A magnetic tape drive has a transport speed of 200 inches per second and a…
2022
A magnetic tape drive has a transport speed of 200 inches per second and a recording density of 1600 bytes per inch. The time required to write 600000 bytes of data grouped in 100-character records with a blocking factor of 10 and an inter-block gap of 0.25 inch is approximately
Answer: B. 2.6251sec — ConceptFor blocked magnetic-tape storage, total writing time equals the total tape length occupied by the data and the inter-block gaps divided by the tape…
- A.
2.0625sec
- B.
2.6251sec
- C.
2.0062sec
- D.
2.6150sec
Attempted by 109 students.
Show answer & explanation
Correct answer: B
Concept
For blocked magnetic-tape storage, total writing time equals the total tape length occupied by the data and the inter-block gaps divided by the tape transport speed.
Data length is total bytes divided by recording density. Block size equals record size multiplied by the blocking factor, and adjacent blocks contribute one inter-block gap each.
Application
Block size = 100 bytes per record × 10 records per block = 1000 bytes per block.
Number of blocks = 600000 bytes ÷ 1000 bytes per block = 600 blocks.
Data length = 600000 bytes ÷ 1600 bytes per inch = 375 inches.
There are 599 gaps between 600 consecutive blocks, so gap length = 599 × 0.25 inch = 149.75 inches.
Total tape length = 375 + 149.75 = 524.75 inches.
Writing time = 524.75 inches ÷ 200 inches per second = 2.62375 seconds.
Cross-check
If the common exam convention counts a trailing gap after the last block, 600 gaps occupy 150 inches and the time is 525 ÷ 200 = 2.625 seconds.
Both conventions make 2.6251 seconds the closest offered value; its difference from 2.62375 seconds is only 0.00135 second.
Therefore, the best available answer is 2.6251 sec.
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