6. External memory *

1 platter - 2 surface - 1Read/1Write

track/cylinder - number of surfaces

track

cylinder

sector

Calculate

RPM - 转/分

Access Time = Seek + Latency

Seek Time = time it takes to position the head of track 读写磁头从最外侧移动到相应磁道track的时间

(track/platter-1)/2 * time/track traversed

latency/rotational delay - time it takes for the beginning to reach the head 磁头移动到相应扇区sector的时间

1/2r

transfer time = read / write

b/rN

burst rate = MB/s = r/s * sector/r * byte/sector

RAID

redundant array of independent disks

0 - no redundancy/strip

1 - copy

2 small-size strip, hamming code (correct 1 bit, detect 2 bit), parallel access, redundancy - logN, not efficient if many disk errors occur

3 small-size strip, single parity bit, high data transfer rate, one drive fail can be restored, parallel access, suffer for multiple request

4 independent access, strip relatively large, answer seperated i/o request parallelly, write penalty(2 read 2 write) when i/o write request of small size is performed

5 big block, independent access, distribution of parity strips across all disks, round robin, avoid l4 bottleneck

6 2 different parity bit, extreme high data availability, handle 3 drive error, write penalty(2 parity blocks)

Q on disk

5) Consider a single platter disk with the following parameters: rotation speed: 7200rpm; number of tracks on one side of platter: 30 000; number of sectors per track: 600; seek time: one ms for every hundred tracks traversed. Let the disk receive a request to access a random sector on a random track and assume the disk head starts at track 0.

a) What is the average seek time?

(30,000-1)/2/100 = 149.995ms

b) What is the average rotational latency?

1/ (2*7200/60) * 1000 = 4.17ms

c) What is the transfer time for a sector?

1/120/600 = 13.9us

d) What is the total average time to satisfy a request?

149.995+4.17=154.165ms

6) Consider a magnetic disk drive with 16 surfaces, 512 tracks per surface, and 64 sectors per track. Sector size is 1KB. The average seek time is 6 ms, the track-to-track access time is 1ms, and the drive rotates at 3600 rpm. Successive tracks in a cylinder can be read without head movement.

a) What is the disk capacity?

16 * 512 * 64 * 1kb = 524288kb = 512mb = 0.5gb

b) What is the average access time?

60 / (2*3600) * 1000 = 8.33ms

6+8.33=14.33ms

c) Estimate the time required to transfer a 5MB file.

each cylinder 16 tracks * 64 sector/track * 1Kb/sector = 1 Mbytes

5MB = 5 cylinders

transfer time for 16 tracks: 16/r = 267ms

6 + (8.3+267 +1) + 4(8.3+267) =

d) What is the burst transfer rate?

3600/60 * 64 * 1k = 3.84MB/s

RAID 3