Implementation of Relational Operators/Estimated Cost

1.Select2.Join

Sailors(sid, sname, rating, age)– Each Sailors tuple is 50 bytes long (fixed length record

format)– A page size is 4K bytes– #tuples: 40,000– All pages are full, unpacked bitmap; 96 bytes are

reserved for slot directory– How many pages for Sailors?

• One page can contain at most tuples

• Sailors occupies pages

4096 96 8050

40000 50080

Cost of Selection Operators)( op . RvalueAR

Factors to Consider)( op . RvalueAR

• No index• unsorted data• sorted data

• Index• tree index• hash-based index

No index, unsorted data. = (R)R A value

Suppose R is Sailors

Best access path: File ScanI/O Cost: 500 pagesI/O time cost: 500 * time to access each pageComplexity: O(|R|)

Notation: |R| is the number of pages in R

No Index, sorted file on R.A

. = (R)R A valueSuppose R is Sailors

Sorted on R.ABest Access Path:

•Binary search to locate the first tuple with R.A=Value•Scan the remaining records

I/O Cost: log2(|R|)+Cost of scan for remaining tuples (0 ~ |R|)

Tree Index on R.A

Selection Cost = cost of traversing from the root to the leaf + cost of retrieving the pages in the sequence set + cost of retrieving pages containing the data records.

• Need to know– Format of the data entries in the leaf node– Clustered or unclustered – Dense or sparse

. = (R)R A value

Index entries

Data entries

Data Records

Page (node)

Alternatives for Data Entry k* in Index Alternative 1: Data entry is the actual tuple

o Tuples organized according to the search key valueo Allow only one Alternative 1 index per relationA

BCDEFGHIJK

4020335163101527334055

L 97

F,10

G,15

H,27

I,33

J,40 K,

55L,97

B,20

C,33

D,51

E,63

A,40

Root

Alternatives for Data Entry k* in Index Alternative 2: Data entry is <k, rid of data record with search key

value k>o Ex. <50, (1, 10)>: search key value = 50, rid tells that the data is in

page 1, slot 10

10* 15* 20* 27* 33* 33* 40* 40* 51* 55* 63* 97*

20 33 51 63

40Root

AB

CD

EF

GH

IJ

K4020

3351

6310

1527

3340

55L 97

Alternatives for Data Entry k* in Index Alternative 3: Data entry is <k, list of rids with search key k>

o Ex: <50, (1,10), (2,20), (3,1)>search key value = 50; three record IDs

10* 15* 20* 27* 33* 40* 51* 55* 63* 97*

20 33 51 63

40Root

AB

CD

EF

GH

IJ

K4020

3351

6310

1527

3340

55L 97

Index entries

Formats of Data entries

Data Records

Pros and Cons Data entries are typically much smaller than data records. So,

Alternatives 2 and 3 are better than Alternative 1 with large data records, especially if search keys are small.

If more than one index is required on a given file, at most one index can use Alternative 1; the rest must use Alternatives 2 or 3.

Alternative 3 is more compact than Alternative 2, but leads to variable sized data entries even if search keys are of fixed length.

Index entries

Data entries

direct search for

(Index File)(Data file)

Data Records

data entries Data entries

Data Records

DENSE INDEX SPARSE INDEX

Dense or Sparse

• Dense: At least one data entry per data record• Sparse: At least one data entry per block/page

Pros and Cons:• Dense: less space-efficient, but great for both equality and range

search• Sparse: more space-efficient, but need sequential search within a

page

Dense or Sparse

10* 15* 20* 27* 33* 33* 40* 40* 51* 55* 63* 97*

20 33 51 63

40Root

AB

CD

EF

GH

IJ

K4020

3351

6310

1527

3340

55L 97

Dense or Sparse

20 33 51 63

40Root

AB

C

D EFG H

I J

K4020

33

51 631015 27

33 40

55 L 97

Index entries

Data entries

direct search for

(Index File)(Data file)

Data Records

data entries Data entries

Data Records

CLUSTERED INDEX UNCLUSTERED INDEX

Clustered or Unclustered

Clustered Index : • The ordering of data records is organized the same as or

close to the ordering of data entries in the index

• Sparse index is always clustered (why?)• A clustered index does not have to be sparse (why?)Pros and Cons:

• Clustered: maintenance cost high, but great for range search• Unclustered: low maintenance cost, but high retrieval cost

• Retrieving one record may need to load one page

Index Classification• Primary vs. secondary: If search key contains

primary key, then called primary index.– Unique index: Search key contains a candidate key.– Non-unique index: Search key is a non-candidate

attributeo Multiple records may be associated with a same

attribute value

• Support equality and range-searches efficiently• Balanced tree• Search/Insert/delete at log F N cost (F = fanout, N = # leaf

pages)• F => 100• The cost of traversing from root to data entries is usually

regarded as 4• Minimum 50% occupancy (except for root). • Each node except root contains d <= m <= 2d entries.

The root node contains 1<= m <= 2d entries.• The parameter d is called the order of the tree.

Index Entries

Data Entries("Sequence set")

(Direct search)

B+Tree: the Most Widely Used Index for RDBMS

doubly linked (why?)

B+ Tree Example• Search begins at root, and key comparisons direct it to a leaf.• Search for tuples whose search key = 15

• Follow the left pointer if the desired value is less than the value in the node

• Otherwise, follow the right pointerRoot

17 24 30

2* 3* 5* 7* 14* 16* 19* 20* 22* 24* 27* 29* 33* 34* 38* 39*

13

B+ Tree Example

Root

17 24 30

2* 3* 5* 7* 14* 16* 19* 20* 22* 24* 27* 29* 33* 34* 38* 39*

13

dense sparse

Dense/SparseClustered/Unclustered

Root

17 24 30

2* 3* 5* 7* 14* 16* 19* 20* 22* 24* 27* 29* 33* 34* 38* 39*

13

B+Tree Index on R.A

– Format of data entry: alternative 2– Size of data entry = 20 bytes; – Page size=4K bytes; 96 bytes are reserved– Total number of records = 100,000; record size = 40

bytes– Reduction Factor = 0.1

• #matching entries/#total entries

Total Cost = Cost of traversing from the root to the leaf (assume 4 I/Os) +Cost of retrieving the pages in the sequence set + the cost of retrieving pages containing the data records

. = (R)R A value

• B+tree, Alternative 2, dense, unclustered• I/O cost of retrieving pages of qualifying data

entries– Matching data entries: 0.1*100000=10,000

entries– #Date entries per page: – Pages of matching data entries = 10000/200 =

50 pages• I/O cost of retrieving qualifying tuples

– 10,000 pages since the index is unclustered, the qualifying tuples are not always in the same order as the data entries.

– In the worst case, for each qualifying data entry, one I/O is needed

• Total I/O Cost = 4+ 50+10,000 pages

4096 96 20020

Data Records

UNCLUSTERED INDEX

Data entries

• Format of data entry: alternative 2• Size of data entry = 20 bytes; • Page size=4K bytes; 96 bytes are reserved• Total number of records = 100,000; record size = 40 bytes• Reduction Factor = 0.1

• B+tree, Alternative 2, dense, clustered• I/O cost of retrieving pages of qualifying data

entries– Matching data entries: 0.1*100000=10000 entries– #Date entries per page: – #Pages of matching data entries =

• I/O cost of retrieving qualifying tuples– #Matching tuples: 10000 – Since the index is dense and clustered, the

qualifying tuples are also clustered– # pages: 10000/100=100 due to

(4096-96)/40=100 tuples per page• Total I/O Cost = 4+ 50+100 pages

4096 96 20020

10000 50

200

Data entries

Data Records

CLUSTERED INDEX

• B+tree, Alternative 2, sparse (must be clustered)• I/O cost of retrieving qualifying tuples

– #Matching tuples: 0.1*100,000=10,000 – Since the index is clustered, the qualifying tuples are also

clustered– # pages: due to 100 tuples per page

• I/O cost of retrieving pages of qualifying data entries– Matching data pages: 100 – #Data entries per page: – #Pages of matching data entries = =1 page

• Total I/O Cost = 4+ 1+100 pages

4096 96 20020

100200

10000100

Data entries

Data Records

CLUSTERED INDEX

SPARSE INDEXEach entry links to one page

Hash-based Index

• Hash-based index on A is good for equality search on attribute A– Usually cannot support range search

. = (R)R A value

Data File

300030005004

4003200760036003

5004

Ashby, 25, 3000

Smith, 44, 3000

Bristow, 30, 2007Basu, 33, 4003

Cass, 50, 5004

Tracy, 44, 5004

Daniels, 22, 6003Jones, 40, 6003

h2 salh2(sal)=00

h2(sal)=11

Hashed indexData entriesData records

I/O costs1) Cost for retrieving the

matching data entries2) Cost for retrieving the

qualifying tuples

I/O cost = cost for retrieving the matching data entries

+ cost for retrieving the qualifying tuples

• I/O cost of retrieving pages of matching data entries– Matching data entries: 0.1*100000=10000 entries– I/O cost = 10000*1.2 = 12000 I/Os

• I/O cost for retrieving the qualifying tuples = 12000 I/Os

• Total cost = 12000+12000 = 24000 I/Os

Total number of records: 100,000Reduction Factor: 0.1Cost for searching the matching data entry: 1.2 I/OAssume dense and unclustered

Factors to Consider

)( op . RvalueAR

• No index• unsorted data• sorted data

• Index• tree index• hash-based index