The Megatron 777 disk has the following characteristics: There are ten surfaces,
ID: 3937880 • Letter: T
Question
The Megatron 777 disk has the following characteristics: There are ten surfaces, with 100, 000 tracks each. Tracks hold an average of 1000 sectors of 1024 bytes each. 20% of each track is used for gaps. The disk rotates at 10, 000 rpm. The time it takes the head to move n tracks is 1 + 0.0002n milliseconds. Answer the following questions about the Megatron 777. What is the capacity of the disk? If t racks are located on the outer inch of a 3.5-inch-diameter surface, what is the average density of bits in the sectors of a track? What is the maximum seek time? What is the maximum rotational latency? If a block is G5, 54G bytes (i.e., 64 sectors), what is the transfer time of a block?Explanation / Answer
a)
The capacity of the disk is calculated as follows:
Number of tracks=number of surfaces* number of tracks
The disk has 10*100,000=100,0000 tracks.
The average track has 1000*1024=1024,000 bytes. Thus, the capacity is 1.024 Gigabytes.
b)
If the tracks are located on the outer inch of a 3.5-inch diameter surface, the average density of bits in the sectors of a track is calculated as follows:
Average Density of bits in the sector of a track = (1000*1024*10) / ((3.5/2) * 3.14*80%)
= (1024 * 10^5) / (5.495*8)
= (1024 * 10^5) / (43.96)
= 23.2939*10^5
= 2.32939 M bits/ inch
c)
The maximum seek time:
Maximum Seek Time occurs when the heads must move across all the tracks. In the given formula 1+0.0002n, substitute n value = 100,000
Maximum Seek time = 1+0.0002(100,000/3)
Maximum Seek time = 1+6.66666666= 7.6666666ms
d)
Maximum rotational latency:
Maximum Rotational Latency is one full revolution. Here the disk rotates 10,000 rpm so maximum rotational latency is 1/ 10,000 of a minute.
Maximum Rotational Latency = 60 / 10,000
Maximum Rotational Latency = 6 milli second
e)
Transfer time of a block:
20% of each track is used for gap.64 sectors have 63 gaps
degrees = ((63/1000) *20% + (64/1000) *80%)*3600
= (0.126+0.0512) *3600 = 63.7920
Transfer time of a block = (degrees/ 360) *10
= (63.7920 /360) *10
= 1.772 ms