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 total offered traffic is (150X16.1) = 2.4 Erl. SINGLE SERVICE DIMENSIONING (2) The ‘Erlang B’ formula illustrated below may be used to calculate the Blocking Probability (PB) or Grade of Service (GoS) of the network. 171! exceeds the largest double-precision number which is about 1. .8*10308 . Applications like call centers often require values of N > 171. When the number of lines N is large. This formula can cause numerical problems on a computer. For example. terms like N! can exceed computer limits. 2684 26. . which gives Blocking equal to 0.84 The calculations are based on α=2.SINGLE SERVICE DIMENSIONING (3) In this example the Blocking Probability is calculated below.415 erl.049%. However.27049 or 27. the accurate value of α = 2. 0.4 erl. A recurrent formula for Erlang B (1) The following recurrent formula is convenient for computation: aEs−1 (a) Es (a) = . E0 (a) = 1 s + aEs−1 (a) Proof (1) A formula used in xls calculations! α E S ( α ) = s! i = S α ∑ i! i=0 1 S −1 S αα S −1 s(s − 1)! S −1 i=0 ∑ α α + i! s! i S = αα S −1 s(s − 1)! S −1 i=0 ∑ α αα + i! s(s − 1)! i S −1 = = i=0 ∑ αi i! 1 αi i! αα S −1 s(s − 1)! αα S −1 s(s − 1)! 1+ S −1 i=0 ∑ . A recurrent formula for Erlang B (2) Proof (2) Since: E S −1 ( α ) = α S −1 (s − 1)! S −1 i=0 ∑ αi i! the recurrent equation results. E0 (a) =1 s + aEs−1(a) . aEs−1(a) Es (a) = . 2 0.0 0.Blocking Probability vs Number of Servers The following graph shows the call blocking probability for various traffic loads (1 to 20 erl) and different number of servers. 1.4 15 servers 0. 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.8 Call Blocking Probability 5 servers 0.0 0 2 4 6 8 10 12 14 16 18 20 Offered Traffic Load (erl) .6 10 servers 0. 01 n:αριθµός πηγών n=50 0 10 n= n: number of users n=40 προσφερόµενο φορτίο α (erl) 40 Offered traffic (erl) 30 n=30 20 n=20 10 n=10 Erlang B n=∞ 0 0 10 20 30 40 50 Number servers Αριθµός of Servers . are called traffic load curves (see fig).01 B=0.Traffic load curves 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. 50 B=0. Performance Measures of Erlang’s loss model N = ∑ jP( j ) = ∑ j =0 j =0 s s s aj a j −1 j P(0) = a∑ P(0) = a(1 − Es (a)) j! j =1 ( j − 1)! N = a(1 − Es (a)) = acarried Average number of calls in the system Average number of busy servers Server Utilization = a (1 − E s ( a )) s where Es(α) is the Erlang B formula Increase in carried traffic when the number of servers is increased from s to s+1: a ( E s ( a ) − E s +1 ( a )) . Since C1 requires 1 channel and C2 requires two. .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. The blocking probability for each service can be calculated by adding the probabilities of each blocking combination. there could never be more than five C1 connected or more than two C2 connected users. MULTI-SERVICE DIMENSIONING (2) Bandwidth sharing policy: Complete Sharing policy Link of C= 5 1st service: b1=1 2nd service: b2=2 1st service calls C=5 lost traffic available bandwidth unit offered traffic 2nd service calls ί carried traffic time . nK). b2 = 2 n2 − n2 12 states (n1. αk=λk/ µκ (erl) and n ⎛ K ak k ⎞ ⎜ ⎟ G ≡ G(Ω) =∑ ⎜ ∏ n ! ⎟ n∈Ω ⎝ k =1 k ⎠ . n2) n2 µ2 λ2 λ1 ( n1 + 1) µ 1 Local balance Ω 2 1 n1− λ1 n1 µ1 n n1+ (n2 + 1)µ2 λ2 n1 0 1 2 3 −1 ⎜ + n2 4 ⎛ ⎜ ⎝ 5 n ak k ⎞ ⎟ nk ! ⎟ ⎠ Product form solution P (n ) = G ∏ k =1 K where n = (n1. b1 = 1. n2.….…nk.MULTI-SERVICE DIMENSIONING (3) State space – Product form solution – Local Balance C = 5. Blocking for Blocking for of 1st class of 2nd class j=n1*b1+n2*b2=n1+2n2 1st class 2nd class n1 n2 j=n1*b1+n2*b2 0 0 0 1 1 1 2 2 3 3 4 5 0 1 2 0 1 2 0 1 0 1 0 0 0 2 4 1 3 5 2 4 3 5 4 5 .MULTI-SERVICE DIMENSIONING (4) State space – Blocking states for our example In-service In-service Occupied bandwidth in calls calls the system. ) local balance λk q( j − bk ) = yk ( j )µk q( j ) Kaufman (1981) y k (j) µ k Call blocking probability: P k = ∑ G q( j ) b j =C −bk +1 Accurate and easy calculation! C −1 G= j =0 ∑ q( j ) C .KAUFMAN-ROBERTS ALGORITHM (1) C = 5. b1 = 1. on Commun. b2 = 2 λ2 λ1 λ1 j=2 λ2 λ1 λ2 λ1 j=3 µ1y1(3) µ1y1(4) λ2 λ1 j=4 µ1y1(5) λ2 j=5 j=0 j=1 µ1y1(1) µ1y1(2) (Note that the formulas use integer values for bandwidth) µ2y2(2) µ2y2(3) µ2y2(4) λk µ2y2(5) Formula Kaufman / Roberts (1981) j j-b k (IEEE Trans. 21765.75. q(2) = 0.1979.2902. q(4) = 0. q(2) = 0.4166. q(5)=0.023579 Pb2 = q(4)+q(5)=0. q(3)= 0. q(1)=1.023579 Blocking probabilities: Pb1 = q(5) = 0. q(5)=0. q(4) = 0.4458 (normalization constant) Normalized values of q(j)’s for our example: The previous values are divided by G q(0)=0.057436.08125 G = 3.KAUFMAN-ROBERTS ALGORITHM (2) Unnormalized values of q(j)’s for our example: q(0)=1.08102 The values in Figure 4-16 (page 106) and Fig 4-17 (page 107) are wrong! .12092. q(3)= 0.2902. q(1)=0. 5625 erl (values chosen so that α1b1 = α2b2).00002) Results for the 2nd service (α2=1.u. and Pb1 = 0.00053) This value should be multiplied by b2=16 b. So Ctotal = 41 + 112 = 153 b. Let α1=25 erl and α2=1.u. > 146 b. 1st approach: Use the Kaufman-Roberts formula Results: C = 146 b.00002 and Pb2 =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.u and b2=16 b.u.u.00099 (smaller than GOS=0.): C = 41 b.KAUFMAN-ROBERTS ALGORITHM (3) Why not to use ErlangB instead of Kaufman-Roberts? Simple Example: Consider two services with b1=1 b.u.u.001. and therefore C = 7*16= 112 b. Find the capacity C.u.u.5625 erl and b2 = 1 b. (Pb1=0. (Pb2 = 0. System should be dimensioned so that GOS = 0. given by Kaufman-Roberts formula .): C = 7 b.u. u.u.KAUFMAN-ROBERTS ALGORITHM (4) 3rd approach: Use the Erlang B formula assuming that αtotal = α1b1 + α2b2 = 50 erl This gives C = 71 b. . given by Kaufman-Roberts formula). (much lower compared to 146 b. The Bandwidth Reservation Policy Random arriving calls QoS guarantee ON constant bit rate Link of Capacity C = 8 1st Service-class: b1=1 2nd Service-class: b2=2 While in service: Traffic Loss Free Bandwidth Unit Reserved Bandwidth Unit (to benefit the 2nd service-class) fixed bandwidth requirement upon arrival 1st Service-class calls C=8 Offered traffic Carried traffic fixed bandwidth requirement upon arrival 2nd Service-class calls time Exponentially Distributed Interarrival Time Bandwidth Reservation Policy .