1. Files and access methods
Layers of a DBMS
Query
Query optimization
Execution engine
Files and access methods
Buffer management
Disk space management
Query
Processor
Query
execution
plan

2. The Memory Hierarchy Main Memory Disk Tape 5-10 MB/S
transmission rates
2-10 GB storage
average time to
access a block:
10-15 msecs.
Need to consider
seek, rotation,
transfer times.
Keep records “close”
to each other.
1.5 MB/S transfer rate
280 GB typical
capacity
Only sequential access
Not for operational
data
Volatile
limited address
spaces
expensive
average access
time:
nanoseconds

3. Disk Space Manager
Task: manage the location of pages on disk (page = block)
Provides commands for:
allocating and deallocating a page
on disk
reading and writing pages.
Why not use the operating system
for this task?
Portability
Limited size of address space
May need to span several
disk devices.
Spindle
Tracks
Disk head
Sector
Arm movement
Platters
Arm assembly

4. Buffer Management in a DBMS
Page Requests from Higher Levels
BUFFER POOL
disk page
free frame
MAIN MEMORY
DISK DB
choice of frame dictated
by replacement policy
Data must be in RAM for DBMS to operate on it!
Table of <frame#, pageid> pairs is maintained. 4

5. Buffer Manager
Manages buffer pool: the pool provides space for a limited
number of pages from disk.
Needs to decide on page replacement policy.
Enables the higher levels of the DBMS to assume that the
needed data is in main memory.
Why not use the Operating System for the task??
- DBMS may be able to anticipate access patterns
- Hence, may also be able to perform prefetching
- DBMS needs the ability to force pages to disk.

6. Record Formats: Fixed Length
Base address (B)
Address = B+L1+L2
Information about field types same for all records in a file; stored in system catalogs.
Finding i’th field requires scan of record.
Note the importance of schema information! 9

7. Files of Records
Page or block is OK when doing I/O, but higher levels of DBMS operate on records, and files of records.
FILE: A collection of pages, each containing a collection of records. Must support:
insert/delete/modify record
read a particular record (specified using record id)
scan all records (possibly with some conditions on the records to be retrieved) 13

8. File Organizations
Heap files: Suitable when typical access is a file scan retrieving all records.
Sorted Files: Best if records must be retrieved in some order, or only a `range’ of records is needed.
Hashed Files: Good for equality selections.
File is a collection of buckets. Bucket = primary page plus zero or more overflow pages.
Hashing function h: h(r) = bucket in which record r belongs. h looks at only some of the fields of r, called the search fields. 2

9. Cost Model for Our Analysis
As a good approximation, we ignore CPU costs:
B: The number of data pages
R: Number of records per page
D: (Average) time to read or write disk page
Measuring number of page I/O’s ignores gains of pre-fetching blocks of pages; thus, even I/O cost is only approximated. 3

10. Sorting
Illustrates the difference in algorithm design when your data is not in main memory:
Problem: sort 1Gb of data with 1Mb of RAM.
Arises in many places in database systems:
Data requested in sorted order (ORDER BY)
Needed for grouping operations
First step in sort-merge join algorithm
Duplicate removal
Bulk loading of B+-tree indexes. 4

11. 2-Way Sort: Requires 3 Buffers
Pass 1: Read a page, sort it, write it.
only one buffer page is used
Pass 2, 3, …, etc.:
three buffer pages used.
INPUT 1
OUTPUT
INPUT 2
Main memory buffers
Disk
Disk 5

12. Two-Way External Merge Sort
3,4
6,2
9,4
8,7
5,6
3,1 2
Input file
Each pass we read + write each page in file.
N pages in the file => the number of passes
So toal cost is:
Idea: Divide and conquer: sort subfiles and merge
PASS 0
3,4
2,6
4,9
7,8
5,6
1,3 2
1-page runs
PASS 1
2,3
4,7
1,3
2-page runs
4,6
8,9
5,6 2
PASS 2
2,3
4,4
1,2
4-page runs
6,7
3,5
8,9 6
PASS 3
1,2
2,3
3,4
8-page runs
4,5
6,6
7,8 9 6

13. General External Merge Sort
More than 3 buffer pages. How can we utilize them?
To sort a file with N pages using B buffer pages:
Pass 0: use B buffer pages. Produce sorted runs of B pages each.
Pass 2, …, etc.: merge B-1 runs.
INPUT 1
. . .
INPUT 2
. . .
. . .
OUTPUT
INPUT B-1
Disk
Disk
B Main memory buffers 7

14. Cost of External Merge Sort
Number of passes:
Cost = 2N * (# of passes)
E.g., with 5 buffer pages, to sort 108 page file:
Pass 0: = 22 sorted runs of 5 pages each (last run is only 3 pages)
Pass 1: = 6 sorted runs of 20 pages each (last run is only 8 pages)
Pass 2: 2 sorted runs, 80 pages and 28 pages
Pass 3: Sorted file of 108 pages 8

15. Number of Passes of External Sort9

16. Cost Model for Our Analysis
As a good approximation, we ignore CPU costs:
B: The number of data pages
R: Number of records per page
D: (Average) time to read or write disk page
Measuring number of page I/O’s ignores gains of pre-fetching blocks of pages; thus, even I/O cost is only approximated.
Average-case analysis; based on several simplistic assumptions. 3

17. Assumptions in Our Analysis
Single record insert and delete.
Heap Files:
Equality selection on key; exactly one match.
Insert always at end of file.
Sorted Files:
Files compacted after deletions.
Selections on sort field(s).
Hashed Files:
No overflow buckets, 80% page occupancy. 4

18. Cost of Operations 5

19. Cost of Operations 6

20. Indexes
An index on a file speeds up selections on the search key fields for the index.
Any subset of the fields of a relation can be the search key for an index on the relation.
Search key is not the same as key (minimal set of fields that uniquely identify a record in a relation).
An index contains a collection of data entries, and supports efficient retrieval of all data entries k* with a given key value k. 7

21. Alternatives for Data Entry k* in Index
Three alternatives:
Data record with key value k
<k, rid of data record with search key value k>
<k, list of rids of data records with search key k>
Choice of alternative for data entries is orthogonal to the indexing technique used to locate data entries with a given key value k.
Examples of indexing techniques: B+ trees, hash-based structures 8

22. Alternatives for Data Entries (2)
If this is used, index structure is a file organization for data records (like Heap files or sorted files).
At most one index on a given collection of data records can use Alternative 1. (Otherwise, data records duplicated, leading to redundant storage and potential inconsistency.)
If data records very large, # of pages containing data entries is high. Implies size of auxiliary information in the index is also large, typically. 9

23. Alternatives for Data Entries (3)
Alternatives 2 and 3:
Data entries typically much smaller than data records. So, 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; rest must use Alternatives 2 or 3.
Alternative 3 more compact than Alternative 2, but leads to variable sized data entries even if search keys are of fixed length. 10

24. Index Classification
Primary vs. secondary: If search key contains primary key, then called primary index.
Clustered vs. unclustered: If order of data records is the same as, or `close to’, order of data entries, then called clustered index.
Alternative 1 implies clustered, but not vice-versa.
A file can be clustered on at most one search key.
Cost of retrieving data records through index varies greatly based on whether index is clustered or not! 11

25. Clustered vs. Unclustered Index
Data entries
Data entries
(Index File)
(Data file)
Data Records
Data Records
CLUSTERED
UNCLUSTERED 12

26. Index Classification (Contd.)
Dense vs. Sparse: If there is at least one data entry per search key value (in some data record), then dense.
Alternative 1 always leads to dense index.
Every sparse index is clustered!
Sparse indexes are smaller;
Ashby, 25, 3000 22
Basu, 33, 4003 25
Bristow, 30, 2007 30
Ashby 33
Cass
Cass, 50, 5004
Smith
Daniels, 22, 6003 40
Jones, 40, 6003 44 44
Smith, 44, 3000 50
Tracy, 44, 5004
Sparse Index
Dense Index on on
Data File
Name
Age 13

27. Index Classification (Contd.)
Composite Search Keys: Search on a combination of fields.
Equality query: Every field value is equal to a constant value. E.g. wrt <sal,age> index:
age=20 and sal =75
Range query: Some field value is not a constant. E.g.:
age =20; or age=20 and sal > 10
Examples of composite key
indexes using lexicographic order.
11,80 11
12,10 12
name
age
sal
12,20 12
13,75
bob 12 10 13
<age, sal>
cal 11 80
<age>
joe 12 20
10,12
sue 13 75 10
20,12
Data records
sorted by name 20
75,13 75
80,11 80
<sal, age>
<sal>
Data entries in index
sorted by <sal,age>
Data entries
sorted by <sal> 13

28. Tree-Based Indexes ``Find all students with gpa > 3.0’’
If data is in sorted file, do binary search to find first such student, then scan to find others.
Cost of binary search can be quite high.
Simple idea: Create an `index’ file.
Index File k1 k2 kN
Data File
Page 1
Page 2
Page 3
Page N
Can do binary search on (smaller) index file! 3

29. Tree-Based Indexes (2) Root index entry P K P K P K P 1 1 2 m 2 m 10*1 1 2 m 2 m
10*
15*
20*
27*
33*
37*
40*
46*
51*
55*
63*
97* 20 33 51 63 40
Root

30. B+ Tree: The Most Widely Used Index
Insert/delete at log F N cost; keep tree height-balanced. (F = fanout, N = # leaf pages)
Minimum 50% occupancy (except for root). Each node contains d <= m <= 2d entries. The parameter d is called the order of the tree.
Root
Index Entries
Data Entries 9

31. Example B+ Tree
Search begins at root, and key comparisons direct it to a leaf.
Search for 5*, 15*, all data entries >= 24* ... 13 17 24 30 2* 3* 5* 7*
14*
16*
19*
20*
22*
24*
27*
29*
33*
34*
38*
39* 10

32. B+ Trees in Practice Typical order: 100. Typical fill-factor: 67%.
average fanout = 133
Typical capacities:
Height 4: 1334 = 312,900,700 records
Height 3: 1333 = 2,352,637 records
Can often hold top levels in buffer pool:
Level 1 = page = Kbytes
Level 2 = pages = Mbyte
Level 3 = 17,689 pages = 133 MBytes

33. Inserting a Data Entry into a B+ Tree
Find correct leaf L.
Put data entry onto L.
If L has enough space, done!
Else, must split L (into L and a new node L2)
Redistribute entries evenly, copy up middle key.
Insert index entry pointing to L2 into parent of L.
This can happen recursively
To split index node, redistribute entries evenly, but push up middle key. (Contrast with leaf splits.) 6

34. Inserting 8* into Example B+ Tree
Note:
why minimum occupancy is guaranteed.
Difference between copy-up and push-up.
Entry to be inserted in parent node. 5
(Note that 5 is
s copied up and
continues to appear in the leaf.) 2* 3* 5* 7* 8* 5 24 30 17 13
Entry to be inserted in parent node.
(Note that 17 is pushed up and only
this with a leaf split.)
appears once in the index. Contrast 12

35. Example B+ Tree After Inserting 8*
Root 17 5 13 24 30 2* 3* 5* 7* 8*
14*
16*
19*
20*
22*
24*
27*
29*
33*
34*
38*
39*
Notice that root was split, leading to increase in height.
In this example, we can avoid split by re-distributing entries; however, this is usually not done in practice. 13

36. Deleting a Data Entry from a B+ Tree
Start at root, find leaf L where entry belongs.
Remove the entry.
If L is at least half-full, done!
If L has only d-1 entries,
Try to re-distribute, borrowing from sibling (adjacent node with same parent as L).
If re-distribution fails, merge L and sibling.
If merge occurred, must delete entry (pointing to L or sibling) from parent of L.
Merge could propagate to root, decreasing height. 14

37. Example Tree After (Inserting 8*, Then) Deleting 19* and 20* ...
Root 17 5 13 27 30 2* 3* 5* 7* 8*
14*
16*
22*
24*
27*
29*
33*
34*
38*
39*
Deleting 19* is easy.
Deleting 20* is done with re-distribution. Notice how middle key is copied up. 15

38. ... And Then Deleting 24* Must merge.
Observe `toss’ of index entry (on right), and `pull down’ of index entry (below). 30
22*
27*
29*
33*
34*
38*
39* 5 13 17 30 2* 3* 5* 7* 8*
14*
16*
22*
27*
29*
33*
34*
38*
39* 16