1 CS222: Principles of Database Management Fall 2010 Professor Chen Li Department of Computer Science University of California, Irvine Notes 01

CS222Notes 012 Topic 1: Data Storage and Record- Oriented File Systems Data Storage –Storage hierarchy –Disks Record-oriented file systems

CS222Notes 013 Storage hierarchy CPU MemoryController Disk/tape... cache

CS222Notes 014 Storage Media Cache: inside/outside CPU –CPU: becoming faster and faster (>=3 GHz now) Main Memory –costs $100/Mbyte -- reduces every year –‘volatile’ -- does not survive system failures –random I/O very fast –data can be processed by CPU directly –capacity limited to orders of magnitude lower than what database needs.

CS222Notes 015 Storage Media: secondary storage Disks (floppy disks, hard disks, CD) –Cheap, and price reduces each year –Non-volatile (except when disk crashes) –Random I/O slow –Data needs to be transferred to memory to be processed by CPU Tape –Cheaper but slower than disks. –Sequential I/O devices. –Handy for backups, sometimes for archival.

CS222Notes 016 Databases and Storage Devices Due to capacity, cost, volatility factors, DBs usually stored in disks. Data brought to main memory for processing from disks There are many ways to interface memory with disk resident data E.g., virtual memory: –VM size limited to max address generated by CPU –Existing VM does not support durability File system provides a more powerful mapping between memory and disk storage A bunch of tricks used ensure that high latency of secondary storage does not impact application response time and system throughput –access disks asynchronously with active applications –prefetch data before application needs it –intelligent caching techniques

CS222Notes 017 Disk Storages -- Outline Disk mechanics Access times (random, sequential) Examples Optimization Other topics

CS222Notes 018 Terms: Spindle, Platters, Magnetic surfaces, Disk head, Disk controller, … … Disk mechanics

CS222Notes 019 Top Views Tracks Sectors Gaps Cylinders

CS222Notes 0110 Characteristics Diameter: 1 inch inches Cylinders: Surfaces:1 (CDs) -- many Tracks/Cyl: 2 (floppies) Sector Size:512B -- 50K Capacity:360 KB (old floppy) -- >=200GB

CS222Notes 0111 “Block” Corresponds to 1 or multiple sectors Its address consists of: –Physical device # (in case of multi disks) –Cylinder # –Surface # –Sector #

CS222Notes 0112 block x in memory I want block X Random disk access time Time = Seek Time + Rotational Delay + Transfer Time + Other time 1 time 2 time 3 time 4

CS222Notes or 5x x 1N Cylinders Traveled Time Time 1: seek time

CS222Notes 0114 Average Random Seek Time   SeekTime(Track i  Track j) S = N(N-1) N N i=1 j=1 j  i Assumptions: –Each track has the same probability to be accessed. –Each track has the probability to jump to another track. Typical S value: 10 ms – 50 ms

CS222Notes 0115 Time 2: Rotational Delay Initial Head Block Wanted Average delay: –R = 1/2 revolution –If disk speed 3600 RPM, then R = 8.33 ms

CS222Notes 0116 Complication May have to wait for start of track before we can read desired block Head Here Block We Want Track Start

CS222Notes 0117 Time 3: Transfer time Transfer time: block size/transfer rate Typical transfer rate:1  3 MB/sec

CS222Notes 0118 Time 4: Other Delays CPU time to issue I/O Contention for controller Contention for bus, memory, etc. Typical value: “0”

CS222Notes 0119 Reading “Next” block Additional time = Block size/transfer rate Other time negligible: –skip gaps –once in a while, next cylinder Sequential disk access

CS222Notes 0120 Average sequential IO time much smaller than random IO time –Random I/O:  20 ms (most time on the initial delay) –Sequential I/O:  1 ms. When designing a structure, try to use sequential IOs. –Data layout on disk becomes critical –Do not just look at the number of IOs Random I/O vs Sequential I/O

CS222Notes 0121 Modify blocks Read block Modify in memory Write block Verify –Optional –If so, the access time needs to add: full rotation + block size/transfer rate

CS222Notes 0122 Disk Specs: 3.5 in diameter 3600 RPM 1 surface Usable capacity: 16 MB = 2 24 # of cylinders: 128 = block = 1 sector = 1 KB 10% overhead between blocks (gaps) seek time: –average = 25 ms. –adjacent cyl = 5 ms. Example 1

CS222Notes 0123 bytes/cyl = 2 24 /2 7 = 2 17 = 128 KB blocks/cyl = 128 KB / 1 KB = 128 Cylinder

CS222Notes 0124 One track... Track Speed: –3600 RPM  60 revolutions / sec  ms/rev In each revolution: –Time over useful data: * 0.9=14.99 ms –Time over gaps: * 0.1 = 1.66 ms –Transfer time 1 block = 14.99/128 = ms –Trans. time 1 block + gap = 16.66/128 = 0.13ms

CS222Notes 0125 Bandwidths Burst bandwidth: –No time on gaps (10%) –1 KB in ms. BB=1KB / 0.117ms = 8.54 KB/ms = 8.33MB/sec Sustained bandwidth: –Including time on gaps –128 KB in ms. SB=128KB /16.66ms = 7.68 KB/ms = 7.50 MB/sec

CS222Notes 0126 Time of random block access Time to read one random block T1 T1 = seek time + rotational delay + Transfer time –Assume we do not have to wait for track start –Seek time = 25ms –Rotational delay = 16.66ms /2 = 8.33 ms –Transfer time =.117 ms –Total = 25 ms ms ms= ms Most of the time is on “seek time” and “rotational delay”!

CS222Notes 0127 Larger blocks? Suppose OS deals with 4 KB blocks We need to include the time of reading 1 block (without gap) and 3 blocks (with gaps) T4 = 25ms + (16.66ms/2) + (.117) x 1 + (.130) * 3 = ms Compare to T1 = ms – not much difference –That’s why we want to use sequential IOs! block

CS222Notes 0128 Reading a track T T = Time to read a full track (start at any block) T T = 25ms (seek time) + (0.13ms / 2) (rotational delay, half of a block) ms (transfer time) = ms The time could be a bit less by ignoring the last gap. Question: what if we need to wait for the start of a track?