排队论的matlab仿真(包括仿真代码).doc

完整版)排队论的matlab仿真(包括仿真代码)
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Experiment Three:
Queuing Theory

ABSTRACT
This experiment is designed to learn the fundamentals of the queuing theory。 Mainly about the M/M/S and M/M/n/n queuing models.

KEY WORDS： queuing theory， M/M/s， M/M/n/n, Erlang B， Erlang C。

INTRODUCTION
A queue is a waiting line and queueing theory is the mathematical theory of waiting lines。 More generally, queueing theory is concerned with the mathematical modeling and analysis of systems that provide service to random demands。 In communication networks, queues are encountered everywhere。 For example， the incoming data packets are randomly arrived and buffered, waiting for the router to deliver。 Such situation is considered as a queue. A queueing model is an abstract description of such a system。 Typically， a queueing model represents （1) the system's physical configuration， by specifying the number and arrangement of the servers， and (2) the stochastic nature of the demands, by specifying the variability in the arrival process and in the service process.
The essence of queueing theory is that it takes into account the randomness of the arrival process and the randomness of the service process。 The most common assumption about the arrival process is that the customer arrivals follow a Poisson process, where the times between arrivals are exponentially distributed. The probability of the exponential distribution function is 。

l Erlang B model
One of the most important queueing models is the Erlang B model （i.e。, M/M/n/n）。 It assumes that the arrivals follow a Poisson process and have a finite n servers. In Erlang B model， it assumes that the arrival customers are blocked and cleared when all the servers are busy. The blocked probability of a Erlang B model is given by the famous Erlang B formula，

where n is the number of servers and A= is the offered load in Erlangs， is the arrival rate and is the average service time. Formula （1.1） is hard to calculate directly from its right side when n and A are large. However, it is easy to calculate it using the following iterative scheme:

l Erlang C model
The Erlang delay model （M/M/n） is similar to Erlang B model, except that now it assumes that the arrival customers are waiting in a queue for a server to become available without considering the length of the queue。 The probability of blocking (all the servers are busy） is given by the Erlang C formula,

Where if and if 。 The quantity indicates the server utilization. The Erlang C formula (1.3） can be easily calculated by the following iterative scheme

where is defined in Eq。（1.1）.

DESCRIPTION OF THE EXPERIMENTS

1. Using the formula （1。2）, calculate the blocking probability of the Erlang B model。 Draw the relationship of the blocking probability PB（n，A) and offered traffic A with n = 1，2， 10， 20， 30， 40， 50, 60， 70， 80， 90， 100. Compare it with the table in the text book （P。281, table 10.3)。
From the introduction, we know that when the n and A are large， it is easy to calculate the blocking probability using the formula 1.2 as follows.

it use the theory of recursion for the calculation。 But the denominator and the numerator of the formula both need to recurs（) when doing the matlab calculation, it waste time and reduce the matlab calculation efficient. So we change the formula to be :

Then the calculation only need recurs once time and is more efficient.

The matlab code for the formula is： erlang_b.m

％＊＊＊＊*＊**＊*＊*＊＊******＊＊*＊*＊＊＊***＊*＊**＊＊
％ File： erlanb_b.m
％ A = offered traffic in Erlangs。
％ n = number of truncked channels.
％ Pb is the result blocking probability.
%＊＊＊*＊＊***＊**＊*****＊*＊*＊****＊*＊*＊*＊＊＊**
function [ Pb ] = erlang_b（ A,n )
if n==0
Pb=1； % P（0,A)=1
else
Pb=1/（1+n/（A*erlang_b(A，n—1))); % use recursion "erlang(A,n—1)"
end
end

As we can see from the table on the text books， it uses the logarithm coordinate, so we also use the logarithm coordinate to plot the result。 We divide the number of servers(n) into three parts， for each part we can define a interval of the traffic intensity（A) based on the figure on the text books ：
1. when 0〈n〈10， 0。1〈A〈10.
2。 when 10〈n〈20, 3〈A<20.
3。 when 30igure the logarithm coordinate.

The matlab code is ：
%＊***＊＊＊*＊***＊**＊＊＊***＊＊**＊＊＊******＊*****＊
% for the three parts。
％ n is the number servers.
％ A is the traffic indensity。
% P is the blocking probability。
%**＊**＊＊＊**＊*＊*＊**＊**＊＊＊＊＊*＊**＊＊*＊*＊**＊＊＊＊
n_1 = ［1:2];
A_1 = linspace（0。1,10,50 % 50 points between 0。1 and 10。
n_2 = ［10：10:20］;
A_2 = linspace（3，20，50）;
n_3 = ［30:10：100]；
A_3 = linspace（13，120，50）；

％＊**＊＊**＊＊＊**＊＊***＊＊***＊＊*＊＊*＊＊＊＊**＊＊＊＊*＊＊
％ for each part， call the erlang_b（) function。
％*＊＊＊*＊＊＊*＊**＊***＊******＊＊*＊＊＊*＊＊＊*＊＊＊***＊
for i = 1:length(n_1）
for j = 1：length(A_1）
p_1（j，i) = erlang_b（A_1（j)，n_1（i））；
end
end
for i = 1：length（n_2）
for j = 1：length(A_2)
p_2（j,i) = erlang_b（A_2（j)，n_2(i));
end
end
for i = 1：length（n_3)
for j = 1:length(A_3)
p_3（j，i) = erlang_b（A_3（j），n_3（i））;
end
end

%**＊＊***＊*＊＊*＊*＊＊＊*＊＊*＊＊**＊*＊**＊＊＊＊**＊**＊＊
％ use loglog to figure the result within logarithm coordinate.
%**＊＊＊＊＊*＊*＊**＊*＊＊**＊＊**＊*********＊*＊＊**＊*
loglog(A_1，p_1，'k—’，A_2，p_2，’k—’，A_3，p_3，’k-')；

xlabel（’Traffic indensity in Erlangs （A）'）
ylabel（’Probability of Blocking （P）’)
axis(［0。1 120 0.001 0.1］)
text（。115， .115，'n=1'）
text(。6， 。115，'n=2’）
text（7， 。115,'10’)
text（17， .115,'20’)
text（27, 。115,’30’）
text（45, .115，’50’）
text(100， .115,’100’）

The figure on the text books is as follow:

We can see from the two pictures that, they are exactly the same with each other except that the result of the experiment have not considered the situation with n=3,4,5,…，12，14，16,18。

2. Using the formula (1.4）， calculate the blocking probability of the Erlang C model。 Draw the relationship of the blocking probability PC（n,A) and offered traffic A with n = 1,2, 10, 20, 30, 40， 50, 60， 70， 80， 90, 100。
From the introduction， we know that the formula 1。4 is :

Since each time we calculate the ， we need to recurs n times， so the formula is not efficient. We change it to be：

Then we only need recurs once. is calculated by the “erlang_b” function as step 1。

The matlab code for the formula is ： erlang_c。m

％＊＊*＊＊**＊*＊＊＊＊＊*＊＊＊*＊＊＊＊*＊*＊＊***＊＊＊＊**＊
% File: erlanb_b.m
% A = offered traffic in Erlangs.
% n = number of truncked channels。
％ Pb is the result blocking probability。
％ erlang_b(A,n) is the function of step 1.
％*＊＊****＊＊＊＊*＊****＊*******＊*＊**＊＊＊＊**＊＊
function ［ Pc ] = erlang_c（ A，n )
Pc=1/（（A/n）+（n—A）/（n*erlang_b（A，n)））;
end

Then to figure out the table in the logarithm coordinate as what shown in the step 1。

The matlab code is ：
%**＊**＊*＊*＊*＊**＊＊＊＊＊＊**＊***＊**＊**＊＊**＊*＊＊*
％ for the three parts.
％ n is the number servers.
% A is the traffic indensity。
% P_c is the blocking probability of erlangC model。
％*＊＊＊*＊*＊*＊*＊*＊*＊＊*＊**＊＊＊＊**＊**＊*＊＊＊＊****＊
n_1 = ［1：2］;
A_1 = linspace(0.1，10，50); % 50 points between 0.1 and 10.
n_2 = [10：10：20］；
A_2 = linspace（3，20，50)；
n_3 = [30:10：100］；
A_3 = linspace(13,120,50）;

%*＊＊＊*＊＊＊*＊**＊**＊＊**＊****＊**＊＊***＊*＊*****＊
% for each part, call the erlang_c（） function。
％＊*＊＊*＊＊****＊＊＊＊＊**＊*＊＊*＊*＊＊*＊＊＊＊**＊*＊＊***
for i = 1:length（n_1）
for j = 1:length(A_1）
p_1_c(j,i） = erlang_c（A_1(j),n_1（i））;
%µ÷Óú¯Êýerlang_c
end
end
for i = 1：length（n_2)
for j = 1：length（A_2)
p_2_c(j，i） = erlang_c(A_2（j),n_2(i));
end
end
for i = 1:length（n_3)
for j = 1：length(A_3）
p_3_c(j,i) = erlang_c(A_3（j），n_3(i））；
end
end

%*＊*＊＊＊＊**＊***＊*＊*＊***＊＊＊＊**＊***＊***＊＊＊**
% use loglog to figure the result within logarithm coordinate。
%＊＊**＊＊*＊＊**＊＊**＊**＊*＊＊*＊＊＊*＊*＊＊＊＊*＊*＊＊＊＊＊
loglog（A_1，p_1_c，’g＊—',A_2，p_2_c，’g＊-’,A_3，p_3_c,'g＊—’）；

xlabel（'Traffic indensity in Erlangs （A）’)
ylabel('Probability of Blocking （P）’)
axis([0.1 120 0.001 0.1])
text（。115, .115,'n=1’)
text（.6， .115,’n=2')
text(6， 。115，'10’）
text(14， 。115,'20’)
text（20, 。115,'30')
text（30， .115,’40’)
text（39, 。115，'50')
text（47, 。115，'60’)
text（55， .115，'70’）
text(65， .115,’80')
text(75, 。115，’90'）
text(85, 。115，'100’）
The result of blocking probability table of erlang C model。

Then we put the table of erlang B and erlang C in the one figure, to compare their characteristic。

The line with ‘ * ’ is the erlang C model， the line without ‘ * ’ is the erlang B model. We can see from the picture that， for a constant traffic intensity (A）, the erlang C model has a higher blocking probability than erlang B model. The blocking probability is increasing with traffic intensity。 The system performs better when has a larger n.

ADDITIONAL BONUS

Write a program to simulate a M/M/k queue system with input parameters of lamda， mu, k.
In this part， we will firstly simulate the M/M/k queue system use matlab to get the figure of the performance of the system such as the leave time of each customer and the queue length of the system。
About the simulation, we firstly calculate the arrive time and the leave time for each customer。 Then analysis out the queue length and the wait time for each customer use “for” loops。
Then we let the input to be lamda = 3, mu = 1 and S = 3, and analysis performance of the system for the first 10 customers in detail.
Finally, we will do two test to compared the performance of the system with input lamda = 1, mu = 1 and S = 3 and the input lamda = 4， mu = 1 and S = 3.

The matlab code is: mms_function.m

function[block_rate,use_rate］=MMS_function(mean_arr，mean_serv,peo_num,server_num）
％%％％％%％%％％%％％%％%％％％%%%％%％％％％%%％%%%％％%％%%%％％%％%%％%％%％%％%％％%%％%％％%%%％%%%%
%first part： compute the arriving time interval, service time
%interval,waiting time, leaving time during the whole service interval
％%％%％％％%％％%%％%％%%%%％％%％%%％%％%%％％%％%％％％%％%%％％%%%％％%%％％%％%％％%％%%%%％%%%%％％％

state=zeros（5,peo_num)；
%represent the state of each customer by
%using a 5*peo_num matrix
％the meaning of each line is: arriving time interval, service time
％interval， waiting time， queue length when NO。ncustomer
％arrive， leaving time
state（1，:）=exprnd（1/mean_arr,1,peo_num）;
％arriving time interval between each
%customer follows exponetial distribution
state(2，:）=exprnd（1/mean_serv,1,peo_num)；
％service time of each customer follows exponetial distribution
for i=1：server_num
state（3,1:server_num)=0；

end
arr_time=cumsum（state（1,：））;
%accumulate arriving time interval to compute
％arriving time of each customer
state（1,:)=arr_time;
state（5，1：server_num)=sum(state（:，1:server_num));
％compute living time of first NO。server_num
%customer by using fomular arriving time + service time
serv_desk=state(5,1：server_num）；
%create a vector to store leaving time of customers which is in service
for i=（server_num+1）：peo_num
if arr_time（i)〉min（serv_desk)
state（3，i)=0；
else
state(3，i)=min(serv_desk)—arr_time(i);
％when customer NO.i arrives and the
％server is all busy， the waiting time can be compute by
％minus arriving time from the minimum leaving time

end
state（5,i)=sum(state(：,i)
for j=1:server_num
if serv_desk（j）==min(serv_desk）
serv_desk（j）=state(5,i）；
break
end
%replace the minimum leaving time by the first waiting customer’s leaving time
end
end

％％%%%％％％%%％%％％%%％％％％%％％%%%％％%%%％％％%％％％％%%%%％％％%%％％％%％％%％%％%%%％％%%％％%％％％％
％second part: compute the queue length during the whole service interval
％%％%％％%％％％%％%％%%％%％％%％%％％％%％％%％%%％％％％％％％%%%%％％%%％％%%％%%％％％％％％%%%%%％％％%％

zero_time=0;
%zero_time is used to identify which server is empty

serv_desk(1:server_num）zero_time；
block_num=0;
block_line=0；
for i=1:peo_num
if block_line==0
find_max=0;
for j=1:server_num
if serv_desk（j）==zero_time
find_max=1； ％means there is empty server
break
else continue
end
end
if find_max==1
％update serv_desk
serv_desk(j)=state(5，i)；
for k=1：server_num
if serv_desk（k)〈arr_time(i） ％before the NO.i customer actually arrives there maybe some customer leave
serv_desk（k)=zero_time；
else continue
end
end
else
if arr_time(i)〉min(serv_desk）
%if a customer will leave before the NO.i
%customer arrive
for k=1:server_num
if arr_time(i)〉serv_desk(k)
serv_desk（k）=state(5,i）；
break
else continue
end
end
for k=1：server_num
if arr_time（i）>serv_desk(k）
serv_desk（k）=zero_time；
else continue
end
end
else %if no customer leave before the NO。i customer arrive
block_num=block_num+1;
block_line=block_line+1；
end
end
else %the situation that the queue length is not zero
n=0；
％compute the number of leaing customer before the NO.i customer arrives
for k=1：server_num
if arr_time（i）>serv_desk(k）
n=n+1；
serv_desk（k)=zero_time；
else continue
end
end
for k=1：block_line
if arr_time（i）>state(5，i—k)
n=n+1;
else continue
end
end
if nh is still not zero
block_num=block_num+1;
for k=0:n-1
if state(5,i—block_line+k)〉arr_time（i）
for m=1：server_num
if serv_desk(m）==zero_time
serv_desk（m)=state（5，i-block_line+k）