SINGLE SERVICE DIMENSIONING (1)

In the illustration below, three lines (3 Erl) are used to provide traffic for 150 speech users with an offered traffic requirement of 16.1 mErl. The totaloffered traffic is (150X16.1) = 2.4 Erl.

SINGLE SERVICE DIMENSIONING (2)

The ‘Erlang B’ formula illustrated below may be used to calculate theBlocking Probability (PB) or Grade of Service (GoS) of the network.

This formula can cause numerical problems on a computer.

When the number of lines N is large, terms like N! can exceed computer limits.

For example, 171! exceeds the largest double-precision number which is about 1,8*10308 .

Applications like call centers often require values of N > 171.

In this example the Blocking Probability is calculated below.

0.2684

SINGLE SERVICE DIMENSIONING (3)

26.84

The calculations are based on α=2.4 erl. However, the accurate value of α = 2.415 erl, which gives Blocking equal to 0.27049 or 27.049%.

A recurrent formula for Erlang B (1)The following recurrent formula is convenient for computation:

1)(,)(

)()( 0

1

1 =+

=−

− aEaaEs

aaEaEs

ss

Proof (1)

1)!s(sαα

i!α

11

1)!s(sαα

i!α

1

1)!s(sαα

i!α

1)!s(sαα

s!α

i!α

1)!s(sαα

i!α

s!α

)α(E

1S

1S

0i

i

1S

1S

0i

i

1S

0i

1Si

1S

1S

0i

Si

1S

S

0i

i

S

S

−+

−

=

=

−+

−=

+

−==

−

−

=

−

−

=

−

=

−

−

−

=

−

=

∑

∑

∑∑∑

A formula usedin xls calculations!

A recurrent formula for Erlang B (2)

Proof (2)

Since:

∑−

=

−

−−

=1S

0i

i

1S

1S

i!α1)!(s

α

)α(E

the recurrent equation results.

1)(,)(

)()( 0

1

1 =+

=−

− aEaaEs

aaEaEs

ss

The following graph shows the call blocking probability for various traffic loads (1 to 20 erl) and different number of servers.

According to the results we see a decrease in call blocking when we add servers (this is expected) which becomes substantial in the case of 15 servers.

Offered Traffic Load (erl)0 2 4 6 8 10 12 14 16 18 20

Cal

l Blo

ckin

g Pr

obab

ility

0.0

0.2

0.4

0.6

0.8

1.0

5 servers

10 servers

15 servers

Blocking Probability vs Number of Servers

In practice it is often necessary to determine the number of trunks required for carrying a given traffic load at a given GOS. Curves indicating the relation between α and the number of trunks s providing a given GOS, are called traffic load curves (see fig).

Αριθµός servers

0 10 20 30 40 50

προσφ

ερόµενο φο

ρτίο

α (e

rl)

0

10

20

30

40

50

n=10

n=20

n=30

n=40

n=50

n=100

Erlang B

B=0.01n:αριθµός πηγών

n=∞Offered traffic

(erl)

Number of Servers

B=0.01n: number of users

Traffic load curves

Performance Measures of Erlang’s loss model

1

0 0 1( ) (0) (0) (1 ( ))

! ( 1)!

(1 ( ))

j js s s

sj j j

s carried

a aN jP j j P a P a E aj j

N a E a a

−

= = =

= = = = −−

= − =

∑ ∑ ∑

Average number of calls in the systemAverage number of busy servers

(1 ( ))sa E aServer Utilizations

−=

Increase in carried traffic when the number of servers is increased from s to s+1:

1( ( ) ( ))s sa E a E a+−

where Es(α) is the Erlang B formula

MULTI-SERVICE DIMENSIONING (1)A network with five channels (5 E) serving 50 users with two data services C1 and C2 requiring 1 and 2 channels respectively is illustrated below.

Since C1 requires 1 channel and C2 requires two, there could never be morethan five C1 connected or more than two C2 connected users.

The blocking probability for each service can be calculated by adding theprobabilities of each blocking combination.

MULTI-SERVICE DIMENSIONING (2)Bandwidth sharing policy: Complete Sharing policy

available bandwidth unit

C=5

time

1st servicecalls

2nd servicecalls

ί

Link ofC= 5 1st service: b1=1 2nd service: b2=2

carried traffic

lost traffic

offered traffic

(n1, n2)

n−1n

+2n

+1n

−2n

22µn

1λ

11µn

2λ

1)11( µ+n

22 )1( µ+n

1λ

2λ

Localbalance

⎟⎟⎠

⎞⎜⎜⎝

⎛∏

=

−K

k k

nk

na k

1

1

!where n = (n1, n2,…nk,…,nK), αk=λk/ µκ (erl) andProduct

formsolution G ≡ G(Ω) =∑ ∏

∈ =⎟⎟⎠

⎞⎜⎜⎝

⎛

Ωn

K

k k

nk

na k

1 !

( )nP = G

12 states

C = 5, b1 = 1, b2 = 2

n2

n1

1

2

1 2 3 4 50

Ω

MULTI-SERVICE DIMENSIONING (3)State space – Product form solution – Local Balance

MULTI-SERVICE DIMENSIONING (4)State space – Blocking states for our example

In-service calls

of 1st class

In-service calls

of 2nd class

Occupied bandwidth in the system,

j=n1*b1+n2*b2=n1+2n2Blocking for

1st classBlocking for

2nd classn1 n2 j=n1*b1+n2*b2

0 0 00 1 20 2 41 0 11 1 31 2 52 0 22 1 43 0 33 1 54 0 45 0 5

KAUFMAN-ROBERTS ALGORITHM (1)

Formula Kaufman / Roberts (1981)

(IEEE Trans. on Commun.)

λk

yk(j) µk

j-bk j

)()()( jqjybjq kkkk µλ =−

local balance

∑∑=+−=

− ==C

j

C

kbCjkb jqGjqGP

01

1 )()(Call blocking probability:

Accurate and easy calculation!

C = 5, b1 = 1, b2 = 2

Kaufman (1981)

µ2y2(5)

λ2

µ1y1(5)µ1y1(3)

λ1

µ1y1(1)

λ2

µ2y2(2)

j = 0 j = 1 λ1

j = 2

λ2

j = 4λ1

j = 5

λ2

µ2y2(4)

λ2

µ2y2(3)

j = 3λ1

µ1y1(2)

λ1

µ1y1(4)

(Note that the formulas use integer values for bandwidth)

KAUFMAN-ROBERTS ALGORITHM (2)

Unnormalized values of q(j)’s for our example:q(0)=1, q(1)=1, q(2) = 0.75, q(3)= 0.4166, q(4) = 0.1979, q(5)=0.08125G = 3.4458 (normalization constant)

Normalized values of q(j)’s for our example:The previous values are divided by Gq(0)=0.2902, q(1)=0.2902, q(2) = 0.21765, q(3)= 0.12092, q(4) = 0.057436, q(5)=0.023579

Blocking probabilities:Pb1 = q(5) = 0.023579 Pb2 = q(4)+q(5)=0.08102

The values in Figure 4-16 (page 106) and Fig 4-17 (page 107) are wrong!

KAUFMAN-ROBERTS ALGORITHM (3)Why not to use ErlangB instead of Kaufman-Roberts?

Simple Example: Consider two services with b1=1 b.u and b2=16 b.u. Let α1=25 erl and α2=1.5625 erl (values chosen so that α1b1 = α2b2).

System should be dimensioned so that GOS = 0.001. Find the capacity C.

1st approach: Use the Kaufman-Roberts formula Results: C = 146 b.u. and Pb1 = 0.00002 and Pb2 =0.00099 (smaller than GOS=0.001)

2nd approach: Use the Erlang B formula for each service-class

Results for the 1st service (α1=25 erl and b1 = 1 b.u.): C = 41 b.u. (Pb1=0.00002)

Results for the 2nd service (α2=1.5625 erl and b2 = 1 b.u.): C = 7 b.u. (Pb2 = 0.00053)This value should be multiplied by b2=16 b.u. and therefore C = 7*16= 112 b.u.

So Ctotal = 41 + 112 = 153 b.u. > 146 b.u. given by Kaufman-Roberts formula

3rd approach: Use the Erlang B formula assuming that αtotal =α1b1 + α2b2 = 50 erl

This gives C = 71 b.u. (much lower compared to 146 b.u. given by Kaufman-Roberts formula).

KAUFMAN-ROBERTS ALGORITHM (4)

The Bandwidth Reservation Policy

Free Bandwidth Unit

C=8

time

1st Service-class calls

Link of Capacity C = 8 1st Service-class: b1=1 2nd Service-class: b2=2

Carried traffic

Traffic Loss

Offered traffic

Exponentially Distributed Interarrival Time

2nd Service-class calls

Reserved Bandwidth Unit (to benefit the 2nd service-class)

fixed bandwidth requirement upon arrival

ON While in service: constant bit rate

Random arriving calls

fixed bandwidth requirement upon arrival

QoSguarantee

BandwidthReservationPolicy