4. Some results on the knock-on delays

a variance small enough and to use the property that the class of normal distributions is closed

Example 4. Let τ has the density (Eq. (8)), and all μ<sup>j</sup> have the same gamma density

<sup>ψ</sup>ðÞ¼ <sup>t</sup> I tð Þ <sup>&</sup>gt; <sup>0</sup> <sup>e</sup>�t=<sup>β</sup><sup>t</sup>

<sup>0</sup> x<sup>α</sup>�<sup>1</sup>e�xdx is gamma function. Put

One can show that in the example under consideration it follows from Eqs. (15) and (16) that

<sup>þ</sup> ae<sup>λ</sup>ð Þ <sup>t</sup>�t0þ<sup>b</sup> <sup>1</sup>

<sup>λ</sup>ð Þ <sup>t</sup>�t<sup>0</sup> <sup>1</sup>

" !#

1 þ λβ

α�1

1 þ λβ

<sup>y</sup> x<sup>α</sup>�<sup>1</sup>e�<sup>x</sup> dx is incomplete gamma function. Graphs of the distribution func-

" #

λ ¼ 0:3, t<sup>0</sup> ¼ 5, α ¼ 14, β ¼ 0:5: (18)

<sup>Γ</sup>ð Þ <sup>α</sup> <sup>β</sup><sup>α</sup> , (17)

� �ð Þ <sup>k</sup>�<sup>1</sup> <sup>α</sup> <sup>Γ</sup> <sup>α</sup>; <sup>1</sup> <sup>þ</sup> λβ � �ð Þ <sup>t</sup> � <sup>t</sup><sup>0</sup> <sup>þ</sup> <sup>b</sup> <sup>=</sup><sup>β</sup> � �

� �ð Þ <sup>k</sup>�<sup>1</sup> <sup>α</sup> <sup>Γ</sup> <sup>α</sup>; <sup>1</sup> <sup>þ</sup> λβ � �ð Þ <sup>t</sup> � <sup>t</sup><sup>0</sup> <sup>=</sup><sup>β</sup> � �

Γð Þ α

Γð Þ α

,

with respect to the convolution operation.

where <sup>α</sup> <sup>&</sup>gt; <sup>0</sup>, <sup>β</sup> <sup>&</sup>gt; 0, <sup>Γ</sup>ð Þ¼ <sup>α</sup> <sup>Ð</sup> <sup>∞</sup>

178 Probabilistic Modeling in System Engineering

where <sup>Γ</sup>ð Þ¼ <sup>α</sup>; <sup>y</sup> <sup>Ð</sup> <sup>∞</sup>

Wkð Þj <sup>t</sup> <sup>a</sup>¼1, <sup>b</sup>¼<sup>0</sup>

lim k!∞

¼ lim

WkðÞ¼ <sup>t</sup> I tð Þ <sup>&</sup>gt; <sup>t</sup><sup>0</sup> <sup>1</sup> � <sup>Γ</sup> <sup>α</sup>;ð Þ <sup>t</sup> � <sup>t</sup><sup>0</sup> <sup>þ</sup> <sup>b</sup> <sup>=</sup><sup>β</sup> � �

<sup>k</sup>!<sup>∞</sup> I tð Þ <sup>&</sup>gt; <sup>t</sup><sup>0</sup> <sup>1</sup> � <sup>Γ</sup> <sup>α</sup>;ð Þ <sup>t</sup> � <sup>t</sup><sup>0</sup> <sup>=</sup><sup>β</sup> � �

<sup>¼</sup> <sup>W</sup>∞ð Þ<sup>t</sup> <sup>≔</sup> I tð Þ <sup>&</sup>gt; <sup>t</sup><sup>0</sup> <sup>1</sup> � <sup>Γ</sup> <sup>α</sup>;ð Þ <sup>t</sup> � <sup>t</sup><sup>0</sup> <sup>=</sup><sup>β</sup> � �

Figure 5. Behavior of the functions Wkð Þt , k = 2, 3, 4, 5.

Γð Þ α

tions Wkð Þt , 2 ≤ k ≤ 5, with the parameters (Eq. (18)) are depicted in Figure 5.

Γð Þ α

Γð Þ α � �

It is not difficult to verify that for Wkð Þt from Example 4 the following formula holds:

þ e

:

Denote by N the random number of knock-on delays (within the framework of the model under consideration).

Lemma 1. For each fixed integer m, 1 ≤ m ≤ n � 1,

$$\mathbb{P}(N \ge m) = \mathbb{P}\left(\tau > \sum\_{j=2}^{m+1} \mu\_j\right). \tag{19}$$

Proof. Easily seen: f g <sup>N</sup> <sup>¼</sup> <sup>0</sup> <sup>¼</sup> <sup>t</sup><sup>0</sup> <sup>≤</sup> <sup>τ</sup> <sup>þ</sup> <sup>t</sup><sup>0</sup> <sup>≤</sup> <sup>T</sup>ð Þ<sup>2</sup> n o, Nf g <sup>¼</sup> <sup>m</sup> <sup>¼</sup> <sup>T</sup>ð Þ <sup>m</sup>þ<sup>1</sup> <sup>&</sup>lt; <sup>τ</sup> <sup>þ</sup> mt<sup>0</sup> <sup>≤</sup> <sup>T</sup>ð Þ <sup>m</sup>þ<sup>2</sup> � <sup>t</sup><sup>0</sup> n o,

<sup>m</sup> = 1, 2, …, <sup>n</sup> – 2, f g <sup>N</sup> <sup>¼</sup> <sup>n</sup> � <sup>1</sup> <sup>¼</sup> <sup>τ</sup> <sup>þ</sup> ð Þ <sup>n</sup> � <sup>1</sup> <sup>t</sup><sup>0</sup> <sup>&</sup>gt; <sup>T</sup>ð Þ <sup>n</sup> n o: This implies that

$$\mathbb{P}(N \ge m) = \mathbb{P}\left(\tau + mt\_0 > T^{(m+1)}\right) = \mathbb{P}\left(\tau > \sum\_{j=2}^{m+1} \mu\_j\right). \tag{7}$$

Here and below, the sign □ denotes the end of the proof.

The corollaries of this lemma are given below. Their proofs are simple and therefore we do not present them.

Corollary 5. If μ<sup>j</sup> ¼ T is a constant value, 2 ≤ j ≤ n, then for every fixed integer m, 1 ≤ m ≤ n � 1, we have the equality Ρð Þ¼ N ≥ m G mT ð Þ.

Corollary 6. If μ<sup>j</sup> ¼ T is a constant value, 2 ≤ j ≤ n, and τ is exponentially distributed with parameter λ, then for every fixed integer m, 1 ≤ m ≤ n � 1, the following equality holds,

$$\mathbf{P}(\mathbf{N} \ge m) = e^{-\lambda m T}.$$

Corollary 7. If μ2, …, μ<sup>n</sup> are independent identically distributed random variables with a density function ψ, then for every fixed integer m, 1 ≤ m ≤ n � 1, we have the equality

$$\mathbb{P}(N \succeq m) = \int\_{-\infty}^{\infty} \overline{G}(\mu) \psi^{\ast m}(\mu) d\mu.$$

In what follows, τ<sup>1</sup> ¼ τ is the delay duration of the first train, τk, k = 2, …, n, is the knock-on delay of the k-th train. The problem is to find the distribution functions GkðÞ¼ t Ρð Þ τ<sup>k</sup> < t , k = 2, 3, …, n. Note that the solution of this problem, which we call by the second problem, allows us to find the distribution of the deviations of the real arrival times from the planned ones.

In what follows, we will use the notation a ∨ b instead of maxð Þ a; b .

Theorem 3. The following formula holds:

$$
\tau\_k = \left(\tau\_{k-1} - \mu\_k\right) \lor 0, \ 2 \le k \le n. \tag{20}
$$

The dotted lines (lines 1 and 2) represent the scheduled trajectories of trains 1 and 2, solid lines (1 and 2) depict the real trajectories taking into account the delays. It can be seen that the

μ2, …, μ<sup>k</sup> have the same distribution function Ψð Þt . They are mutually independent. The ran-

GkðÞ¼ t I tð Þ > 0 G t þ μ<sup>k</sup>

ð∞ �∞

τ<sup>2</sup> ¼ τ � μ<sup>2</sup>

Proof. Let t > 0 be the time spent by the train on the path length (distance to the place, where an unplanned stop of the train 1 occurred). We show the equality τ<sup>2</sup> ¼ 0 holds under the condition τ ≤ μ2. The departure time of the train 1 after stopping is t þ τ. The time point when train 2 reaches s can be written as μ<sup>2</sup> þ t<sup>0</sup> þ t. The knock-on delay of train 2 will not occur, i.e., τ<sup>2</sup> ¼ 0, in the case, when the indicated time points are separated by the value r ≥ t0,

, 2 ≤ k ≤ n. As it follows from the assumption that the random variables

GkðÞ¼ <sup>t</sup> I tð Þ <sup>&</sup>gt; <sup>0</sup> <sup>Ρ</sup> <sup>τ</sup> � <sup>μ</sup><sup>k</sup> <sup>&</sup>lt; <sup>t</sup> � �, <sup>2</sup> <sup>≤</sup> <sup>k</sup> <sup>≤</sup> <sup>n</sup>: (22)

GkðÞ¼ t I tð Þ > 0 G tð Þ þ ð Þ k � 1 T : (24)

, 2 ≤ j ≤ n be independent identically distributed random variables with a continu-

, 2 ≤ j ≤ n be independent identically distributed random variables with a density

� �: (23)

Probabilistic Model of Delay Propagation along the Train Flow

http://dx.doi.org/10.5772/intechopen.75494

181

G tð Þ <sup>þ</sup> <sup>y</sup> <sup>d</sup>Ψ<sup>∗</sup>ð Þ <sup>k</sup>�<sup>1</sup> ð Þ<sup>y</sup> : (25)

, 2 ≤ j ≤ n and has a density function g tð Þ. Then GkðÞ¼ t

� � ∨ 0: (26)

arrival time of the train 1 differs from the schedule at τ and the train 2 on the τ2.

Denote <sup>μ</sup><sup>k</sup> <sup>¼</sup> <sup>P</sup><sup>k</sup>

Corollary 12. Let μ<sup>j</sup>

Corollary 13. Let μ<sup>j</sup>

�∞

Ð<sup>t</sup>þ<sup>y</sup>

I tð Þ <sup>&</sup>gt; <sup>0</sup> <sup>Ð</sup> <sup>∞</sup>

function ψð Þt . Let τ be independent of μ<sup>j</sup>

Lemma 2. The following formula is valid:

�<sup>∞</sup> g zð Þdz � �ψ<sup>∗</sup>ð Þ <sup>k</sup>�<sup>1</sup> ð Þ<sup>y</sup> dy.

5. Proof of Theorem 3 and its corollaries

<sup>j</sup>¼<sup>2</sup> <sup>μ</sup><sup>j</sup>

dom variable <sup>μ</sup><sup>k</sup> has the distribution function <sup>Ψ</sup><sup>∗</sup>ð Þ <sup>k</sup>�<sup>1</sup> ð Þ<sup>t</sup> .

Corollary 9. The distribution function of τ<sup>k</sup> has the following form:

Corollary 10. Let μ<sup>j</sup> > 0, 2 ≤ j ≤ n be some constant values. Then

Corollary 11. Let μ<sup>j</sup> ¼ T > 0, 2 ≤ j ≤ n be a constant value. Then

ous distribution function Ψð Þt . Let τ be independent of μ2, …, μn. Then

GkðÞ¼ t I tð Þ > 0

The next Corollaries 10 and 11 follow from Corollary 9 in an obvious way.

Corollary 8. The following formula holds:

$$
\pi\_k = \left(\tau - \sum\_{j=2}^k \mu\_j\right) \vee 0, \ 2 \le k \le n. \tag{21}
$$

It should be noted that within the framework of our model the deviation of the real arrival time from the planned one for k-th train coincides with τk, 1 ≤ k ≤ n. Figure 6 illustrates this statement.

Figure 6. Deviation arrival times from the schedule: delays τ<sup>1</sup> and τ2.

The dotted lines (lines 1 and 2) represent the scheduled trajectories of trains 1 and 2, solid lines (1 and 2) depict the real trajectories taking into account the delays. It can be seen that the arrival time of the train 1 differs from the schedule at τ and the train 2 on the τ2.

Denote <sup>μ</sup><sup>k</sup> <sup>¼</sup> <sup>P</sup><sup>k</sup> <sup>j</sup>¼<sup>2</sup> <sup>μ</sup><sup>j</sup> , 2 ≤ k ≤ n. As it follows from the assumption that the random variables μ2, …, μ<sup>k</sup> have the same distribution function Ψð Þt . They are mutually independent. The random variable <sup>μ</sup><sup>k</sup> has the distribution function <sup>Ψ</sup><sup>∗</sup>ð Þ <sup>k</sup>�<sup>1</sup> ð Þ<sup>t</sup> .

Corollary 9. The distribution function of τ<sup>k</sup> has the following form:

Ρð Þ¼ N ≥ m

τ<sup>k</sup> ¼ τ<sup>k</sup>�<sup>1</sup> � μ<sup>k</sup>

<sup>τ</sup><sup>k</sup> <sup>¼</sup> <sup>τ</sup> �<sup>X</sup>

0 @

k

1

j¼2 μj

It should be noted that within the framework of our model the deviation of the real arrival time from the planned one for k-th train coincides with τk, 1 ≤ k ≤ n. Figure 6 illustrates this statement.

In what follows, we will use the notation a ∨ b instead of maxð Þ a; b .

Theorem 3. The following formula holds:

180 Probabilistic Modeling in System Engineering

Corollary 8. The following formula holds:

Figure 6. Deviation arrival times from the schedule: delays τ<sup>1</sup> and τ2.

ð∞ �∞

In what follows, τ<sup>1</sup> ¼ τ is the delay duration of the first train, τk, k = 2, …, n, is the knock-on delay of the k-th train. The problem is to find the distribution functions GkðÞ¼ t Ρð Þ τ<sup>k</sup> < t , k = 2, 3, …, n. Note that the solution of this problem, which we call by the second problem, allows us to find the distribution of the deviations of the real arrival times from the planned ones.

G uð Þψ<sup>∗</sup><sup>m</sup>ð Þ <sup>u</sup> du:

� � ∨ 0, 2 ≤ k ≤ n: (20)

A ∨ 0, 2 ≤ k ≤ n: (21)

$$\mathcal{G}\_k(t) = I(t > 0)\mathcal{P}\left(\tau - \overline{\mu}\_k < t\right), \ 2 \le k \le n. \tag{22}$$

The next Corollaries 10 and 11 follow from Corollary 9 in an obvious way.

Corollary 10. Let μ<sup>j</sup> > 0, 2 ≤ j ≤ n be some constant values. Then

$$G\_k(t) = I(t>0)G\left(t+\overline{\mu}\_k\right). \tag{23}$$

Corollary 11. Let μ<sup>j</sup> ¼ T > 0, 2 ≤ j ≤ n be a constant value. Then

$$\mathbf{G}\_k(t) = I(t>0)\mathbf{G}(t+(k-1)T). \tag{24}$$

Corollary 12. Let μ<sup>j</sup> , 2 ≤ j ≤ n be independent identically distributed random variables with a continuous distribution function Ψð Þt . Let τ be independent of μ2, …, μn. Then

$$\mathcal{G}\_k(t) = I(t > 0) \int\_{-\infty}^{\infty} G(t + y) d\Psi^{\*(k-1)}(y). \tag{25}$$

Corollary 13. Let μ<sup>j</sup> , 2 ≤ j ≤ n be independent identically distributed random variables with a density function ψð Þt . Let τ be independent of μ<sup>j</sup> , 2 ≤ j ≤ n and has a density function g tð Þ. Then GkðÞ¼ t I tð Þ <sup>&</sup>gt; <sup>0</sup> <sup>Ð</sup> <sup>∞</sup> �∞ Ð<sup>t</sup>þ<sup>y</sup> �<sup>∞</sup> g zð Þdz � �ψ<sup>∗</sup>ð Þ <sup>k</sup>�<sup>1</sup> ð Þ<sup>y</sup> dy.

#### 5. Proof of Theorem 3 and its corollaries

Lemma 2. The following formula is valid:

$$
\pi\_2 = \left(\pi - \mu\_2\right) \lor 0. \tag{26}
$$

Proof. Let t > 0 be the time spent by the train on the path length (distance to the place, where an unplanned stop of the train 1 occurred). We show the equality τ<sup>2</sup> ¼ 0 holds under the condition τ ≤ μ2. The departure time of the train 1 after stopping is t þ τ. The time point when train 2 reaches s can be written as μ<sup>2</sup> þ t<sup>0</sup> þ t. The knock-on delay of train 2 will not occur, i.e., τ<sup>2</sup> ¼ 0, in the case, when the indicated time points are separated by the value r ≥ t0,

Figure 7. Two traffic scenarios: (a) lack of the knock-on delay; (b) knock-on delay occurs.

i.e., r ¼ μ2þ t<sup>0</sup> þ t � ð Þ t þ τ ≥ t0, or, which is the same thing, τ ≤ μ2. The considered case is illustrated in Figure 7a.

The knock-on delay of the duration τ<sup>2</sup> ¼ τ � μ<sup>2</sup> will occur when τ > μ2. Indeed, since trains after a random stop depart simultaneously, then the equality t þ τ ¼ μ<sup>2</sup> þ t<sup>0</sup> þ ð Þþ t � t<sup>0</sup> τ<sup>2</sup> holds, i.e., τ ¼ μ<sup>2</sup> þ τ2. The case under consideration is illustrated in Figure 7b. Thus, the validity of Eq. (26) is shown. □

Proof of Theorem 3. We shall use the method of mathematical induction. The equality (Eq. (20)) for k = 2 is established by Lemma 2. Let Eq. (20) be satisfied. We show that:

$$
\pi\_{k+1} = \left(\pi\_k - \mu\_{k+1}\right) \lor 0, 2 \le k+1 \le n. \tag{27}
$$

Case 2. If the k-th train is delayed, then (k + 1)-th one will not be delayed (τ<sup>k</sup>þ<sup>1</sup> ¼ 0) only if <sup>τ</sup><sup>k</sup> <sup>≤</sup> <sup>μ</sup><sup>k</sup>þ<sup>1</sup>. Case 2 is illustrated in Figure 9b. Note that if the knock-on delay of the <sup>k</sup>-th train occurs, a conflict of the k-th train with (k + 1)-th is described similar to the interaction of trains 1 and 2 (see Lemma 2). All described cases lead to Eq. (20). □ Proof of Corollary 8. We indicate that Eq. (21) is similar to Eq. (20). According to the statement of

Using the method of mathematical induction and taking into account that <sup>μ</sup><sup>k</sup>�<sup>1</sup> <sup>þ</sup> <sup>μ</sup><sup>k</sup> <sup>¼</sup> <sup>μ</sup>k, we obtain Eq. (21) from Eq. (28). □

Figure 9. Two traffic scenarios: (a) the case, when <sup>τ</sup><sup>k</sup>�<sup>1</sup> <sup>&</sup>gt; <sup>μ</sup>k, <sup>τ</sup><sup>k</sup> <sup>&</sup>gt; <sup>μ</sup><sup>k</sup>þ<sup>1</sup>: all trains up to (<sup>k</sup> + 1)-th are detained; (b) the case,

<sup>∨</sup> <sup>0</sup>, <sup>τ</sup><sup>k</sup> <sup>¼</sup> <sup>τ</sup><sup>k</sup>�<sup>1</sup> � <sup>μ</sup><sup>k</sup>

∨ 0: (28)

Probabilistic Model of Delay Propagation along the Train Flow

http://dx.doi.org/10.5772/intechopen.75494

183

Theorem 3, we have:

τ<sup>2</sup> ¼ τ � μ<sup>2</sup>

when <sup>τ</sup><sup>k</sup>�<sup>1</sup> <sup>&</sup>gt; <sup>μ</sup>k, <sup>τ</sup><sup>k</sup> <sup>≤</sup> <sup>μ</sup><sup>k</sup>þ<sup>1</sup>: all trains from (<sup>k</sup> + 1)-th up to <sup>n</sup>-th are not delayed.

<sup>∨</sup> <sup>0</sup>, <sup>τ</sup><sup>3</sup> <sup>¼</sup> <sup>τ</sup><sup>2</sup> � <sup>μ</sup><sup>3</sup>

Figure 8. The case, when τ<sup>k</sup>�<sup>1</sup> ≤ μk: knock-on delays of k-th and consecutive trains are not observed.

It follows from the inductive hypothesis that τ<sup>k</sup> ¼ 0 under the condition τ<sup>k</sup>�<sup>1</sup> ≤ μk. But if the delay of the k-th train is 0, then the next train does not undergo any delay, that is, τ<sup>k</sup>þ<sup>1</sup> ¼ 0. The present case is illustrated in Figure 8.

In the case, when τ<sup>k</sup>�<sup>1</sup> > μk, a knock-on delay of the k-th train occurs and equals to τ<sup>k</sup> ¼ τ<sup>k</sup>�<sup>1</sup> � μ<sup>k</sup> (according to the inductive hypothesis). Further, two cases are possible: either (1) a delay τ<sup>k</sup> entails a delay τ<sup>k</sup>þ1, or (2) τ<sup>k</sup>þ<sup>1</sup> ¼ 0.

Case 1. If the <sup>k</sup>-th train is delayed, then (<sup>k</sup> + 1)-th one will be delayed only if <sup>τ</sup><sup>k</sup> <sup>&</sup>gt; <sup>μ</sup><sup>k</sup>þ<sup>1</sup>, and its delay duration is <sup>τ</sup><sup>k</sup>þ<sup>1</sup> <sup>¼</sup> <sup>τ</sup><sup>k</sup> � <sup>μ</sup><sup>k</sup>þ<sup>1</sup> (this fact follows from the equality of the moments of departure of the <sup>k</sup>-th and (<sup>k</sup> + 1)-th trains after an unscheduled stop: <sup>T</sup>ð Þ<sup>k</sup> <sup>þ</sup> <sup>t</sup> � ð Þ <sup>k</sup> � <sup>1</sup> <sup>t</sup><sup>0</sup> <sup>þ</sup> <sup>τ</sup><sup>k</sup> <sup>¼</sup> <sup>T</sup>ð Þ <sup>k</sup>þ<sup>1</sup> <sup>þ</sup> <sup>t</sup> � kt<sup>0</sup> <sup>þ</sup> <sup>τ</sup><sup>k</sup>þ1). Case 1 is illustrated in Figure 9a.

Figure 8. The case, when τ<sup>k</sup>�<sup>1</sup> ≤ μk: knock-on delays of k-th and consecutive trains are not observed.

i.e., r ¼ μ2þ t<sup>0</sup> þ t � ð Þ t þ τ ≥ t0, or, which is the same thing, τ ≤ μ2. The considered case is

The knock-on delay of the duration τ<sup>2</sup> ¼ τ � μ<sup>2</sup> will occur when τ > μ2. Indeed, since trains after a random stop depart simultaneously, then the equality t þ τ ¼ μ<sup>2</sup> þ t<sup>0</sup> þ ð Þþ t � t<sup>0</sup> τ<sup>2</sup> holds, i.e., τ ¼ μ<sup>2</sup> þ τ2. The case under consideration is illustrated in Figure 7b. Thus, the validity of Eq. (26) is shown. □ Proof of Theorem 3. We shall use the method of mathematical induction. The equality (Eq. (20))

It follows from the inductive hypothesis that τ<sup>k</sup> ¼ 0 under the condition τ<sup>k</sup>�<sup>1</sup> ≤ μk. But if the delay of the k-th train is 0, then the next train does not undergo any delay, that is, τ<sup>k</sup>þ<sup>1</sup> ¼ 0. The

In the case, when τ<sup>k</sup>�<sup>1</sup> > μk, a knock-on delay of the k-th train occurs and equals to τ<sup>k</sup> ¼ τ<sup>k</sup>�<sup>1</sup> � μ<sup>k</sup> (according to the inductive hypothesis). Further, two cases are possible: either (1) a

Case 1. If the <sup>k</sup>-th train is delayed, then (<sup>k</sup> + 1)-th one will be delayed only if <sup>τ</sup><sup>k</sup> <sup>&</sup>gt; <sup>μ</sup><sup>k</sup>þ<sup>1</sup>, and its delay duration is <sup>τ</sup><sup>k</sup>þ<sup>1</sup> <sup>¼</sup> <sup>τ</sup><sup>k</sup> � <sup>μ</sup><sup>k</sup>þ<sup>1</sup> (this fact follows from the equality of the moments of departure of the <sup>k</sup>-th and (<sup>k</sup> + 1)-th trains after an unscheduled stop: <sup>T</sup>ð Þ<sup>k</sup> <sup>þ</sup> <sup>t</sup> � ð Þ <sup>k</sup> � <sup>1</sup> <sup>t</sup><sup>0</sup> <sup>þ</sup> <sup>τ</sup><sup>k</sup> <sup>¼</sup>

<sup>∨</sup> <sup>0</sup>, <sup>2</sup> <sup>≤</sup> <sup>k</sup> <sup>þ</sup> <sup>1</sup> <sup>≤</sup> <sup>n</sup>: (27)

for k = 2 is established by Lemma 2. Let Eq. (20) be satisfied. We show that:

Figure 7. Two traffic scenarios: (a) lack of the knock-on delay; (b) knock-on delay occurs.

<sup>τ</sup><sup>k</sup>þ<sup>1</sup> <sup>¼</sup> <sup>τ</sup><sup>k</sup> � <sup>μ</sup><sup>k</sup>þ<sup>1</sup>

illustrated in Figure 7a.

182 Probabilistic Modeling in System Engineering

present case is illustrated in Figure 8.

delay τ<sup>k</sup> entails a delay τ<sup>k</sup>þ1, or (2) τ<sup>k</sup>þ<sup>1</sup> ¼ 0.

<sup>T</sup>ð Þ <sup>k</sup>þ<sup>1</sup> <sup>þ</sup> <sup>t</sup> � kt<sup>0</sup> <sup>þ</sup> <sup>τ</sup><sup>k</sup>þ1). Case 1 is illustrated in Figure 9a.

Case 2. If the k-th train is delayed, then (k + 1)-th one will not be delayed (τ<sup>k</sup>þ<sup>1</sup> ¼ 0) only if <sup>τ</sup><sup>k</sup> <sup>≤</sup> <sup>μ</sup><sup>k</sup>þ<sup>1</sup>. Case 2 is illustrated in Figure 9b. Note that if the knock-on delay of the <sup>k</sup>-th train occurs, a conflict of the k-th train with (k + 1)-th is described similar to the interaction of trains 1 and 2 (see Lemma 2). All described cases lead to Eq. (20). □

Proof of Corollary 8. We indicate that Eq. (21) is similar to Eq. (20). According to the statement of Theorem 3, we have:

$$
\tau\_2 = \left(\tau - \mu\_2\right) \lor 0, \ \tau\_3 = \left(\tau\_2 - \mu\_3\right) \lor 0, \ \tau\_k = \left(\tau\_{k-1} - \mu\_k\right) \lor 0. \tag{28}
$$

Using the method of mathematical induction and taking into account that <sup>μ</sup><sup>k</sup>�<sup>1</sup> <sup>þ</sup> <sup>μ</sup><sup>k</sup> <sup>¼</sup> <sup>μ</sup>k, we obtain Eq. (21) from Eq. (28). □

Figure 9. Two traffic scenarios: (a) the case, when <sup>τ</sup><sup>k</sup>�<sup>1</sup> <sup>&</sup>gt; <sup>μ</sup>k, <sup>τ</sup><sup>k</sup> <sup>&</sup>gt; <sup>μ</sup><sup>k</sup>þ<sup>1</sup>: all trains up to (<sup>k</sup> + 1)-th are detained; (b) the case, when <sup>τ</sup><sup>k</sup>�<sup>1</sup> <sup>&</sup>gt; <sup>μ</sup>k, <sup>τ</sup><sup>k</sup> <sup>≤</sup> <sup>μ</sup><sup>k</sup>þ<sup>1</sup>: all trains from (<sup>k</sup> + 1)-th up to <sup>n</sup>-th are not delayed.

Proof of Corollary 9. It follows from Corollary 8 that τ<sup>k</sup> ¼ 0 if τ ≤ μ<sup>k</sup> (see, e.g., Figure 8), and τ<sup>k</sup> ¼ τ � μ<sup>k</sup> if τ > μ<sup>k</sup> (see, e.g., Figure 9a). Using the law of total probability, we obtain the following chain of equalities:

$$\begin{split} \mathcal{G}\_{k}(t) &= \mathcal{P}(\tau\_{k} < t) = I(t > 0)\mathcal{P}(\tau\_{k} < t) \\ &= I(t > 0) \left( \mathcal{P}\{\tau\_{k} < t | \tau \le \overline{\mu}\_{k}\} \mathcal{P}\{\tau \le \overline{\mu}\_{k}\} + \mathcal{P}\{\tau\_{k} < t | \tau > \overline{\mu}\_{k}\} \mathcal{P}\{\tau > \overline{\mu}\_{k}\} \right) \\ &= I(t > 0)\mathcal{P}\{\tau - \overline{\mu}\_{k} \le 0\} + I(t > 0)\mathcal{P}\{0 < \tau - \overline{\mu}\_{k} < t\} = I(t > 0)\mathcal{P}\{\tau - \overline{\mu}\_{k} < t\}. \qquad \bigtriangleup \end{split}$$

where 0 ≤ a ≤ 1, b ≥ 0, and λ > 0 are some parameters. Such distribution function is considered, for example, in [8]. It is easy to see that G xð Þ¼ ð Þ 1 � a G0ð Þþ x � b aG xð Þ � b; λ , where G0ð Þx is the distribution function of the degenerate distribution concentrated at the point x ¼ 0,

Let us find out the form of the distribution functions (Eqs. (13) and (14)) in the case of Eq. (33),

ð∞ b e ð∞

t�t0þb e �λz dΨð Þz

> ð∞ t�t<sup>0</sup> e �λz dΨð Þz

Probabilistic Model of Delay Propagation along the Train Flow

http://dx.doi.org/10.5772/intechopen.75494

, k ≥ 3:

(35)

185

�<sup>λ</sup>udΨ<sup>∗</sup>ð Þ <sup>k</sup>�<sup>2</sup> ð Þ <sup>u</sup>

�

�<sup>∞</sup> G zð Þ � <sup>t</sup> <sup>þ</sup> <sup>t</sup><sup>0</sup> <sup>d</sup>Ψð Þ<sup>z</sup> exists for any continuous distribution

G xð Þ¼ I xð Þþ <sup>&</sup>lt; <sup>b</sup> I xð Þ <sup>≥</sup> <sup>b</sup> ae�λð Þ <sup>x</sup>�<sup>b</sup> , (36)

ð∞

V uð ÞdΨ<sup>∗</sup>ð Þ <sup>k</sup>�<sup>2</sup> ð Þ <sup>u</sup> � �, (38)

V uð Þ¼ V1ð Þþ u V2ð Þ u , (39)

t�t0þb e �λz

dΨð Þz : (37)

<sup>t</sup>�t<sup>0</sup> I zð þ <sup>þ</sup> <sup>u</sup> � <sup>t</sup> <sup>t</sup><sup>0</sup> <sup>&</sup>lt; <sup>b</sup>ÞdΨð Þ<sup>z</sup> .

�

� �, (34)

when the function Ψ is continuous. In what follows, we mean that n ≥ 3.

Lemma 3. Let the function G be defined by Eq. (33), and Ψ be continuous. Then

<sup>W</sup>2ðÞ¼ <sup>t</sup> I tð Þ <sup>&</sup>gt; <sup>t</sup><sup>0</sup> <sup>Ψ</sup>ð Þþ <sup>t</sup> � <sup>t</sup><sup>0</sup> <sup>þ</sup> <sup>b</sup> ae<sup>λ</sup>ð Þ <sup>t</sup>�t0þ<sup>b</sup>

� �

e �λz dΨð Þz � �dΨ<sup>∗</sup>ð Þ <sup>k</sup>�<sup>2</sup> ð Þ <sup>u</sup>

G zð Þ � <sup>t</sup> <sup>þ</sup> <sup>t</sup><sup>0</sup> <sup>d</sup>Ψð Þ¼ <sup>z</sup> <sup>Ψ</sup>ð Þþ <sup>t</sup> � <sup>t</sup><sup>0</sup> <sup>þ</sup> <sup>b</sup> ae<sup>λ</sup>ð Þ <sup>t</sup>�t0þ<sup>b</sup>

In accordance with Eq. (13), the relation (Eq. (34)) is proved.

WkðÞ¼ t I tð Þ > t<sup>0</sup> Ψð Þþ t � t<sup>0</sup>

ð Þ <sup>Ψ</sup>ð<sup>t</sup> � <sup>t</sup><sup>0</sup> � <sup>u</sup> <sup>þ</sup> <sup>b</sup>Þ � <sup>Ψ</sup>ð Þ <sup>t</sup> � <sup>t</sup><sup>0</sup> <sup>d</sup>Ψ<sup>∗</sup>ð Þ <sup>k</sup>�<sup>2</sup> ð Þ <sup>u</sup>

Proof. According to Eq. (33), one may conclude that function G xð Þ has a unique discontinuity

function Ψ. Note that if Ψð Þz had a discontinuity point z ¼ t1, then the function G zð Þ � t þ t<sup>0</sup> would also be discontinuous at the point z ¼ t<sup>1</sup> for t ¼ t<sup>0</sup> þ t<sup>1</sup> � b, and then the considered

> ð∞ �∞

<sup>t</sup>�t<sup>0</sup> G zð Þ <sup>þ</sup> <sup>u</sup> � <sup>t</sup> <sup>þ</sup> <sup>t</sup><sup>0</sup> <sup>d</sup>Ψð Þ<sup>z</sup> . Given Eq. (36), it is easy to see that

<sup>t</sup>�t<sup>0</sup> I zð Þ <sup>þ</sup> <sup>u</sup> � <sup>t</sup> <sup>þ</sup> <sup>t</sup><sup>0</sup> <sup>≥</sup> <sup>b</sup> <sup>e</sup>�<sup>λ</sup>zdΨð Þ<sup>z</sup> , <sup>V</sup>2ð Þ¼ <sup>u</sup> <sup>Ð</sup> <sup>∞</sup>

t�t0�uþb

WkðÞ¼ <sup>t</sup> I tð Þ <sup>&</sup>gt; <sup>t</sup><sup>0</sup> <sup>Ψ</sup>ð Þþ <sup>t</sup> � <sup>t</sup><sup>0</sup> ae<sup>λ</sup>ð Þ <sup>t</sup>�t0þ<sup>b</sup>

þ ðb �∞ e �λu ð∞

þ ðb �∞

point <sup>x</sup> <sup>¼</sup> <sup>b</sup>. Hence, the integral <sup>Ð</sup> <sup>∞</sup>

ð∞ �∞

where V uð Þ¼ <sup>Ð</sup> <sup>∞</sup>

<sup>V</sup>1ð Þ¼ <sup>u</sup> ae�λð Þ <sup>u</sup>�tþt0�<sup>b</sup> <sup>Ð</sup> <sup>∞</sup>

Let k ≥ 3. It follows from Eq. (14) that

then

integral would not exist (see Remark 1). Since

G xð Þ¼ ; <sup>λ</sup> I xð Þ <sup>≥</sup> <sup>0</sup> <sup>1</sup> � <sup>e</sup>�λ<sup>x</sup> � �.

Proof of Corollary 12. Apply the well-known assertion to Eq. (22): if Y<sup>1</sup> and Y<sup>2</sup> are independent random variables, then for any function of two variables fð Þ �; � and any c∈R, the following equality holds: <sup>Ρ</sup>ðf Yð Þ <sup>1</sup>;Y<sup>2</sup> <sup>&</sup>lt; <sup>c</sup>Þ ¼ <sup>Ð</sup> <sup>∞</sup> �<sup>∞</sup> <sup>Ρ</sup>ð Þ f yð Þ ;Y<sup>2</sup> <sup>&</sup>lt; <sup>c</sup> dF1ð Þ<sup>y</sup> , where <sup>F</sup><sup>1</sup> is the distribution function of <sup>Y</sup>1. Consequently, GkðÞ¼ <sup>t</sup> I tð Þ <sup>&</sup>gt; <sup>0</sup> <sup>Ð</sup> <sup>∞</sup> �<sup>∞</sup> <sup>Ρ</sup>ð Þ <sup>τ</sup> � <sup>y</sup> <sup>&</sup>lt; <sup>t</sup> <sup>d</sup>Ψ<sup>∗</sup>ð Þ <sup>k</sup>�<sup>1</sup> ð Þ<sup>y</sup> . This implies Eq. (25). □

Proof of Corollary 13. The assertion follows from Eq. (25). □

Note that the function Gkð Þt has a jump at zero which is equal to:

$$G\_k(0+) = \int\_{-\infty}^{\infty} G^+(y) d\Psi^{\*(k-1)}(y), \text{ where } G^+(y) = \lim\_{t \to 0+} G(t+y).$$

In the case, when τ and μ<sup>j</sup> are absolutely continuous, it follows from Eq. (25) that

$$\mathcal{G}\_k(t) = I(t>0) \int\_{-\infty}^{\infty} \left( \int\_{-\infty}^{t+y} g(z) dz \right) \psi^{\*(k-1)}(y) dy,\tag{29}$$

where <sup>g</sup>ð Þ� and <sup>ψ</sup>ð Þ� are the density functions of <sup>τ</sup> and <sup>μ</sup>1, respectively, <sup>ψ</sup><sup>∗</sup><sup>j</sup> ð Þy is the j-fold convolution of the density ψð Þ� . In this case, we also have

$$\mathcal{G}\_k(t) \coloneqq I(t>0)G\_k'(t) = I(t>0) \int\_{-\infty}^{\infty} \mathbf{g}(t+y)\psi^{\*(k-1)}(y)dy. \tag{30}$$

If we assume that τ ≥ 0, then we deduce from Eq. (29) that

$$G\_k(t) = I(t>0) \int\_{-t}^{\infty} \left( \int\_0^{t+y} g(z) dz \right) \psi^{\*(k-1)}(y) dy,\tag{31}$$

and we get from Eq. (30),

$$\mathcal{g}\_k(t) = I(t>0) \int\_{-t}^{\infty} \mathbf{g}(t+y)\psi^{\*(k-1)}(y) dy. \tag{32}$$
