Response to J. Richard Guadagno's comments on Jörg Schweizer's InTransSys Critique

by Jörg Schweizer

March, 2000 

I would like to thank J. Richard Guadagno for his response and additional data on the Integrated Transportation System (InTranSys). Nevertheless I see the need for clarification of a couple of issues, such that the interested reader can better spot common points and differences between my opinion and the one of Mr. Guadagno. In particular, I am still not sure whether the logistic problem is solved. Therefore, I propose a simple model in order to estimate the probability that a local system failure propagates across the network.

Let me first mention the common points:

The core of my critique can be summarized as follows:

There may be a couple of other problems that I cannot further consider because I have not enough information about the InTranSys design or, with which assumptions certain quantities have been determined. Two of these problems are: the high required construction energy of the entire system, air friction of the vehicles because of the large vehicle cross section. It further appears that InTranSys does not satisfy the brickwall criteria for speeds at 200 kilometers per hour with a throughput of 36,000 vehicles per hour. And if so, what is the maximum failure deceleration? I do not insist on the brickwall criteria but in order to compare capacities of different transportation systems, one has to apply the same safety criteria.

I will now present a simple model that allows one to quantify the logistics problem of InTranSys, such that the interested readers can make themselves a picture of how likely it is that a local failure can propagate and block large parts of the network.

First I define the traffic density on the InTranSys by means of the probability p that a time slot is occupied by a carrier. A time slot at a certain point in the network, is the time that a moving block (with or without carrier) needs to pass that point. A moving block includes the space for one carrier plus the necessary safety margins. If the system is operating at its capacity limits then all time slots are occupied by a carrier and p=1. If the system is operating at 50% of its capacity then, on the average, only half of the time slots are occupied and p=0.5. It is further assumed that the timeslots are occupied independently. This assumption is reasonable since the starting point and departure time are chosen independently by the system users. The system could pre-allocate timeslots for certain origin-destination pairs (whether they are used or not) but this would only result in an a priori limitation of capacity. Therefore, the possibility to occupy time-slots independently is an adequate assumption and, in fact, the logistics of InTranSys seems to take advantage of an independent slot allocation. If this should not be the case I wonder what the alternatives are. For sake of simplicity it is assumed that the time-slot occupation is equally distributed over the entire network, which is unlikely, if not impossible. However, it is straight forward to extend the presented theory and to use different probabilities pm for each considered merge process. The equal distribution assumption is actually influencing the results in favour of InTranSys.

Before the carrier can start (or resume) a trip, it needs to allocate such a time slot for all points along the desired path. Or, in other words, to book a block on the guideway that moves from origin to destination. The allocation is successful if it is assured that no other carrier will use this time slot during the entire trip. In the case where the allocation is not successful, the carrier must wait and check if a later time-slot is available. Whether the allocation is successful depends on the number of merge points on the trip. At a simple merge, the time-slots of two input-lines are merged into one time-slot on the output-line. Obviously, only one time-slot at one of the input lines can be occupied. When a carrier is approaching a merge point then the probability that the other input line is not occupied equals 1-p. At the second merge point, the probability that the time-slot of the other input line is not occupied again equals 1-p. Because the two merge events are independent, the probability that the time-slot at the first and the second merge points are free equals  (1-p)*(1-p)=(1-p)^2. Therefore, if a trip encounters m merge points, the probability of a successful allocation equals (1-p)^m.

This means in practice that one has to wait in average 1/(1-p)^m time slots before a trip across m merge points can be allocated. If the system is operating at 50% of its capacity and the planned trip crosses m=5 merge points then one has to wait an average of 32 time-slots. With m=10 merge points the average waiting time is already at 1024 time-slots. At traveling speeds of 100 kilometers per hour, a capacity of 14,400 vehicles per hour has been claimed, which is equivalent to a time slot of 0.25 seconds. Thus, one must expect a waiting time of 8 seconds for a trip with 5 merge points and 256 seconds (=4.27 minutes) for a trip with 10 merge points. This may be at the limits of acceptability but so far, there is no logistical problem.

The allocation problem may become more serious in the following situation: one branch of the network breakes down or gets blocked (in the following called the dead branch). As this event occurs almost every day with the railway network it may also happen to a nation-wide InTranSys network from time to time. There is nothing to worry about as long as the failure remains restricted to the dead branch of the network. This is why I try to determine the probability that the failure in the dead branch will propagate to other brances of the network. 

Let me first introduce n as the total number of vehicles that have already left their station at the time when the failure occured and that allocated time-slots within the dead branch. In other words, all vehicles upstream of the dead branch that are already on their way to pass through the dead branch. These are all the vehicles that need to be ``shunted off to stations'' or whose path needs to be rerouted (reallocated), where ``shunting off'' means simply a rerouting to the nearest station with available parking. Considering just one vehicle, vehicle i, we know from previous results that the probability of a successful allocation equals (1-p)^mi where mi is the number of merge points that vehicle i passes before reaching its desination station. Therefore, the probability P that the path for all vehicles (i=1..n) can be reallocated equals

P=(1-p)^m1*(1-p)^m2*... *(1-p)^mn

Please note that in the present case none of the vehicles can wait for the next free time slot. Either there are free time slots for all concerned vehicles or at least one vehicle needs to be stopped before it enters a merge with an already occupied time-slot. Clearly, P is the probability that none of the concerned vehicles needs to be stopped and 1-P is the probability that at least one vehicle needs to be stopped on the track, blocking another branch of the network. This, in return, would be a necessary condition for a propagation of the initial local failure.

Now, to get a "feeling" for the meaning of this equation, one can replace the variables by numbers to calculate the probability that another branch of the network will ``die''. If we take for example again p=0.5 (system is operating at 50% of its capacity) then 7200 vehicles would pass through a dead branch, which is blocked for one hour. Let us assume only 10% of these vehicles began their trip before the failure occurred (this is the amount of vehicles, coming from remoter origins; the other 90% would not start but remain in their stations until they find another time-slot with an alternative path). Anyway, 720 vehicles need to be rerouted on the fly or shunted off to an alternative station. Let us further assume that even 90% of these vehicle can be shunted of to a station without passing a further merge point, and the other 72 vehicles find a free station after passing only one merge point. Finally, with n=72 vehicles, p=0.5, mi=1 for all i=1...72 vehicles, the probability that all 72 vehicles find a free slot to arrive at the stations is

P= (0.5^1)^72 = 0.5^72 = 2.12*10^(-22)

which is appoximately zero, and the probability 1-P is quasi one. Therefore, it is almost sure that a failure in one branch will block at least one other branch (which will certainly block at least one other and so forth). This probability of failure propagation is only reasonably low if the traffic density is very low (p close to zero). This means InTranSys has a huge (theoretical) capacity, but for logistical reasons, it cannot take full advantage of it.

In conclusion, InTranSys, with its present design, may be efficient as container, car or bus transportation system in single line operation or with only few interchanges.


Last modified: August 13, 2002