Question

Consider a disk with a sector size of 512 bytes, 2000 tracks per surface, 50 sectors per track, five double-sided platters, and average seek time of 10 msec

Suppose that a block size of 1024 bytes is chosen. Suppose that a file containing 100,000 records of 100 bytes each is to be stored on such a disk and that no record is allowed to span two blocks

1. Suppose the interleaving factor is 3. What time is required to read a file containing 100,000 records of 100 bytes each sequentially?

2. What is the time required to read a file containing 100,000 records of 100 bytes each in a random order? To read a record, the block containing the record has to be fetched from disk. Assume that each block request incurs the average seek time and rotational delay

Answer 1

I have first answered the second question and then with first question, kindly don’t confuse, comment me if you have any clarification.

Solution for second answer:

Let me first define these factors so that you can understand vividly:

Access time is defined as the time taken to approach a block and then seek time is defined as the time consumed to transfer the disk head for locating the block, the rotational delay is regarded as the waiting time taken by the block to revolve around under the disk head while transfer time is defined as the time consumed for write and read, in other words, time consumed for disk to rotate. Therefore, we can derive that, access time can be gained by adding seek time with rotational delay and transfer time together. It is evident that the seek time is 10msec , rotational is 6msec and transfer time is 1k/2,250 where 2,250 is obtained by capacity of a track which is 25K bytes, 0.11 is the maximum rotational delay , therefore, 25K/ 0.011= 2, 250Kbytes/sec. Now we can find the access time which is 16.44 msec and since the question says 10,000 blocks, the answer would be 164.4 seconds.

Solution for first answer:

Transfer Time is obtained by 400 tracks multiplied with maximum rotational delay which is 0.011, therefore the total transfer time is 4.4 seconds and seek time is obtained by 40 cylinders multiplied with seek time of 0.01 about 100,000 records of 100 bytes, therefore, the total seek time is 0.4 seconds. Now by adding the seek time with transfer time we obtain the total access time which is 4.8 seconds.