Magnetic tape

  1. 1. The problem statement, all variables and given/known data

    Assume that a 2400 foot magnetic tape has recording density of 6400 Bpi. Data (logical) records are 100 bytes, and the memory buffer is 10,000 bytes. What is the largest IRG that will allow 80 percent of the tape to be data?

    2. Relevant equations and 3. The attempt at a solution


    Tape length = 2400 ft = 2400×12 in=28800 inches
    Recording density = 6400 bytes/in
    Each record has 100 bytes = 64 inches
    Memory buffer = 10,000 bytes
    Largest IRG that allows 80% of tape to be data

    Amount data = 80% of 2400 ft = 1920 ft consists of data.
    Since each record has 100 bytes for 64 inches, then, 1 inch has 6400 bytes of data.
    Total amount of data = 28800 inches × 6400 bytes = 184320000 bytes
    = 184320000 bytes ÷ 220
    = 175.78125 MBytes
     
  2. jcsd
  3. lewando

    lewando 1,077
    Gold Member

    This has not been answered. Is this what you are having trouble with?
    If the density is 6400 bytes/in, then 6400 bytes = 1 in, and then 100 bytes should be much smaller than an inch.
    Somehow your bad premise resulted in a correct statement.
    True for 100% utilization (no IRG).
    You have been pretty good at documenting your steps. Where does 220 come from?
    How does 184,320,000/220 = 175,781,250? (I get the distinction you are using with "M", but still.../220?)
     
    Last edited: Sep 15, 2011
  4. It is not two hundred and twenty(220)...it suppose to be 220 which gives the answer in megabytes same as multiplying 1024 by 1024..
     
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