I am looking at the following problem in operations research:
Suppose that a train is operated over two identical consecutive trips, where on each trip the train incurs an exponentially distributed disturbance with average $1/\lambda$. In order to cope with these disturbances, a time supplements $s_1$ and $s_2$ are to be allocated to the two trips and we denote the total supplement over both trips by $S=s_1 + s_2$. This problem is taken out of "Stochastic Improvement of Cyclic Railway Timetables" by Kroon et. al. 2008.
In the paper, we first calculate the probability that after the first trip, the train delay $D_1$ does not exceed d. This is straightforward, as this probability corresponds to the probability that the disturbance of the first trip does not exceed $s_1 + d$, which is just $$ P(D_1 \leq d) = 1 - e^{-\lambda(s_1 + d)}. $$
Now we want to give the probability that after the second trip, the delay $D_2$ of the train does not exceed $d$. According to the paper, this probability is given by: $$ P(D_2 \leq d) = 1 - e^{-\lambda(s_2 + d)} - \lambda(s_2+d)e^{-\lambda(S + d)}. $$ However, I have not been able to prove this formula.
I think that the probability that the sum of both disturbances does not exceed $S+d$ is given (per convolution of the two exponential distributions) by $$ P(Disturbance 1 + Disturbance 2 \leq S+d) = 1 - e^{-\lambda(S + d)} - \lambda(S+d)e^{-\lambda(S + d)}. $$ This is pretty close to the formula for $P(D_2 \leq d)$, so I might be on the right track, but does not correspond to $P(D_2 \leq d)$ because the delay after the second trip can be greater than $d$ even if the sum of the disturbances is not greater than $S+d$ (if the first trip has no disturbance, the second journey will still begin only after the first time supplement has passed, so if the disturbance of the second trip is bigger than $s_2+d$ but smaller than $S+d$, the delay $D_2$ will be greater than $d$ while the sum of the disturbances is smaller than $S+d$).
So, my idea for the formula was something like $$ P(D_2 \leq d) = 1 - P(Disturbance2 > s_2 + d) - P(Disturbance2 \leq s_2 + d \quad \wedge \quad Disturbance1 + Disturbance2 > S + d) $$ as the delay will always be greater than $d$ if the second trip disturbance is greater than $s_2 + d$, and if it isn't, but the sum of the disturbances exceeds $S+d$, the delay $D_2$ will also be greater than $d$. However, I have not been able to give an analytical expression of the last term here and I am not convinced that this would result in the formula even if I had.
Can anyone verify the formula of the paper and maybe give an idea on how to prove it / where I'm going wrong?
Thank you.
