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Optimal aggregation factor and clustering under delay constraints in aggregate sequential group paging schemes

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Abstract

This paper considers several optimization problems of sequential paging with aggregation mechanism which has been shown to reduce significantly the paging cost of a wireless communication system. An important problem is to find the optimal aggregation factor subject to a constraint on the average paging delay. Another problem is, given a cost function that depends on both paging cost and paging delay, how to find the optimal aggregation factor to minimize that cost function. We have formulated and shown that these can be solved nicely due to the monotonicity and convexity of the average paging cost function and paging delay function. We demonstrate that the optimization problems of the aggregate factor and subnet clustering are not separable. This leads to joint optimization problems of aggregation factor and clustering that are investigated in this paper. The paper presents different algorithms to solve these joint optimization problems using the monotonicity in the aggregation factor and the number of clusters of the average paging cost and delay with the unconstrained optimal clustering and the structures of the constrained optimal clustering.

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Correspondence to Ushio Yamamoto.

Appendices

Appendices

1.1 Proof of proposition 2

The problem to be considered is as follows.

Suppose we have two partition vectors v = (|g 1|,…,|g m |) and v  = (|g′ 1|,…,|g′ m |) in which g i  ≡ g′ i for i  j, l such that 1 ≤ jl ≤ m and P G (g j ) ≤ P G (g l ). In partition v , g j and g l are swapped, all other elements of v and v are the same. We want to see if C(k,v) ≥ C(k,v ) and L(k,v) ≥ L(k,v ).

1.1.1 Average paging cost function

$$ \begin{aligned} C(k,{\user2{v}}) \ge & C(k,{\user2{v}}\prime ) \\ \Leftrightarrow & \sum\limits_{i = 1}^{m} {\left( {A_{i}^{k} - A_{i - 1}^{k} } \right)B_{i} } \le \sum\limits_{i = 1}^{m} {\left( {A_{i}^{\prime k} - A_{i - 1}^{\prime k} } \right)B_{i} } \\ \Leftrightarrow & \sum\limits_{i = 1}^{j - 1} {\left( {A_{i}^{k} - A_{i - 1}^{k} } \right)B_{i} } + \left( {A_{j}^{k} - A_{j - 1}^{k} } \right)B_{j} + \sum\limits_{i = j + 1}^{l - 1} {\left( {A_{i}^{k} - A_{i - 1}^{k} } \right)B_{i} } + \left( {A_{l}^{k} - A_{l - 1}^{k} } \right)B_{l} + \sum\limits_{i = l + 1}^{m} {\left( {A_{i}^{k} - A_{i - 1}^{k} } \right)B_{i} } \\ \le & \sum\limits_{i = 1}^{j - 1} {\left( {A_{i}^{\prime k} - A_{i - 1}^{\prime k} } \right)B_{i} } + \left( {A_{j}^{\prime k} - A_{j - 1}^{\prime k} } \right)B_{j} + \sum\limits_{i = m + 1}^{l - 1} {\left( {A_{i}^{\prime k} - A_{i - 1}^{\prime k} } \right)B_{i} } + \left( {A_{l}^{\prime k} - A_{l - 1}^{k} } \right)B_{l} \\ & + \sum\limits_{i = l + 1}^{m} {\left( {A_{i}^{'k} - A_{i - 1}^{'k} } \right)B_{i} } \\ \Leftrightarrow & \left( {A_{j}^{k} - A_{j - 1}^{k} } \right)B_{j} + \sum\limits_{i = j + 1}^{l - 1} {\left( {A_{i}^{k} - A_{i - 1}^{k} } \right)B_{i} } + \left( {A_{l}^{k} - A_{l - 1}^{k} } \right)B_{l} \\ \le & \left( {A_{j}^{'k} - A_{j - 1}^{'k} } \right)B_{j} + \sum\limits_{i = j + 1}^{l - 1} {\left( {A_{i}^{'k} - A_{i - 1}^{'k} } \right)B_{i} } + \left( {A_{l}^{'k} - A_{l - 1}^{'k} } \right)B_{l} \\ \Leftrightarrow & \left( {A_{j}^{k} - A_{j}^{'k} } \right)B_{j} + \left( {A_{l - 1}^{'k} - A_{l - 1}^{k} } \right)B_{l} + \sum\limits_{i = j + 1}^{l - 1} {\left( {\left( {A_{i}^{k} - A_{i - 1}^{k} } \right) - \left( {A_{i}^{'k} - A_{i - 1}^{'k} } \right)} \right)B_{i} } \ge 0 \\ \end{aligned} $$
(16)

Note that in the above derivation, we use B i  = B i for i and Aj−1 = A′ j−1 , A l  = A′ l .

For k = 1, we have A i   A i−1=P G (g i ) = P G (g′ i ) = A′ i−1, so the above (16) becomes simply the following:

$$ \begin{array}{ll} &\left( {A_{j} - A\prime_{j} } \right)B_{j} + \left( {A\prime_{l - 1} - A_{l - 1} } \right)B_{l} \ge 0 \hfill \\ &\quad\Leftrightarrow \left( {P_{G} (g_{j} ) - P_{G} (g_{l} )}\right)B_{j} + \left( {P_{G} (g_{l} ) - P_{G} (g_{j} )} \right)B_{l} \ge 0 \hfill \\&\quad\Leftrightarrow \left( {P_{G} (g_{l} ) - P_{G} (g_{j} )} \right)\left( {B_{l} - B_{j} } \right) \ge 0 \hfill \\\end{array} $$

This inequality obviously holds.

In general, the inequality is not simplified as for k = 1, and hence more difficult to verify.

Instead of considering the above inequality, we consider the case where l = j + 1, i.e. two groups are adjacent. All unordered partition can be transformed to the descent order partition using this basic swap operation as in the bubble-sorting algorithm. Thus, it suffices to show the inequality for adjacent groups.

Notice that we can write A i  = a i  + P G (g j ), A′ i  = a i  + P G (g l ) for j ≤ i ≤ l − 1 where

$$ a_{i} = A_{j - 1} + \sum\nolimits_{t = j + 1}^{i} {P_{G} (g_{t} )} . $$
$$ \begin{gathered} \Leftrightarrow \left( {\left( {a_{j} + P_{G} (g_{j} )} \right)^{k} - \left( {a_{j} + P_{G} (g_{l} )} \right)^{k} } \right)B_{j} + \left( {\left( {a_{l - 1} + P_{G} (g_{l} )} \right)^{k} - \left( {a_{l - 1} + P_{G} (g_{j} )} \right)^{k} } \right)B_{l} \hfill \\ + \sum\limits_{t = j + 1}^{l - 1} {\left( {\left( {\left( {a_{t} + P_{G} (g_{j} )} \right)^{k} - \left( {a_{t - 1} + P_{G} (g_{j} )} \right)^{k} } \right) - \left( {\left( {a_{t} + P_{G} (g_{l} )} \right)^{k} - \left( {a_{t - 1} + P_{G} (g_{l} )} \right)^{k} } \right)} \right)B_{t} } \ge 0 \hfill \\ \end{gathered} $$

For l = j + 1, the above becomes:

$$ \begin{gathered} \left( {\left( {a_{j} + P_{G} (g_{j} )} \right)^{k} - \left( {a_{j} + P_{G} (g_{l} )} \right)^{k} } \right)B_{j} + \left( {\left( {a_{j} + P_{G} (g_{l} )} \right)^{k} - \left( {a_{j} + P_{G} (g_{j} )} \right)^{k} } \right)B_{l} \ge 0 \hfill \\ \Leftrightarrow \left( {\left( {a_{j} + P_{G} (g_{l} )} \right)^{k} - \left( {a_{j} + P_{G} (g_{j} )} \right)^{k} } \right)\left( {B_{l} - B_{j} } \right) \ge 0 \hfill \\ \end{gathered} $$

This also obviously holds.

1.1.2 Average paging delay function

The average paging delay function consists of an affine function of k, which is not dependent on partition v, and a function dependent on v, but independent of k. The proof is essentially the same as for k = 1.

$$ \begin{aligned} L(k,v) = & \sum\limits_{i = 1}^{m} {P_{G} (g_{i} )i} + {\frac{{L_{\text{aggr}} }}{2}} = \sum\limits_{i = 1}^{m} {\left( {A_{i} - A_{i - 1} } \right)i} + {\frac{{L_{{_{\text{aggr}} }} }}{2}} \\ \ge & L(k,v') = \sum\limits_{i = 1}^{m} {P_{G} (g'_{i} )i} + {\frac{{L'_{\text{aggr}} }}{2}} = \sum\limits_{i = 1}^{m} {\left( {A'_{i} - A'_{i - 1} } \right)i} + {\frac{{L'_{{_{\text{aggr}} }} }}{2}} \\ \Leftrightarrow & \left( {A_{j} - A_{j - 1} } \right)j + \left( {A_{l} - A_{l - 1} } \right)l \ge \left( {A'_{j} - A'_{j - 1} } \right)j + \left( {A'_{l} - A'_{l - 1} } \right)l \\ \Leftrightarrow & \left( {A_{j} - A'_{j} } \right)j + \left( {A'_{l - 1} - A_{l - 1} } \right)l \ge 0 \\ \Leftrightarrow & \left( {P_{G} (g_{l} ) - P_{G} (g_{j} )} \right)\left( {l - j} \right) \ge 0 \\ \end{aligned} $$

Proof of Proposition 3

Consider the function C(x), where x is a positive real number, we will show that it is convex in x.

Let x 1, x 2 be positive real numbers and x = αx 1 + (1 − α)x 2, where α ∈ [0,1]. We need to show that C(x) ≤ α C(x 1) + (1 − α) C(x 2).

$$ \begin{aligned} C(x) = & C(\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} ) \\ = & {\frac{1}{{\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} }}}\left( {\sum\limits_{i = 1}^{m} {\left( {A_{i}^{{\alpha x_{1} + (1 - \alpha )x_{2} }} - A_{i - 1}^{{\alpha x_{1} + (1 - \alpha )x_{2} }} } \right)B_{i} } } \right) \\ \le & {\frac{\alpha }{{x_{1} }}}\sum\limits_{i = 1}^{m} {\left( {A_{i}^{{x_{1} }} - A_{i - 1}^{{x_{1} }} } \right)B_{i} } + {\frac{1 - \alpha }{{x_{2} }}}\sum\limits_{i = 1}^{m} {\left( {A_{i}^{{x_{2} }} - A_{i - 1}^{{x_{2} }} } \right)B_{i} } \\ \Leftrightarrow & LHS \le {\frac{1}{{x_{1} x_{2} }}}\left( {\sum\limits_{i = 1}^{m} {\left( {A_{i}^{{x_{1} }} - A_{i - 1}^{{x_{1} }} } \right)\alpha x_{2} B_{i} + \sum\limits_{i = 1}^{m} {\left( {A_{i}^{{x_{2} }} - A_{i - 1}^{{x_{2} }} } \right)\left( {1 - \alpha } \right)x_{1} B_{i} } } } \right) \\ = & {\frac{1}{{x_{1} x_{2} }}}\sum\limits_{i = 1}^{m} {\left( {\left( {\alpha x_{2} A_{i}^{{x_{1} }} + \left( {1 - \alpha } \right)x_{1} A_{i}^{{x_{2} }} } \right) - \left( {\alpha x_{2} A_{i - 1}^{{x_{1} }} + (1 - \alpha )x_{1} A_{i - 1}^{{x_{2} }} } \right)} \right)} B_{i} \\ \Leftrightarrow & x_{1} x_{2} \sum\limits_{i = 1}^{m} {\left( {A_{i}^{{\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} }} - A_{i - 1}^{{\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} }} } \right)B_{i} } \\ \le & \left( {\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} } \right)\sum\limits_{i = 1}^{m} {\left( {\left( {\alpha x_{2} A_{i}^{{x_{1} }} + (1 - \alpha )x_{1} A_{i}^{{x_{2} }} } \right) - \left( {\alpha x_{2} A_{i - 1}^{{x_{1} }} + (1 - \alpha )x_{1} A_{i - 1}^{{x_{2} }} } \right)} \right)} B_{i} \\ \Leftrightarrow & \sum\limits_{i = 1}^{m} {x_{1} x_{2} A_{i}^{{\alpha x_{1} + (1 - \alpha )x_{2} }} B_{i} } - \sum\limits_{i = 1}^{m} {x_{1} x_{2} A_{i - 1}^{{\alpha x_{1} + (1 - \alpha )x_{2} }} B_{i} } \\ \le & \sum\limits_{i = 1}^{m} {\left( {\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} } \right)\left( {\alpha x_{2} A_{i}^{{x_{1} }} + (1 - \alpha )x_{1} A_{i}^{{x_{2} }} } \right)B_{i} } - \sum\limits_{i = 1}^{m} {(\alpha x_{1} + (1 - \alpha )x_{2} )\left( {\alpha x_{2} A_{i - 1}^{{x_{1} }} + (1 - \alpha )x_{1} A_{i - 1}^{{x_{2} }} } \right)B_{i} } \\ \Leftrightarrow & \sum\limits_{i = 1}^{m} {\left( {x_{1} x_{2} A_{i}^{{\alpha x_{1} + (1 - \alpha )x_{2} }} - \left( {\alpha x_{1} + (1 - \alpha )x_{2} } \right)\left( {\alpha x_{2} A_{i}^{{x_{1} }} + \left( {1 - \alpha } \right)x_{1} A_{i}^{{x_{2} }} } \right)} \right)} B_{i} \\ \le & \sum\limits_{i = 1}^{m} {\left( {x_{1} x_{2} A_{i - 1}^{{\alpha x_{1} + (1 - \alpha )x_{2} }} - \left( {\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} } \right)\left( {\alpha x_{2} A_{i - 1}^{{x_{1} }} + (1 - \alpha )x_{1} A_{i - 1}^{{x_{2} }} } \right)} \right)B_{i} } \\ \end{aligned} $$
(17)

Let \( h\left( {A_{i} } \right) = x_{ 1} x_{ 2} A_{i}^{{\alpha x 1 + (1 - \alpha )x^{2} }} - \left( {\alpha x_{ 1} ( 1- \alpha )x_{2} } \right)\left( {\alpha x_{ 2} A_{i}^{{x^{1} }} + \left( { 1- \alpha } \right)x_{ 1} A_{i}^{{x^{2} }} } \right), \) then it is easy to see that the following is a sufficient condition for the above (17):

$$ \sum\limits_{i = 1}^{m} {\left( {h(A_{i} ) - h(A_{i - 1} )} \right)} \le 0 $$
(18)

Again we investigate the first derivative of h(A i ):

$$ \begin{aligned} {\frac{{{\text{d}}h(A_{i} )}}{{{\text{d}}A_{i} }}} = & x_{1} x_{2} \left( {\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} } \right)A_{i}^{{\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} - 1}} - \alpha x_{1} x_{2} \left( {\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} } \right)A_{i}^{{x_{1} - 1}} \\ & - \left( {\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} } \right)\left( {1 - \alpha } \right)x_{1} x_{2} A_{i}^{{x_{2} - 1}} \\ = & {\frac{{x_{1} x_{2} \left( {\alpha x_{1} + \left( {1 - \alpha } \right)x_{2} } \right)}}{{A_{i} }}}\left( {A_{i}^{{\theta x_{1} + \left( {1 - \theta } \right)x_{2} }} - \left( {\alpha A_{i}^{{x_{1} }} + \left( {1 - \alpha } \right)A_{i}^{{x_{2} }} } \right)} \right) \\ \end{aligned} $$

Note that the function a x is convex in x for a ∈ (0,1], so by definition, we have

$$ A_{i}^{{\alpha x_{1} + (1 - \alpha )x_{2} }} \le \alpha A_{i}^{{x_{1} }} + \left( {1 - \alpha } \right)A_{i}^{{x_{2} }} $$

This implies that \( {\frac{{{\text{d}}h(A_{i} )}}{{{\text{d}}A_{i} }}} \le 0 \). Consequently, h(A i ) is non-increasing function. As A i−1 < A i , we have (18). This implies that (17) holds, and consequently, C(x) is convex in x.

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Do, H.T., Onozato, Y. & Yamamoto, U. Optimal aggregation factor and clustering under delay constraints in aggregate sequential group paging schemes. Wireless Netw 16, 1427–1446 (2010). https://doi.org/10.1007/s11276-009-0212-z

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