Consider a generalization of an M/M/k/k system to the case where the arrival process is not Poisson. As mentioned in the previous section, one way non-Poisson traffic occurs is when we consider a secondary server group to which traffic overflows from a primary server group. If we know the offered traffic to the primary server group (say A) and the number of servers in the primary and secondary groups are given by k1 and k2, respectively, then the blocking
probability of the secondary server groups is obtain by
Pb(secondary) =
Ek1+k2(A)
Ek1(A)
In this case we also know the mean and variance of the traffic offered to the secondary server group which is readily obtained by Equations (342) and (343).
However, the more challenging problem is the following: given a multi-server loss system with
k2 servers loaded by non-Poisson offered traffic with meanM and variance V, find the blocking probability. This offered traffic could have come from or overflowed from various sources, and unlike the previous problem, here we do not know anything about the original offered traffic streams or the characteristics of any primary systems. All we know are the values of M and
V. This problem does not have an exact solution, but reasonable approximations are available. We will now present two approximations:
1. Equivalent Random Method (ERM) 2. Hayward Approximation.
Equivalent Random Method (ERM)
We wish to estimate the blocking probability for a system with k2 servers loaded by non-pure
chance offered traffic with mean M and variance V. We know that if this traffic is offered to a system with infinite number of servers (instead ofk2), the mean and variance of the number of busy servers will beM and V, respectively.
Under the ERM, due to [89], we model the system as if the traffic was the overflow traffic from a primary system with Neq circuits and offered traffic Aeq that follows Poisson process. If we find such Aeq and Neq then by Eq. (344), the blocking probability in our k2-server system,
denoted P Bk2(Neq, Aeq), will be estimated by:
P Bk2(Neq, Aeq) =
ENeq+k2(Aeq)
ENeq(Aeq)
.
To approximate Aeq and Neq, we use the following:
Aeq =V + 3Z(Z−1); (345)
Neq =
Aeq(M +Z)
M +Z −1 −M−1. (346)
Note that Equation (345) is an approximation, but Equation (346) is exact and it results in an approximation only because Equation (345) is an approximation.
Hayward Approximation
The Hayward approximation [30] is based on the following result. A multi-server system with
k servers fed by traffic with meanM and varianceV has a similar blocking probability to that of an M/M/k/k system with offered load MZ and Zk servers, hence Erlang B formula that gives
Ek Z
M Z
Homework 8.20
The XYZ Corporation has measured their offered traffic to a set of 24 circuits and found that it has a mean of M = 21 and a variance of V = 31.5. Arriving calls cannot be queued and delayed before they are served. This means that if a call arrives and all the circuits (servers) are busy, the call is blocked and cleared from the system.
Use the Hayward Approximation as well as the Equivalent Random Method to estimate the blocking probability.
Solution
M = 21; V = 31.5;
Peakedness: Z = 3121.5 = 1.5.
Hayward Approximation
The mean offered traffic in the equivalent system = 21 / 1.5 = 14. The number of servers in the equivalent system = 24/1.5 = 16. Blocking probability =E16(14) = 0.1145.
Equivalent Random Method
Aeq=V + 3Z(Z−1) = 31.5 + 3x1.5x0.5 = 33.75. Neq= Aeq(M +Z) M +Z−1 –M −1 = 33.75(21 + 1.5) 21 + 1.5−1 −21−1 = 13.32. We will useNeq= 13 conservatively.
From Erlang B:
E13(33.75) = 0.631. E13+24(33.75) = 0.075.
Then, the blocking probability is obtained by
E13+24(33.75) E13(33.75) = 0.07536 0.631 = 0.1194. Homework 8.21
Assume that non-Poisson traffic with mean = 65 Erlangs and variance = 78 is offered to a loss system. Use both Hayward Approximation and the Equivalent Random Method to estimate the minimal number of circuits required to guarantee that the blocking probability is not more than 1%.
Solution
Let us use the notation N∗ to represent the minimal number of circuits required to guarantee that the blocking probability is not more than 1%. Previously, we use k2 to represent the givennumber of servers in the secondary system. Now we use the notation N∗ to represent the desired number of servers in the system.
Given,M = 65 and V = 78, the peakedness is given by
Z = 78
65 = 1.2.
Equivalent Random Method
By (345): Aeq = 78 + 3×1.2×0.2 = 78.72. By (346): Neq =
78.72(65+1.2)
65+1.2−1 −65−1 = 13.92736 = 14 approx.
A conservative rounding would be to round it down to Neq = 13. This will result in a more conservative dimensioning because lower value for Neq will imply higher loss in the primary system, which leads to higher overflow traffic from the primary to the secondary system. This in turn will require more servers in the secondary system to meet a required blocking probability level.
In the present case, because the result is 13.92736 (so close to 14), we round it up toNeq= 14. In any case, we need to be aware of the implication of our choice. We will repeat the calculation using the more conservative choice ofNeq = 13.
The blocking probability in the primary equivalent system is given by,
E14(78.72) = 0.825.
Next, find the minimal value forN∗+ 14 such that
EN∗+14(78.72)
0.825 ≤0.01, or,
EN∗+14(78.72)≤0.00825.
By Erlang B formula: N∗+ 14 = 96, so the number of required servers is estimated by N∗ = 82.
Notice that the choice ofN∗+ 14 = 96 is made because
E96(78.72) = 0.0071
satisfies the requirement and
E95(78.72) = 0.0087,
Now we consider Neq = 13.
The blocking probability in the primary equivalent system is given by,
E13(78.72) = 0.8373.
Next, find minimal value forN∗+ 13 such that
EN∗+13(78.72)
0.8373 ≤0.01, or,
EN∗+13(78.72)≤0.008373.
By Erlang B formula: we choose N∗+ 13 = 96 because as above E96(78.72) = 0.0071 satisfies
the requirement and E95(78.72) = 0.0087 does not.
Because of our conservative choice ofNeq = 13, the more conservative choice for the number of required servers is N∗ = 83.
Hayward Approximation
Mean offered traffic in the equivalent system = 65/1.2 = 54.16667.
By Erlang B formula, the number of required servers in the equivalent system for 1% blocking is 68.
Then, 68×1.2 = 81.6.Rounding up conservatively, we obtain that 82 servers are required. Now the designer will need to decide between 82 servers based on the Hayward approximation and based on a reasonable but less conservative approach according to Equivalent Random Method, or 83 according to the very conservative approach based on the Equivalent Random Method.
Homework 8.22
Consider again the two loss systems the primary and the secondary and use them to compare numerically between:
1. the exact solution;
2. the Hayward approximation;
3. the Equivalent Random Method approximation;
4. a 3rd approximation that is based on the assumption that the arrival process into the secondary system follows a Poisson process. For this approximation assume that the traffic lost in the primary system is offered to the secondary system following a Poisson process.
Guide
For the comparison, at first assume that you know A, k1 and k2 and compute M and V, i.e.,
the mean and variance of the offered load to the secondary system as well as the blocking probability of traffic in the secondary system.
Next, assume thatA andk1 are not known butk2 is known; also known areM and V, i.e., the
mean and variance of the offered load to the secondary system that you computed previously. And evaluate the blocking probability using both Hayward, the Equivalent Random Method and the Poisson approximations.
Compare the results for a wide range of parameters.
9
M/M/k
The M/M/k queue is a generalization of the M/M/1 queue to the case of k servers. As in M/M/1, for an M/M/k queue, the buffer is infinite and the arrival process is Poisson with rate
λ. The service time of each of the k servers is exponentially distributed with parameterµ. As in the case of M/M/1 we assume that the service times are independent and are independent of the arrival process.
9.1
Steady-State Equations and Their Solution
Letting A = λ/µ, and assuming the stability condition λ < kµ, or A < k, the M/M/k queue gives rise to the following steady-state equations:
π1 =Aπ0 π2 =Aπ1/2 = A2π0/2 π3 =Aπ2/3 = A3π0/(3!) . . . πk=Aπk−1/k =Akπ0/(k!) πk+1 =Aπk/k =Ak+1π0/(k!k) πk+2 =Aπk+1/k =Ak+2π0/(k!k2) . . . πk+j =Aπk+j−1/k =Ak+jπ0/(k!kj) forj = 1, 2, 3, . . . and in general: πn = Anπ 0 n! forn = 0, 1, 2, . . . , k−1 (347) and πn= Anπ 0 k!kn−k for n=k, k+ 1, k+ 2, . . . . (348) These balance equations can also be described by the following state transition diagram of M/M/k: 0 λ ,, 1 µ l l λ ,, 2 2µ l l λ ** · · · 3µ l l λ ,, k kµ j j λ -- k+ 1 kµ l l λ -- k+ 2 kµ m m λ ** · · · kµ m m
To obtain π0, we sum up both sides of Eqs. (347) and (348), and because the sum of the πns equals one, we obtain an equation forπ0, which its solution is
π0 = k−1 X n=0 An n! + Ak k! k (k−A) !−1 . (349)
Substituting the latter in Eqs. (347) and (348), we obtain the steady-state probabilitiesπn, n= 0, 1, 2, . . . .