ВКР - Математическая модель задержек поездов (1222229), страница 10

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Proof. We have

Observe that departure time of the first train is . Therefore, according to order of the train departures, if , then and, consequently, . Otherwise if , then , i.e. . Hereof we get

and thus we obtain the following equality

which is equivalent to (1.11). 

Let is fixed. We consider the random events

(1.28)

Note that these events are mutually exclusive and .

Notice that if, for instance, n = 3, then , and if n = 4, then , i.e. events , in general, depend on n.

Lemma 4. Let . The following formula holds

(1.29)

Proof. Suppose for clarity that n = 6. We have

(1.30)

By Theorem 1, item 1, the event entails the following equality

thus,

(1.31)

The event implies that the inequality is satisfied. Then, from (1.25), we obtain

hence

(1.32)

The event entails inequality (1.22) as follows

According to Theorem 1 from this inequality we deduce that . Thus,

(1.33)

Similarly, we obtain

(1.34)

Gathering (1.30) – (1.34) we arrive at the equality

(1.35)

Obviously,

(1.36)

Next, since , then

(1.37)

Note that . Consequently,

(1.38)

From (1.35) – (1.38) we obtain

(1.39)

Equality (1.39) is equivalent to (1.29). Indeed, given the implication holds. Therefore

.

Thus,

hence

The proof in the case an arbitrary remains unchanged. For n = 3 the proof only simplified. 

Proof of Theorem 2. The formula (1.11) is proved in Lemma 3 and the formula (1.12) for k = 3 is proved in Lemma 4.

The proof of eqn (1.12) for we carry out by analogy with the proof of Lemma 4.

We have

(1.40)

By Theorem 1, item 1, the event entails the following equality

hence

(1.41)

If the event occurred, then due to (1.26) the following relation holds

from which we get

Likewise, for all

(1.42)

From (1.41) and (1.42), bearing in mind that

we obtain

(1.43)

Furthermore, by using (1.25) and relation

we find that

(1.44)

Let us assume that the event occurred. Then inequality (1.22) holds. Hence according to (1.7) we deduce that . Thus,

(1.45)

Gathering (1.40), (1.43) – (1.45) we arrive at the equality

(1.46)

Arguing as at the end of the proof of Lemma 4, we obtain

Therefore

It follows that (1.46) entails (1.12). 

1.4 Proof of corollaries from the theorem 2

Proof of Corollary 2. Let . Taking into account (1.13), we denote

(1.47)

From Corollary 1 and (1.12) it follows that for

(1.48)

where

We have

(1.49)

and

(1.50)

Taking into account the continuity of random variable , from (1.48) – (1.50) we obtain (1.14). 

Proof of Corollary 3. When we replace in (1.14) with T, we obtain eqn (1.15). 

Proof of Corollary 4. It is well-known, that if and are two independent random variables, and besides has the density function , then for any measurable function of two variables и and for each

(1.51)

Formula (1.51) entails the following form of :

(1.52)

Formulas (1.11) and (1.52) lead to (1.17).

Let . Consider the formula (1.12). We use the notation (1.47). Note that
, and are mutually independent, has the density function . Therefore

(1.53)

Next, similarly to (1.51), we get

(1.54)

In turn,

Consequently,

(1.55)

From (1.54) and (1.55) we obtain

(1.56)

According to (1.12), the relations (1.53) and (1.56) are to be added. As a result, we arrive at (1.18). 

1.5 Proof of Lemma 1

We use the events, similar to (1.28). It is easy to see that

Hence



1.6 The case when the time intervals T_j are constants

Lemma 5. Let all T_j be constants and be equal to some . If

(1.57)

with , then for

(1.58)

Moreover,

(1.59)

(1.60)

Proof. We use the result of Corollary 3. From (1.57) it follows that

(1.61)

Using this equation, the formula (1.58) for k = 2 is obtained from (1.16), а for it is obtained from (1.15).

Let us calculate the mean and variance of the random variables . Since

then has one point of discontinuity , and , , has two points of discontinuity and (see Figure 1.3). Moreover, for

Therefore, for every bounded continuous function , we have

(1.62)

. It is easily seen that (1.62) is also true when k = 2. Using well known equations

then, for we obtain

(1.63)

(1.64)

From (1.62) and (1.63) it follows that

This proves formula (1.59).

In turn, from (1.62) and (1.64) the following relation implies

Consequently,

(1.65)

Transforming the right hand side of (1.65), we arrive at (1.60). 

Figure 1.3 shows the graphs of functions from (1.58) with k = 2, 3 and the parameters

(1.66)

Figure 1.3 – The behavior of functions and

With the aid of equations (1.59) and (1.60) we get with precision to

Remark 3. It is easily seen that the greater k, the function from (1.58) is closer to . This is consistent with Figure 1.3, and formulas (1.59) and (1.60), whereby , when . 

Remark 4. If maximum probability p that occurs at least m of secondary delays is given, and has the density function (1.57), then by Corollary 5 the parameter T satisfies the following condition

(1.67)

(see also [6]). We denote by the minimum T, that satisfies to the inequality (1.67). Consider the following example. Let . Figure 1.4 shows the behavior of as a function of a continuous parameter m when p = 0.1 and
p = 0.05. It is clear that with a decrease of p the value of increases. Exact calculations can be performed according to the following formula

(1.68)

Figure 1.4 – The behavior of when p = 0.1 and p = 0.05

Now we fix p = 0.1. Figure 1.5 shows the behavior of as a function of a continuous parameter m for and . In accordance with equation (1.68) the value increases, when decreases. Since random variable is exponentially distributed and the equality holds, then the reduction of leads to an increase in the mean value of primary delay, which in turn leads to an increase in the intervals between trains (if we want to reduce the number of secondary delays).

Figure 1.5 – The behavior of when and

All parameters, which we use in this example, are valid for the main railway sites (directions) of Russian railways.

The calculation shows that additional (to the value of T) interval is equal to 4.5 minutes for m = 2. Total duration of the minimum interval between trains is 9.5 minutes. It should be noted that the headway of heavy freight trains at the Russian railways lies in the range from 10 to 14 minutes. Obviously, this decision is due to the need to obtain a small amount of unplanned delays. Indeed, when m_knock is equal to 1, the interval, which is calculated by our technique, represents 13 minutes.

1.7 The case when the time intervals T_j have the normal distribution

Let us define

(1.69)

where , T and are some positive parameters. Mark that since random intervals T_j cannot be negative numbers, the assumption of a density function in (1.69) is incorrect. However, by choosing the parameter (when T is known), we can ensure that the probability of the event was closer to 0. We assume that

. (1.70)

We consider the standard normal random variable Y. Its distribution function is equal to

.

From (1.70), in particular, it follows that . Then

.

Note that the choice of the density function from (1.69) is caused by the following property: for any integer the following equality holds

. (1.71)

Also, when , we denote

.

Lemma 6. Suppose that all T_j are independent identically distributed random variables. If the density functions g and are defined in (1.69), then

(1.72)

(1.73)

Moreover,

(1.74)

(1.75)

(1.76)

(1.77)

Proof. We use the result of Corollary 4. According to (1.17) we get

.

Thus

, (1.78)

where

.

Let Y be standard normal random variable. Then

. (1.79)

Let us calculate J₂. It is easy to check that for any and

,

whence

. (1.80)

Consequently,

, (1.81)

where

.

Thus, for any

(1.82)

From (1.82) it follows that

(1.83)

Equation (1.72) follows from (1.17), (1.78), (1.79) and (1.83).

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