Three-detector problem and Newell's method

Template:New unreviewed article Three-detector problem is the problem how to predict vehicle counts at some intermediate points based on the data from downstream and upstream detectors. Using the predicted counts, one can estimate the traffic state between two detectors in the condition of lack of middle detectors, but also can improve the method of incident diagnosis by comparing the observed and predicted data. So a realistic solution to this problem is important. Newell G.F. gave out a simplified method to solve this problem. In this Newell's method, one can get the cumulative counts curve (N-curve) of any intermediate points just by shifting N-curves of upstream and downstream. This method is further explained by Daganzo C.F.'s variational theory

A special case to demonstrate Newell's method

Assumption. In this special case, we use the triangular fundamental diagram (TFD) with three parameters: free flow speed $v_{f}$, wave velocity -w and maximum density $k_{j}$ (see Figure 1). Additionally, we consider a long study period where traffic past upstream detector (U) is unrestricted and traffic path downstream detector (D) is restricted so that waves from both boundaries point into the (t,x) solution space (see Figure 2).

File:Three detector problem 2.jpg

Figure 1 Triangular fundamental diagram

File:Three detector problem 1.jpg

Figure 2 Three detector problem

The goal of three-detector problem is calculating the vehicle at a generic point (P) on the "world line" of detector M (See Figure 2). Upstream. Since the upstream state is uncongested, there must be a characteristic with slope $v_{f}$ that reaches P from the upstream detector. Such a wave must be emitted $\tau _{1}=L_{U}/v_{f}$ times unit earlier, at point B on the figure. And since the vehicle number does not change along this characteristic, we see that the vehicle number at the M-detector calculated from conditions upstream is the same as that observed at the upstream detector $\tau _{1}$ time units earlier.Since $\tau _{1}$ is independent of the traffic state (it is a constant), this result is equivalent to shifting the smoothed N-curve of the upstream detector (curve U of Figure 3) to the right by an amount $\tau _{1}$.

Downstream. Likewise, since the state over the downstream detector is queued, there will be a wave reaching P from a location C and wave velocity $-w<0$. The change in vehicular label along this characteristic can be obtained with the moving observer construction of Figure 4, for an observer moving with the wave. In our particular case, the slanted line corresponding to the observer is parallel to the congested part of TFD. This means that the observer flow is independent of the traffic state and takes on the value: $k_{j}(-w)$. Therefore, in the time that it takes for the wave to reach the middle location, $\tau _{2}=L_{D}/(-w)$, the change in count is $\delta =k_{j}(-w)\tau _{2}=k_{j}L_{D}$; i.e., the change in count equals the number of vehicles that fit between M and D at jam density. This result is equivalent to shifting the D-curve to the right $\tau _{2}$ units and up $\delta$ units.

Actual count at M. In view of the Newell-Luke Minimum Principle, we see that the actual count at M should be the lower envelope of the U'- and D'-curves. This is the dark curves, M(t). The intersections of the U'- and D'- curves denote the shock's passages over the detector; i.e., the times when transitions between queued and unqueued states take place as the queue advances and recedes over the middle detector. The area between the U'- and M-curves is the delay experienced upstream of location M, trip times are the horizontal separation between curves U(t), M(t) and D(t), accumulation is given by vertical separations, etc.

Mathematical expression. In terms of the function N(t,x) and the detector location ($x_{u}$, $x_{m}$, $x_{d}$) as follows:

$N(t,x_{m})=min\{\ N(t-L_{U}/v_{f},x_{u})\ ,\ N(t+L_{D}/w,x_{d})+k_{j}L_{D}\ \}\qquad (1)$

where $L_{U}=x_{m}-x_{u}$ and $L_{D}=x_{d}-x_{m}$.

Basic principles of variational theory (VT)

Goal. Suppose we know the number of vehicles (N) along a boundary in a time-space region and we are looking for the number of vehicles at a generic point P (denoted as $N_{p}$) beyond that boundary in the direction of increasing time(see Figure 5).

Suppose an observer that starts moving from the boundary to point P along path L. We know the vehicle number the observer sees, $N_{L}$. We then break the path of the observer into small sections (such as the one show between A and B) and note that we also know the maximum number of vehicles that can pass the observer along that small section is, $C_{AB}$. The relative capacity formula tells us that it is: $C_{AB}=r(v^{0})\Delta {t}$. For TFD and using $v_{AB}$ for the slope of segment AB, $C_{AB}$ can be written as:

$C_{AB}=r(v_{AB})\Delta {t}=q_{0}\Delta {t}-k_{0}\Delta {x}=q_{0}(t_{B}-t_{A})-k_{0}(x_{B}-x_{A});for\ v_{AB}\in [-w,v_{f}]\qquad (2)$

So, if we now add the vehicle number on the boundary to the sum of all C_{AB} along path L we get an upper bound for N_p. This upper bound applies to any observer that moves with speeds in the range $[-w,v_{f}]$. Thus we can write:

$N_{P}\leq N_{L}+\sum _{L}(C_{AB}),\ v_{AB}\in [-w,v_{f}]\qquad (3)$

Equations (1) and (2) are based on the relative capacity constraint which itself follows from the conservation law.

Maximum principle. It states that $N_{P}$ is the largest possible value, subject to the capacity constraints. Thus the VT recipe is:

$N_{P}={min}_{L}\{N_{L}+\sum _{L}(C_{AB})\}\qquad (4)$

Equation (4) is a shortest path(i.e., calculus of variations) problem with $C_{AB}=r(v_{AB})$ as the cost function. It turns out that it produces the same solution as Kinematic wave theory.

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Figure 3 Newell's solution method

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Figure 4 Geometrical depiction of the relative flow seen by a moving observer

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Figure 5 Variational theory

Generalized solution

 Three steps:
 1. Find the minimum upstream count, $N_{U}$
 2. Find the minimum downstream count, $N_{D}$
 3. Choose the lower of the two, $N_{P}=min\{N_{U}\ ,\ N_{D}\}$

Step 1

All possible observer straight lines between the upstream boundary and point P have to constructed with observer speeds smaller than free flow speed:

$C_{QP}=q_{0}\Delta {t}-k_{0}\Delta {x}=q_{0}(t_{P}-t_{Q})-k_{0}(x_{P}-x_{Q})\qquad (5)$

where $\Delta {t}=t_{P}-t_{Q}={\frac {(x_{M}-x_{U})}{v_{QP}}}$ for $v_{QP}\in [0,v_{f}]$ and $\Delta {x}=x_{M}-x_{U}$

Thus we need to minimize ${N_{Q}+q_{0}(t_{P}-t_{Q})-k_{0}(x_{M}-x_{U})}$; i.e.,

$N_{U}={min}_{t_{Q}}\{N_{Q}+q_{0}(t_{P}-t_{Q})-k_{0}(x_{M}-x_{U})\}\qquad (6)$

Since $dN_{Q}/dt\leq q_{0}$, we see that the objective function is non-incresing and therefore $t_{Q}^{*}=t_{P_{1}}$. So Q should be placed at $P_{1}$ and we have:

$C_{QP}=C_{P_{1}P}=q_{0}({\frac {x_{M}-x_{U}}{v_{f}}})-k_{0}(x_{M}-x_{U})=0\qquad (7)$

Thus, $N_{U}=N_{P_{1}}$

Step 2

We have:$N_{D}=min_{Q^{'}}\{N_{Q^{'}}+C_{Q^{'}P}\}=min_{Q^{'}}\{N_{Q^{'}}+q_{0}\Delta {t}-k_{0}\Delta {x}\}$ So repeat the same steps we find that $N_{D}$ is minimized when $Q^{'}=P_{2}$. And at point $P_{2}$ we get:

$N_{D}=N_{P_{2}}+q_{0}({\frac {x_{D}-x_{M}}{w}})-k_{0}(x_{D}-x_{M})\qquad (8)$

Since the FD is triangular, ${\frac {q_{0}}{w}}+k_{0}=k_{j}$. Therefore (8) reduces to:

$N_{D}=N_{P_{2}}+(x_{D}-x_{M})k_{j}\qquad (9)$

Step 3

To get the solution we now choose the lower of $N_{U}$ and $N_{D}$.

$N_{P}=min\{N_{U}\ ,\ ND\}=min\{N_{P_{1}}\ ,\ N_{P_{2}}+(x_{D}-x_{M})k_{j}\}\qquad (10)$

This is Newell's the recipe for the 3-detector problem.

External links

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