# quiz1 - Total number of blocks that can be stored = 2400 12...

This preview shows page 1. Sign up to view the full content.

CS351 - QUIZ 1 Q: Assume that we have 400 byte records Tape density = 1600 bpi Tape length = 2400 feet (1 foot = 12 inches) IBG = 0.5 inch How many records can be stored on this tape: a) With blocking factor = 3 ? b) No blocking ? Solution: a) If blocking factor = 3, the number of bytes in a block = 400*3 = 1200 bytes. That means we the length spent for the records in each block is 1200 bytes = 0.75” 1600 bytes/inch As an IBG follows each block of data, for each block (data + IBG) we spend 0.75 + 0.5 = 1.25”.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Total number of blocks that can be stored = 2400 * 12 / 1.25 = 23040 Thus the total number of records that can be stored = 3 * 23040 = 69120 records. b) With no blocking, 400 / 1600 = 0.25” is spent for the record in each block and in this case the number of interblock gaps is equal to the number of blocks. Thus, for each block (record + IBG) 0.25 + 0.5 = 0.75” is spent. number of records that can be stored = 2400 * 12 / 0.75 = 38400 records....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online