In reality, there could well be a different distribution for each stop and for different times of the day. For our deterministic model, we can take a constant mean value for all passengers, say three stops. So, when N e ( I, J ) passengers board bus I at stop J, this gives rise to N 1 ( I, J + 3) = N e ( I, J ). The total number of passengers on board the bus at any time will be given by the sum Σ J N 1 ( I, J ) and we need to check that this does not exceed 60. We now have the beginnings of a simple model. We see that we need two arrays, namely T d ( I, J ) and N 1 ( I, J ), and clearly

For our convenience, we have assumed that no passengers arrive while a bus is actually at a stop. We have also assumed that we do not have buses overtaking each other. Since a study of overtaking was one of our objectives, we must build this feature into the model. As we have assumed a constant travelling time, any overtaking that occurs must happen at a stop. Once overtaking has occurred, the bus numbers become mixed up. We need an array P ( K ) such that P ( K ) is the original number of the bus which is currently K th in the sequence. Initially, we have P ( K ) = K for K = 1 to N b . The K th bus in the sequence will arrive at stop J before the ( K − 1)th bus has left if

Table 9.16

We call this ‘bunching’ and we can use a variable B to record the number of occurrences. When it happens, we can assume for convenience that all waiting passengers continue to board the bus in front so that N e ( P ( K ), J ) = 0. The condition for overtaking is that T d ( P ( K ), J ) < T d ( P ( K − 1), J ). (We must define P (0) = 0.) When it happens, we must interchange the number labels of the K th and ( K − 1)th buses, i.e. R = P ( K ), P ( K ) = P ( K − 1) and P ( K − 1) = R . Note that we assume no ‘double overtaking’.

In spite of all our simplifications, our model is already getting rather complicated and we have not yet considered the passengers! We need variables such as T ar ( I, J ) which is the time of arrival of the I th passenger currently at stop J, W ( I, J ) which is the waiting time of the I th passenger at stop J, and T j ( I, J ) which is the total journey time of the I th passenger at stop J and equals W ( I, J ) + boarding time + travel time + disembarking time.

For a more realistic stochastic model, we can use a Poisson process to generate the arrivals of passengers at a stop and to sample the journey lengths from a distribution such as that mentioned earlier. We can also introduce random delays into the journey times between stops.

Obtain the mathematical solution

A run of the simple deterministic model on a microcomputer using only four stops and four buses with intervals of 500 s between starting times from the depot and 60 s between passenger arrivals at any stop