Ideal Grading Interconnection In Telecommunication Traffic Engineer

A grading can be said ideal if total group that made at beam in is maximal and each group has access (only can access) to k, with connection:

æ N ö = N!
è k ø k ! (N – k) !

Where:
Line connection = k
Total line out = N, and N > k ® there is interconnection

Example:

N = 4
k = 2

Total maximal group that must be made:

= æ 4 ö = 4! = 4.3 = 6 group
è 2 ø 2 ! (4 – 2) ! 2

The figure as follows:

Figure 5.13 configurations for ideal grading.

So the problem, from (total group) x k = 6 x 2 = 12 line out ® interconnecting so that be 4 line out.

Interconnection Techniques In Telecommunication Traffic Engineer

Cascade Interconnection (grading)

Figure 5.8 Cascade Interconnection.

Enclose interconnection

Figure 5.9 Enclose interconnection (1).

Enclose interconnection

Figure 5.10 Enclose interconnection (2).

Enclose interconnection

Figure 5.11 Enclose interconnection (3)

Combination Interconnection

Figure 5.12 Combination Interconnection

Best Grading In Telecommunication Traffic Engineer

The best grading is grading that give where does searching line out, from 1 line out to the other line out, smoothest progression and line out load is equal. In figure 5.6 showed a system with beginning condition: k = 10, n = 31 line and total group = 6.
















Figure 5.6 System with 31 line out (a, b, c, and d).

Assumed grading with N = 31 line out (server) not yet optimum and be looked for best grading form. When assumed best grading is achieved in N = 27 line out, so grading form looked for in the following way:
Make equation based on system condition in the beginning situation:

a + b + c + d = 10

6a + 3b + 2c + d = 27

From both above equation will get:

5a + 2b + c = 17

Do predictions (approximately) from a, b, c, d, values and so on, and these values must fit if putted into to all above equation. then count the absolute difference value (NSM) that is total from difference (absolute value) a, b, c, d, value in a series with equation:

NSM = S ( |a-b| + |b-c| + |c-d| )

Do predictions and calculation several times, example 10 times, and the result table likes in table 5.1.


Individual a 3 3 2 2 2 2 1 1 1 0 0
Pairs b 1 0 0 1 2 3 4 5 6 8 6
Three c 0 2 7 5 3 1 4 2 0 1 3
Common d 6 5 1 2 3 4 1 2 3 1 0
NSM 9 8 15 8 1 6 6 7 14 1 14

Table 5.1 best grading calculation table.

Based on calculation data result, take values a, b, c, and d, that gives minimum NSM. From calculation, got value a, b, c, and d, that give minimum NSM and fulfill equation: a = 2, b = 2, c = 3, d = 3.

Got final configuration as best grading:














Figure 5.7 The Best grading configuration.

Grading Interconnection In Telecommunication Traffic Engineer

Traffic units many used in traffic engineer analysis are offered traffic (A), grade of service (B), and device capacity (N). This third unit is connect very tight, so that when two unit already known, then third unit countable. Usually B value is determined, so that if A is fluctuates then N will follow that change.

Generally companies make device with N standard, so that when A is big and b is certain, then will emerge how if N that need is bigger than N standard? (N > Nstd ?)

Example:
















Figure 5.1 Example system divide in to several group.

Like showed in figure 5.1, for example showed a switching system that has line out (selector or server) is bigger than N standard and used to serve 1000 customers. To handle that condition, so system is divided in to 5 sub-system with each serve 200 customers and line capacity n1, n2, ……, n5, each £ total line out (server) standard.

In next example, see figure 5.2, showed complete calculation so that seen change A and N value for appointed B value.

In example is showed switching system that used to process traffic as big as 6,71 Erlang with desirable grade of service 0,2 %. Based on calculation, for the mentioned is need 15 line out (server). The bundle is perfect bundle.

When obvious there is no N standard that have 15 line out, so the solution with divide to be 2 sub group as follows:

Sub group I : A = 6,71 / 2 = 3,355 Erlang
B = 0,2 %
® needed n1 = 10 line

Sub group II : A = 6,71 / 2 = 3,355 Erlang
B = 0,2 %
® needed n2 = 10 line

The result from that technique showed in figure 5.2 (B), where seen:
- A is total and B is constant
- Each sub group is perfect bundle
- Need 2 x 10 = 20 line out (server)
- In its entirety not perfect bundle again






















Figure 5.2 Solving example for N > N standard.

Conclusion:
Perfect bundle gives:
Total line out is minimum
Efficiency per line out optimum

By becoming imperfect bundle then efficiency is decreased. To increase the efficiency, can be done by multiplication process with interconnection, grading, or increasing switching level. Example configuration that simplified from multiplication is showed in figure 5.3 and 5.4.

Interconnection :








Figure 5.3 Multiplication configurations with interconnection.

Increasing Switching Level:








Figure 5.4 multiplication configurations with increasing switching level.

Interconnection example:
- Selector system that used has line connection k = 5
- N = 8 line out (server)
- Found 2 groups that will interconnected





K = 5 ü before interconnection, total line out (server) = 10
ý
2 group þ After interconnection N = 8 line

Interconnection can be done with several ways, usually done with cascade interconnection (grading). Example from cascade interconnection form, showed in figure 5.5, with:
k = 10
total group = 4
N = 23










Figure 5.5 Example cascade interconnection (grading) configuration.

Before interconnection: Total line out (server) : N = 4 x 10 = 40 line
After interconnection: N = 23 line (12 individual, 8 pair, 3 four).

In this case explanation from interconnection = grading.

Engset and Bernoulli Distribution In Telecommunication Traffic Engineer

Engset and Bernoulli distribution got from condition as follows:

1. Calling source (total customer) is limited.
2. Total server (line) that processing is limited.
3. Incoming call rate depend on not busy total customer.
4. Customer that made a success of occupation server doesn't do calling again.

If S declares full scale customer and N declare total server, so when:
S > N: got Engset distribution (operative Engset traffic or Purechance traffic (type 2)
S £ N: got Bernoulli or Binominal distribution, (operative Bernoulli traffic)

Because incoming call rate or offered traffic to system is proportional with not busy total customer, so for n condition can be declared with equation:

Cn = (S – n) l

Where:
l : Incoming call rate per customer and is call intensities free calling source.
(S – n) : is total calling source (costumer) that still free.

Cn can be assumed as birth coefficient in n(=bn) condition, so that condition diagram for Engset and Bernoulli distribution can be described as follows:













Figure 4.3 Engset and Bernoulli distribution condition diagram.

Balance equation:

(S-n)l.P(n) = (n+1)s.P(n+1) n = 0, 1, 2, ….. , (N-1 atau S-1)

So that completion for:
n = 0: Sl.P(0) = s.P(1)
® P(1) = S(l/s).P(0)

n = 1: (S-1)l.P(1) = 2s.P(2)
® P(2) = S(S-1)(l/s)2.(1/2).P(0)

n = 2: (S-2)l.P(2) = 3s.P(3)
® P(3) = S(S-1)(S-2)(l/s)3.(1/3.2.1).P(0)
So on.

Got:

Pn = { S ! / n ! (S-n) ! }.(l/s)n.P(0)

Or:

Pn = æ S ö . (l/s)n.P (0)
è n ø

Where:

æ S ö is binomial coefficient = S!
è n ø n!(S – n)

Equation is demoted from diagram condition, and good operative for Engset also Bernoulli:

If Engset : end condition = N (S>N)
Binomial : end condition = S (S£N)

Completion for Engset Distribution : (S>N)
Blocking probability (n=N)

æ S öæ l öN
PN = è Nøè s ø
N
∑ æ S öæ l öj
j=0 è Nøè sø

PN = all busy/occupied line probability
= Time Congestion

Average offered rate:

N
C = l (S - S n Pn)
N=0

Value S n Pn in equation above represent average total busy server. Traffic that proceed by network is total average accepted calling during average service time period. This is equal to total average busy server in a certain time that is given. Thereby equation above can be made to be:

C = l (S – Y)

So offered traffic to be:

A = C tr = l tr (S – Y)

When system on N condition, offered traffic rate is (S-N)l, and all incoming call on that condition will refused, so that lost traffic is:

A – Y = (S – N) l tr PN

Thereby, now the GOS to be:

GOS = S – N . PN
S – Y

Indicating that is for Engset traffic, blocking probability and unequal grade of service, in this case time congestion and call congestion has different value.

Completion for Bernouli distribution: (S £ N)
With using equation:

Pn = æ S öæ l ön
è n øè s ø

where :

(l/s)n = {(l+s)/s}S.{1-l/(l+s)}S-n.{ l/(l+s)}n

Then got:

Pn = æ l+s öS . P0æ S ö.æ l ön æ1 - l öS-n
è s ø è n ø èl+sø è l+sø

Bernoulli formula:

Pn = æ S ö.pn æ1 - pöS-n
è n ø

From both above equation got:

{(l+s)/s}S.P0 = 1 and {l /(l+s)} = p

So:

P0 =. 1 = æ s öS
æl+söS èl+sø
è s ø

Equation above known as Bernoulli or Binomial equation, that explain servers not interdependent one with another.

Erlang Distribution In Telecommunication Traffic Engineer

Generally telecommunication system can classify as loss system or waits system (delay system). This matter happens because total customer that connected to network usually more than total server, so that shaping network that causes the happening of blocking. As finally, be produced overflow traffic in network.

In loss system, overflow traffic will refuse. In this case, there is 3 point may be happen in overflow traffic, that is:
Overflow traffic that refuse by a switching device, there is possibility can be accommodated by other switching device that found in network. This condition is called as LCC (Lost Call Cleared).
Refuse traffic will try to do recalling after a few moments. This condition is called LCR (Lost Call Returned).
Traffic that held by device and then served (proceed) after there is empty server and can be occupied. This condition is called as LCH (Lost Call Held).

Erlang distribution, this distribution is got from condition as follows:
- Infinite calling source
- Average incoming call rate: a (constant)
- Total server (line) that processing is limited
- Incoming calls when all servers busy, refused (LCC system).

Distribution with above condition usually used to analyze traffic in trunk transmission system. Usually in a network found several group of trunk that connects between one switching device (central) with another switching device.



When a group trunk that connect direct between two busy switching devices (all occupied), so make possible to shift traffic pass other switching device that use different group trunk. In this way, callings that experience blocking at a group trunk the service is shifted to other group trunk

LCC traffic analysis condition, first time is done by A.K. Erlang in the year 1917. The aim principal from that analysis is to estimate blocking probability and grade of service (GOS). In this case, to express offered traffic, can be used equation:

A = a.tr

Where a, be Poisson average incoming call rate that the value is equivalent with C (average incoming call rate).

When all servers in a busy state, incoming traffic will be refused by system. Because the traffic loss is occurring, so be produced different incoming rate, called as effective incoming call rate. We are notation average effective incoming call rate as C0, and effective incoming rate in I condition is as Ci. System stay in j condition when is there j busy server. Along all idle servers, so every incoming traffic that can process by network. When all servers is busy, so incoming traffic will be refused. Traffic at one network that follow the limitation, is known as Erlang traffic or pure chance traffic (type 1). In this case at can:

Ci = a untuk 0 £ I < cn =" 0" c0 =" ∑" i="0" pn =" 1" c0 =" a" y =" C0" tr =" a" a =" a." y =" A" pn =" A" gos =" Pn," dn =" n.s" pn =" 0" 1 =" (A"> 0
P1 = A P0 for n = 0

for n = 1, got :

P2 = (A P1 + P1 – AP0) / 2

Substitution for P1

P2 = A2 P0 / 2

With same technique for n = 2, got :

P3 = (A P2 + 2P2 – AP1) / 3 = A3 P0 / (3x2) = A3 P0 / 3!

So that generally got,

Pj = Aj P0 / j!
P0 + AP0 + ….. + AN P0 / N! = 1
P0 = 1 / {1 + A + A2/2! + ….. + AN/N!}

And

Pn = An/n!
1 + A + A2/2! + ….. + AN/N!

For n = N :

Pn = AN/N!
1 + A + A2/2! + ….. + AN/N!

Above equation known as Erlang Loss Equation (Erlang B formula). Where Pn be probability all busy servers and also is blocking probability from system (GOS).

Condition diagram for Erlang distribution showed in figure 4.2, where balanced still occur until n = N - 1.








Figure 4.2 Condition Diagram for Erlang distribution.

Poisson Distribution In Telecommunication Traffic Engineer

This distribution got from condition as follows:

1. Incoming call is randomly, with incoming call rate = a (constant, doesn't depending total occupation) because total calling source is infinite (big).
2. There's only birth process, there is no death process.
3. Total server (channel) that accommodates (to cultivate) is infinite (big), so that that incoming call always can be served by servers.

For condition there is no death, so dn = 0 and equation condition birth-death process be

dPn(t) / dt = Pn-1(t).an-1 – an.Pn(t) for n ³ 1
dP0(t) / dt = - a0.P0(t) for n = 0

Observation assumption is begun in t = 0, that is when does system present in condition to zero where there is no birth occur. Got equation:

ì 1 for n = 0
Pn(0) = í
0 for n ¹ 0

With that condition, than got equation solution:

P0(t) = e-at

From equation above will got (for n = 1) :

dP1(t) / dt = ae-at – aP1(t)

the solving :

P1(t) = at.e-at

for n = 2,

P2(t) = (at)2 e-at / 2!

for n = 3,

P3(t) = (at)3 e-at / 3!

So on, ….. , so that with induction get general resolve as,

Pn(t) = (at)n e-at
n!

The above equation is known as Poisson Distribution Equation or Poisson Arrival Process Equation. This equation expresses probability system with total occupation as much as n on time t., this represent existence probability n arrival in time interval t. Equation also derivable with pay attention condition equation so that furthermore can see the condition diagram form, that is:

dPn(t) = - (bn + dn).Pn(t) + bn-1(t) + dn+1.Pn+1(t)
dt

Whit b1, b2, ….. , bn and d1, d2, ….. , dn, where:
bn : Birth Coefficient
For monstrous calling source, this mean that incoming call rate is constant and hasn't depending on how big calling that success occupy device (server)
so : b1 = b2 = ….. = bn = a (constant)
dn : Death Coefficient

In condition there is no death, so dn = 0, while bn = a, because constant calling rate (doesn't depending n), so that birth-death process equation condition be:

dPn(t) / dt = Pn-1(t).an-1 – an Pn(t) for n ³ 1
dPn(t) / dt = - an Pn(t) for n = 0

To finish this equation, observation assumption is begun in t = 0, that is when does system present in condition to zero where there is no birth that occur. The equation:

ì 1 for n = 0
Pn(0) = í
0 for n ¹ 0

With that condition, for n = 0, the solution is:

Pn(t) = e-at

for n = 1

dp1(t) / dt = ae-at – aP1(t)

The solution:

P1(t) =at e-at

for n = 2

P2(t) = (at)2 e-at / 2!

for n = 3

P3(t) = (at)3 e-at / 3!

So on, ….. , so that with induction is got general resolve as,

Pn(t) = (at)n e-at / n!

The result is Poisson distribution equation. In this case (at) be incoming call average rate multiply with long time occupation average rate = traffic = A, so that equation can be written as,

Pn = An e-A
n!

Condition diagram:









Figure 4.1 Condition diagram for Poisson distribution

Birth coefficient bn = a
Taken from incoming 1 calling probability during dt time = a. dt, (a: incoming call average rate in 1 hour (busy).
Death coefficient dn = n.s
Taken from the end probability just any 1 occupation during dt time = n. s.dt ( n. s : end occupation average rate in condition n in 1 hour (busy rate)).