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1860: Computer Systems & Architecture > External Memory > Flashcards

External Memory Flashcards

Week 2.8 (99 cards)

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1
Q

write mechanism of magnetic disks

A
  • magnetic patterns recorded on platter
  • current direction detemines magnetism
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2
Q

read mechanism of magnetic disks

A

same or seperate read heads

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3
Q

organisation of magnetic disk

A
  1. head
  2. tracks
  3. intertrack gaps
  4. sectors
  5. intersector gaps
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4
Q

what does the head do in a magnetic disk

A

reads from or writes to a portion of the rotating platter

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5
Q

what is a track

A

concentric rings where data is stored

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6
Q

what do intertrack gaps do

A

minimise errors due to misalignment or interference

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7
Q

what are sectors

A

smallest unit of data transfer

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8
Q

what are do intersector gaps do

A

prevent precision-related errors

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9
Q

what is the main problem with magnetic disks

A

outer tracks move faster under the read/write head

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10
Q

what are the 2 solutions to the magnetic disk rotational problem

A
  1. constant angular velocity (CAV)
  2. multiple zone recording (MZR)
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11
Q

define CAV

A
  • entire disk spins as one unit at a fixed speed
  • outer tracks have a larger circunmference but store the same amount of data
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12
Q

advantage of CAV

A

faster & simpler data access - fixed sector addressing

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13
Q

disadvantage of CAV

A

wasted storage space

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14
Q

define MZR

A
  • disk is divided into multiple zones, each with a difference number of sectors per track
  • outer zones have more sectors - ^ storage efficiency
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15
Q

advantage of mZR

A

^ storage capacity

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16
Q

disadvantage of MZR

A

requires complex timing adjustments

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17
Q

what are the 4 physical characteristics of magnetic disks

A
  1. disk portability
  2. disk sides
  3. single/multiple platter disks
  4. disk head mechanism
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18
Q

describe disk portability

A
  • nonremovable disks - fixed in the drive
  • removable disks - allows unlimited storage capacity with limited drives & enables portability
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19
Q

describe difference in disk sides

A
  • double-sided = ^ capacity
  • single-sided = v cost
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20
Q

describe single/multiple platter disks

A
  • multiple platters stacked together - each with their own read/write head
  • set of track at same position = cylinder
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21
Q

describe disk head mechanisms

A
  • fixed gap heads = hover at fixed distance above the platter
  • contact heads = head touches disk surface
  • winchester heads = operates very close to the disk for higher data density
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22
Q

what are the 5 measurements involved in disk I/O operation & timing

A
  1. seek time
  2. rotational delay
  3. access time
  4. transfer time
  5. queuing delays
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23
Q

define seek time

A

time to move head to the correct track
- start-up time + track traversal time
- + settling time to confirm track positioning

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24
Q

define rotational latency

A

time for the sector to align with the head

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25
define access time
seek time + rotational latency
26
define transfer time
time taken to read/write as the sector moves under the head
27
define queuing delays
process waits for device availability & I/O channel assignment
28
describe rotational position sensing
- after seek command, the I/O channel is released for other operations - disk must reconnect to its I/O module and the I/O channel when the desired sector approaches - if channel is busy, the disk must wait a full rotation (RPS miss) - RPS misses introduce additional delays in disk access
29
transfer time formula
T = b/rN where: - b = number of bytes to be transferred - N = number of bytes per track - r = revolutions per second
30
total read/write time formula
Ttotal = Ts + 1/2r + b/rN where: - b = number of bytes to be transferred - N = number of bytes per track - r = revolutions per second - 1/2r = average rotational delay - Ts = seek time
31
average rotational delay formula
1/2r
32
one full disk rotation forumla
1/r
33
why is more platters not the best solution
- mechanical limitations = ^ weight = v speed of head - all heads move together = limits independent reads - heat & power issues
34
what is a better solution than more disk platters
multiple drives in parallel - RAID
35
describe redundant array of independent disks (RAID)
1. a set of physical disk drives viewed by the operation system as a single logical drive 2. data is distributed across the drives using a method called striping 3. redundant disk capacity stored parity information for data recovery in case of failure
36
disks required for RAID 0
N
37
describe RAID 0
- no redundancy - when performance + capacity has a greater need than reliability - data is stripped across all disks in the array = parallel access - multiple I/O requests handled simultaneously - ideal for large file transfers, database queries
38
what is array management software
- maps between logical & physical disk space - can run in the disk subsystem or the host computer
39
what are 2 requirements for RAID 0 to have high data transfer capacity
1. high transfer capacity across entire path 2. large, continuous I/O requests
40
why does RAID 0 have high data transfer capacity
- depends on patterns & data layout - no redundancy overhead - large I/O requests enable parallel transfers from multiple disks - if small, RAID 0 similar speed to a single disk
41
why does RAID 0 have a high I/O request rate
- when response time matters more than transfer rate - small I/O requests are dominated by seek time and rotational latency - RAID 0 improves I/O execution rates by balancing the load across multiple disks - high performance = multiple I/O requests outstanding at the same time
42
differences in size of strip and their impact on RAID 0's high I/O request rate
1. large = each I/O goes to a single disk - parallel execution 2. small = multiple disks involved in a single request, reducing parallelism
43
effect of small strip size on RAID 0 configuration
distributes large files across disks for parallel access - ^ sequential transfer speed
44
effect of large strip size on RAID 0 configuration
allows different disks to handle speperate requests independently - ^ responsiveness
45
RAID 1 name
mirror for redundancy
46
how many disks are need for RAID 1
2N
47
describe RAID 1
- duplicates all data across 2 disks, providing full redundancy - read requests can be handled by either disk - reduced access time - write requests must updated both disks but can be done in parallel - simple recovery
48
disadvantage of RAID 1
costly - 2x storage capacity needed
49
use of RAID 1
best suited for critical data storage
50
RAID 2 name
parallel access with error correction
51
how many disks needed for RAID 2
N + m
52
describe RAID 2
- each bit is stored on a different disk at the same relative position - uses parallel access, every disk participates in every I/O request - synchronised spindle = all disk heads are aligned - hamming code used for error correction, stored in separate parity disks - parity disks do not store data - only detect and correct
53
disadvantage of RAID 2
high redundancy cost - number of parity disks grow with data disks
54
use of RAID 2
rarely used today due ti the high reliability of modern disks
55
RAID 3 name
parallel access with single parity disk
56
how many disks are needed for RAID 3
N + 1
57
describe RAID 3
- data striped at byte level across multiple disks - parallel access - single parity disk = XOR-based parity for fault tolerance - fault recovery: if a disk fails, missing data is reconstructed using the XOR of the remaining disks and parity disks
58
disadvantage of RAID 3
limited independent disk operations
59
use of RAID 3
best for high-speed sequential data transfers
60
RAID 4 name
independent access with single parity disk
61
how many disks are needed for RAID 4
N + 1
62
describe RAID 4
- independent access - separate I/O requests - striped at block-level - single parity disk provides redundancy using XOR calculations - efficient for reads
63
disadvantages of RAID 4
- write penalty: small writes require reading old data and parity before updating - parity disk bottleneck: all writes must update the parity disk - limiting performance
64
use of RAID 4
best for high IOPS workloads
65
RAID 5 name
distributed parity for better performance
66
how many disks needed for RAID 5
N + 1
67
describe RAID 5
- distributed parity strips across all disks - uses a round-robin allocation to evenly spread parity - eliminates parity disk bottleneck - provides redundancy with improved write performance
68
use of RAID 5
best for balanced workloads with frequent reads and writes
69
RAID 6 name
dual parity for high availability
70
how many disks are needed for RAID 6
N + 2
71
describe RAID 6
- two different parity calculations - N + 2 disks - parity blocks stored separately - can recover from two simultaneous disk failures - extreme high data availability - read performance similar to RAID 5
72
disadvantage of RAID 6
write penalty: each write affects 2 parity blocks - reducing performance
73
what is RAID 10 best for
mirroring & striping = high performance & redundancy
74
6 advantages of using solid state storage
1. ^ IOPS 2. durability 3. lifespan 4. v power consumption 5. quieter 6. v latency
75
system architecture of solid state storage
- file system software and I/O drivers - interface component connects SSD to host system - components
76
5 components of solid state storage
1. controller - operations and firmware 2. addressing - locations 3. data buffer/cache - RAM to improve throughput 4. error correction 5. flash memory - NAND
77
what is the main problem with SSDs
performance degradation over time
78
why does SSD performance degrade over time
store data in pages but pages are grouped into larger blocks - writing a new page = ○ read entire block to RAM ○ modify required page ○ erase entire block ○ write updated block back to flash memory - fragmentation Issue: ○ new data stored contiguously ○ over time, files are deleted/modified, pages become scattered across multiple blocks ○writing a new file may require modifying multiple fragmented blocks - slowing down performance
79
what are the 7 solutions to SSD degradation over time
1. overprovisionising 2. TRIM command 3. wear levelling 4. caching mechanism 5. bad block management 6. RAID for SSDs 7. self-monitoring
80
what is overprovisionising
extra storage space is reserved to optimise write operations and defragment during idle time
81
what is the TRIM command
allows OS to instruct SSD to erase deleted blocks in advance, making them ready for future writes
82
what is wear leveling
distributes writes evenly across memory cells to extend SSD lifespan
83
what are caching mechanisms for SSDs
delay and group writes to reduce wear
84
what is bad block management
detects and avoids failing memory cells
85
what is RAID for SSDs
increases reliability by reducing data loss
86
what is self-monitoring for SSDs
SSDs estimate their lifespan, enabling proactive failure prevention
87
describe CD-ROM
- better error correction - pits & lands - laser creates master disk - used to create copies - reflective layer for readability
88
how does CD-ROM read data
- laser - disk spins while laser moves across - pits scatter light, low-intensity reflection - lands reflect with higher intensity - photosensor detects changes - transitions: 1. pits - lands = 1 2. no change = 0
89
describe the spiral track
- single spiral track - track starts at the centre & spirals outwards - outer tracks have greater capacity but ^ rotational delay
90
describe constant linear velocity (CLV)
- data evenly placed across the disk - disc spins faster near the centre & slows down near the edge - ensure data is read at a steady rate
91
advantages of CD-ROM
- low-cost in comparison to magnetic disks - removable storage for archival purposes
92
disadvantages of CD-ROM
- read-only format - v access time compared to magnetic disks
93
describe CD-R
write once, read-many
94
describe CD-RW
rewriteable
95
describe DVDs
- ^ picture quality & random access - ^ capacity - shorter-wavelength laser - ^ data pack density - dual-layer & double-sided
96
5 reasons why optical discs have declined
1. space constraint - too bulky 2. v capacity 3. digital distribution - streaming & downloads 4. cloud storage - enables online data access & sharing 5. smartphone integration
97
describe magnetic tape storage
- digital: stores 0s and 1s for archival and large-scale storage applications - analog: stores continuous waveforms, allowing audio playback in cassette players - sequential access makes tapes slower than HDDs and SSDs
98
use of magnetic tapes
backing up data that doesn't require frequent access
99

1860: Computer Systems & Architecture (22 decks)

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