We call a controller causal if at point t = tk its decision depends only on information from t = tk, non-causal otherwise.
Proposition: For the two objectives an optimal controller is non-causal.
Proof (waiting time): the proof is given by an example, by contradiction. Let us construct a case to show that the future should be known to the deterministic controller. The situation is given in figure 8. Note that the controller only knows if there are customers in a queue or not.
Figure 8: diagram of arriving customers
A causal controller decides to give first green to approach road A. The corresponding waiting time is w1. A non-causal controller would result in a waiting time w2.
Figure 9: different decisions
The following variables are introduced: a length of the queue of customers arriving together at side A; b the same as a for side B; z intermediate time (z = zAB = zBA here); d time difference between arrival times of customers A and B (d = a + z); k proportionality constant for scaling the waiting times
w1 = kb ( a + z – d )
w2 = ka ( d + b + z )
If the causal controller's decision is optimal, w2 > w1 must hold.
a (d + z) > b (z – d)
d (b + a) > z (b – a)
We choose b as a multiple of a, thus b = na with n >1. Then we obtain the solution, expressed by a quotient
Because d is not known, nor n, the inequality cannot hold in general
henceforth. 
Proof (throughput): also by an example; the duration v is calculated as follows (zAB ? zBA) :
v1 = a + b + zAB
v2 = a + b + zBA + d
Again we demand v2 > v1 to
be true for a causal decision to be optimal. It follows that zBA
+ d > zAB. The inequality cannot hold in general.
It can be shown [4] that for an optimal decision, also for d = 0 both a and b have to be known to the controller.
We have seen so far that it can be sufficient to know only a "bit" of the future for an optimal control, but in general the future over infinite time must be known. What remedy should be applied to tackle for the non-causality problem?
We will limit to a finite time horizon. The application shows that often the finite time horizon is not reached and that a deterministic decision can be made.
3.2 Lining upAs known in queuing theory, a server (a crossing) can become saturated (unstable for a finite time). The queue length of at least one approach road will increase. Saturation of a crossing leads to another problem: how to control a crossing such that the cycle time is minimal. It will be discussed in section 4.5.
Only during green and red phases customers can be processed. Each time an intermediate time is used, productive time is lost.
Proposition: changing phases must be avoided as much as possible.
Figure 10: a situation
Proof (waiting time): by contradiction. Light A is green and a block of customers with length a is passing it. It has already accumulated a certain amount kA of waiting time before green. At the same time a queue with length b waits in front of light B, having already a waiting time kB. Figure 10 shows the situation. We denote the waiting time of the first case with w1 (splitting up: interrupting processing a after the time e) and the other case with w2 (sequential: processing b after a). The variables are defined in the same way as before.
Figure 11: splitting up a queue or not
For the waiting times it holds (see figure 11):
w1 = kA + kB + zb + (a – e) (2z + b)
w2 = kA + kB + b (a – e + z)
Of course 0 = e < d + z and z > 0 hold. If breaking up queues is better, the following equations apply:
w1 < w2
kA + kB + zb + (a – e) (2z + b) < kA + kB + b (a – e + z)
zb + (a – e) (2z + b) < b (a – e) + zb
2z + b < b
z < 0
This contradicts the definition of z > 0.
Proof (throughput): the equations are:
v1 = e + z + b + z + (a - e)
v2 = a + z + b
Again if case 1 should be preferred, v1 < v2 holds and therefore 2 z < z. Only a z < 0
could meet this.
As a consequence herefrom green phases will become excessively long. To avoid this the Generating Strategy must be restricted by maximal green durations.