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Congestion probabilities in a batched Poisson multirate loss model supporting elastic and adaptive traffic

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Abstract

The ever increasing demand of elastic and adaptive services, where in-service calls can tolerate bandwidth compression/expansion, together with the bursty nature of traffic, necessitates a proper teletraffic loss model which can contribute to the call-level performance evaluation of modern communication networks. In this paper, we propose a multirate loss model that supports elastic and adaptive traffic, under the assumption that calls arrive in a single link according to a batched Poisson process (a more “bursty” process than the Poisson process, where calls arrive in batches). We assume a general batch size distribution and the partial batch blocking discipline, whereby one or more calls of a new batch are blocked and lost, depending on the available bandwidth of the link. The proposed model does not have a product form solution, and therefore we propose approximate but recursive formulas for the efficient calculation of time and call congestion probabilities, link utilization, average number of calls in the system, and average bandwidth allocated to calls. The consistency and the accuracy of the model are verified through simulation and found to be quite satisfactory.

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Notes

  1. When no bandwidth compression takes place, i.e., \({\textbf{\textit{nb}}}+b_{k} \le C\), then r ≡ r(n) = 1.

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Authors and Affiliations

  1. Department of Telecommunications Science and Technology, University of Peloponnese, 221 00, Tripolis, Greece

    Ioannis Moscholios & Anthony Boucouvalas

  2. Department of Electrical & Computer Engineering, University of Patras, 265 04, Patras, Greece

    John Vardakas & Michael Logothetis

Authors
  1. Ioannis Moscholios
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  2. John Vardakas
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  3. Michael Logothetis
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  4. Anthony Boucouvalas
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Corresponding author

Correspondence to Michael Logothetis.

Appendices

Appendix 1

1.1 Tutorial example

Consider again the example of Section 2.1. A link of capacity C = 3 b.u. and T = 5 b.u. accommodates calls of two service classes. The first service class is elastic and the second service class is adaptive. Remind that b 1 = 1 b.u. and b 2 = 2 b.u., while \({\mu_1}^{-1}={\mu_2}^{-1}= 1\) time unit. Furthermore, we assume that the batch size of both service classes follows the geometric distribution with parameters β 1 = 0.2 and β 2 = 0.5, which corresponds to 1.25 and 2.0 mean number of call arrivals per batch, respectively. The system has 12 states \(\textbf{\textit{n}} = (n_1,n_2)\), which are presented in Table 8, together with the occupied link bandwidth j in each state, and the state-dependent factors \(r(\textbf{\textit{n}})=C/j\), \(\phi_1(\textbf{\textit{n}})\), and \(\phi_2(\textbf{\textit{n}})\). The values of \(\phi_1(\textbf{\textit{n}})\) and \(\phi_2(\textbf{\textit{n}})\) are calculated through Eqs. 13 and 14.

Table 8 State space, occupied link bandwidth, and state-dependent factors of the tutorial example
Full size table

The global balance equation for a state \(\textbf{\textit{n}} = (n_1, n_2)\) is of the form

$$\begin{array}{lll} \sum \limits_{k=1}^{K=2} (R^{(\rm up)} (L_{\textbf{\textit{n}}}^{k} ) -R^{(\rm up)} (L_{{\textbf{\textit{n}}}_{{\it k}}^{{\it -}{\rm 1}} }^{k} )) \nonumber\\ \quad\sum \limits_{k=1}^{K=2} (R^{(\rm down)} (L_{\textbf{\textit{n}}}^{k} ) -R^{(\rm down)} (L_{{\textbf{\textit{n}}}_{{\it k}}^{{\it -}{\rm 1}} }^{k} )). \end{array}$$
(50)

When we use the state-dependent factor \(r(\textbf{\textit{n}})\) (irreversible Markov chain) in Eq. 50, we get the following system of equations:

$$\begin{array}{rll}{\rm State}\,\, \textbf{\textit{n}}&=&(0,0): P(0,1) + P(1,0) - 2.0P(0,0) =0\\ {\rm State}\,\, \textbf{\textit{n}}&=&(0,1): 0.5P(0,0) + 2.0P(0,2) + P(1,1) \\ &&-\, 3.0P(0,1) =0\\ {\rm State}\,\, \textbf{\textit{n}}&=&(0,2): 0.5P(0,0) + P(0,1) + 0.6P(1,2) \\&&-\, 3.0P(0,2) =0\\ {\rm State}\,\, \textbf{\textit{n}}&=&(1,0): 0.8P(0,0) + P(1,1) + 2.0P(2,0) \\&&-\, 3.0P(1,0) =0\\ {\rm State}\,\, \textbf{\textit{n}}&=&(1,1): 0.8P(0,1) + 0.5P(1,0) + 2.0P(1,2)\\&& +\, 1.5P(2,1) - 4.0P(1,1) =0\\ {\rm State}\,\, \textbf{\textit{n}}&=&(1,2): P(0,2) + 0.5P(1,0) + P(1,1) \\&&-\, 2.6P(1,2) =0\end{array}$$
$$\begin{array}{rll}{\rm State}\,\, \textbf{\textit{n}}&=&(2,0): 0.16P(0,0) + 0.8P(1,0) + 3.0P(3,0) \\[1pt]&& +\, P(2,1) - 4.0P(2,0) =0\\[1pt] {\rm State}\,\, \textbf{\textit{n}}&=&(2,1): 0.16P(0,1) + 0.8P(1,1) + P(2,0)\\ [1pt]&&+\, 1.8P(3,1) - 3.5P(2,1) =0\\[1pt] {\rm State}\,\, \textbf{\textit{n}}&\!=\!&(3,0): 0.032P(0,0) \!+\! 0.16P(1,0) \!+\! 0.8P(2,0)\\[1pt]&& +\, P(3,1) + 3.0P(4,0) - 5.0P(3,0) =0\\[1pt] {\rm State}\,\, \textbf{\textit{n}}&=&(3,1): 0.04P(0,1) + 0.2P(1,1) + P(2,1) \\[1pt]&&+\, P(3,0) - 2.8P(3,1) = 0\\[1pt] {\rm State}\,\, \textbf{\textit{n}}&=&(4,0): 0.0064P(0,0)+ 0.032P(1,0) \\[1pt]&&+\, 0.16P(2,0) + 0.80P(3,0) + 3.0P(5,0) \\[1pt]&&- 4.0P(4,0) =0\\[1pt] {\rm State}\,\, \textbf{\textit{n}}&=&(5,0): 0.0016P(0,0) + 0.008P(1,0) \\[1pt]&&+\, 0.04P(2,0) + 0.20P(3,0) + P(4,0) \\[1pt]&&- 3.0P(5,0) =0. \end{array}$$

Its solution is

$$\begin{array}{lll} P(0,0) &=& 0.1250 \;\;\; P(0,1) = 0.1166\;\;\; \\[3pt]P(0,2) &=& 0.0807 \;\;\; P(1,0) = 0.1334 \\[3pt] P(1,1) &=& 0.1259 \;\;\; P(1,2) = 0.1051\;\;\; \\[3pt]P(2,0) &=& 0.0871 \;\;\; P(2,1) = 0.0889 \\[3pt] P(3,0) &=& 0.0442 \;\;\; P(3,1) = 0.0582\;\;\; \\[3pt]P(4,0) &=& 0.0227 \;\;\; P(5,0) = 0.0121. \end{array}$$

Based on the values of \(P(\textbf{\textit{n}} )\)’ s, the exact values of G(j)’ s are G(0) = 0.1250, G(1) = P(1,0) = 0.1334, G(2) = P(0,1) + P(2,0) = 0.2037, G(3) = P(1,1) + P(3,0) = 0.1701, G(4) = P(0,2) + P(2,1) + P(4,0) = 0.1923, and G(5) =  P(1,2) + P(3,1) + P(5,0) =  0.1754.

Thus, the exact values of TC and CC probabilities are \(P_{b_1}=0.1754\), \(P_{b_2}=0.3677\), \(C_{b_1}=0.2225\), and \(C_{b_2}=0.6192\).

When we use the state-dependent factors \(\phi_1(\textbf{\textit{n}})\) and \(\phi_2(\textbf{\textit{n}})\) (reversible Markov chain) in Eq. 50, we get the following system of equations:

$$\begin{array}{rll}{\rm State}\,\, \textbf{\textit{n}}&=& (0,0): P(0,1)+P(1,0)-2.0P(0,0)= 0\\ {\rm State}\,\, \textbf{\textit{n}}&=& (0,1): 0.5P(0,0)+2.0P(0,2)+P(1,1)\\&&-\,3.0P(0,1) = 0\\ {\rm State}\,\, \textbf{\textit{n}}&=& (0,2): 0.5P(0,0)+P(0,1)+0.88235P(1,2)\\&&-\,3.0P(0,2) = 0\\ {\rm State}\,\, \textbf{\textit{n}}&=& (1,0): 0.8P(0,0)+P(1,1)+2.0P(2,0)\\&&-\,3.0P(1,0) = 0\\ {\rm State}\,\, \textbf{\textit{n}}&=& (1,1): 0.8P(0,1)\!+\!0.5P(1,0)\!+\!1.7647P(1,2)\\&&+1.71428P(2,1)-4.0P(1,1) = 0\\ {\rm State}\,\, \textbf{\textit{n}}&=& (1,2): P(0,2)+0.5P(1,0)+P(1,1)\\&&-\,2.64705P(1,2) = 0\\ {\rm State}\,\, \textbf{\textit{n}}&=& (2,0): 0.16P(0,0)+0.8P(1,0)+3.0P(3,0)\\&&+\, 0.85714P(2,1)-4.0P(2,0)= 0\\ {\rm State}\,\, \textbf{\textit{n}}&=& (2,1): 0.16P(0,1)+0.8P(1,1)+P(2,0)\\&&+\,2.23404P(3,1) -3.57142P(2,1) = 0\\ {\rm State}\,\, \textbf{\textit{n}}&\!=\!& (3,0): 0.032P(0,0)\!+\!0.16P(1,0)\!+\!0.8P(2,0)\\&&+\, 0.63829P(3,1)\!+\!3.0P(4,0)\!-\!5.0P(3,0) = 0\\ {\rm State}\,\, \textbf{\textit{n}}&=& (3,1): 0.04P(0,1)+0.2P(1,1)+P(2,1)\\&&+\,P(3,0)- 2.87233P(3,1)= 0\\ {\rm State}\,\, \textbf{\textit{n}}&=& (4,0): 0.0064P(0,0)+0.032P(1,0)\\&&+\,0.16P(2,0)+ 0.80P(3,0)+3.0P(5,0)\\&&-4.0P(4,0) = 0\\ {\rm State}\,\, \textbf{\textit{n}}&=& (5,0): 0.0016P(0,0)+0.008P(1,0)\\&&+\,0.04P(2,0)+ 0.20P(3,0)+P(4,0)\\&&-\,3.0P(5,0)= 0. \end{array}$$

The solution of this system is

$$\begin{array}{lll} P(0,0) &=& 0.1277 \;\;\; P(0,1)= 0.1278 \;\;\; \\P(0,2) &=& 0.0958 \;\;\;P(1,0)= 0.1276 \\ P(1,1) &=& 0.1278 \;\;\; P(1,2)= 0.1086 \;\;\; \\P(2,0) &=& 0.0764 \;\;\;P(2,1)= 0.0897 \\ P(3,0) &=& 0.0354 \;\;\; P(3,1)= 0.0542 \;\;\; \\P(4,0) &=& 0.0189 \;\;\;P(5,0)= 0.0101. \end{array}$$

Based on the values of \(P(\textbf{\textit{n}})\)’ s, the approximate values of \(G(j)\)’ s are \(G()\), \(G()P(,)\) \(0\), \(G()P(,)P(,)\), \(G()\) P(1,1) + P(3,0) = 0.1632, \(G(4) = P(0,2) + P(2,1) + P(,)\), and \(G()P(,)P(,)\) P(5,0) = 0.1729.

The same values of G(j)’s can also be obtained through Eq. 29. Thus, the approximate values of TC and CC probabilities (based on Eqs. 41 and 43, respectively) are \(P_{b_1}=0.1729\), \(P_{b_2}=0.3773\), \(C_{b_1}=0.2222\), and \(C_{b_2}=0.6248\). These results are very close to the corresponding exact values given above and show that the proposed formula can give quite good results even in the case of small examples.

Appendix 2

2.1 Notion of state-dependent multipliers φ k (n) and their calculation

Consider a link that accommodates calls of two service classes. The first service class is elastic and the second is adaptive.

When \(\textbf{\textit{nb}}\le {\it C}\), the bandwidth b k and the service rate μ k , k = 1,2, of calls are not affected. In this case, \(\phi _{k}(\textbf{\textit{n}})=1\), k = 1,2, where n = (n 1 ,n 2 ) and b = (b 1, b 2).

When \(T \ge \textbf{\textit{nb}} > C\), the bandwidth b k and the service rate μ k , of calls are decreased by \(\phi _{k}(\textbf{\textit{n}})\), k = 1,2, so that the link operates at its full capacity:

$$\label{EQ51} n_{1} b_{1}^{\rm compressed} +n_{2} b_{2}^{\rm compressed}=C. $$
(51)

Furthermore, from Eqs. 16 and 17, we have

$$\label{EQ52} b_{1}^{\rm compressed} =b_{1} \phi _{1} (\textbf{\textit{n}}) ~~~~{\rm (elastic~service~class)} $$
(52)
$$\label{EQ53} b_{2}^{\rm compressed} =b_{2}^{} \phi _{2} (\textbf{\textit{n}})r(\textbf{\textit{n}}) ~~~~{\rm (adaptive~service~class)}. $$
(53)

Based on Eqs. 52 and 53, Eq. 51 becomes

$$\label{EQ54} n_{1}b_{1}\phi _{1}(\textbf{\textit{n}})+n_{2}b_{2}\phi _{2}(\textbf{\textit{n}})r(\textbf{\textit{n}})=C. $$
(54)

Consider now an excerpt of the state transition diagram of our system that consists of four adjacent states as shown in Fig. 7.

Fig. 7
figure 7

State transition diagram of four adjacent states

Full size image

According to Kolmogorov’ s criterion [41], the system Markov chain (Fig. 7) becomes reversible if

$$ \textit{Flow clockwise} = \textit{Flow counter-clockwise} $$

By applying this criterion to our Markov chain (Fig. 7), we have

$$\begin{array}{lll} \lambda _{1} \phi _{2} (n_{1} ,n_{2} )n_{2} \mu _{2} \phi _{1} (n_{1} ,n_{2}-1)n_{1} \mu _{1} \lambda _{2} \nonumber\\ \quad \phi _{1} (n_{1} ,n_{2} )n_{1} \mu _{1} \phi _{2} (n_{1}-1,n_{2} )n_{2} \mu _{2} \lambda _{1} \lambda _{2}. \end{array}$$
(55)

In Eq. 55, we notice that the Kolmogorov’ s criterion holds by a proper selection of the state-dependent multipliers \(\phi _{k}(\textbf{\textit{n}})\), which affect the bandwidth and service rate requirements (not the arrival rates).

Equation 55 is simplified to

$$\label{EQ56} \phi _{2} (n_{1} ,n_{2} )\phi _{1} (n_{1} ,n_{2}-1)=\phi _{2} (n_{1}-1,n_{2} )\phi _{1} (n_{1} ,n_{2} ). $$
(56)

By choosing

$$\label{EQ57} \phi _{k}(\textbf{\textit{n}})=\left\{\begin{array}{l} {{\rm 1}\, \, \, \, \, \, \, \, \, \, \, \, \, \, ,\, \, {\rm when}\, \, \textbf{\textit{nb}}\le C\, \, {\rm and}\, \, \textbf{\textit{n}} \in \, \Omega\, \, \, } \\ {\frac{x( \textbf{\textit{n}}_{k}^{-1} )}{ x(\textbf{\textit{n}})} ,\, \, \, {\rm when}\, C< \textbf{\textit{nb}}\le T\, \, {\rm and}\, \, \textbf{\textit{n}} \in \, \Omega \; \; } \\ {{\rm 0}\, \, \, \, \, \, \, \, \, \, \, \, \, ,\, \, \, {\rm otherwise\; \; \; \; \; }} \end{array}\right. $$
(57)

Equation 56 holds and the Markov chain becomes reversible.

Equation 54, based on Eq. 57, is written as

$$\label{EQ58} x(\textbf{\textit{n}})=\frac{1}{C} \left[n_{1} b_{1}x(\textbf{\textit{n}}_{{\rm 1}}^{{\rm -1}} )+r(\textbf{\textit{n}})n_{2} b_{2}^{} x(\textbf{\textit{n}}_{2}^{{\rm -1}} )\right] $$
(58)

where \(r(\textbf{\textit{n}})=C/(\textbf{\textit{nb}})\).

Equation 58 is Eq. 14 for two service classes (elastic and adaptive).

To determine \(\phi _{1}(\textbf{\textit{n}})\) and \(\phi _{2}(\textbf{\textit{n}})\), we proceed as follows. Based on Eqs. 54 and 56, we get

$$\begin{array}{lll} {\phi _{1} (n_{1} ,n_{2} )=\left\{\begin{array}{l} {\frac{C}{n_{1} b_{1} } ,\, \, \, \, \,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~{\rm if}\, \, n_{1} \ge 1\, \, {\rm and}\, n_{2} =0\, } \\ {\frac{C\phi _{1} (n_{1} ,n_{2} -1)}{n_{1} b_{1} \phi _{1} (n_{1} ,n_{2} -1)+r(n_{1} ,n_{2} )n_{2} b_{2} \phi _{2} (n_{1} -1,n_{2} )} ,\, \, \, \, \, {\rm if}\, \, n_{1} \ge 1\, \, {\rm and}\, n_{2} \ge 1\, } \end{array}\right. } \\ {\phi _{2} (n_{1} ,n_{2} )=\left\{\begin{array}{l} {\frac{C}{r(n_{1} ,n_{2} )n_{2} b_{2} } ,\, \, \, \, \, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~{\rm if}\, \, n_{2} \ge 1\, \, {\rm and}\, n_{1} =0\, } \\ {\frac{C\phi _{2} (n_{1} -1,n_{2} )}{n_{1} b_{1} \phi _{1} (n_{1} ,n_{2} -1)+r(n_{1} ,n_{2} )n_{2} b_{2} \phi _{2} (n_{1} -1,n_{2} )} ,\, \, \, \, \,~ {\rm if}\, \, n_{1} \ge 1\, \, {\rm and}\, n_{2} \ge 1} \end{array}\right. } \end{array}. $$
(59)

For K service classes, when n k  ≥ 1, k ∈ K, we have

$$ \label{EQ60} \begin{array}{l} {\phi _{k}(n_{1},...,n_{k},...,n_{K} )} ={\frac{C\phi _{k} (n_{1} -1,...,n_{k-1}-1,n_{k} ,n_{k+1}-1,...,n_{K}-1)}{\sum \limits_{i\in \! K_{\rm e}}n_{i} b_{i}\phi _{i} (n_{1}-1,...,n_{i-1}-1,n_{i},n_{i+1}-1,...,n_{K_{\rm e}}-1)+r(\textbf{\textit{n}})\sum \limits_{i\in K_{\rm a}}n_{i} b_{i} \phi _{i} (n_{1}-1,...,n_{i-1}-1,n_{i},n_{i+1}-1,...,n_{K_{\rm a}}-1)}}. \end{array} $$
(60)

Equation 60 can also be used in the case of the E-EMLM [25]. If we consider only the case of elastic service classes [24], then Eq. 60 becomes

$$\label{EQ61} \phi _{k} (n_{1} ,...,n_{k},...,n_{K})=\frac{C\phi _{k}(n_{1}-1,...,n_{k-1}-1,n_{k},n_{k+1}-1,...,n_{K}-1)}{\sum \limits_{i \in K} n_{i} b_{i} \phi _{i}(n_{1}-1,...,n_{i-1}-1,n_{i},n_{i+1}-1,...,n_{K}-1)}, \, \, \, ~{\rm if}\, \, n_{k} \ge 1,\,~ k\in K. $$
(61)

Although the Markov chain becomes reversible by using \(\phi _{k}(\textbf{\textit{n}})\)’s, the existence of summations in Eqs. 60 and 61 reveals that no PFS exists for the calculation of steady-state distribution P(n).

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Moscholios, I., Vardakas, J., Logothetis, M. et al. Congestion probabilities in a batched Poisson multirate loss model supporting elastic and adaptive traffic. Ann. Telecommun. 68, 327–344 (2013). https://doi.org/10.1007/s12243-012-0326-7

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