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A certain moving arm disk-storage device has the following specifications:

  • Number of tracks per surface $= 404$
  • Track storage capacity $=130030$ bytes.
  • Disk speed $=3600$ rpm
  • Average seek time $=30$ m secs.

Estimate the average latency, the disk storage capacity, and the data transfer rate.

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  1. Avg Latency $= \frac{1}{2}\times \frac{60}{R}$ =$\frac{1}{2}\times \frac{60}{3600} = 8.33$ ms
  2. Disk Storage Capacity = (We need a number of surface to calculate it) $404 \times  130030 \;\text{Bytes} \simeq 50$ MB per surface (approx)
  3. Data transfer rate $=$ Track capacity $\times \frac{R}{60} = 130030 \times \frac{3600}{60} = 7801.8$ kBps
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  1. 3600 rotations = 1 minutes = 60 secs => 1 rotation = (60/3600) secs => for average latency = (1/2)*(60/3600) = 8.33 ms
  1. Disk Storage Capacity = (No. of platters)*(No. of surfaces)*(No. of Tracks)*(No. of sectors)*(Data)

                                                 =1*1*404*130030                     

                          /* Track storage capacity = (no. of tracks*no of sectors*data) */

                                                 = close to 50 MB

  1. Data Transfer Rate = (No. of surfaces)*(Track Capacity)*(No of rotations per second)

                                            = 1*130030*(3600/60)

                                            = 7801.8 Kb/sec

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