\beta$.

Bottleneck model Nash equilibrium

Equilibrium conditions ： No traveler can reduce the travel cost by changing their departure time .

Travelers' travel results are early or late , Under different travel results , The travel cost is expressed in different forms . because $T_0$ Is a fixed constant , It doesn't work for subsequent analysis , For the sake of simplicity , Set up $T_0=0$. Suppose the traveler starts from the starting point and directly enters the bottleneck section .

The departure rate of early travelers

For early travelers ,$max\{0,t^{\ast }-t-T(t)\}=t^{\ast }-t-T(t)$, $max\{0,t+T(t)-t^{\ast }\}=0$, The formula （5） It can be specifically expressed as
$$c(t)=\alpha T(t)+\beta (t^{\ast }-t-T(t)).(6)$$ When equilibrium is reached , The traveler changes the departure time , Travel costs will not change again , So in equilibrium there is
$$\frac{dc(t)}{dt}=0$$
The specific derivation process is ：
$$\begin{array}{rl}\dot{c}(t)& =\alpha \dot{T}(t)-\beta -\beta \dot{T}(t)\\ & =\frac{\alpha }{s}\dot{D}(t)-\beta -\frac{\beta }{s}\dot{D}(t)\\ & =\frac{\alpha -\beta }{s}(r(t)-s)-\beta \\ & =0\end{array}$$ Thus, the equilibrium departure rate can be obtained $r^{\ast }(t)$ by
$$r^{\ast }(t)=s+\frac{\beta s}{\alpha -\beta }(7)$$

The departure rate of late travelers

For late travelers ,$max\{0,t^{\ast }-t-T(t)\}=0$, $max\{0,t+T(t)-t^{\ast }\}=t+T(t)-t^{\ast }$, The formula （5） It can be specifically expressed as
$$c(t)=\alpha T(t)+\gamma (t+T(t)-t^{\ast }).(8)$$
It is similar to the derivation process of the equilibrium departure rate of early travelers , We can know the equilibrium departure rate of late arrival $r^{\ast }(t)$ by
$$r^{\ast }(t)=s-\frac{\gamma s}{\alpha +\gamma }(9)$$

Queue formation time and dispersion under the condition of equilibrium departure rate , Calculate the departure time according to the expected time

In the above simple bottleneck model, the cumulative departure and arrival conditions are shown in the following figure ：

among $B$ The point is that the last early traveler can be on time at the expected time $t^{\ast }$ terminus ad quem , The total waiting time is BC The length of the line segment . At the moment $\tilde{t}$ Travelers only have waiting time and no utility cost of being late or leaving early .
Whether you arrive early or late , The total number of travelers is $N$, The total length of the formed queue is equal to the total length of the dispersed queue , And $\tilde{t}$ Time is defined as from $\tilde{t}$ Set out at all times , Just in time to arrive at the destination , Then there are

$$(\tilde{t}-t_{qb})(s+\frac{\beta s}{\alpha -\beta })+(t_{qe}-\tilde{t})(s-\frac{\gamma s}{\alpha +\gamma })=N(10.1)$$$$(\tilde{t}-t_{qb})\frac{\beta s}{\alpha -\beta }=(t_{qe}-\tilde{t})\frac{\gamma s}{\alpha +\gamma }(10.2)$$$$\tilde{t}+\frac{\beta }{\alpha -\beta }(\tilde{t}-t_{qb})=t^{\ast }(10.3)$$
among
The formula (10.1) Indicates that the total number of travelers who can arrive early plus travelers who will arrive late is $N$;
The formula (10.2) It means that $\tilde{t}$ The maximum queue length is reached at any time , The length of the formed queue should be equal to the length of the dispersed queue ;
The formula (10.3) It means that $\tilde{t}$ To go through a period of waiting time , Just arrived at the destination .
stay $\tilde{t}$ The total number of vehicles entering at any time is $(s+\frac{\beta s}{\alpha -\beta })(\tilde{t}-t_{qb})$, The total number of vehicles driven out is $s(\tilde{t}-t_{qb})$, Therefore, the accumulated number of vehicles $$D(\tilde{t})=(s+\frac{\beta s}{\alpha -\beta })(\tilde{t}-t_{qb})-s(\tilde{t}-t_{qb})=\frac{\beta s}{\alpha -\beta }(\tilde{t}-t_{qb})$$
Then the waiting time $$T^v(t)=\frac{D(\tilde{t})}{s}=\frac{\beta }{\alpha -\beta }(\tilde{t}-t_{qb}),$$ There's a formula (10.3) establish . The simultaneous formula (10.1)-(10.3) It can be solved
$$t_{qb}=t^{\ast }-(\frac{\gamma }{\beta +\gamma })(\frac{N}{s})(11.1)$$$$t_{qe}=t^{\ast }+(\frac{\gamma }{\beta +\gamma })(\frac{N}{s})(11.2)$$$$\tilde{t}=t^{\ast }-(\frac{\beta \gamma }{\alpha (\beta +\gamma )})(\frac{N}{s})(11.3)$$

The equilibrium travel cost of travelers

Due to the arbitrariness of the traveler in the above expression , Therefore, the travel cost of all travelers is the same when the equilibrium state is reached , Take the travelers at the head of the queue as an example , At the moment $t_{qb}$ set out , Because it's in front of the queue , So it didn't go through the bottleneck period , Therefore, the waiting time is 0, The travel cost is only the utility cost of arriving early , namely
$$c(t)=\beta (t^{\ast }-t_{qb}).(12)$$ Bring in the formula (11.1) The equilibrium travel cost is
$$c^{\ast }(t)=(\frac{\beta \gamma }{\beta +\gamma })(\frac{N}{s}).(13)$$