1. Suppose you have a hard disk with 2200 tracks per surface, each track divided
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Question
1. Suppose you have a hard disk with 2200 tracks per surface, each track divided into 110 sec- tors, six platters and a blocksize of 512 bytes (G.e.. kilobyte), what is the total raw capacity of the disk drive? 2. A program may be written in a language that is compiled to native binary (to the instruc- tion opcodes the actual CPU knows how to do).orit may be interpreted (translated to native binary statement by statement as the program runs.) Suppose part of a program initializes a 500,000 element array to all o (i.e., each of 500.000 elements is set to 0). A compiled ver- sion performs this action in 0.008 seconds: an interpreted version takes 0.608 seconds. Why is the compiled version so much faster? 3. Suppose a disk drive in your system has a rotation speed of 7 200 rpm. and seek times of 0.75 ms track-to-track, or 22 ms full 'stroke (inner edge to outer edge). (a) What is the average latency to access this drive? (b) If the sustained read rate is 36 Mbits/sec, how much more time after the average la- tency (you calculated above) will it take to read 750 Kbytes (assuming all those bytes are in sequential sectors on the same track)? How long to write that same amount of data if the sustained write speed is 24 Mbits/sec? 4. Suppose you replace the hard disk in the previous question with an optical drive like a DVD drive. The drive you use has an average access time of 185 ms, and can do sustained reads at 1600 Kbytes per second. How long will it take, including access time, to read the same 750 Kbytes off this optical disk (again, assuming all the data is in sequential sectors on same track)? 5. Unimpressed with your experiment with the optical drive. you now replace it with a solid- state drive. This dive has a sustained read rate of 580 Mbps, sustained write speed of 520 Mbps. For the 750 Kbytes of data, including average access time (latency), how long does it take to read the data? write the data?Explanation / Answer
Solution:
1 and 2 are solved, please repost others.
1)
Number of track= 2200
Number of sectors on each track= 110
The sector size= 512 bytes
The total capacity of the hard drive= 2200*110*512
= 123904000 bytes
= 123.904000 MB
2)
The compiled version executes faster because it is executing on the low level where the extra overhead of interpreting the language written code to machine level or low level code is saved, that is why compiled version is faster.
For example, if you have a .exe file compiled in C language, running that will be faster compared to compile and run.
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