Performance analysis of the cumulative ARQ in IEEE 802.16 networks

Abstract

In this paper, we study the performance of the cumulative Automatic Repeat reQuest (ARQ) in IEEE 802.16 networks. An analytical model is developed to investigate some important performance metrics, such as protocol data unit (PDU) delivery delay, service data unit (SDU) delivery delay, and goodput. A general scheduling scheme and the flexible retransmission of lost PDUs are jointly considered in the analytical model, which provides a more valuable and practical guideline for the system design and performance evaluation. Extensive simulations are conducted to demonstrate the impacts of different operational parameters on the performance metrics and verify the accuracy of the analytical model.

Appendix: Analyses of service probability and inter-service time for SSs

In the Appendix, we discuss the service probability and inter-service time for SSs given that the general scheduling scheme proposed in [19] is adopted.

1.1 Analysis of service probability

The service probability is defined as the probability that an SS obtains the transmission opportunities at a DL sub-frame. Let the SS under discussion be referred to as the tagged SS. In order to analyze the service probability for the tagged SS, Firstly, we classify all SSs into three groups based on the channel state of the tagged SS at a specific MAC frame. Given the channel state of the tagged SS is at state ‘n’, the three groups is composed of the SSs with channel conditions better than, same as, and worse than the state ‘n’, respectively, which is denoted as the group G1, G2, and G3, respectively. Let k ₁, k ₂ and k ₃ denote the number of SSs in the group G1, G2, and G3, respectively. The tagged SS, which belongs to G2, obtains the chance of transmission only when the condition k ₁ < h holds. Otherwise all the selected SSs should come from the group G1. When the condition k ₁< h is satisfied, the probability that the tagged SS obtains the chance of transmission is derived based on the value of k ₁ and k ₂.

Since the total number of selected SSs is h, and k ₁ SSs are at the channel state better than state ‘n’, h − k ₁ is the quota left for the SSs at G2 and G3. When the left quota, h − k ₁ , is larger than k2, all the SSs at the group G2 are selected. That is, the tagged queue obtains the chance of transmission at this DL sub-frame with a probability 1. On the other hand, when $k_{2} > h - k_{1} ,$ a tie occurs. In this case, the BS randomly selects (h − k ₁) out of k ₂ SSs in G2. Therefore, the tagged queue obtains the chance of transmission with a probability$\left( {\frac{{h - k_{1} }}{{k_{2} }}} \right).$ To take these situations into account, we define the function ξ (.), as given in (25). It is concluded that k ₁ is a value between the set [0, h − 1] such that the tagged SS can obtain the chance of transmission. Under a specific value of k ₁ , k ₂ is in the set of [1, M − k ₁], where M is the total number of SSs in the system. The set begins with 1 since at least the tagged SS is in the group G2. When the values of k ₁ and k ₂ are given, k ₃ is M − k ₁ − k ₂.

$$ \xi \left( {\frac{{h - k_{1} }}{{k_{2} }}} \right) = \left\{ \begin{gathered} 1\quad \;\quad \quad h - k_{1} \ge k_{2} \hfill \\ \frac{{h - k_{1} }}{{k_{2} }}\quad \;h - k_{1} < k_{2} \hfill \\ \end{gathered} \right. $$

(25)

Let $\Upomega \left( {j_{1} } \right),$ $\Upomega \left( {j_{2} } \right),$ and $\Upomega \left( {j_{3} } \right)$ denote the set of SSs in groups G1, G2 and G3, respectively. Thus, the probability that the tagged SS obtains the chance of service and stays at the state “n” along with the specific $\Upomega \left( {j_{1} } \right),$ $\Upomega \left( {j_{2} } \right)$and $\Upomega \left( {j_{3} } \right)$ can be expressed as

$$ \xi \left( {\frac{{h - k_{1} }}{{k_{2} }}} \right)\left[ {\prod\limits_{{i_{1} \; \in \;\Upomega \;\left( {j_{1} } \right)}} {Pr\left( {S_{{i_{1} }} > n} \right)} \prod\limits_{{i_{2} \in \;\Upomega \;\left( {j_{2} } \right)}} {Pr\left( {S_{{i_{2} }} = n} \right)} \prod\limits_{{i_{3} \; \in \;\Upomega \;\left( {j_{3} } \right)}} {Pr\left( {S_{{i_{3} }} < n} \right)} } \right] $$

(26)

where the function ξ(.) is defined as (25), while $\left[ {\prod\limits_{{i_{1} \; \in \;\Upomega \;\left( {j_{1} } \right)}} {Pr\left( {S_{{i_{1} }} > n} \right)} ,\;\prod\limits_{{i_{2} \in \;\Upomega \;\left( {j_{2} } \right)}} {Pr\left( {S_{{i_{2} }} = n} \right)} ,\;{\text{and}}\;\prod\limits_{{i_{3} \; \in \;\Upomega \;\left( {j_{3} } \right)}} {Pr\left( {S_{{i_{3} }} < n} \right)} } \right]$ denote the probability that all SSs in groups G1, G2, and G3 are at the channel states better than, same as, and worse than the state n, respectively. For instance, the system is composed of SS1, SS2, SS3, SS4, SS5, and SS6. The tagged SS is SS1, which stays at channel state n = 3. Let h = 3, k ₁ = 2, and k ₂ = 3. The probability that SS1 obtains the chance of transmission with the channel state 3 while $\Upomega \left( {j_{1} } \right),$ $\Upomega \left( {j_{2} } \right),$ and $\Upomega \left( {j_{3} } \right)$ are {SS2,SS3}, {SS1, SS4, SS5}, and {SS6}, respectively, is given by

$$ \frac{1}{3}\left[ {\prod\limits_{{i_{1} \in \left\{ {SS2,SS3} \right\}}} {Pr\left( {S_{{i_{1} }} > 3} \right)\prod\limits_{{i_{2} \in \left\{ {SS4,SS5} \right\}}} {Pr\left( {S_{{i_{2} }} = 3} \right)} \prod\limits_{{i_{3} \in \left\{ {SS6} \right\}}} {Pr\left( {S_{{i_{3} }} < 3} \right)} } } \right] $$

(27)

Equation 27 is for the specific $\Upomega \left( {j_{1} } \right),$ $\Upomega \left( {j_{2} } \right),$ and $\Upomega \left( {j_{3} } \right).$ In the following, the number of all possible $\Upomega \left( {j_{1} } \right),\,\,\Upomega \left( {j_{2} } \right)$ and $\Upomega \left( {j_{3} } \right)$ are taken into consideration. Let a ₁ represents the number of possible combinations for selecting k ₁ SSs out of (M − 1)SSs to construct the group G1, where M is the total number of SSs in the system. After SSs in G1 are selected, there are M − k ₁ SSs left. Let a ₂ represent the number of possible combinations for selecting (k ₂ − 1) SSs out of the left (M − k ₁ − 1) SSs to construct the group G2. At last, the left M−k ₁−k ₂ SSs consist of the group G3. We have $\;a_{{_{1} }} = \left( {\begin{array}{*{20}c} {M - 1} \\ {k_{1} } \\ \end{array} } \right)$ and $a_{{_{2} }} = \left( {\begin{array}{*{20}c} {M - k_{2} - 1} \\ {k_{2} - 1} \\ \end{array} } \right).$ In other words, given a k ₁, the total number of possible $\Upomega \left( {j_{1} } \right)$ is a ₁, and the set of all possible $\Upomega \left( {j_{1} } \right)$ is represented by $\left\{ {\Upomega \left( {j_{1} } \right),\;\;j_{1} = 1,2, \ldots a_{1} } \right\}.$ Given a $\Upomega \left( {j_{1} } \right),$ the total number of possible $\Upomega \left( {j_{2} } \right)$ is a ₂ , and the set of all possible $\Upomega \left( {j_{2} } \right)$ is represented by $\left\{ {\Upomega \left( {j_{2} } \right),\;\;j_{2} = 1,2, \ldots a_{2} } \right\}.$ Note that given a $\Upomega \left( {j_{1} } \right)$ and $\Upomega \left( {j_{2} } \right),$ the number of possible $\Upomega \left( {j_{3} } \right)$ is 1 since the group G3 is composed of all the left SSs that belong to neither G2 nor G3. In other words, j ₃ is always 1. Let the vector $\Upxi = \sigma_{S0} \,\,\sigma_{S1} \,\,\,\sigma_{S2}\,\,\sigma_{S3} \,\, \cdots \,\,\sigma_{S7}$ be the probability for the tagged SS to obtain the transmission opportunity when the channel state of the tagged SS is ‘n’ (n = 0,1,…,7). The service probability of the tagged SS with the channel state ‘n’ is given as

$$ \sigma_{{_{{S_{n} }} }} = \left\{ \begin{gathered} \sum\limits_{{k_{1} = 1}}^{h - 1} {Q\left( {\frac{{h - k_{1} }}{{k_{2} }}} \right)\sum\limits_{{j_{1} = 1}}^{{a_{1} }} {\prod\limits_{{i_{1} \in \Upomega \left( {j_{1} } \right)}} {Pr\left( {S_{{i_{1} }} > n} \right)} } } \sum\limits_{{k_{2} = 1}}^{{M - k_{1} }} {\sum\limits_{{j_{2} = 1}}^{{a_{2} }} {\prod\limits_{{i_{2} \in \Upomega \left( {j_{2} } \right)}} {Pr\left( {S_{{i_{2} }} = n} \right)} \prod\limits_{{i_{3} \in \Upomega \left( {j_{3} } \right)}} {Pr\left( {S_{{i_{3} }} < n} \right)} } } \quad \quad n = 1, \ldots ,7 \hfill \\ 0\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\,\quad n = 0 \hfill \\ \end{gathered} \right. $$

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Note that $\sigma_{{_{S0} }} = 0$ since the tagged queue is not allowed to transmit when the channel state of the tagged SS is ‘0’, considering a higher error bit rate at such a poor channel condition.

1.2 Analysis of the inter-service time

By jointly considering ARQ and the general scheduling scheme, a Markov model can be constructed according to the channel states of the tagged SS. Each state in the Markov model represents the current channel state of the tagged SS and whether the tagged queue obtains the chance of transmission in the current DL sub-frame. Since the channel state includes (N + 1) states (N = 7 in the study), and each state, excluding state ‘0’, may represent either one of the following two situations: the tagged SS either is selected or not, which equivalently means that the tagged queue either obtains or losses the chance of transmission at the current DL sub-frame; when the channel state of the tagged SS is at the state “0”, the tagged queue keeps silent, which equivalently means that the tagged queue always lose the chance of transmission. Thus, the combined Markov model consists of 2N + 1 states as shown in Fig. 16, where (n, s) and (n, w) represent that the tagged queue obtains and loses the chance of transmission with the channel state of ‘n’, respectively.

Fig. 16

The Markov model for the tagged SS

The transmission probability matrix is given by

$$ \underline{\underline{Q}} = \left[ {\begin{array}{*{20}c} {p_{{_{00} }} } & {p_{{_{01} \quad }} p_{{_{01} }} } & {} & {p_{{_{0N} \quad }} p_{{_{0N} }} } \\ \begin{gathered} p_{{_{10} }} \hfill \\ p_{{_{10} }} \hfill \\ \end{gathered} & \begin{gathered} p_{{_{11} }} \quad p_{{_{11} }} \hfill \\ p_{{_{11} }} \quad p_{{_{11} }} \hfill \\ \end{gathered} & \begin{gathered} \ldots \hfill \\ \ldots \hfill \\ \end{gathered} & \begin{gathered} p_{{_{1N} \quad }} p_{{_{1N} }} \hfill \\ p_{{_{1N} \quad }} p_{{_{1N} }} \hfill \\ \end{gathered} \\ \begin{gathered} \vdots \hfill \\ \vdots \hfill \\ \end{gathered} & \begin{gathered} \vdots \hfill \\ \vdots \hfill \\ \end{gathered} & \begin{gathered} \vdots \hfill \\ \vdots \hfill \\ \end{gathered} & \begin{gathered} \vdots \hfill \\ \vdots \hfill \\ \end{gathered} \\ \begin{gathered} p_{{_{N0} }} \hfill \\ p_{{_{N0} }} \hfill \\ \end{gathered} & \begin{gathered} p_{{_{N1} }} \quad p_{{_{N1} }} \hfill \\ p_{{_{N1} }} \quad p_{{_{N1} }} \hfill \\ \end{gathered} & \begin{gathered} \ldots \hfill \\ \ldots \hfill \\ \end{gathered} & \begin{gathered} p_{{_{NN} \quad }} p_{{_{NN} }} \hfill \\ p_{{_{NN} \quad }} p_{{_{NN} }} \hfill \\ \end{gathered} \\ \end{array} } \right] \cdot \left[ {\begin{array}{*{20}c} {1 - \sigma_{{s_{0} }} } & 0 & {\quad 0 \cdots } & {0\quad \quad 0} \\ \begin{gathered} 0 \hfill \\ 0 \hfill \\ \vdots \hfill \\ \end{gathered} & \begin{gathered} \sigma_{{s_{1} }} \hfill \\ 0 \hfill \\ \;\; \vdots \hfill \\ \end{gathered} & \begin{gathered} \quad 0 \cdots \hfill \\ 1 - \sigma_{{s_{1} }} \hfill \\ \quad \vdots \hfill \\ \end{gathered} & \begin{gathered} 0\quad \quad 0 \hfill \\ 0\quad \quad 0 \hfill \\ \vdots \hfill \\ \end{gathered} \\ \begin{gathered} \vdots \hfill \\ 0 \hfill \\ \end{gathered} & \begin{gathered} \vdots \hfill \\ 0 \hfill \\ \end{gathered} & \begin{gathered} \vdots \hfill \\ 0 \hfill \\ \end{gathered} & \begin{gathered} \vdots \quad \quad \vdots \hfill \\ \sigma_{{s_{N} }} \quad 0 \hfill \\ \end{gathered} \\ 0 & 0 & 0 & {\quad 0\;\quad 1 - \sigma_{{s_{N} }} } \\ \end{array} } \right] $$

(29)

In order to derive the inter-service time between two adjacent transmissions, states in Fig. 16 can be grouped into two states shown in Fig. 17, denoted as ‘S’ and ‘W’, respectively. The states ‘S’ and ‘W’ represent the states where the tagged queue obtains and losses the chance of transmission, respectively.

Fig. 17

The grouped Markov model for the tagged queue

The transition probabilities of the grouped Markov model are given by

$$ p_{sw} = \frac{{\sum\limits_{n = 0}^{7} {\left[ {\theta \left( {n,s} \right)\sum\limits_{j = 0}^{7} {p_{ns,jw} } } \right]} }}{{\sum\limits_{n = 0}^{7} {\theta \left( {n,s} \right)} }};\quad p_{ss} = 1 - p_{sw} $$

(30)

$$ p_{ws} = \frac{{\sum\limits_{n = 0}^{7} {\left[ {\theta \left( {n,w} \right)\sum\limits_{j = 0}^{7} {p_{nw,js} } } \right]} }}{{\sum\limits_{n = 0}^{7} {\theta \left( {n,w} \right)} }};\;\quad p_{ww} = 1 - p_{ws} $$

(31)

where $\theta \left( {n,s} \right)$is the steady state probability of the state (n,s), and $p_{ns,jw}$ is the one-step transition probability from the state (n,s) to the state (j,w) (n,j = 1,2,…,7).

Let m denote the inter-service time, which is defined as the duration (in unit of frame) between two successive transmission chances of the tagged SS. The probability mass function of m, is given by

$$ pr\left[ {m = i} \right] = \left\{ \begin{array} {ll} p_{ss} & i = 0 \\ p_{sw} \left( {p_{ww} } \right)^{i - 1} p_{ws} & i > 0 \\ \end{array} \right. $$

(32)

The mean of m is given by

$$ \begin{gathered} E\left[ m \right] = \sum\limits_{i = 1}^{\infty } {i\;p_{sw} \left( {p_{ww} } \right)^{i - 1} p_{ws} } \hfill \\ \quad \quad = p_{sw} p_{ws} \sum\limits_{i = 1}^{\infty } {i\;\left( {p_{ww} } \right)^{i - 1} } = \frac{{p_{sw} }}{{p_{ws} }} \hfill \\ \end{gathered} $$

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