data storage – numerical problems

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Data Storage

Problem:
Consider a 5.25 inch disk with 16 double-surfaced platters rotating at 5280 rpm. It has a usable capacity of 16 gigabytes (2**34 bytes) stored on 1024 cylinders. Assume 12% of each track is used as overhead.

    1. What is the burst bandwidth this disk could support reading a single block from one track?
    2. What is the sustained bandwidth this disk could support reading an entire track?
    3. What is the average rotational latency, assuming it is not necessary to start aboutt the beginning of the track?
    4. Assuming the average seek time is 10 ms, what is the average time to fetch a 4-Kbyte (2**12 bytes) sector?

    Solutions:
    1. What is the burst bandwidth this disk could support reading a single block from one track?

    #Bytes/Track = Capacity / (#cylinders * #platters * 2)
    = 2**34 / (2**10 * 2**4 * 2)
    = 2**19 Bytes
    Time/revolution = 1/5280 minute = 1/88 second

    Suppose a sector has X bytes.
    #sector/track = 2**19 / X
    total time spent per sector + gap = (1/88 second) / (2**19/X)
    = X / (88 * 2**19) second
    time spent per sector = 0.88 * total time for a sector and a gap
    = X / (100 * 2**19) second
    BB = bytes per sector / time spent per sector
    = X / (X / (100 * 2**19))
    = 100 * 2**19 B/s
    = 100 * 2**19 / 2**20 MB/s
    = 50 MB/s

    2. What is the sustained bandwidth this disk could support reading an entire track?

    SB (Sustained Bandwidth) = bytes per track / time per revolution
    = 2**19 bytes / (1/88) seconds
    = 2**19 * 88 B/s
    = (2**19 * 88) / (2**20) MB/s
    = 44 MB/s

    3. What is the average rotational latency, assuming it is not necessary to start abt the beginning of the track?

    Avg. rotational latency = 0.5 * time per revolution
    = 0.5 * (1/88) seconds
    = 5.68ms

    4. Assuming the average seek time is 10 ms, what is the average time to fetch a 4-Kbyte (2**12 bytes) sector?

    Transfer time = (2**12)/(100 * 2**19) seconds = 0.078 ms

    Fetch time = seek time + avg. rotational time + transfer time
    = 10 + 5.68 + 0.078
    = 15.758 ms


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