Exploiting replication on dependent data allocation for ordered queries over multiple broadcast channels

Abstract

Data broadcasting has been recognized as an important means for information dissemination in mobile computing environments. In some mobile applications, the data items broadcast are dependent upon one another. However, most prior studies on broadcasting dependent data do not employ replication in broadcast program generation. In view of this, we explore in this paper the problem of broadcasting dependent data in multiple broadcast channels, and explicitly investigate the effect of data replication. After analyzing the model of dependent data broadcasting, we derive several theoretical properties to formulate the average access time of broadcast programs. In light of the theoretical results, we develop an algorithm to exploit replication on broadcast program generation. Our experimental results show that the proposed algorithm is able to generate broadcast programs of very high quality. In addition, the results also show that broadcast programs with replication is more robust than those without replication in error-prone environments.

Appendix: Proof of all Lemmas

Proof of Lemma 1

Consider a broadcast program with replication and a query Q _i . Suppose that the first required data item of Q _i is D _k (i.e., k = q ⁱ(1)). Since each data item is of multiple replicas, the mobile device will retrieve the nearest copy of D _k in order to minimize the access time of the Q _i . Without loss of generality, we order each copy of D _k by its position in the broadcast program, and let a(j) be position(D _k (j)).

Suppose that a mobile user submits Q _i in the m-th broadcast cycle and the time interval between the start time of the m-th broadcast cycle and the moment when the user submits Q _i to be x. As shown in Fig. 11, the broadcast program can be divided into b + 1 parts, where b is the number of replicas of D _k , by the start time of each copy of D _k . It is obvious that the mobile device will read D _k (j) when x lies on part j, 1 ≤ j ≤ b. In addition, the mobile device will read D _k (1) in (m + 1)-st broadcast cycle when x lies on part (b + 1).

Fig. 10

Effect of data error rate

Recall that the time interval between the start time of the m-th broadcast cycle and the moment of the appearance of the j-th copy of D _k is $(\rm position(D_{k}(j))-1)\times{\frac{s}{B}}=(a(j)-1)\times{\frac{s}{B}}.$ Suppose that x is a uniform distribution over (0, L ^*). The average startup time of Q _i on the broadcast program with replication can be formulated as below.

$$ \begin{aligned} T_{\rm Startup}(Q_{i}) = & {\frac{1}{L^{*}}}\times\left\{\int_{x=0}^{a(1)-1}\left[a(1)-1-x\right]dx +\int_{x=a(1)-1}^{a(2)-1}\left[a(2)-1-x\right]dx\right. \cr & \left.+\cdots+\int_{x=a(b-1)-1}^{a(b)-1}\left[a(b)-1-x\right]dx +\int_{x=a(b)-1}^{L^{*}}\left[L^{*}-x+a(1)-1\right]dx\right\}\times{\frac{s}{B}} \cr = & {\frac{s}{L^{*}\times B}}\times\left\{\sum_{i=1}^{b-1}{\frac{\left[a(i+1)-a(i)\right]^{2}}{2}} +{\frac{\left[L^{*}-a(b)+a(1)\right]^{2}}{2}}\right\} \end{aligned} $$

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Proof of Lemma 2

Consider the same scenario in the Proof of Lemma 1 and Fig. 11. Suppose that the first required data item of Q _i is D _k (i.e., k = q ⁱ(1)) and the mobile device will try its best to minimize the access time. It is obvious that the mobile device will retrieve the j-th copy of D _k , j = 2, 3,…, b, as the first required data item when x lies on part j. In addition, the mobile device will retrieve the first copy of D _k when x lies on part 1 and part b + 1. Suppose that x is a uniform distribution over (0, L ^*). The probability that x lies on part j is equal to the ratio of the length of part j over the length of the broadcast program. Therefore, we can formulate p _i (j) as follows:

$$ p_{i}(j)=\left\{ \begin{array}{lll} {\frac{L^{*}-a(b)\,+\,a(1)}{L^{*}}} & : & \hbox{if}\,j=1, \cr {\frac{a(j)\,-\,a(j-1)}{L^{*}}} & : & \hbox{otherwise}. \end{array}\right. $$

Since x is a uniform distribution over (0, L ^*), for j = 1, T _Startup(Q ⁱ(j)) can be formulated as

$$ \begin{aligned} T_{\rm Startup}(Q^{i}(j)) &= {\frac{1}{L^{*}-a(b)+a(1)}}\times\left(\int_{x=a(b)}^{L^{*}}(L^{*}-x+a(1))dx + \int_{x=0}^{a(1)}(a(1)-x)\right)dx\times{\frac{s}{B}} \cr &= {\frac{1}{L^{*}-a(b)+a(1)}}\times\left[(L^{*}-a(1))\times\left({\frac{1}{2}}L^{*}+a(1)-{\frac{1}{2}}a(b)\right)+{\frac{1}{2}}a(1)^{2}\right]\times{\frac{s}{B}} \cr &= {\frac{L^{*}-a(b)+a(1)}{2}}\times{\frac{s}{B}}. \end{aligned} $$

Similarly, for j = 2, 3,…, b, T _Startup(Q ⁱ(j)) can be formulated as

$$ \begin{aligned} T_{\rm Startup}(Q^{i}(j)) &= {\frac{1}{a(j)-a(j-1)}}\times\int_{x=a(j-1)}^{a(j)}(a(j)-x)dx\times{\frac{s}{B}} \cr &= {\frac{1} {a(j)-a(j-1)}}\times\left(a(j)\int_{x=a(j-1)}^{a(j)}1dx-\int_{x=a(j-1)}^{a(j)}xdx\right)\times{\frac{s}{B}} \cr& = {\frac{a(j)-a(j-1)}{2}}\times{\frac{s}{B}}. \end{aligned} $$

To summarize the above results, we have

$$ T_{\rm Startup}(Q^{i}(j))=\left\{ \begin{array}{lll} {\frac{L^{*}-a(b)+a(1)}{2}}\times{\frac{s}{B}} & : & \hbox{if}\,j=1, \cr {\frac{a(j)-a(j-1)}{2}}\times{\frac{s}{B}} & : & \hbox{otherwise}. \end{array}\right. $$

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Proof of Lemma 3

We prove this lemma by mathematical induction on the value of b.

1.1 Basis

We assume that the first data item of Q _i is D _k (that is, $D_{k}=D_{q^{i}(j)}$) and the number of the replicas D _k is b. Consider the case that b = 1. In this case, there is the only one copy of D _k in the broadcast program. On average, the user has to wait for a half broadcast cycle to retrieve D _k . Thus, $T_{\rm Startup}(Q_{i})={\frac{L^{*}}{2}}\times{\frac{s}{B}}= {\frac{L^{*}}{2b}}\times{\frac{s}{B}},$ and Lemma 3 is true when b = 1.

Consider the case that b = 2. Without loss of generality, we reorder the replicas of D _k based on their positions in ascending order and assume that the first copy of D _k is placed in the first slot of the broadcast program (i.e., a(1) = 1). Thus, the T _Startup(Q _i ) can be formulated as

$$ \begin{aligned} & T_{\rm Startup}(Q_{i}) \cr = & \sum_{j=1}^{b}(p_{i}(j)\times Q^{i}(j)) \cr = & \left[{\frac{a(2)-a(1)}{L^{*}}}\times{\frac{a(2)-a(1)}{2}} +{\frac{L^{*}-(a(2)-a(1))}{L^{*}}}\times{\frac{L^{*}-(a(2)-a(1))}{2}}\right]\times{\frac{s}{B}} \cr = & {\frac{1}{2\times L^{*}}}\times (a(2)^{2}-2a(2)+1)+{\frac{1}{2\times L^{*}}}\times(L^{*2}-2\times L^{*}\times (a(2)-1)+(a(2)-1)^{2})\times{\frac{s}{B}} \cr = & {\frac{1}{2\times L^{*}}}\times \left( 2a(2)^{2}-(2L^{*}+4)a(2)+L^{*2}+2L^{*}+2\right)\times{\frac{s}{B}}. \end{aligned} $$

(10)

Since only a(2) is unknown in the above equation, we have $T_{\rm Startup}^{''}(Q_{i})={\frac{2}{L^{*}}}\times{\frac{s}{B}}>0.$ Thus, T _Startup(Q _i ) has a minimal value and the minimal value is in the point that T ^′_Startup (Q _i ) = 0. By the following derivation, the minimal value of T _Startup(Q _i ) is in the point that $a(2)={\frac{L^{*}}{2}}+1.$

$$ \begin{aligned} T_{\rm Startup}^{\prime}(Q_{i})&=0 \cr 4a(2)-(2L^{*}+4)&=0 \cr a(2)&={\frac{L^{*}}{2}}+1 \end{aligned} $$

Then, we have $a(2)-a(1)={\frac{L^{*}}{2}}$ and $L^{*}-a(b)+a(1)={\frac{L^{*}}{2}}.$ In addition, by substituting $a(2)={\frac{L^{*}}{2}}+1$ into Eq. (10), we obtain that the lower bound of T _Startup(Q _i ) is ${\frac{L^{*}}{4}}\times{\frac{s}{B}}={\frac{L^{*}}{2b}}\times{\frac{s}{B}}.$ Thus, Lemma 3 is true when b = 2.

1.2 Induction step

Assume that Lemma 3 is true for b = 1, 2,…, k. We would like to use this hypothesis to prove that Lemma 3 is also true when b = k + 1. Without loss of generality, we assume that a(1) = 1. To facilitate the following derivation, we reorder the replicas of D _k based on their positions in ascending order. In addition, as shown in Fig. 12, we divide the broadcast program into two parts according to the position of the b-th replica of D _k (i.e., a(b)).

Fig. 11

The illustration for the Proof of Lemma 1 and 2

Fig. 12

The illustration for the Proof of Lemma 3

Part 1 is equivalent to a broadcast program of length a(b) − 1 and with b − 1 replicas of D _k . Based on the hypothesis, the lower bound of the average startup time of part 1 is ${\frac{a(b)-1}{2(b-1)}}={\frac{a(b)-1}{2k}}.$ In addition, we also have

$$ a(j+1)-a(j)={\frac{a(b)-1}{b}}={\frac{a(b)-1}{k}},\,\ \hbox{for } j\,=\,1,2,\ldots, b-2,\hbox{and} $$

(11)

$$ a(b-1)={\frac{a(b)-1}{k}}\times (b-2)+1={\frac{k-1}{k}}\times (a(b)-1)+1. $$

(12)

Thus, T _Startup(Q _i ) can be formulated as

$$ \begin{aligned} & T_{\rm Startup}(Q_{i}) \cr = & \left[{\frac{a(b)-1}{L^{*}}}\times{\frac{a(b)-1}{2k}}+{\frac{L^{*}-a(b) +1}{L^{*}}}\times{\frac{L^{*}-a(b)+1}{2}}\right]\times{\frac{s}{B}} \cr = & {\frac{1}{2\times L^{*}}} \times\left[\left(1+{\frac{1}{k}}\right)a(b)^{2}-\left({\frac{2}{k}} +2L^{*}+2\right)a(b)\,+\,{\frac{1}{k}}+L^{*2}+2L^{*}+1\right]\times{\frac{s}{B}} \end{aligned} $$

(13)

Similarly, only a(2) is unknown in the above equation. Since $T_{\rm Startup}^{''}(Q_{i})={\frac{1+{\frac{1}{k}}}{L^{*}}}\times{\frac{s}{B}}>0.$ the minimal value of T _Startup(Q _i ) is in the point that T ^′_Startup (Q _i ) = 0. By the following derivation, the minimal value of T _Startup(Q _i ) is in the point that $a(b)={\frac{k}{k+1}}L^{*}+1.$

$$ \begin{aligned} T_{\rm Startup}^{\prime}(Q_{i})&=0 \cr {\frac{1}{2\times L^{*}}} \times\left[\left(2+{\frac{2}{k}}\right)a(b)-\left({\frac{2}{k}} +2L^{*}+2\right)\right]\times{\frac{s}{B}}&=0 \cr a(b)&={{\frac{\frac{2}{k}}+2L^{*}+2}{2+{\frac{2}{k}}}}={\frac{k}{k+1}}L^{*}+1 \end{aligned} $$

By substituting the value of a(b) into Eq. (12), we have $a(b-1)={\frac{k-1}{k}}\times (a(b)-1)+1={\frac{k-1}{k+1}}L^{*}+1$ and $a(b)-a(b-1)={\frac{1}{k+1}}L^{*}.$ The following equations can be obtained by summarizing this result and Eq. (11).

$$ \begin{aligned} a(j+1)-a(j)={\frac{a(b)-1}{k}}&={\frac{L^{*}}{k+1}}, j=1,2,\ldots b-1 \cr L^{*}-a(b)+a(1)={\frac{L^{*}}{k+1}}&={\frac{L^{*}}{b}} \end{aligned} $$

By substituting $a(b)={\frac{k}{k+1}}L^{*}+1$ into Eq. (13), we obtain that the lower bound of T _Startup(Q _i ) is ${\frac{L^{*}}{2(k+1)}}\times{\frac{s}{B}}={\frac{L^{*}}{2b}}\times{\frac{s}{B}}.$ To summarize the above results, Lemma 3 is true for b = k + 1 when Lemma 3 is true for b = 1, 2,…, k.

Finally, Lemma 3 is proven by mathematical induction.