erlangb kaufman
TRANSCRIPT
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