1. CPSC-608 Database Systems
Fall 2017
Instructor: Jianer Chen
Office: HRBB 315C
Phone:
Notes #7

2. Graduate Database DBMS lock table DDL language DDL complier file
administrator
DDL
complier
lock table
DDL
language
file
manager
logging &
recovery
concurrency
control
transaction
manager
database
programmer
index/file
manager
buffer
manager
query
execution
engine
DML
complier
main
memory
buffers
DML (query)
language
secondary
storage
(disks)
DBMS
Graduate Database

3. Graduate Database DBMS lock table DDL language DDL complier file
administrator
DDL
complier
lock table
DDL
language
file
manager
logging &
recovery
concurrency
control
transaction
manager
database
programmer
index/file
manager
buffer
manager
query
execution
engine
DML
complier
main
memory
buffers
DML (query)
language
secondary
storage
(disks)
DBMS
Graduate Database

4. Computer Memory Hierarchy

5. A Typical Computer ... CPU main memory Secondary Storage bus Disk
controller
Secondary
Storage
disks

6. Main Memory
fast
small capacity (gigabytes)
volatile
Disks
slow
large capacity (100’s gigabytes)
non-volatile

7. Typical Disk …
Terms: Platter, Head, Cylinder,
Track, Sector, Gap

8. Top View
Track
Sector
Gap

9. Original image © IBM Corporation
Top view of a 36 GB, 10,000 RPM, IBM SCSI server hard disk, with its top cover removed. Note the height of the drive and the 10 stacked platters. (The IBM Ultrastar 36ZX.)
Original image © IBM Corporation
Video show at

10. A “typical” disk 5 platters (thus 10 surfaces)
A surface has 20,000 tracks
A track has 500 sectors (million bytes)
A sector has several thousand bytes
Disk makes 5000 revolutions per minute (so about 10 millisecond per rotation)

11. Blocks A (logic) block = one or several sectors (typical size 16KB)
Block address
Physical device
Cylinder #
Surface #
Sector

12. ? Disk Access Time I want block X block X in memory
Time = Seek Time + Rotational Delay +
Transfer Time + Other

13. Seek Time
3 or 5x x 1 N
Cylinders Traveled
Time

14. Average Random Seek Time
  SEEKTIME (i  j)
S =
N(N-1)
i=1
j=1
ji
typical seek time: 10 ms  40 ms

15. Rotational Delay Average Rotational Delay R = 1/2 revolution
Head here
Block I want
Average Rotational Delay
R = 1/2 revolution
typical rotational delay = 8 ms

16. Transfer Rate: typical: t = 80 MB/second = 80 KB/millisecond
transfer time: block size / t
~ 10/80 < 1 ms

17. Other Delays Typical value: ≈ 0 CPU time to issue I/O
Contention for controller
Contention for bus, memory
Typical value: ≈ 0

18. Thus, reading a block of 16K bytes:
Time = Seek Time + Rotational Delay +
Transfer Time + Other
~ 30 ms + 8 ms + 16/80 ms + 0
~ 40 ms

19. slow (read/write: 1~40 millisecond) large capacity (100’s gigabytes)
Disks
slow (read/write: 1~40 millisecond)
large capacity (100’s gigabytes)
non-volatile
Main Memory
fast (read/write: nanosecond)
small capacity (gigabytes)
volatile
Disks are about 105~106 times slower than
main memory

20. I/O Model of Computation
Dominance of I/O cost: if a block needs to be moved between disk and main memory, then the time taken to perform the read/write is much larger than the time likely to be used to manipulate that data in main memory.
The number of disk block reads/writes is a good approximation to the entire computation.

21. Disk I/O Optimization: Example I

22. Disk I/O Optimization: Example I
Optimizing Disk Seek Time (by disk controller):

23. Disk I/O Optimization: Example I
Optimizing Disk Seek Time (by disk controller):
On a (dynamic) sequence of disk I/O requests, how do we order the requests to minimize the seek time?

24. Disk I/O Optimization: Example I
Optimizing Disk Seek Time (by disk controller):
On a (dynamic) sequence of disk I/O requests, how do we order the requests to minimize the seek time?
(seeking tracks is the most time consuming component of disk I/O. Moving to a nearer track takes less time.)

25. Disk I/O Optimization: Example I
Optimizing Disk Seek Time (by disk controller):
On a (dynamic) sequence of disk I/O requests, how do we order the requests to minimize the seek time?
(seeking tracks is the most time consuming component of disk I/O. Moving to a nearer track takes less time.)
Elevator Algorithm:
Let the head move along its current direction, process each encountered request on the way until no request is ahead. Then reverse the direction.

26. Elevator Algorithm
Keep an UpperQ and a LowerQ, and elevator’s current Direction and Position.
Repeat
1. If Direction = Up Then
If UpperQ  
Then x = Min(UpperQ); Position = x;
Delete(UpperQ, x);
Else Direction = Down;
2. Else If LowerQ  
Then x = Max(LowerQ); Position = x;
Delete(LowerQ, x);
Else Direction = Up.

27. Disk I/O Optimization: Example II
Reducing the number of disk I/Os.
Example. Sorting on disk
Each tuple (with a key) takes 160 bytes
Each block holds 100 tuples (16KB)
A relation R has 10M tuples
(1.6 GB, 100K blocks)
Main memory has 100MB (6400 blocks)
A disk read/write: 40 ms

28. Main memory sorting algorithms
disk read/write: 40 ms a tuple: 160 bytes a block: 16KB (100 tuples)
a relation R: 1.6 GB (10M tuples, 100K blocks) main memory: 100MB (6400 blocks)
Main memory sorting algorithms
heap sort:
10M * log2 (10M)
= 230M disk block read/write
= 9200M ms = seconds > 100 day
quick sort and merge sort:
2 * 100K (blocks) * log2 (10M)
= 4.6M disk block read/write
= 184M ms = seconds > 2 day

29. Two-phase Multiway MergeSort
disk read/write: 40 ms a tuple: 160 bytes a block: 16KB (100 tuples)
a relation R: 1.6 GB (10M tuples, 100K blocks) main memory: 100MB (6400 blocks)
Two-phase Multiway MergeSort
Phase 1. making sorted sublist
repeat
fill the main memory with remaining tuples in
R and sort them;
write the sorted sublist (of 6400 blocks) back to disk
Phase 2. Merging
bring in a block from each of the sorted sublist;
merge them and put in an “output” block;
write the “output” block back to disk when it is full

30. Two-phase Multiway MergeSort
Main memory
Two-phase Multiway MergeSort
Disk

31. Two-phase Multiway MergeSort
First Phase
Main memory
Two-phase Multiway MergeSort
Disk

32. Two-phase Multiway MergeSort
First Phase
Sort it
Main memory
Two-phase Multiway MergeSort
Disk

33. Two-phase Multiway MergeSort
First Phase
Main memory
Two-phase Multiway MergeSort
Disk

34. Two-phase Multiway MergeSort
First Phase
Main memory
Two-phase Multiway MergeSort
Disk

35. Two-phase Multiway MergeSort
First Phase
Sort it
Main memory
Two-phase Multiway MergeSort
Disk

36. Two-phase Multiway MergeSort
First Phase
Main memory
Two-phase Multiway MergeSort
Disk

37. Two-phase Multiway MergeSort
First Phase
Main memory
Two-phase Multiway MergeSort
Disk

38. Two-phase Multiway MergeSort
First Phase
Sort it
Main memory
Two-phase Multiway MergeSort
Disk

39. Two-phase Multiway MergeSort
First Phase
Main memory
Two-phase Multiway MergeSort
Disk

40. Two-phase Multiway MergeSort
First Phase
Main memory
Two-phase Multiway MergeSort
Disk

41. Two-phase Multiway MergeSort
First Phase
Sort it
Main memory
Two-phase Multiway MergeSort
Disk

42. Two-phase Multiway MergeSort
First Phase
Main memory
Two-phase Multiway MergeSort
Disk

43. Second Phase
Main memory
Disk

44. Two-phase Multiway MergeSort
Second Phase
Main memory
One block per sublist
Two-phase Multiway MergeSort
Disk

45. Two-phase Multiway MergeSort
Main memory
One block per sublist
Two-phase Multiway MergeSort
Disk

46. Two-phase Multiway MergeSort
Main memory
One block per sublist
Two-phase Multiway MergeSort
Disk

47. Two-phase Multiway MergeSort
Main memory
One block per sublist
Two-phase Multiway MergeSort
Disk

48. Two-phase Multiway MergeSort
Main memory
One block per sublist
Two-phase Multiway MergeSort
Disk

49. Two-phase Multiway MergeSort
Main memory
One block per sublist
Two-phase Multiway MergeSort
Disk

50. Two-phase Multiway MergeSort
Main memory
Two-phase Multiway MergeSort
Disk

51. Two-phase Multiway MergeSort
disk read/write: 40 ms a tuple: 160 bytes a block: 16KB (100 tuples)
a relation R: 1.6 GB (10M tuples, 100K blocks) main memory: 100MB (6400 blocks)
Two-phase Multiway MergeSort
# sublists = 100K/6400 = 16
thus, in phase 2, we can easily hold a block for each sublist in the main memory
Disk block read/write:
100K (blocks) * 4
= 400K disk block read/write
= 16M ms = seconds < 4.5 hours

52. Graduate Database DBMS lock table DDL language DDL complier file
administrator
DDL
complier
lock table
DDL
language
file
manager
logging &
recovery
concurrency
control
transaction
manager
database
programmer
index/file
manager
buffer
manager
query
execution
engine
DML
complier
main
memory
buffers
DML (query)
language
secondary
storage
(disks)
DBMS
Graduate Database

53. Read Chapter 13 for more details on
memory structures