Expected Delay for Various Access Probabilities Access Probability Expected Delay (in broadcast units) A B C Flat (a) Skewed (b) Multi-disk (c)

Transcription

1 Organization of Broadcast Data Source: [2]: S. Acharya, et al. Broadcast disks: data management for asymmetric communication environments," ACM SIGMOD International Conference on Management of Data (SIGMOD 1995),

2 Expected Delay for Various Access Probabilities Access Probability Expected Delay (in broadcast units) A B C Flat (a) Skewed (b) Multi-disk (c)

3 Broadcast Program Generation Broadcast Program Generation (For simplicity, assume that data items are pages", that is, they are of a uniform, fixed length.) 1. Order the pages from hottest (most popular) to coldest. 2. Partition the list of pages into multiple range, where each range contains pages with similar access probabilities. These ranges are referred to as disk. 3. Choose the relative frequency of broadcast for each of the disks. The only restriction on the relative frequencies is that they must be integers. For example given two disks, disk 1 could be broadcast three times for every two times that disk 2 is broadcast, thus, rel freq(1)=3, and rel freq(2)=2. 3 (Next page to continue )

4 Broadcast Program Generation (con't) 4. Split each disk into a number of smaller units. These units are called chunks (C ij refers to the j th chunk in disk i). First, calculate max chunks as the Least Common Multiple (LCM) of the relative frequencies. Then, split each disk i into number chunks(i) = max chunks / rel freq(i) chunks. In the previous example, number chunks(1) would be 2, while number chunks(2) would be Create the broadcast program by interleaving the chunks of each disk in the following manner: 4 (Next page to continue )

5 Broadcast Program Generation (con't) 01 for i = 0 to max chunks 1 02 for j = 1 to num disks 03 k = i mod num chunks(j) 04 Broadcast chunk C j;k 05 endfor 06 endfor 5

6 Database (pages) HOT COLD Deriving a Server Broadcast Program Tracks T 1 T 2 T 3 Chunks C 1,1 C 2,1 C 2,2 C 3,1 C 3,2 C 3,3 C 3,4 Major Cycle Broadcast Disk program C 1,1 C 2,1 C 3,1 C 1,1 C 2,2 C 3,2 C 1,1 C 2,1 C 3,3 C 1,1 C 2,2 C 3,4 Minor Cycle 6

7 Organization of Broadcast Data Source: M.S. Chen, K.L. Wu and P.S. Yu, Indexed sequential data broadcasting in a wireless computing environment," 17th IEEE International Conference on Distributed Computing Systems. 1997, pp

8 ffl To achieve energy saving Index Broadcasting: ffl When a palmtop is actively listening to a channel, it CPU must be in an active mode to examine data packets so as to determine if they match the predefined patterns. ffl Therefore, to achieve energy saving it is highly desirable to let the palmtops stay in the doze mode most of the time and only come to the active mode when it is necessary. 8

9 Index Broadcasting: 9 I a1 a2 a3 R1 R2 R3 R4 R5 R6 R7 R8 R9 (a) an example index tree I a1 R1 R2 R3 a2 R4 R5 R6 a3 R7 R8 R9 (b) index probing scenario to data record R5

10 ffl Where does the system arrange the index? The system broadcasts the index and broadcast data in a channel. The system broadcasts the index and broadcast data in two different channels. ffl Here, we only discussed the former case, i.e., the index and broadcast data are in the same channel. 10

11 ffl The main architecture of index design in this subsection. Design an index tree which can minimize the average cost of index probes. Two cases are considered: one for fixed index fanouts and the other for variant index fanouts. Λ For the case of fixed index fanouts, in lights of Huffman code, CF (constant fanout) is derived for the optimal index tree construction that minimizes the average number of index probes. Λ For the case of variant index fanouts, VF (variant fanout) is derived and shows that the average cost of index probes can be reduced not only by employing an imbalanced index tree that is designed in accordance with data access skew, but also by exploiting variant fanouts for index notes. 11

12 Imbalanced Index Tree Construction for Fixed Fanouts The access probability for each data record and the number of index probes required to reach each each record by T B d=3 and T I d=3. Data record R1 R2 R3 R4 R5 R6 R7 R8 R9 Pr(Ri) Ipb(Ri) in T B d= Ipb(Ri) in T I d=

13 Alogrithm CF : Use access frequencies to build an index tree with a fixed fanout d. Step 1: Initially, we have a forest of n subtrees, each of which is a single node. Step 2: Attach the d subtrees with the smallest labels to a new node. Label the resulting subtrees with the sum of all labels from its d subtrees. Step 3: n = n d + 1. (Then, n is the number of remaining subtrees.) If n = 1 stop else go to Step 2. Theorem 1: Given a fixed index fanout, the average number of index probes, i.e., P 1»i»n Pr(R i)ipb(ri), is minimized by using the index tree constructed by algorithm CF. 13

14 Illustration of an imbalanced index tree I a1 R1 R2 I R1 R2 a1 R3 a2 R4 R5 R6 a3 R7 R8 R a2 a (b) a corresponding data broadcast sequence R3 R4 R5 R6 R7 R8 R (a) an index tree built by CF 14

15 ffl Denote the cost of probing from an index ai as f(ai). Then, the average cost of locating a record by probing indexes can be expressed as: X X f(aj)) (Pr(Ri) j 2P ath(ri) a 1»i»n ffl The index probe cost required to reach each record by different index tree built by CF Data record R1 R2 R3 R4 R5 R6 R7 R8 R Pr(Ri) j 2P ath(r i ) d(a j ) in T I d= Pa Pa j 2P ath(r i ) d(a j ) in T I d= Pa j 2P ath(r i ) d(a j ) in T I d=

16 An index tree T I d=2 built by algorithm CF I 0.4 a1 0.6 R a2 R a3 a a a R3.05 a R4.05 R5.02 R6.02 R7.02 R8.02 R

17 An index tree T I d=4 built by algorithm CF I a1 R1 R a2 R3.05 R4.05 R5.02 R6 R R8.02 R

18 Imbalanced Index Tree Construction for Variant Fanouts ffl Basically, algorithm VF is greedy in nature and builds the index tree in a top down manners ffl VF starts with attaching all data records to the root node. Then, after some evaluation, VF groups nodes with small access frequencies and moves them to one level lower so as to minimize the average index probe cost. 18

19 Illustration of VF idea r r h1 h2 h1 h2 hi hx hm hi+1 hi+2 hm 19

20 Lemma: Suppose that node r has m child nodes, h1, h2,, hm, which are sorted according to descending order of Pr(hj), 1» j» m, i.e., Pr(hj) Pr(hk) iff j» k. Then, the advantage cost of index probes can be reduced by grouping nodes hi+1, hi+2,, hm and attaching them under a new child node if and only if 1) X (m Pr(hj) > X Pr(hj) i i+1»j»m 1»j»i 20

21 Alogrithm VF : Step 1: Assume that R1, R2,, Rn have been sorted according to descending order of Pr(Rj). Step 2: P artition(r1; R2; ; Rn). Step 3: Report the resulting index tree. 21

22 Procedure P artition(h1; h2; ; hm): 1. Let y(i) = (m i 1) P 1»j»i Pr(h j) P i+1»j»m Pr(h j). Determine i Λ such that y(i Λ ) = max 8i2f1;m 2g fy(i)g. 2. If y(i Λ )» 0 then return. 3. Attach nodes hi Λ +1, hi Λ +2,, hm under a new index node hx in the index tree. 4. P artition(hi Λ +1;hi Λ +2; ;hm). (h1, h2,, hi Λ +1) according to descending order of Pr(hj), 1» j» i Λ P artition(h1;h2; ;hi Λ +1). 6. Return. 22

23 Example: We use an example to illustrate algorithm VF. Data record R1 R2 R3 R4 R5 R6 R7 R8 R9 R10 R11 Pr(Ri)

24 Illustration of VF I I a1.12 R1 R2 R3 R4 R5 R6 R7 R8 R9 R10 R11 R1 R2 R3 R4 (a) R5 R6 R7 R8 R9 R10 R11 (b) I I a a3.52 a2 a1 a1 R1 R2 R3 R4 R1 R2 R3 R4 (c) (d) 24

25 ffl Determine the set of nodes to be grouped together in (a), where m = i i) X (10 1<j<iPr(hj ) 9*0.28 8*0.48 7*0.68 6*0.88 5*0.92 4*0.96 3*0.98 2* X Pr(hj ) i+1<j<11 y(i) ffl Determine the set of nodes to be grouped together in (b), where m = 5. i i) P 1<j<i Pr(h j ) 3*0.28 2* (4 Pr(h j ) Pi+1<j<5 y(i)

26 Illustration of VF Further Partitions of Nodes Under Index a1 a1 a1.08 a4 R5 R6 R7 R8 R9 R10 R11.02 a5 R5 (a) a1 R6 R7.08 a4 R8 R9 R10 R11 (c) R5 R6 R7 R8 R9 R10 R11 (b) 26

27 Illustration of VF the Resulting Index Tree I.48 a3.52 a2.12 a1 R1 R2 R3 R4.08 a4.02 a5 R5 R6 R7 R8 R9 R10 R11 27

28 Interleaving the broadcasting sequence with index nodes R1 R2 R3 R4 R5 R6 R7 R8 R9 R10 R11 I, a3 a2 a1 a4 a5 (a) I a3 R1 R2 a2 R3 R4 a1 R5 a4 R6 R7 a5 R8 R9 R10 R11 (b) 28