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ACTIVITY 1

Simulation of random events with a given distribution law

Playing out a discrete random variable

Let it be required to play a discrete random variable, i.e. get the sequence of its possible values ​​x i (i = 1,2,3,...n), knowing the distribution law X:

Denote by R a continuous random variable. The value of R is distributed uniformly in the interval (0,1). Denote by r j (j = 1,2,...) the possible values ​​of the random variable R. Let us divide the interval 0< R < 1 на оси 0r точками с координатами на n частичных интервалов.

Then we get:

It can be seen that the length of the partial interval with index i is equal to the probability Р with the same index. Length

Thus, when a random number r i falls into the interval, the random variable X takes on the value x i with probability P i .

There is the following theorem:

If each random number that fell into the interval is assigned a possible value x i , then the played value will have a given distribution law

Algorithm for playing a discrete random variable given by the distribution law

1. It is necessary to break the interval (0,1) of the 0r axis into n partial intervals:

2. Select (for example, from a table of random numbers, or in a computer) a random number r j .

If r j fell into the interval, then the discrete random variable being played took the possible value x i .

Playing a Continuous Random Variable

Let it be required to play a continuous random variable X, i.e. get the sequence of its possible values ​​x i (i = 1,2,...). In this case, the distribution function F(X) is known.

Exists next theorem.

If r i is a random number, then the possible value x i of the played continuous random variable X with a known distribution function F(X) corresponding to r i is the root of the equation

Algorithm for playing a continuous random variable:

1. It is necessary to choose a random number r i .

2. Equate the selected random number of the known distribution function F(X) and get the equation.

3. Solve this equation for x i . The resulting value x i will simultaneously correspond to a random number r i . and given distribution law F(X).

Example. Play 3 possible values ​​of a continuous random variable X distributed uniformly in the interval (2; 10).

The distribution function of X has the following form:

By condition, a = 2, b = 10, therefore,

In accordance with the algorithm for playing a continuous random variable, we equate F(X) to the chosen random number r i .. We obtain from this:

Substitute these numbers into equation (5.3). We obtain the corresponding possible values ​​of x:

Problems for modeling random events with a given distribution law

1. It is required to play 10 values ​​of a discrete random variable, i.e. get a sequence of its possible values ​​x i (i=1,2,3,…n), knowing the distribution law Х

Let's choose from the table of random numbers a random number r j: 0.10; 0.12; 0.37; 0.09; 0.65; 0.66; 0.99; 0.19; 0.88; 0.59; 0.78

2. The frequency of receipt of applications for service is subject to the exponential distribution law () , x, the parameter l is known (hereinafter l = 1/t - intensity of receipt of applications)

l=0.5 requests/hour. Determine the sequence of values ​​for the duration of the intervals between receipts of requests. The number of realizations is equal to 5. Number r j: 0.10; 0.12; 0.37; 0.09; 0.65; 0.99;

ACTIVITY 2

Queuing system

Systems in which, on the one hand, there are mass requests for the performance of any types of services, and on the other hand, these requests are satisfied, are called queuing systems. Any QS serves to fulfill the flow of requests.

QS include: a source of requirements, an incoming flow, a queue, a service device, an outgoing flow of requests.

SMOs are divided into:

QS with losses (failures)

CMO with waiting (unlimited queue length)

QS with limited queue length

CMO with limited waiting time.

According to the number of channels or service devices, QS are single-channel and multi-channel.

According to the location of the source of requirements: open and closed.

By the number of service elements per requirement: single-phase and multi-phase.

One of the forms of classification is the classification of D. Kendall - A / B / X / Y / Z

A - determines the distribution of time between arrivals;

B - determines the distribution of service time;

X - determines the number of service channels;

Y - determines the system throughput (queue length);

Z - determines the order of service.

When the system capacity is infinite and the service order is first come first served, the Y/Z parts are omitted. The first digit (A) uses the following symbols:

The M-distribution has an exponential law,

G - the absence of any assumptions about the service process, or it is identified with the symbol GI, meaning a recurrent service process,

D- deterministic (service time is fixed),

Е n - Erlangian of the n-th order,

NM n - hyper-Erlangian of the n-th order.

The second digit (B) uses the same symbols.

The fourth digit (Y) shows the capacity of the buffer, i.e. the maximum number of seats in the queue.

The fifth digit (Z) indicates the method of choosing from the queue in a waiting system: SP-equiprobable, FF-first in-first out, LF-last in-first out, PR-priority.

For tasks:

l - the average number of applications arriving per unit of time

µ is the average number of requests served per unit of time

Channel 1 load factor, or percentage of time that a channel is busy.

Main characteristics:

1) P ref - the probability of failure - the probability that the system will refuse service and the requirement is lost. This happens when the channel or all channels are busy (PSTN).

For a multi-channel QS R otk = R n , where n is the number of service channels.

For a QS with a limited queue length Р otk =Р n + l , where l is the allowable queue length.

2) Relative q and absolute A throughput of the system

q \u003d 1-P otk A \u003d ql

3) The total number of requirements in the system

L sys = n - for QS with failures, n is the number of channels occupied by the service.

For QS with waiting and limited queue length

L sys \u003d n + L cool

where L exp is the average number of requests waiting for service to start, etc.

The remaining characteristics will be considered in the course of solving problems.

Single-channel and multi-channel queuing systems. Failure systems.

The simplest single-channel model with a probabilistic input flow and a service procedure is a model characterized by an exponential distribution of both the durations of the intervals between arrivals of claims and the durations of servicing. In this case, the distribution density of the durations of the intervals between arrivals of claims has the form

Service duration distribution density:

The flows of requests and services are the simplest. Let the system work with failures. This type of QS can be used in modeling transmission channels in local networks. It is necessary to determine the absolute and relative throughput of the system. Let's represent this queuing system as a graph (Figure 2), which has two states:

S 0 - the channel is free (waiting);

S 1 - the channel is busy (the request is being serviced).

Figure 2. Graph of states of a single-channel QS with failures

Let us designate the probabilities of states: P 0 (t) - the probability of the state "channel is free"; P 1 (t) - the probability of the state "channel is busy". Based on the labeled state graph, we compose the system differential equations Kolmogorov for state probabilities:

The system of linear differential equations has a solution subject to the normalization condition P 0 (t) + P 1 (t) = 1 . The solution of this system is called non-stationary, since it directly depends on t and looks like this:

P 1 (t) = 1 - P 0 (t) (3.4.3)

It is easy to see that for a single-channel QS with failures, the probability P 0 (t) is nothing but the relative capacity of the system q. Indeed, P 0 is the probability that at time t the channel is free and the claim that has arrived at time t will be serviced, and, therefore, for this moment time t, the average ratio of the number of serviced requests to the number of received ones is also equal to P 0 (t), i.e., q = P 0 (t).

After a long time interval (at), a stationary (steady-state) mode is reached:

Knowing the relative throughput, it is easy to find the absolute one. Absolute throughput (A) - the average number of applications that the queuing system can serve per unit of time:

The probability of refusal to service the request will be equal to the probability of the "channel is busy" state:

This value P otk can be interpreted as the average share of unserved requests among those submitted.

In the overwhelming majority of cases, in practice, queuing systems are multichannel, and, therefore, models with n serving channels (where n>1) are of undoubted interest. The queuing process described by this model is characterized by the intensity input stream l, while no more than n clients (requests) can be served in parallel. The average service time for one request is 1/m. The input and output streams are Poisson. The mode of operation of one or another service channel does not affect the mode of operation of other service channels of the system, and the duration of the service procedure for each of the channels is a random variable subject to an exponential distribution law. The ultimate goal of using n service channels connected in parallel is to increase (compared to a single-channel system) the speed of servicing requests by serving n clients simultaneously. The state graph of a multichannel queuing system with failures has the form shown in Figure 4.

Figure 4. Graph of states of a multichannel QS with failures

S 0 - all channels are free;

S 1 - one channel is busy, the rest are free;

S k - exactly k channels are occupied, the rest are free;

S n - all n channels are occupied, the rest are free.

The Kolmogorov equations for the probabilities of the system states P 0 , ... ,P k , ... P n will have the following form:

The initial conditions for solving the system are as follows:

P 0 (0) = 1, P 1 (0) = P 2 (0) = ... = P k (0) = ... = P 1 (0) = 0 .

The stationary solution of the system has the form:

Formulas for calculating the probabilities P k (3.5.1) are called Erlang formulas.

Let us determine the probabilistic characteristics of the functioning of a multichannel QS with failures in a stationary mode:

1) failure probability:

since the request is rejected if it arrives at the moment when all n channels are busy. The value of P otk characterizes the completeness of the service of the incoming stream;

2) the probability that the application will be accepted for service (it is also the relative throughput of the system q) complements P otk to unity:

3) absolute bandwidth

4) the average number of channels occupied by the service () is the following:

The value characterizes the degree of loading of the QS.

Tasksto lesson 2

1. The communication branch, which has one channel, receives the simplest flow of messages with an intensity of n = 0.08 messages per second. The transmission time is distributed according to the exp law. Servicing of one message occurs with the intensity µ=0.1. Messages arriving at times when the serving channel is busy transmitting a previously received message receive a transmission failure.

Coeff. Relative channel load (probability of channel being busy)

P

Q is the relative capacity of the internodal branch

And the absolute bandwidth of the communication branch.

2. The communication branch has one channel and receives messages every 10 seconds. The service time for one message is 5 seconds. The message transmission time is distributed exponentially. Messages arriving at times when the channel is busy are denied service.

Define

Р zan - the probability of occupancy of the communication channel (factor of relative load)

Q- relative bandwidth

A is the absolute bandwidth of the communication branch

4. The internodal branch of the secondary communication network has n = 4 channels. The flow of messages arriving for transmission over the channels of the communication branch has a rate of = 8 messages per second. The average transmission time of one message is t = 0.1 seconds. A message arriving at the moment when all n channels are busy receives a transmission failure along the communication branch. Find CMO characteristics:

ACTIVITY 3

Single-channel system with waiting

Consider now a single-channel QS with expectation. The queuing system has one channel. The incoming flow of service requests is the simplest flow with intensity. The intensity of the service flow is equal (i.e., on average, a continuously busy channel will issue serviced requests). Service duration is a random variable subject to an exponential distribution law. Service flow is the simplest Poisson flow of events. A request that arrives at a time when the channel is busy is queued and awaits service. This QS is the most common in modeling. With one or another degree of approximation, it can be used to simulate almost any node of a local area network (LAN).

Let us assume that no matter how many demands enter the input of the serving system, this system(queue + served clients) can not accommodate more than N-requirements (applications), i.e., clients who do not fall into the waiting period are forced to be served elsewhere. M/M/1/N system. Finally, the source that generates service requests has an unlimited (infinitely large) capacity. The QS state graph in this case has the form shown in Figure 3

Figure 3. Graph of states of a single-channel QS with waiting (death and reproduction scheme)

QS states have the following interpretation:

S 0 - "channel is free";

S 1 - "channel is busy" (there is no queue);

S 2 - "channel is busy" (one application is in the queue);

S n - "the channel is busy" (n -1 applications are in the queue);

S N - "the channel is busy" (N - 1 applications are in the queue).

The stationary process in this system will be described by the following system of algebraic equations:

where p=load factor

n - state number.

The solution of the above system of equations for our QS model has the form:

The initial value of the probability for a QS with a limited queue length

For a QS with an infinite queue H =? :

P 0 \u003d 1- s (3.4.7)

It should be noted that the fulfillment of the stationarity condition for this QS is not necessary, since the number of applications admitted to the serving system is controlled by introducing a restriction on the queue length, which cannot exceed (N - 1), and not by the ratio between the intensities of the input stream, i.e. not the ratio c=l/m.

In contrast to the single-channel system, which was considered above and with an unlimited queue, in this case, the stationary distribution of the number of requests exists for any finite values ​​of the load factor c.

Let us determine the characteristics of a single-channel QS with waiting and a limited queue length equal to (N - 1) (M/M/1/N), as well as for a single-channel QS with an unlimited capacity buffer (M/M/1/?). For a QS with an infinite queue, the condition with<1, т.е., для того, чтобы в системе не накапливалась бесконечная очередь необходимо, чтобы в среднем запросы в системе обслуживались быстрее, чем они туда поступают.

1) the probability of refusal to service the application:

One of the most important characteristics of systems in which requests can be lost is the probability P loss that an arbitrary request will be lost. In this case, the probability of losing an arbitrary request coincides with the probability that at an arbitrary moment of time all waiting places are occupied, i.e. the formula P from k \u003d P H is valid

2) relative throughput of the system:

For CMO with unlimitedth queue q=1, because all applications will be served

3) absolute bandwidth:

4) the average number of applications in the system:

L S with unlimited queue

5) average residence time of an application in the system:

For unlimited queue

6) the average duration of the client's (application's) stay in the queue:

With unlimited queue

7) the average number of applications (clients) in the queue (queue length):

with unlimited queue

Comparing the expressions for the average waiting time in the queue T pt and the formula for the average queue length L pt, as well as the average residence time of requests in the system T S and the average number of requests in the system L S , we see that

L och \u003d l * T och L s \u003d l * T s

Note that these formulas are also valid for many queuing systems more general than the M/M/1 system under consideration and are called Little's formulas. The practical significance of these formulas lies in the fact that they eliminate the need to directly calculate the values ​​of T och and T s with a known value of the values ​​of L och and L s and vice versa.

Tasks for single-channel CMOwith expectation, Withanticipation andlimited queue length

1. Given a single-line QS with an unlimited queue accumulator. Applications arrive every t =14 seconds. The average transmission time of one message is t=10 seconds. Messages arriving at times when the serving channel is busy are received into the queue without leaving it until service begins.

Determine the following performance indicators:

2. The internodal branch of communication, which has one channel and a queue drive for m=3 waiting messages (N-1=m), receives the simplest flow of messages with a rate of n=5 messages. in sec.. The time of message transmission is distributed according to the exponential law. The average transmission time for one message is 0.1 seconds. Messages arriving at times when the serving channel is busy transmitting a previously received message and there is no free space in the drive are rejected.

Р otk - the probability of failure to receive a message

L syst - the average total number of messages in the queue and transmitted along the communication branch

T och - the average time the message stays in the queue before the start of transmission

T syst - the average total time spent by a message in the system, the sum of the average waiting time in the queue and the average transmission time

Q- relative bandwidth

A is the absolute throughput

3. The internodal branch of the secondary communication network, which has one channel and a queue store for m = 4 (N-1=4) waiting messages, receives the simplest message flow with a rate of = 8 messages per second. The message transmission time is distributed exponentially. The average transmission time for one message is t = 0.1 second. Messages arriving at times when the serving channel is busy transmitting a previously received message and there is no free space in the drive are denied in the queue.

P otk - the probability of failure to receive a message for transmission over the communication channel of the internodal branch;

L och - the average number of messages in the queue to the communication branch of the secondary network of the queue;

L syst - the average total number of messages in the queue and transmitted through the communication branch of the secondary network;

T och - the average time the message stays in the queue before the start of transmission;

Р zan - the probability of occupancy of the communication channel (coefficient of relative channel load);

Q is the relative capacity of the internodal branch;

A is the absolute capacity of the internodal branch;

4. The internodal communication branch, which has one channel and a queue drive for m=2 waiting messages, receives the simplest flow of messages with an intensity of n=4 messages. in sec.. The time of message transmission is distributed according to the exponential law. The average transmission time for one message is 0.1 seconds. Messages arriving at times when the serving channel is busy transmitting a previously received message and there is no free space in the drive are rejected.

Determine the following performance indicators of the communication branch:

Р otk - the probability of failure to receive a message

L och - the average number of messages in the queue to the communication branch

L syst - the average total number of messages in the queue and transmitted along the communication branch

T och - the average time the message stays in the queue before the start of transmission

T syst - the average total time spent by a message in the system, the sum of the average waiting time in the queue and the average transmission time

Р zan - the probability of occupancy of the communication channel (coefficient of relative channel load c)

Q- relative bandwidth

A is the absolute throughput

5. The internodal branch of the secondary communication network, which has one channel and an unlimited storage queue of waiting messages, receives the simplest flow of messages with an intensity of n = 0.06 messages per second. Average transmission time of one message t =10 seconds. Messages arriving at times when the communication channel is busy are received into the queue and do not leave it until the start of service.

Determine the following performance indicators of the communication branch of the secondary network:

L och - the average number of messages in the queue to the communication branch;

L syst - the average total number of messages in the queue and transmitted along the communication branch;

T och - the average time spent by a message in the queue;

T syst is the average total time spent by a message in the system, which is the sum of the average waiting time in the queue and the average transmission time;

Р zan - the probability of occupancy of the communication channel (the coefficient of the relative load of the channel);

Q is the relative capacity of the internodal branch;

A - absolute throughput of the internodal branch

6. Given a one-line QS with an unlimited queue accumulator. Applications arrive every t =13 seconds. Average transmission time per message

t=10 seconds. Messages arriving at times when the serving channel is busy are received into the queue without leaving it until service begins.

Determine the following performance indicators:

L och - the average number of messages in the queue

L syst - the average total number of messages in the queue and transmitted along the communication branch

T och - the average time the message stays in the queue before the start of transmission

T syst - the average total time spent by a message in the system, the sum of the average waiting time in the queue and the average transmission time

Р zan - probability of occupancy (coefficient of relative channel load c)

Q- relative bandwidth

A is the absolute throughput

7. A specialized diagnostic post is a single-channel QS. The number of parking lots for cars waiting for diagnostics is limited and equals 3 [(N - 1) = 3]. If all parking lots are occupied, i.e. there are already three cars in the queue, then the next car that arrived for diagnostics does not get into the service queue. The flow of cars arriving for diagnostics is distributed according to the Poisson law and has an intensity = 0.85 (cars per hour). The time of car diagnostics is distributed according to the exponential law and equals 1.05 hours on average.

It is required to determine the probabilistic characteristics of the diagnostic post operating in stationary mode: P 0 , P 1 , P 2 , P 3 , P 4 , P open, q, A, L och, L sys, T och, T sis

ACTIVITY 4

Multi-channel QS with waiting, with waiting and limited queue length

Consider a multichannel queuing system with waiting. This type of QS is often used when modeling groups of LAN subscriber terminals operating in an on-line mode. The queuing process is characterized by the following: the input and output flows are Poisson with intensities and respectively; no more than n clients can be served in parallel. The system has n service channels. The average service time per client is 1/m for each channel. This system also refers to the process of death and reproduction.

с=l/nm - the ratio of the intensity of the incoming flow to the total intensity of service, is the system load factor

(With<1). Существует стационарное распределение числа запросов в рассматриваемой системе. При этом вероятности состояний Р к определяются:

where Р 0 is the probability of a free state of all channels with an unlimited queue, k is the number of applications.

if we accept c=l / m, then P 0 can be determined for an unlimited queue:

For limited queue:

where m is the length of the queue

With unlimited queue:

Relative throughput q=1,

Absolute bandwidth A \u003d l,

Average number of occupied channels Z=A/m

With limited queue

1 The inter-nodal branch of the secondary communication network has n = 4 channels. The flow of messages arriving for transmission over the channels of the communication branch has a rate of = 8 messages per second. The average time t = 0.1 for the transmission of one message by each communication channel is t/n = 0.025 seconds. The waiting time for messages in the queue is unlimited. Find CMO characteristics:

R otk - the probability of failure to transmit messages;

Q is the relative throughput of the communication branch;

A is the absolute bandwidth of the communication branch;

Z is the average number of busy channels;

L och - the average number of messages in the queue;

T exp - average waiting time;

T syst - the average total time spent by messages in the queue and transmission along the communication branch.

2. The mechanical workshop of the plant with three posts (channels) performs repairs of small-scale mechanization. The flow of faulty mechanisms arriving at the workshop is Poisson and has an intensity of = 2.5 mechanisms per day, the average repair time for one mechanism is distributed according to the exponential law and is equal to = 0.5 days. Suppose that there is no other workshop in the factory, and, therefore, the queue of mechanisms in front of the workshop can grow almost indefinitely. It is required to calculate the following limit values ​​of the probabilistic characteristics of the system:

Probabilities of system states;

The average number of applications in the service queue;

The average number of applications in the system;

The average duration of the application in the queue;

The average duration of an application's stay in the system.

3. The internodal branch of the secondary communication network has n=3 channels. The flow of messages arriving for transmission through the channels of the communication branch has an intensity of n=5 messages per second. The average transmission time of one message is t=0.1 , t/n=0.033 sec. There can be up to m= 2 messages in the pending message queue drive. A message that arrives at a time when all places in the queue are occupied receives a transmission rejection on the communication branch. Find the characteristics of the QS: P ref - probability of message transmission failure, Q - relative throughput, A - absolute throughput, Z - average number of busy channels, L och - average number of messages in the queue, T exp - average waiting time, T system - the average total time spent by a message in the queue and its transmission along the communication branch.

ACTIVITY 5

Closed QS

Let's consider the model of servicing the machine park, which is a model of a closed queuing system. Until now, we have considered only such queuing systems for which the intensity of the incoming flow of requests does not depend on the state of the system. In this case, the source of claims is external to the QS and generates an unlimited flow of claims. Consider queuing systems for which depends on the state of the system, where the source of requirements is internal and generates a limited flow of requests. For example, a machine park consisting of N machines is maintained by a team of R mechanics (N > R), and each machine can be serviced by only one mechanic. Here machines are sources of requirements (requests for service), and mechanics are service channels. A failed machine after service is used for its intended purpose and becomes a potential source of service requirements. Obviously, the intensity depends on how many cars are currently in operation (N - k) and how many cars are being serviced or standing in line waiting for service (k). In the model under consideration, the capacity of the source of requirements should be considered limited. The incoming flow of requirements comes from a limited number of machines in operation (N - k), which at random times fail and require maintenance. Moreover, each machine from (N - k) is in operation. Generates a Poisson demand flow with intensity X regardless of other objects, the total incoming flow has intensity. A request that enters the system at the moment when at least one channel is free is immediately sent for servicing. If a requirement finds all channels busy serving other requirements, then it does not leave the system, but queues up and waits until one of the channels becomes free. Thus, in a closed queuing system, the incoming flow of requirements is formed from the outgoing one. The state S k of the system is characterized by the total number of requests being serviced and in the queue, equal to k. For the considered closed system, obviously, k = 0, 1, 2, ... , N. Moreover, if the system is in the state S k , then the number of objects in operation is (N - k). If - the intensity of the flow of requirements per machine, then:

The system of algebraic equations describing the operation of a closed QS in a stationary mode is as follows:

Solving this system, we find the probability of the k-th state:

The value of P 0 is determined from the condition of normalizing the results obtained by the formulas for P k , k = 0, 1, 2, ... , N. Let us define the following probabilistic characteristics of the system:

Average number of requests in the service queue:

Average number of requests in the system (in service and in queue)

average number of mechanics (channels) "idle" due to lack of work

The downtime ratio of the serviced object (machine) in the queue

Utilization rate of objects (machines)

Downtime ratio of service channels (mechanics)

Average waiting time for service (time waiting for service in queue)

Closed QS problem

1. Let two engineers of the same productivity be allocated to service ten personal computers (PCs). The flow of failures (malfunctions) of one computer is Poisson with intensity = 0.2. The service time of a PC obeys an exponential law. The average maintenance time for one PC by one engineer is: = 1.25 hours. The following service organization options are available:

Both engineers serve all ten computers, so if the PC fails, one of the free engineers serves it, in this case R = 2, N = 10;

Each of the two engineers maintains five PCs assigned to him. In this case, R = 1, N = 5.

It is necessary to choose the best option for organizing PC maintenance.

It is necessary to define all the probabilities of states P k: P 1 - P 10, given that and using the results of calculating P k, we calculate P 0

ACTIVITY 6

Traffic calculation.

The theory of teletraffic is a section of the theory of queuing. The foundations of the theory of teletraffic were laid by the Danish scientist A.K. Erlang. His works were published in 1909-1928. Let us give important definitions used in the theory of teletraffic (TT). The term "traffic" (English, traffic) corresponds to the term "telephone load". It implies the load created by the flow of calls, requirements, messages arriving at the inputs of the QS. The volume of traffic is called the value of the total, integral time interval missed by one or another resource, during which this resource was occupied for the analyzed period of time. A unit of work can be considered as a second occupation of a resource. Sometimes you can read about hours, and sometimes just seconds or hours. However, the ITU recommendations give the dimension of traffic volume in erlango hours. To understand the meaning of such a unit of measurement, one more traffic parameter must be considered - traffic intensity. In this case, they often talk about the average intensity of traffic (load) on a given pool (set) of resources. If at each moment of time t from a given interval (t 1 ,t 2) the number of resources from this set occupied by servicing traffic is equal to A(t), then the average traffic intensity will be

The value of the traffic intensity is characterized as the average number of resources occupied by the traffic in a given time interval. The unit of measurement of the intensity of the load is one Erlang (1 Erl, 1 E), i.e. 1 erlang is the amount of traffic that requires the full employment of one resource, or, in other words, at which the work of one second is performed by the resource - occupation for the time of one second. In American literature, you can sometimes find another unit of measurement called CCS- Centrum (or hundred) Calls Second (hectosecond occupations). The CCS number reflects the time the servers are occupied in 100 second intervals in 1 hour. The intensity measured in CCS can be converted to Erlangs using the formula 36CCS=1 Erl.

The traffic generated by one source and expressed in hour-sessions is equal to the product of the number of call attempts c for a certain time interval T and the average duration of one attempt t: y = c t (h-h). Traffic can be calculated in three different ways:

1) let the number of calls c per hour be 1800, and the average duration of the lesson t = 3 minutes, then Y = 1800 calls. /h 0.05 h = 90 Erl;

2) let the durations t i of all n occupations of the outputs of a certain bundle be fixed during time T, then the traffic is determined as follows:

3) let during the time T, observation is carried out at regular intervals over the number of simultaneously occupied exits of a certain beam, according to the results of the observations, a step function of time x(t) is built (Figure 8).

Figure 8. Counts of simultaneously occupied beam exits

Traffic during time T can be estimated as the average value of x(t) over this time:

where n is the number of samples of simultaneously occupied outputs. The value of Y is the average number of simultaneously occupied beam exits during time T.

Traffic fluctuations. The traffic of secondary telephone networks fluctuates significantly over time. During the working day, the traffic curve has two or even three peaks (Figure 9).

Figure 9. Fluctuations in traffic during the day

The hour of the day during which long-term traffic is at its most significant is called the busy hour (BUSH). Knowledge of the traffic in the CNN is fundamentally important, since it determines the number of channels (lines), the amount of equipment of stations and nodes. The traffic of the same day of the week has seasonal fluctuations. If the day of the week is a pre-holiday day, then the NPV of this day is higher than the day after the holiday. If the number of services supported by the network grows, then so does the traffic. Therefore, it is problematic to predict with sufficient certainty the occurrence of traffic peaks. Traffic is closely monitored by the network administration and design organizations. Traffic measurement rules are developed by ITU-T and are used by national network administrations in order to meet the quality of service requirements for both subscribers of their own network and subscribers of other networks connected to it. The theory of teletraffic can be used for practical calculations of losses or equipment volume of a station (node) only if the traffic is stationary (statistically steady). This condition is approximately satisfied by the traffic in the CNN. The amount of load received per day on the PBX affects the prevention and repair of equipment. The unevenness of the load on the station during the day is determined by the concentration coefficient

A more rigorous definition of NNN is as follows. Recommendation ITU E.500 prescribes to analyze intensity data for 12 months, select the 30 busiest days from them, find the busiest hours on these days and average the intensity measurement results over these intervals. This calculation of the traffic intensity (load) is called the normal estimate of the traffic intensity in the busy hour or level A. A more stringent estimate can be averaged over the 5 busiest days of the selected 30-day period. Such an assessment is called an increased or an assessment of the level of B.

The process of creating traffic. As every user of the telephone network knows, not all attempts to establish a connection with the called subscriber end successfully. Sometimes you have to make several unsuccessful attempts before the desired connection is established.

Figure 10. Diagram of events when a connection is established between subscribers

Let's consider possible events when simulating the establishment of a connection between subscribers A and B (Figure 10). Statistical data on calls in telephone networks are as follows: the share of completed calls is 70-50%, the share of failed calls is 30-50%. Any attempt by the subscriber occupies the entrance of the QS. With successful attempts (when the conversation took place), the occupation time of the switching devices that establish connections between inputs and outputs is longer than with unsuccessful attempts. The subscriber can interrupt connection attempts at any time. Retries can be caused by the following reasons:

Number dialed incorrectly;

Assumption of an error in the network;

The degree of urgency of the conversation;

Unsuccessful previous attempts;

Knowing the habits of subscriber B;

Doubt about correct dialing.

A retry may be attempted depending on the following circumstances:

Degrees of urgency;

Estimates of the reasons for failure;

Estimates of the expediency of repeating attempts,

Estimates of the acceptable interval between attempts.

Refusal to retry may be associated with a low degree of urgency. There are several types of traffic generated by calls: incoming (offered) Y p and missed Y p. Traffic Y p includes all successful and unsuccessful attempts, traffic Y p, which is part of Y p, includes successful and part of unsuccessful attempts:

Y pr \u003d Y p + Y np,

where Y p - conversational (useful) traffic, and Y np - traffic created by unsuccessful attempts. Equality Y p = Y p is possible only in the ideal case, if there are no losses, errors of callers and no responses of called subscribers.

The difference between the incoming and missed loads for a certain period of time will be the lost load.

Traffic forecasting. Limited resources lead to the need for a phased expansion of the station and the network. The network administration makes a forecast of an increase in traffic during the development phase, considering that:

Income is determined by the part of the passed traffic Y p, - costs are determined by the quality of service at the highest traffic;

A large proportion of losses (low quality) occurs in rare cases and is typical for the end of the development period;

The largest volume of missed traffic falls on periods when there are practically no losses - if the losses are less than 10%, then subscribers do not respond to them. When planning the development of stations and the network, the designer must answer the question, what are the requirements for the quality of service provision (for losses). To do this, it is necessary to measure traffic losses according to the rules adopted in the country.

An example of traffic measurement.

First, consider how you can display the operation of a QS that has several resources that serve some traffic at the same time. We will further talk about such resources as servers that serve the flow of applications or requirements. One of the most visual and commonly used ways to depict the process of servicing requests by a pool of servers is a Gantt chart. This chart is a rectangular coordinate system, the abscissa of which represents time, and the ordinate represents discrete points corresponding to the pool servers. Figure 11 shows a Gantt chart for a system with three servers.

In the first three time intervals (we consider them a second), the first and third servers are occupied, the next two seconds - only the third, then the second works for one second, then the second and the first for two seconds, and the last two seconds - only the first.

The constructed diagram allows you to calculate the amount of traffic and its intensity. The diagram shows only served or missed traffic, since it does not say anything about whether requests entered the system that could not be served by the servers.

The volume of the passed traffic is calculated as the total length of all segments of the Gantt chart. Volume in 10 seconds:

Associate with each time interval plotted along the abscissa, an integer equal to the number of servers occupied in this single interval. This value A(t) is the instantaneous intensity. For our example

A(t)= (2, 2, 2, 1, 1, 1, 2, 2, 1, 1)

Let us now find the average traffic intensity over a period of 10 seconds

Thus, the average intensity of traffic passed by the considered system of three servers is equal to 1.5 Erl.

Main load parameters

Telephone communication is used by various categories of subscribers, which are characterized by:

the number of load sources - N,

the average number of calls from one source in a certain time (HNN usually) - s,

the average duration of one occupation of the switching system when servicing one call is t.

The load intensity will be

Let's define different call sources. For example,

Average number of calls per office phone per office phone;

The average number of calls from one apartment individual device; random event queuing teletraffic

with a count - the same from the apparatus for collective use;

with ma - the same from one coin machine;

with sl - the same from one connecting line.

Then the average number of calls from one source is:

There are approximate data for the average number of calls from one source of the corresponding category:

3.5 - 5, \u003d 0.5 - 1, with count \u003d 1.5 - 2, with ma \u003d 15 - 30, with sl \u003d 10 - 30.

There are the following types of connections, which, depending on the outcome of the connection, create a different telephone load at the station:

k p - coefficient showing the proportion of connections that ended in a conversation;

k c - connections that did not end with a conversation due to the busyness of the called subscriber;

k but - coefficient expressing the proportion of connections that did not end with a conversation due to the non-response of the called subscriber;

k osh - connections that did not end with a conversation due to the caller's errors;

k of those - calls that did not end with a conversation for technical reasons.

During normal operation of the network, the values ​​of these coefficients are equal to:

k p =0.60-0.75; k c =0.12-0.15; k but =0.08-0.12; k osh =0.02-0.05; k those = 0.005-0.01.

The average duration of a lesson depends on the types of connections. For example, if the connection ended with a conversation, the average duration of the occupation of devices t state will be equal to

where is the duration of connection establishment;

t cond. - the conversation that took place;

t in - the duration of sending a call to the telephone set of the called subscriber;

t p - duration of the conversation

where t co - station response signal;

1.5n - dialing time of the called subscriber (n - number of characters in the number);

t with - the time required to establish a connection by switching mechanisms and disconnect the connection after the end of the conversation. Approximate values ​​of the considered quantities:

t co \u003d 3 sec., t c \u003d 1-2.5 sec., t in \u003d 8-10 sec., t p \u003d 90-130 sec.

Calls that do not end with a conversation also create a telephone load.

The average occupancy time of devices when the called subscriber is busy is equal to

where t is set. determined by (4.2.3)

t buzzer - time of listening to the busy buzzer, t buzzer =6sec.

The average duration of the occupation of devices when the called subscriber does not answer is equal to

where t pv is the time of listening to the ringback control signal, t pv = 20 sec.

If there was no conversation due to subscriber errors, then on average t osh = 30 sec.

The duration of the sessions that did not end with a conversation for technical reasons is not defined, since the percentage of such sessions is small.

From all of the above, it follows that the total load created by a group of sources for the NTT is equal to the sum of the loads of individual types of occupations.

where is a coefficient that takes into account the terms as shares

On a telephone network with a seven-digit numbering, an automatic telephone exchange was designed, the structural composition of subscribers of which is as follows:

N chr \u003d 4000, N ind \u003d 1000, N count \u003d 2000, N ma \u003d 400, N sl \u003d 400.

The average number of calls coming from one source in a busy hour is

By formulas (4.2.3) and (4.2.6) we find the load

1.10.62826767 sec. = 785.2 hz.

Average lesson duration t from the formula Y=Nct

t= Y/Nc= 2826767/7800*3.8=95.4 sec.

Load task

1. On a telephone network with seven-digit numbering, an automatic telephone exchange was designed, the structural composition of subscribers of which is as follows:

N uchr \u003d 5000, N ind \u003d 1500, N count \u003d 3000, N ma \u003d 500, N sl \u003d 500.

Determine the load arriving at the station - Y, the average duration of occupation t, if it is known that

with chr \u003d 4, with ind \u003d 1, with count \u003d 2, with ma \u003d 10, with sl \u003d 12, t p \u003d 120 sec., t in \u003d 10 sec., k p \u003d 0.6, t with \u003d 1 sec., \u003d 1.1.

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LABORATORY WORK MM-03

PLAYING DISCRETE AND CONTINUOUS ROVs

Purpose of the work: study and software implementation of methods for playing discrete and continuous RVs

QUESTIONS TO STUDY FROM LECTURE SUMMARY:

1. Discrete random variables and their characteristics.

2. Playing a complete group of random events.

3. Playing a continuous random variable by the inverse function method.

4. Choice of a random direction in space.

5. Standard normal distribution and its recalculation for given parameters.

6. The method of polar coordinates for playing out the normal distribution.

TASK 1. Formulate (in writing) a rule for playing out the values ​​of a discrete RV, the distribution law of which is given in the form of a table. Compose a subroutine-function for playing the values ​​of CV using the BSV received from the RNG subroutine. Play 50 CB values ​​and display them on the screen.

Where N is the variant number.

TASK 2. The distribution density function f(x) of a continuous random variable X is given.

In the report, write down the formulas and calculation of the following values:

A) normalization constant;

B) distribution function F(x);

C) mathematical expectation M(X);

D) dispersion D(X);

E) a formula for playing out the values ​​of CB using the inverse function method.

Compose a function subroutine to play the given CV and get 1000 values ​​of this CV.

Construct a histogram of the distribution of the obtained numbers over 20 segments.

TASK 3. Write a procedure that allows you to play the parameters of a random direction in space. Play 100 random directions in space.

Use the built-in pseudo-random number generator.

The written report on laboratory work should contain:

1) Name and purpose of the work, group, surname and number of the student's option;

2) For each task: -condition, -necessary formulas and mathematical transformations, -name of the program file that implements the algorithm used, -calculation results.

The debugged program files are handed over together with the written report.

APPLICATION

Variants of the distribution density of continuous SW

Var-t	SW distribution density	Var-t	SW distribution density

Definition 24.1.random numbers name possible values r continuous random variable R, distributed uniformly in the interval (0; 1).

1. Playing a discrete random variable.

Let it be required to play a discrete random variable X, that is, to obtain a sequence of its possible values, knowing the distribution law X:

x x 1 X 2 … x n

p p 1 R 2 … r p .

Consider a random variable uniformly distributed in (0, 1) R and split the interval (0, 1) by points with coordinates R 1, R 1 + R 2 , …, R 1 + R 2 +… +r p-1 on P partial intervals whose lengths are equal to the probabilities with the same indices.

Theorem 24.1. If each random number that falls into the interval is assigned a possible value , then the played value will have a given distribution law:

x x 1 X 2 … x n

p p 1 R 2 … r p .

Proof.

Possible values ​​of the obtained random variable coincide with the set X 1 , X 2 ,… x n, since the number of intervals is P, and when hit rj in the interval, a random variable can take only one of the values X 1 , X 2 ,… x n.

Because R is uniformly distributed, then the probability of its falling into each interval is equal to its length, which implies that each value corresponds to the probability pi. Thus, the random variable being played has a given distribution law.

Example. Play 10 values ​​of a discrete random variable X, whose distribution law has the form: X 2 3 6 8

R 0,1 0,3 0,5 0,1

Solution. Let's break the interval (0, 1) into partial intervals: D 1 - (0; 0.1), D 2 - (0.1; 0.4), D 3 - (0.4; 0.9), D 4 – (0.9; 1). Let's write out 10 numbers from the table of random numbers: 0.09; 0.73; 0.25; 0.33; 0.76; 0.52; 0.01; 0.35; 0.86; 0.34. The first and seventh numbers lie on the interval D 1 , therefore, in these cases, the random variable being played has taken the value X 1 = 2; the third, fourth, eighth and tenth numbers fell into the interval D 2 , which corresponds to X 2 = 3; the second, fifth, sixth and ninth numbers were in the interval D 3 - while X = x 3 = 6; not a single number fell into the last interval. So, the played out possible values X are: 2, 6, 3, 3, 6, 6, 2, 3, 6, 3.

2. Playing out opposite events.

Let it be required to play trials, in each of which the event A appears with a known probability R. Consider a discrete random variable X, which takes the values ​​1 (if the event A happened) with a probability R and 0 (if A did not happen) with a probability q = 1 – p. Then we play this random variable as suggested in the previous paragraph.

Example. Play 10 challenges, each with an event A appears with a probability of 0.3.

Solution. For a random variable X with distribution law X 1 0

R 0,3 0,7

we get the intervals D 1 - (0; 0.3) and D 2 - (0.3; 1). We use the same sample of random numbers as in the previous example, for which the numbers №№1,3 and 7 fall into the interval D 1, and the rest - into the interval D 2 . Therefore, we can assume that the event A happened in the first, third, and seventh trials, but did not happen in the others.

3. Playing a complete group of events.

If events A 1 , A 2 , …, A p, whose probabilities are equal R 1 , R 2 ,… r p, form a complete group, then for playing out (that is, modeling the sequence of their appearances in a series of tests), you can play a discrete random variable X with distribution law X 1 2 … P, doing this in the same way as in paragraph 1. At the same time, we assume that

p p 1 R 2 … r p

If X takes on the value x i = i, then in this trial an event occurred A i.

4. Playing a continuous random variable.

a) Method of inverse functions.

Let it be required to play a continuous random variable X, i.e. get the sequence of its possible values x i (i = 1, 2, …, n), knowing the distribution function F(x).

Theorem 24.2. If r i is a random number, then the possible value x i played continuous random variable X with a given distribution function F(x), corresponding r i, is the root of the equation

F(x i) = r i. (24.1)

Proof.

Because F(x) increases monotonically in the range from 0 to 1, then there is a (and unique) value of the argument x i, at which the distribution function takes the value r i. Hence, equation (24.1) has a unique solution: x i= F -1 (r i), Where F-1 - function inverse to F. Let us prove that the root of equation (24.1) is a possible value of the considered random variable X. Suppose first that x i is a possible value of some random variable x, and we prove that the probability of x falling into the interval ( c, d) is equal to F(d) – F(c). Indeed, due to the monotonicity F(x) and that F(x i) = r i. Then

Therefore, Hence, the probability of x falling into the interval ( c, d) is equal to the increment of the distribution function F(x) on this interval, hence x = X.

Play 3 possible values ​​of a continuous random variable X, distributed uniformly in the interval (5; 8).

F(x) = , that is, it is required to solve the equation Let's choose 3 random numbers: 0.23; 0.09 and 0.56 and substitute them into this equation. Get the corresponding possible values X:

b) Superposition method.

If the distribution function of the random variable being played can be represented as a linear combination of two distribution functions:

then , because at X®¥ F(x) ® 1.

We introduce an auxiliary discrete random variable Z with distribution law

Z 12 . Let's choose 2 independent random numbers r 1 and r 2 and play out the possible

pc 1 C 2

meaning Z by number r 1 (see paragraph 1). If Z= 1, then we are looking for the desired possible value X from the equation, and if Z= 2, then we solve the equation .

It can be proved that in this case the distribution function of the random variable being played is equal to the given distribution function.

c) Approximate simulation of a normal random variable.

Since for R, uniformly distributed in (0, 1), , then for the sum P independent, uniformly distributed in the interval (0,1) random variables . Then, by virtue of the central limit theorem, the normalized random variable at P® ¥ will have a distribution close to normal, with parameters A= 0 and s =1. In particular, a fairly good approximation is obtained for P = 12:

So, to play the possible value of the normalized normal random variable X, you need to add 12 independent random numbers and subtract 6 from the sum.

Of all random variables, it is easiest to play (simulate) a uniformly distributed variable. Let's see how it's done.

Let's take some device, at the output of which the digits 0 or 1 can appear with probability; the appearance of one or another number should be random. Such a device can be a tossed coin, a dice (even - 0, odd - 1) or a special generator based on counting the number of radioactive decays or bursts of radio noise over a certain time (even or odd).

Let's write y as a binary fraction and replace successive digits with numbers generated by the generator: for example, . Since the first digit is equally likely to be 0 or 1, this number is equally likely to lie in the left or right half of the segment. Since 0 and 1 are also equally likely in the second digit, the number lies in each half of these halves with equal probability, and so on. Hence, a binary fraction with random digits really takes any value on the segment with equal probability

Strictly speaking, only a finite number of bits k can be played. Therefore, the distribution will not be completely required; the mathematical expectation will be less than 1/2 by the value (because the value is possible, but the value is impossible). So that this factor does not affect, multi-digit numbers should be taken; True, in the method of statistical testing, the accuracy of the answer usually does not exceed 0.1% -103, and the condition gives that on modern computers it is overfulfilled with a large margin.

pseudo-random numbers. Real random number generators are not free from systematic errors: coin asymmetry, zero drift, etc. Therefore, the quality of the numbers they produce is checked by special tests. The simplest test is to calculate for each digit the frequency of occurrence of zero; if the frequency is noticeably different from 1/2, then there is a systematic error, and if it is too close to 1/2, then the numbers are not random - there is some pattern. More complex tests are the calculation of correlation coefficients of consecutive numbers

or groups of digits within a number; these coefficients should be close to zero.

If any sequence of numbers satisfies these tests, then it can be used in calculations according to the method of statistical tests, without being interested in its origin.

Algorithms for constructing such sequences have been developed; symbolically they are written by recurrent formulas

Such numbers are called pseudo-random and are calculated on a computer. This is usually more convenient than using special generators. But each algorithm has its own limit on the number of sequence members that can be used in calculations; with a larger number of terms, the random character of numbers is lost, for example, periodicity is found.

The first algorithm for obtaining pseudo-random numbers was proposed by Neumann. Let's take a number from digits (decimal for definiteness) and square it. We leave the middle numbers near the square, discarding the last and (or) the first. We square the resulting number again, and so on. The values ​​are obtained by multiplying these numbers by For example, let's set and choose the initial number 46; then we get

But the distribution of Neumann numbers is not uniform enough (values ​​predominate, which is clearly seen in the example above), and now they are rarely used.

The most commonly used now is a simple and good algorithm related to the selection of the fractional part of the product

where A is a very large constant (the curly bracket denotes the fractional part of the number). The quality of pseudo-random numbers strongly depends on the choice of the value A: this number in binary notation must have a sufficiently "random" value, although its last digit should be taken as one. The value has little effect on the quality of the sequence, but it has been noted that some values ​​are unsuccessful.

With the help of experiments and theoretical analysis, the following values ​​have been investigated and recommended: for BESM-4; for BESM-6. For some American computers, these numbers are recommended and are related to the number of digits in the mantissa and the order of the number, so they are different for each type of computer.

Remark 1. In principle, formulas like (54) can give very long good sequences if they are written in a non-recursive form and all multiplications are performed without rounding. Normal rounding on a computer degrades the quality of pseudo-random numbers, but nevertheless, the members of the sequence are usually suitable.

Remark 2. The quality of the sequence improves if small random perturbations are introduced into algorithm (54); for example, after normalizing a number, it is useful to send the binary order of the number to the last binary digits of its mantissa

Strictly speaking, the regularity of pseudo-random numbers should be imperceptible in relation to the required particular application. Therefore, in simple or well-formulated problems, sequences of not very good quality can be used, but special checks are required.

Arbitrary distribution. To play a random variable with non-uniform distribution, you can use formula (52). Play y and determine from equality

If the integral is taken in its final form and the formula is simple, then this is the most convenient way. For some important distributions - Gauss, Poisson - the corresponding integrals are not taken and special ways of playing out have been developed.

Let it be required to play a continuous random variable X, i.e. get the sequence of its possible values ​​(i=1, 2, ..., n), knowing the distribution function F(x).

Theorem. If is a random number, then the possible value of the continuous random variable X being played with a given distribution function F (x), corresponding to , is the root of the equation .

Rule 1 To find a possible value, a continuous random variable X, knowing its distribution function F (x), it is necessary to choose a random number , equate its distribution function and solve the resulting equation .

Remark 1. If it is not possible to solve this equation explicitly, then resort to graphical or numerical methods.

Example 1. Play 3 possible values ​​of a continuous random variable X distributed uniformly in the interval (2, 10).

Solution: Let's write the distribution function of the value X, distributed uniformly in the interval (a, b): .

By condition, a=2, b=10, therefore, .

Using rule 1, we write an equation to find possible values ​​of , for which we equate the distribution function to a random number:

From here .

Let's choose 3 random numbers, for example, , , . Substitute these numbers into the equation, resolved with respect to ; as a result, we obtain the corresponding possible values ​​of X: ; ; .

Example 2. A continuous random variable X is distributed according to an exponential law given by the distribution function (the parameter is known) (x > 0). It is required to find an explicit formula for playing out the possible values ​​of X.

Solution: Using the rule, write the equation .

Let's solve this equation for : , or .

The random number is in the interval (0, 1); hence the number is also random and belongs to the interval (0,1). In other words, R and 1-R are equally distributed. Therefore, to find it, you can use a simpler formula.

Remark 2. It is known that .

In particular, .

It follows that if the probability density is known, then to play out X, instead of the equations, we can solve the equation with respect to .

Rule 2 In order to find the possible value of a continuous random variable X, knowing its probability density , one must choose a random number and solve an equation or equation with respect to , where a is the smallest finite possible value of X.

Example 3. Given the probability density of a continuous random variable X in the interval ; outside this interval. It is required to find an explicit formula for playing out the possible values ​​of X.

Solution: Let's write an equation in accordance with rule 2.

After integrating and solving the resulting quadratic equation relatively , finally we get .

18.7 Approximate play of a normal random variable

Recall first that if a random variable R is uniformly distributed in the interval (0, 1), then its mathematical expectation and variance are respectively equal: М(R)=1/2, D(R)=1/12.

Let us compose the sum of n independent, uniformly distributed in the interval (0, 1) random variables : .

To normalize this sum, we first find its mathematical expectation and variance.

It is known that the mathematical expectation of the sum of random variables is equal to the sum of the mathematical expectations of the terms. The sum contains n terms, the mathematical expectation of each of which, due to M(R)=1/2, is 1/2; therefore, the expectation of the sum

It is known that the variance of the sum of independent random variables is equal to the sum of the variances of the terms. The sum contains n independent terms, the variance of each of which, due to D(R)=1/12, is equal to 1/12; hence the variance of the sum

Hence the standard deviation of the sum

We normalize the sum under consideration, for which we subtract the mathematical expectation and divide the result by the standard deviation: .

By virtue of the central limit theorem at , the distribution of this normalized random variable tends to normal with the parameters a=0 and . For finite n, the distribution is approximately normal. In particular, for n=12 we obtain a fairly good and easy-to-calculate approximation .

The estimates are satisfactory: close to zero, little different from one.

List of sources used

1. Gmurman V.E. Theory of Probability and Mathematical Statistics. - M.: Higher school, 2001.

2. Kalinina V.N., Pankin V.F. Math statistics. - M .: Higher school, 2001.

3. Gmurman V.E. Guide to solving problems in probability theory and mathematical statistics. - M .: Higher school, 2001.

4. Kochetkov E.S., Smerchinskaya S.O., Sokolov V.V. Theory of Probability and Mathematical Statistics. - M.: FORUM: INFRA-M, 2003.

5. Agapov G.I. Problem book on the theory of probability. - M .: Higher School, 1994.

6. Kolemaev V.A., Kalinina V.N. Theory of Probability and Mathematical Statistics. – M.: INFRA-M, 2001.

7. Wentzel E.S. Probability Theory. - M .: Higher school, 2001.