User-equilibrium model

User-equilibrium model of traffic assignment is based on the fact that humans choose a route so as to minimize his / her travel time and on the assumption that such a behaviour on the individual level creates an equilibrium at the system (or network) level. Flows on links (whose travel times are assumed to vary with flow) are said to be in equilibrium when no trip maker can improve his / her travel time by unilaterally shifting to another route. This notion of equilibrium flows is generally referred to as Wardrop's principle. Before presenting the model which can determine such equilibrium flows on a network, the idea of equilibrium flows or the concept of user-equilibrium needs to be explained further.

Consider, the example network shown in Figure (a). Figure (b) gives the travel time function for each of the three single link routes between the origin O and destination D. The total demand from O to D is 110.

図: Example problem on user-equilibrium assignment technique.

In the example, obviously if demand is less than or equal to 100, everybody will travel using Route 2 since the travel time offered by Route 2 (between 30 minutes and 40 minutes) will be less than any other route. However, when demand is in excess of 100, if everybody travels on Route 2, then the travel time on Route 2 will become greater than 40 minutes, implying that Route 1 will become more lucrative (offering about 40 minutes of travel time, since the volume on Route 1 is zero at this stage) and some can benefit by unilaterally shifting to Route 1. On doing so if the the volume carried by Route 2 falls below 100 then again Route 2 will become lucrative and some will shift to Route 2 from Route 1. Given this, and the fact the total demand is 110, it can be said that if 100 use Route 2 and 10 use Route 1 then each one will face a travel time of about 40 minutes and none can improve his / her travel time by unilaterally shifting from one route to another. This then will be the equilibrium flow to which the network has to eventually converge given that every traveler always tries to minimize his / her travel time. Notice that under the equilibrium condition Route 3 is not used as it offers more than 40 minutes of travel time.

The example, in addition to clarifying the concept of equilibrium also illustrates the fact that at equilibrium all routes which are used between a given origin and destination offer the same travel time which is less than all routes which are not used. Another interesting feature is that at equilibrium flow the sum of the areas under the travel time versus flow plots is the least. This feature can be easily seen by visually inspecting the sum of the areas shown in Figure . In the figure, (a) corresponds to the equilibrium flow, and the sum of the areas is equal to area ACDE + area ABFG. In the figure (b), corresponds to an instance of non-equilibrium flow conditions with Route 2 carrying all the flow (i.e., 110); in this case the relevant sum of the areas is simply equal to area HIJK. In the same figure, (c) represents another instance of non-equilibrium flow where Route 2 carries a flow of 90 and the rest 20 are carried by Route 1; in this case the relevant sum of areas is area LMNO + area LPQR. Finally, part (d) of the figure represents another instance of non-equilibrium flow conditions with Routes 1, 2, and 3 carrying flows of 30, 70, and 10, respectively; the relevant sum of areas in this case is area TUVW + area TXYZ + area T $\alpha \beta \gamma$. Note that the sum of the areas is the least in (a) -- the case representing equilibrium flow. The interested reader may try to verify this feature by constructing other such examples or by proving it mathematically.

図: Areas under travel-time versus flow plots under various flow distributions.

The user-equilibrium model formulation is based on this area related observation about equilibrium flows and the fact that the travel time on a route is simply a linear sum of the travel times on the constituent links. The formulation, which is a non-linear programming problem is given in Equation .

$$\min_{x_a} \sum_a \int_0^{x_a} \tau_a(\omega) d\omega$$

Subject to

$$\sum_k \Phi_k^{i,j} = t_{ij} \;\;\;\;\;\;\;\;\;\;\;\;\;\; \forall i, j$$

$$\sum_i \sum_j \sum_k \Phi_k^{i,j} \delta_{a,k}^{i,j} = x_a \;\;\;\;\;\;\;\;\;\;\;\;\;\; \forall a$$ (28)

$$\Phi_k^{i,j} \ge 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\; \forall i, j, k$$

where $\omega$ is a variable denoting flow. Non-linear programming problems of this kind (with linear constraints and a non-linear objective function) can be effectively solved using convex combination algorithms. The interested reader may refer to Taha [#!taha1!#] for more details on convex combination algorithms. Here only the principle is given so that the latter discussions on solving user-equilibrium problems are more meaningful to the reader.

Principles of convex combination methods for solving optimization problems with non-linear objective function and linear constraints

The convex combination algorithm determines the optimum point by proceeding in the following iterative manner.

• First, a feasible solution (i.e., a solution which satisfies all the constraints), say $f_1$, is determined.
• Next, another feasible solution, say $f_2$, is determined.
• Since all the constraints are linear (thereby giving rise to a convex solution space), it can be shown that any point, $f_3$ on a straight line joining $f_1$ and $f_2$ (note $f_3 = \alpha f_1 + (1 - \alpha) f_2$; where $0 \le \alpha \le 1$) is also feasible. The convex combination algorithm then determines the value of $\alpha$ which gives the best value for the objective function. Note that determination of the best value of $\alpha$, say $\alpha^*$ requires an unconstrained optimization of the objective function written in terms of $\alpha$.
• Once the value of $\alpha^*$ is determined, the point $f_3$ is precisely identified by using the value of $\alpha^*$ in place of $\alpha$ in the earlier equation for $f_3$.
• Next $f_3$ is set equal to $f_1$; also another feasible point is determined and $f_2$ is replaced with the new feasible point. The process goes back two steps and continues repeatedly till the objective function value changes little from one iteration to the next. The optimal solution is $f_1$ obtained in the last iteration.

Solving the mathematical programming formulation of the user-equilibrium model

As stated earlier, the convex combination algorithm is used to solve the mathematical programming formulation of the user-equilibrium model. Here the procedure is presented step-wise. For a better understanding of the procedure parallels are drawn with the steps presented in the earlier section.

Step 1:

Set iteration counter $n=1$. Determine a feasible solution by performing all-or-nothing assignment with $\tau_a(0)$ as the travel time on link $a$. Note that the feasibility conditions of the programming formulation will be met by performing the all-or-nothing assignment. Let this solution vector be $s_1^n$.

Step 2:

Update the travel time of the links using the link volumes obtained in the $s_1^n$.

Step 3:

With the updated travel time information perform an all-or-nothing assignment to obtain another feasible solution. Let this solution vector be $s_2^n$.

Step 4:

Determine $\alpha^{*,n}$ by finding that value of $\alpha$ which minimizes the following quantity

$$\sum_a \int_0^{[s_1^n + \alpha (s_2^n - s_1^n)]_a} \tau_a(\omega) d\omega$$

where, $[s_1^n + \alpha (s_2^n - s_1^n)]_a$ indicates the flow value for link ``$a$'' as per the solution vector $[s_1^n + \alpha (s_2^n - s_1^n)]$.

Step 5:

Obtain the new feasible point, $s_1^{n+1} = s_1^n + \alpha^{*,n} (s_2^n - s_1^n)$.

Step 6:

Check whether the value of the objective function (given in Equation  or in Step 4 here) has changed significantly. If it has not, then stop and report $s_1^{n+1}$ as the solution. Otherwise, set $n=n+1$ and continue from Step 2.

An example to illustrate how the solution procedure suggested here works is provided next.

Example

For the network shown in Figure  and 1000 trips per day from Node A to Node B determine the link flows using the user-equilibrium assignment technique. The link travel times, $\tau_a(x_a)$, are given by $\tau_a(x_a) = k_a(1+0.15(x_a/b_a)^4)$. The link number, the $k_a$ value and the $b_a$ value for a particular link are mentioned as $(\alpha , \beta , \gamma)$ on the links. Note that in this case each route has one link.

図: Road network for the example on user-equilibrium method of traffic assignment.

Solution

Table  provides the output of every step at each iteration. As per the table the flow on link $a$, $x_a$ is: $x_1 = 0$, $x_2 = 359$, $x_3 = 470$, and $x_4 = 171$. Note that all links which are used have approximately the same travel time and this is less than the travel time offerred by the unused link (in this case Link 1).

表: Solution of the example on user equilibrium model of traffic assignment.

$n$	Step		Links				Obj. fn.
			1	2	3	4	value
1	1	$\tau_a(0)$	35	10	20	25
		$s_1^n$	0	1000	0	0	1975.0
	2	$\tau_a(s_1^n)$	35	947	20	25
	3	$s_2^n$	0	0	1000	0
	4	$\alpha^{*,n}$	0.596
	5	$s_1^{n+1}$	0	404	596	0	197.0
	6		n=2; Go to Step 2
2	2	$\tau_a(s_1^n)$	35	35	35	25
	3	$s_2^n$	0	0	0	1000
	4	$\alpha^{*,n}$	0.161
	5	$s_1^{n+1}$	0	339	500	161	189.98
	6		n=3; Go to Step 2
3	2	$\tau_a(s_1^n)$	35	22.3	27.3	35.3
	3	$s_2^n$	0	1000	0	0
	4	$\alpha^{*,n}$	0.035
	5	$s_1^{n+1}$	0	362	483	155	189.44
	6		n=4; Go to Step 2
4	2	$\tau_a(s_1^n)$	35	26.1	26.3	25.3
	3	$s_2^n$	0	0	0	1000
	4	$\alpha^{*,n}$	0.020
	5	$s_1^{n+1}$	0	354	473	172	189.33
	6		n=5; Go to Step 2
5	2	$\tau_a(s_1^n)$	35	24.8	25.8	25.4
	3	$s_2^n$	0	1000	0	0
	4	$\alpha^{*,n}$	0.007
	5	$s_1^{n+1}$	0	359	470	171	189.33
	6		Stop; Report $s_1^{n+1}$
			as solution