Disk Management

Question 1

Disk Access Time

Answer 1

For b bytes:
Ta = Ts + 1/(2r) + b/(rN)

• N = bytes per track
• Ts = seek time
• r = spinning speed
• b = bytes to access

Question 2

Seek Times: 4 phases

Answer 2

• speed-up
• coast
• slowdown
• settle

Question 3

Skewing

Answer 3

Goal: keep read speed high
=> shift sectors on consecutive tracks somewhat -> reduce rotational delay

Question 4

Sparing

Answer 4

Map faulty sectors to other sectors via a list passed to the controller

Question 5

Disk Scheduling Algorithm: FIFO

Answer 5

= Process requests in order of arrival

-> fair
-> very efficient for clustered operations
-> inefficient for random reads

Question 6

Disk Scheduling Algorithm: Shortest Seek Time (SSTF)

Answer 6

= process first request with smallest seek time

-> requires estimation of seek time
-> high utilization/efficiency

Question 7

Disk Scheduling Algorithm: SCAN

Answer 7

= SSTF with arm changing direction when reaching the extremes of the disk

-> less starvation than SSTF
-> lower delay variance, higher average delay
-> unfairness center vs. extreme tracks

Question 8

Disk Scheduling Algorithm: C-SCAN

Answer 8

= SCAN but once extreme is reached, continue from the other extreme

-> reduces unfairness

Question 9

Disk Scheduling Algorithm: LOOK

Answer 9

= SCAN but the arm changes direction if there are no more requests in the current direction

-> slightly more efficient than SCAN
-> more complex
-> unfairness center vs. extreme tracks

Question 10

Disk Scheduling Algorithm: C-LOOK

Answer 10

= C-SCAN but changes direction if there are no more requests in the current direction

Question 11

Disk Scheduling Algorithm: V-SCAN

Answer 11

Process first request which is closest located with:
distance = SSTF distance + 1 (if changing direction) * R * width of the disk

• VSCAN(0) = SSTF
• VSCAN(1) = LOOK
• VSCAN(0.2) gives good balance between SSTF and LOOK

Question 12

Disk Scheduling Algorithm: FSCAN

Answer 12

Use a 2-stage buffer to handle I/O
Apply SCAN on the requests in the first buffer, then process requests in the second buffer

Question 13

RAID

Answer 13

= Redundant Arrays of Independent Disks
=> many inexpensive disks form one large, fast and reliable disk

Question 14

RAID 0

Answer 14

Data equally striped over N disks:
Disk i contains blocks i, i+N, i+2N,..

-> poor robustness

effect of stripe length:
- large: random reads simultaneous
- small: sequential read fast

Question 15

RAID 1

Answer 15

Data mirroring over 2 discs
-> all data on both disks

Very high robustness but expensive solution

Profit on random reads but write is slower than with one disk

Question 16

RAID 3+4

Answer 16

= striping over N disks (RAID 0) + 1 extra disk for parity bits
-> RAID 3 = short stripe, synchronized disk heads (fast sequential reads, parity on write mostly no read required)
-> RAID 4 = long stripe (random read fast, 1 write = 2 writes + 2 reads

Question 17

RAID 5+6

Answer 17

= RAID 4 with parity information spread across all disks
-> better random read performance

RAID 6 = one more disk

Question 18

RAID 0+1 & 1+0

Answer 18

RAID 0+1 = mirroring and then striping
-> on read operation: select mirror first, then use striping

RAID 1+0 = striping and then duplicate
-> more robust, can handle many double disk failures
-> on read operation: striping and for each stripe we choose a mirror

Question 19

Combinate RAIDS: X+Y

Answer 19

organize data according RAID Y, then subdivide each of the obtained disks according to X

Question 20

What is Zone Bit Recording?

Answer 20

◦ Modern discs divided into zones (~15)
▪ Within a zone: fixed number of sectors on each track
▪ Across zones: Number of sectors per track varies (decreases towards the inside)
◦ In practice: up to 2 times as many sectors in outer zone (several hundreds)
◦ Result: almost double reading and writing speed on the outside (filling from outside to inside)

Question 21

What are the disk components?

Answer 21

◦ One or more circular (metal) disks
◦ Disk head (per disk)
◦ Fixed rotation speed

Question 22

How is a disk partioned?

Answer 22

◦ Tracks (concentric circles)
◦ Sectors (fixed number of bytes): unit of reading and write operations

Question 23

What is the classical layout of a disk?

Answer 23

◦ Fixed number sectors per track
◦ Compact storage near the center [high density]
◦ Low density on the outside
◦ Fixed read and write speed

Question 24

What is the disk controller?

Answer 24

◦ Connected to the bus
◦ Receives request (via device driver software) and handles them
◦ Can buffer at least one sector (necessary because of faster bus)
◦ Command Queueing:
controller can receive multiple requests and determine the order of processing

Question 25

What is disk caching?

Answer 25

◦ Cache with read-ahead strategy
◦ Read data from cache is replaced immediately
→ file reading is usually sequential
◦ Aggressive read-ahead: over multiple tracks
◦ Old disk: on-arrival read-ahead
▪ Reading begins when the head is in the right track location
▪ Useful if we usually need the largest part of the track (increasingly less the case)
◦ Cache useful for quickly writing the same data, but volatile