
#1
Sep1311, 12:24 AM

P: 4

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
Sep1511, 06:32 AM

PF Gold
P: 1,054





#3
Sep2611, 11:01 PM

P: 4

It is not two hundred and twenty(220)...it suppose to be 2^{20} which gives the answer in megabytes same as multiplying 1024 by 1024..



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