Q. The term "fast wide SCSI-II" signifies a SCSI bus that operates at a data rate of 20 megabytes per second when it moves a packet of bytes among the host and a device. Presume that a fast wide SCSI-II disk drive spins at 7200 RPM has a sector size of 512 bytes and holds 160 sectors per track.
a. Estimate the continued transfer rate of this drive in megabytes per second.
b. Presume that the drive has 7000 cylinders 20 tracks per cylinder a head switch time from one platter to another of 0.5 millisecond as well as an adjacent cylinder seek time of 2 milliseconds. Utilize this additional information to give an accurate estimate of the sustained transfer rate for a huge transfer.
c. Presume that the average seek time for the drive is 8 milliseconds. Evaluate the I/Os per second as well as the effective transfer rate for a random-access workload that reads individual sectors that are scattered across the disk.
d. Compute the random-access I/Os per second and transfer rate for I/O sizes of 4 kilobytes 8 kilobytes and 64 kilobytes.
e. If numerous requests are in the queue a scheduling algorithm such as SCAN should be able to reduce the average seeks distance. Presume that a random-access workload is reading 8-kilobyte pages the average queue length is 10 and the scheduling algorithm decreases the average seek time to 3 milliseconds. Now compute the I/Os per second and the effective transfer rate of the drive.
Answer:
a. The disk spins 120 times per second as well as each spin transfers a track of 80 KB. Therefore the sustained transfer rate can be approximated as 9600 KB/s.
b. Presume that 100 cylinders is a huge transfer. The transfer rate is entire bytes divided through total time. Bytes: 100 cyl * 20 trk/cyl * 80 KB/trk that is 160,000 KB. Time: rotation time + track switch time + cylinder switch time and rotation time is 2000 trks/120 trks per sec, i.e., 16.667 s. Track switch time is 19 switch per cyl * 100 cyl * 0.5 ms that is 950 ms. Cylinder switch time is 99 * 2 ms that is 198 ms. Therefore the total time is 16.667 + 0.950 + 0.198 that is 17.815 s. (We are ignoring any initial seek as well as rotational latency which might add about 12 ms to the schedule that is 0.1%.) therefore the transfer rate is 8981.2 KB/s. The overhead of track in addition to cylinder switching is about 6.5%.
c. The time/per transfer is 8ms to seek + 4.167ms average rotational latency + 0.052 ms (calculated from 1/(120 trk per second * 160 sector per trk)) to rotate one sector past the disk head during reading. We compute the transfers per second as 1/(0.012219), i.e., 81.8. Since every transfer is 0.5 KB the transfer rate is 40.9 KB/s.
d. We ignore track as well as cylinder crossings for simplicity. For reads of size 4 KB and 8 KB and 64 KB the corresponding I/Os per second are calculated from the seek and rotational latency and rotational transfer time as in the previous item giving (respectively) 1/ (0.0126) 1/(0.013), and 1/(0.019). Therefore we get 79.4, 76.9, and 52.6 transfers per second respectively. Transfer rates are acquired from 4, 8 and 64 times these I/O rates giving 318 KB/s, 615 KB/s and 3366 KB/s respectively.
e. From 1/ (3+4.167+0.83) we acquire 125 I/Os per second. From 8 KB per I/O we acquire 1000 KB/s.