Storage Design Based On Application Requirements and Disk Performance
Varez IT
Member Posts: 32 ■■□□□□□□□□
I am currently studying for the EMCISA certification and my study guide is Information Storage and Management Second Edition. I'm currently reading through section 2.10 "Storage Design Based On Application Requirements and Disk Performance" and I have some questions but before asking the questions please allow me to provide some context.
The study guide states that the Disk Service Time is Ts = T + L + X (where T = average seek time in milliseconds, L = rotational latency and X = internal transfer rate). I quote "40 MB/s internal data transfer rate, from which the internal transfer rate time (X) is derived on the block size of the I/O - for example, an I/O with a block size of 32 KB; therefore X = 32 KB/40 MB. Consequently, the time take by the I/O controller to serve an I/O of block size 32 KB is (Ts) = 5 ms + (0.5/250) + 32 KB/40 MB = 7.8 ms."
I understand how to arrive at the average seek time (T) as well as the rotational latency (L) but I am confused by the internal transfer rate (X).
Why is the author using 40 MB? Also, what is the answer to 32 KB/40 MB?
Again, I understand the average seek time (T) as well as the rotational latency (L) but I can't seem to understand how to arrive at the internal transfer rate (X) and get an answer of 7.8 ms.
Any help is much appreciated.
The study guide states that the Disk Service Time is Ts = T + L + X (where T = average seek time in milliseconds, L = rotational latency and X = internal transfer rate). I quote "40 MB/s internal data transfer rate, from which the internal transfer rate time (X) is derived on the block size of the I/O - for example, an I/O with a block size of 32 KB; therefore X = 32 KB/40 MB. Consequently, the time take by the I/O controller to serve an I/O of block size 32 KB is (Ts) = 5 ms + (0.5/250) + 32 KB/40 MB = 7.8 ms."
I understand how to arrive at the average seek time (T) as well as the rotational latency (L) but I am confused by the internal transfer rate (X).
Why is the author using 40 MB? Also, what is the answer to 32 KB/40 MB?
Again, I understand the average seek time (T) as well as the rotational latency (L) but I can't seem to understand how to arrive at the internal transfer rate (X) and get an answer of 7.8 ms.
Any help is much appreciated.
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TheProf
Users Awaiting Email Confirmation Posts: 331 ■■■■□□□□□□
I know what you mean, I was also a bit confused, but I was able to find a site that gave pretty good explanation which I linked below.Calculating the data transfer rate is reasonably simple, provided you know the true specifications of the drive; figuring the transfer rate will show you what design parameters have an impact on this performance measure. The transfer rate is a measure of the amount of data that can be accessed over a period of time. So we need to know how much data is able to pass under the read/write heads in one second. This is dependent on the density of the data (how tightly packed the data is into each linear inch of disk track), and also how fast the disk is spinning. The density of the data can be calculated easily if we know how many sectors are on the track, since we know how many bytes there are in a sector. The speed of the disk is calculated in RPM, so we divide it by 60 to get revolutions per second. This gives us a calculation of the data transfer rate in megabits per second as follows (to get the result in megabytes per second, simply divide by 8- Data Transfer Rate = (Spindle Speed / 60 * Sectors-Per-Track * 512 * 8 / 1,000,000
For more info, take a look on this site -
Varez IT
Member Posts: 32 ■■□□□□□□□□
TheProf,
Thank you! That information was very helpful!
Also, is there a way to mark your answer as 'correct' on this website so that you receive a credit? It doesn't appear that is the case.
