Suppose a large mailing-list file with one million 100-byte records is stored in a tape. Assume that
the files is stored on a 6250-bpi (bits per inch) nine-track (or 6250 bytes per inch) tape,
each data block contains fifty 100-byte records (the blocking factor: 50), and
the tape has an interblock gap of 0.3 inches.
Calculate how much tape is needed:
Space requirement for storing a file
= Number of data blocks ×
(Physical length of a data block + Length of an interblock gap)
= (1000000/50)×[(100×50)/6250+0.3)] = 22000 inches = 1833.33 feet
Estimating Data Transmission Times
Assuming a tape is with 6250-bpi, nine-track, 200-ips (inches per second), and 0.3-inch interblock gap.
Calculate the nominal transmission rate: Nominal transmission rate
= Tape density (bpi) × Tape speed (ips)
= 6250 × 200 = 1250000 bytes/sec = 1250 kilobytes/sec
Calculate the effective transmission rate:
Assume that the blocking factor is 1. Effective transmission rate
= Effective recording density (bpi) × Tape speed (ips)
= [100 / (100/6250+0.3)] × 200
= 63280 bytes/sec = 63.3 kilobytes/sec
Assume that the blocking factor is 50. Effective transmission rate
= Effective recording density (bpi) × Tape speed (ips)
= {(50×100) / [(100×50)/6250 + 0.3]} × 200
= 909090.9 bytes/sec = 909.1 kilobytes/sec
