1) The document describes deriving the probability of blocking for a blocking system using an LCC infinite source traffic model. It shows the derivation of the blocking probability formula using Markov chains. 2) The blocking probability formula is then simulated using MATLAB to generate a graph of blocking probability versus traffic intensity for different numbers of channels. 3) The graph can be used for network planning to determine the maximum traffic or required channels for a given blocking probability goal.

1. BLOCKING PROBABILITY: VERIFYING RESULTS USING COMPUTER SIMULATION
Jasper Valentine Hatilima (MSc Student #: 11042005)
School of Information Science and Technology
Southwest Jiaotong University
Chengdu, Sichuan, PR China
Phone: +8613540757472, email: jasperhatilima@yahoo.com
November 2011
------------------------------------------------------------------------------------------------------------------------------------------
To describe the call handling capacity of a server system, we can define a hypothetical network
with U potential users and is able to handle C simultaneous users (capacity of C channels). If U=<C, then
the system is called a non blocking system; this implies that all calls can be handled at the same time.
Else if U>C, then the system is referred to as a blocking system. For a blocking system, there are two
factors that are used to define the traffic model. Firstly, it is the manner in which a blocked call is
handled: Lost Call delayed (LCD) or Lost Call Cleared (LCC). The second traffic element model is whether
the number of users is assumed infinite or finite. Infinite sources are analytically easier to deal with and
are reasonable when number of sources is 5 to 10 times the capacity.
1. Derivation
Using LCC and Infinite Source model, we will derive the probability of blocking at state k as
below. The system to consider has the following parameters:
 Number of trunked channels = C
 Average mean holding time = H or service rate
 If we take traffic intensity to be A , then A =H [offered Traffic]
We define two statistical components of the trunks:
 Arrival Times
A Poisson distribution is assumed for the arrival of the calls. Therefore from 0 to t seconds, the
probability of n calls arriving is:
e  (t 0) [ (t  0)]n e   t ( t ) n
Pn (t  0)  = Pn (t )  (1)
n! n!
 Service Times
The mean call duration is H and so 1/H=  is the mean number of calls serviced in a unit of time
i.e. mean service time. Here we assume an exponential distribution of the service time.
th
Therefore the probability that the duration S k of the K call will be less than an arbitrary call
duration S is:
Pr{ S k < S }= 1  e  S S >0
This gives a probability density function
p ( S k )   e  Sk (2)
We use the property of Markov chains i.e. the transition from the present state i to the next state i+1
depends only on the state i. The trunk operation is a continuous process but may be analyzed in small

2. discrete small time intervals  . If N z is the number of calls in the system at time z , then N z = N ( z ),
where N is a random process representing the number of occupied channels at discrete times.
Using equation (1) and the small interval of time interval  , then the probability of k arrivals in 
seconds is
e  ( )k
Pr  {k} 
k!
 defines a transition probability to use in the Markov chain i.e. probability of having k channels in
use is equivalent to the probability of having (k-1) channels in use multiplied by transition probability
 i.e.
 Pk 1 = Pk (3)
1
But  is the time equivalent to
k
Therefore using this in equation (3) gives,
 Pk 1  k  Pk
 P0
For k =1, P (4)
1

Equation (3) under steady state condition i.e. a Markov chain over a long period of time becomes:
 Pk 1  k  Pk (5)
 1
For various states up to state k, equation (5) becomes Pk  P0 ( )
k
(6)
 k!
 C
and P0  ( )k Pk k !  1   Pi (7)
 i 1
In equation (6), the probability of blocking for C channels is
 1
PC  P0 ( )C (8)
 C!
1 
( )C
C! 
Substituting equation (7) into (8) yields PC  C (9)
 1
 (  )k k !
k 0

But traffic intensity is A =  H = ; substituting this into equation (9) gives:

1
AC
Probability of Blocking for C servers is PC  C C !
1
 Ak k !
k 0

3. 2. Computer Simulation
The probability of blocking using the above derivation was simulated using Matlab and the following
graph was obtained. Also shown below is the Matlab code used for this graphical analysis.
Figure 1.0 Probability of Blocking Vs Traffic Intensity obtained from Matlab simulation
When C is plotted against Offered Traffic, the graph can be used for two kinds dimensioning cases in
network planning. From the graph a desired probability of blocking (GoS) can be used against the given
number of servers C to determine the maximum traffic that can be carried. Alternatively, this data can
be used to determine the channels required to carry a specific amount of maximum traffic at a given
GoS.
Find below the Matlab M-file used for this analysis. The M-file is also included in the submitted folder
under the file name is Prob_Block_Simulation_Jasper.m:

4. %%Program to simulate blocking probability
%% Name: Jasper Valentine Hatilima
%% Student #: 11042005
%% Southwest Jiaotong University (2011-11-18)
%% ---------------------------Program----------------------------------
clear all;
close all;
C=[1 2 3 4 5 10 20 60]; % The number of Servers
rho=[0.1:0.1:100]; % Traffic intensity rho = lambda/mu
for i=1:length(C),
inc=0;
for q=0:C(i)
inc=inc+rho.^q/factorial(q);
end
pb(i,:)=rho.^C(i)./factorial(C(i))./inc;
end
figure(1)
loglog(rho,pb(1,:),'-b',rho,pb(2,:),'-b',rho,pb(3,:),'-b',...
rho,pb(4,:),'-b',rho,pb(5,:),'-b',rho,pb(6,:),'-b',rho,pb(7,:),'-b',...
rho,pb(8,:),'-b');
%% ----------------------------Plotting---------------------------------
xlabel('Traffic Intensity,rho [Erlangs]');
ylabel('Probability of Blocking');
title('Probability of Blocking Vs Traffic Intensity');
axis([0.1 100 1e-4 1]);
text(2.e-1,0.15,'C=1');
text(3.2e-1,3e-2,'C=2');
text(4.8e-1,1e-2,'C=3');
text(7.2e-1,5e-3,'C=4');
text(1,3e-3,'C=5');
text(3.4,1.5e-3,'C=10');
text(9.8,1.6e-3,'C=20');
text(41,1.6e-3,'C=60');
grid on;
%% ------------------------------END------------------------------------