Given a set of 5 disks organized in RAID4 configuration with 4 data disks and 1

Question

Given a set of 5 disks organized in RAID4 configuration with 4 data disks and 1 parity disk. The
bandwidth of the CPU can sustain data transfer from/to two disks in parallel.
For each disk, we have: seek time = 3ms, rotation speed = 5000 rotations/minute, transfer speed
= 50MB/s, controller overhead and the waiting time is negligible. Each transfer moves a block of
size 512Bytes on each disk.
A read operation moves the data only from the disk that hold the data.
For a write operation, the parity block must be recomputed. Therefore the parity block and
the data disk must be read and the modify values written back.

Compute the data access time to the disk array for a read operation and a disk write operation.

Explanation / Answer

Answer:

Given: seel time = 3 ms, rotation speed = 5000 rpm, transfer speed = 50 MB/s,

controller overhead = 0 (as said to be negligible)

waiting time = 0 (as said to be negligible)

Block size = 512 Bytes

Data access time for a read operation is given by:

data access time = Seek time + Rotation time + Transfer time + Controller overhead time

Seek time is given. Calculating Rotation time:

Rotation time = 5000 rpm = 5000/60 R/s = 83.33 R/s = 1000/83.33 ms/R = 12 ms/R (approx.) = 12/2 msR = 6 ms/R (as we are seeking only on 1 disk for read operation).

Now Transfer time = 512/(50*1024*1024) = 9.77e-6 s = 0.00977 ms = 0.0098 ms

Hence Seek time = 3 + 6 + 0.0098 + 0 = 9.0098 ms

For write operation, two disks are accessed (as parity block is also read and recomputed). It will involve first reading, then writing. For reading time will be same as previously. For writing, time will depend on rotation time and transfer time. We assume rotation time to be same as previously i.e. 6 ms. Hence, total access time for write operation will be 9.0098+6+0.0098 = 15.0196 ms

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