Storage Systems - Systems Architecture - Lecture Slides | CMSC 411, Study notes of Computer Science

Download Storage Systems - Systems Architecture - Lecture Slides | CMSC 411 and more Study notes Computer Science in PDF only on Docsity! CMSC 411 - A. Sussman (from D. O'Leary) 1 Computer Systems Architecture CMSC 411 Unit 6 – Storage Systems Alan Sussman May 2, 2006 CMSC 411 - Alan Sussman 2 Administrivia • HW #5 due today, #6 due Thursday • Quiz 2 scheduled for May 9 – on Units 4 and 4b (ILP) – practice quiz posted by tomorrow • Read Chapter 7, except 7.8, 7.12, 7.13 • Online course evaluation available at https://www.courses.umd.edu/online_evalua tion CMSC 411 - Alan Sussman 3 Last week • Loop unrolling – to minimize pipeline stalls by scheduling multiple loop iterations together, still respecting data dependences – limited by # registers, instruction cache size, data and control dependences (including loop-carried ones) – problems include increased code size, slower compilation, work for programmer or compiler writer – data dependences – GCD test – did you get that? • Conditional instructions – turn control into data dependences – move dependence to end of pipeline – helps with scheduling for superscalar (moving instructions past branches – problems include using extra processor resources, slowing clock rate • Speculative instructions – only serious problem is preserving exception behavior – example is speculative load instruction, that fails if a subsequent store is to the same memory address (before the load check instruction) • IA-64 – supports compiler-based ILP • predicated instructions • deferred exceptions and load speculation CMSC 411 - Alan Sussman 4 Storage systems • We already know about four levels of storage: – registers – cache – memory – disk • but we've been a little vague on how these devices are interconnected • In this unit, we study – input/output units such as disks and tapes – buses to connect storage devices – I/O performance issues – design of file systems (won’t talk much about this) Disk and Tape Technologies CMSC 411 - Alan Sussman 6 (Hard) Disks • What it is: – a collection of 1-20 platters (like 2-sided CD's) – between 1 and 8 inches in diameter – 2.5 & 3.5 inch most common today – rotating on a central spindle – with 500-2500 tracks on each surface – divided into (maybe) 64 sectors • older disks: all tracks have the same number of sectors • current disks: outer tracks have more sectors • larger diameter: best retrieval times • smaller diameter: cheaper and uses less power CMSC 411 - A. Sussman (from D. O'Leary) 2 CMSC 411 - Alan Sussman 7 Disks (cont.) – Fig. 7.1 • Used for – file storage – slowest level of virtual memory during program execution CMSC 411 - Alan Sussman 8 Disks (cont.) • How information is retrieved: – Wait for previous requests to be filled Time = queuing delay – A movable arm is positioned at the correct cylinder Time = seek time – The system waits for the correct sector to appear under the arm Time = rotational latency – Then a magnetic head senses • the sector number • the information recorded in the sector • an error correction code CMSC 411 - Alan Sussman 9 Disks (cont.) – and the information is transferred to a buffer Time = transfer time – The retrieval is handled by a disk controller, which may impose some extra overhead Time = controller time • Because all of this is so expensive, might also read the next sector or two, hoping that the next information needed is located there (prefetch or read ahead) CMSC 411 - Alan Sussman 10 Example • Average disk access time (in millisec): average seek time + average rotational delay + transfer time + controller overhead1024 bytessector size .5 mscontroller overhead 8000 RPMrotation speed 10MB/sectransfer rate 5 msaverage seek time CMSC 411 - Alan Sussman 11 Example (cont.) • average seek time = 5 ms • average rotational delay = • transfer time = • controller overhead = .5 ms • Total: 5 + 3.75 + .1 + .5 = 9.35 ms ms RPSRPM 75.3 )60/000,8( 5.0 000,8 5.0 == ms bytes bytes MB KB 1.sec10 sec/10 10 sec/10 1 4 7 3 === − CMSC 411 - Alan Sussman 12 Technology gap between memory and disk – Fig. 7.5 CMSC 411 - A. Sussman (from D. O'Leary) 5 CMSC 411 - Alan Sussman 25 Failure rate vs. Availability • Failure rate: concerns whether any of the hardware is broken • Availability: concerns whether the system is usable, even if some pieces are broken • Example 1: Your bank can improve the availability of the ATM system by installing two ATM machines so that one is available even if one breaks • Example 2: Your bank can reduce the failure rate of the ATM system by installing a machine that does not break as often – Also increases the availability • Generally, hope that more complicated hardware improves availability and performance, but it also may increase the failure rate CMSC 411 - Alan Sussman 26 Example: Disk arrays • Suppose a machine has an array of 20 disks – Case 1: If distribute the data across the disks (striping), then all 20 disks must be working properly in order to access the data - but throughput can be improved – Case 2: If store 20 copies of the data, one copy per disk, have good availability: can access the data even if some disks fail • But reliability of the 20 disks is less than reliability of a single disk: the probability of one of the 20 disks failing is essentially 20 times the probability that a single disk will fail CMSC 411 - Alan Sussman 27 Disk arrays (cont.) • In Case 2, store multiple copies on multiple disks, called RAID: redundant arrays of inexpensive disks • RAID is actually not inexpensive (because of the cost of the controllers, power supplies, and fans), so often the “I” is said to stand for “independent” – More than 80% of non-PC disk drive sales are now RAID, a $19B industry – Typically store 2 copies, not 20 – Used when availability is critical, in applications such as: • airline reservations • medical records • stock market CMSC 411 - Alan Sussman 28 RAID – Fig. 7.17 • There are various levels of RAID, depending on the relative importance of availability, accuracy, and cost 2826 – P+Q redundancy widely used1815 - 4 w/distributed parity Network Appliance1814 - Block-interleaved parity Storage Concepts1813 - Bit-interleaved parity 4812 - Memory-style ECC EMC, Compaq, IBM8811 - Mirrored widely used0800 - Striped CompaniesCheck disks Example data disks # faults survived RAID level CMSC 411 - Alan Sussman 29 RAID levels 0 & 1 • One copy of data: RAID 0 – Data striped across a disk array • Two full copies of data (mirroring): RAID 1 – If one disk fails, go to other – Can also use this to distribute the load of READs – Most expensive RAID option • RAID 0 and 1 can be combined – 1+0 (or 10) – mirror pairs of disks, then stripe across pairs – 0+1 (or 01) – stripe across one set of half the disks, then mirror writes to both sets CMSC 411 - Alan Sussman 30 RAID 3 • Bit-interleaved parity: RAID 3 – One copy of the data, stored among several disks, and one extra disk to hold a parity bit (checksum) for the others • Example: Suppose have 4 data disks, and one piece of the data looks like this: Disk 1: 0 1 0 1 1 0 0 0 Disk 2: 0 1 1 1 0 1 1 0 Disk 3: 0 1 1 1 1 0 0 0 Disk 4: 0 0 0 1 0 1 0 1 – Then the parity bits are set by taking the sums mod 2: Disk 5: 0 1 0 0 0 0 1 1 CMSC 411 - A. Sussman (from D. O'Leary) 6 CMSC 411 - Alan Sussman 31 RAID 3 (cont.) • So if the data on one of the disks becomes corrupted, the parity bits on Disk 5 will be wrong, so can tell there has been a failure – and be able to fix it if know which disk failed • Disadvantage: Each data access must read from all 5 disks in order to retrieve the data and check for corruption – also can’t always tell where the error is (could even be on the parity disk) CMSC 411 - Alan Sussman 32 RAID 4 • Block-interleaved parity: RAID 4 – Same organization of data as RAID 3 but cheaper reads and writes – Read: Read one sector at a time, and count on the sector's own error detection mechanisms. – Write: In each write, note which bits are changing - this is enough information to change the parity bits without reading from the other disks CMSC 411 - Alan Sussman 33 RAID 4 example • If the original contents are Disk 1: 0 1 0 1 1 0 0 0 Disk 2: 0 1 1 1 0 1 1 0 Disk 3: 0 1 1 1 1 0 0 0 Disk 4: 0 0 0 1 0 1 0 1 Disk 5: 0 1 0 0 0 0 1 1 • And write Disk 2: 0 1 1 1 0 1 1 0 old Disk 2: 1 0 1 1 0 0 1 1 new • Then since bits 0, 1, 5, and 7 changed, need to flip those parity bits: Disk 5: 0 1 0 0 0 0 1 1 old Disk 5: 1 0 0 0 0 1 1 0 new CMSC 411 - Alan Sussman 34 RAID 5 – Fig. 7.19 • Disadvantage of RAID 4: Parity disk is a bottleneck, so it is better to interleave the parity information across all of the disks (RAID 5) CMSC 411 - Alan Sussman 35 RAID 6 • Also called P+Q redundancy – to allow recovery from a second failure, since parity schemes only can recover from one – need a second extra (check) disk – computation is more complicated than simple parity CMSC 411 - Alan Sussman 36 RAID summary • Higher throughput than single disk – in either MB/sec or I/Os/sec • Failure recovery easy • Allows taking advantage of small size and low power requirements of small disks, and still get these advantages – RAIDs now dominate large-scale storage systems • Note: No need to memorize the RAID levels – But you need to be able to explain how the example RAID levels work CMSC 411 - A. Sussman (from D. O'Leary) 7 I/O performance measures CMSC 411 - Alan Sussman 38 I/O performance measures • diversity: which I/O devices can connect to the system? • capacity: how many I/O devices can connect to the system? • bandwidth: throughput, or how much data can be moved per unit time • latency: response time, the interval between a request and its completion • High throughput usually means slow response time! CMSC 411 - Alan Sussman 39 Throughput vs. latency Fig. 7.24 Fig. 7.25 CMSC 411 - Alan Sussman 40 Improving performance (cont.) • Adding another server can decrease response time, if workload is held constant – but keeping work balanced between servers is difficult • To design a responsive system, must understand what the “typical” user wants to do with it • Each transaction consists of three parts: – entry time: the time for the user to make the request – system response time: the latency – think time: the time between system response and the next entry • Key observation is that a faster system produces a lower think time – see Fig. 7.26 CMSC 411 - Alan Sussman 41 Modeling computer performance • The usual way to model computer performance is using queuing theory (mathematics again) • Unfortunately, even queuing theory does not provide a very good model, so more complicated mathematics is now being applied (e.g., stochastic differential equations) • But, H&P only consider queuing models – and we don’t even have time to go into that Data Management Issues