Moving Query Monitoring in Spatial Network Environments

Abstract

Moving queries over mobile objects are an important type of query in moving object database systems. In recent years, there have been quite a few works in this area. Due to the high frequency in location updates and the expensive cost of continuous query processing, server computation capacity and wireless communication bandwidth are the two limiting factors for large-scale deployment of moving object database systems. Many techniques have been proposed to address the server bottleneck including one using distributed servers. To address both scalability factors, distributed query processing techniques have been considered. These schemes enable moving objects to participate in query processing to substantially reduce the demand on server computation, and wireless communications associated with location updates. Most of these techniques, however, assume an open-space environment. Since Euclidean distance is different from network distance, techniques designed specifically for an open space cannot be easily adapted for a spatial network. In this paper, we present a distributed framework which can answer moving query over moving objects in a spatial network. To illustrate the effectiveness of the proposed framework, we study two representative moving queries, namely, moving range queries and moving k-nearest-neighbor queries. Detailed algorithms and communication mechanisms are presented. The simulation studies indicate that the proposed technique can significantly reduce server workload and wireless communication cost.

Appendix: Number of segments included in a monitoring region

In this appendix, we will show that for the n defined as:

$$n = \left\lceil avg\_r / avg\_seg \right\rceil $$

the total number of segments covered by the query’s monitoring region, denoted as f(n), is:

$$f(n) = 4n^2 + 4n - 1 $$

(12)

When n = 1, as shown in Fig. 18a, there are 7 segments in the monitoring region. The solid thick segment in the center is the segment where a query object is on. All the surrounding 6 segments are included in the monitoring region.

Fig. 18

Monitoring regions for n = 1 and n = 2

This means that Eq. 12 holds when n = 1:

$$f(1) = 7 = 4 \times 1^2 + 4 \times 1 - 1 \label{eq:f(1)} $$

(13)

Now, assume that when n = k − 1, with k ≥ 2, we have

$$f(k-1) = 4(k-1)^2 + 4(k-1) -1 \label{eq:f(k-1)} $$

(14)

The next step is to identify the relationship between f(k) and f(k − 1). When n = 2, the monitoring region is drawn in Fig. 18b. As we can see, the same segments inherited from when n = 1 are still included in the monitoring region, represented with solid thick segments. Moreover, there are new ones added into the monitoring region represented in the solid thin segments.

By comparing Figs. 18a and b, we can tell that when n is increased from 1 to 2, there are two groups of new segments added into the monitoring region, as depicted by the solid thin segments in Fig. 19a and b, respectively. The thin solid segments in Fig. 19a are the corner segments to be added into the new monitoring region. When n = 2, there are four such corner segments in each corner. When n = k, there will be 2k of such segments in the corner. Since there are four corners, there are 2k * 4 = 8k such segments, except two segments that are counted twice (the segments on the far left and right). So the number of segments belonging to the first set is 8k − 2. The second group of new segments is depicted in Fig. 19b. There is one solid thin segment that bridges the left-top corner and the right-top corner, similarly, there is another one in the bottom. When n is increased by one, there are always two new segments to be added into the new monitoring region.

Fig. 19

Observations when n = 2

To help verify these two observations, we draw the monitoring region for n = 3 in Fig. 20, which shows that there are 22 segments belonging to the first group and 2 segments from the second group.

Fig. 20

Monitoring region for n = 3

As a conclusion, when n is increased from k − 1 to k, there are two groups of segments added into the new monitoring region. The first group has (8k − 2) segments, while the second group has 2 segments. Then we have

$$\begin{array}{rll} f(k) & = &f(k-1) + (8k - 2) + 2 \nonumber \\ & = &f(k-1) + 8k \label{eq:f(k1)} \end{array}$$

(15)

After plugging Eq. 14 into Eq. 15, we get

$$\begin{array}{rll} f(k) & =& 4(k-1)^2 + 4(k-1) -1 + 8k \nonumber \\ & =& 4k^2 + 4k - 1 \label{eq:f(k2)} \end{array}$$

(16)

By combining Eq. 13, 14, and 16, we prove the correctness of Eq. 12 using mathematical induction.