Question 2.
2.1.
Cylinder capacity = number of surfaces x track capacity
= number of surfaces x number of sector per track x bytes per sector
= 16 x 63 x 512
= 516096 (bytes)
Number of cylinders = (350000 x 80)/516096
» 54.25(cylinders)
2.
6 records per sector
6 x 63 = 378 records per track
378 x 16 = 6048 records per cylinder
Number of cylinders = 350000/6048
» 57.87(cylinders)
3.
Average total time first track per cylinder
= Average seek time + Average rotational delay + Average transfer time for a track
= 12 + 6 + 12
= 30 (msecs)
Average total time rest tracks per cylinder
= Average rotational delay + Average transfer time for a track
= 6 + 12
= 18 (msecs)
Average total time per cylinder except the last cylinder
= 30 + 18 x 15
= 300 (msecs)
From 2, we know the file occupies 57.87 cylinders.
Tracks in the last cylinder
= (350000 – 57 x 6048) / 378
» 13.93 (tracks)
= 13 tracks 59 sectors
Average total time last cylinder
= 30 + 18 x 12 + total time for last track
= 30 + 18 x 12 + (6 + 59 x 12/63)
= 263.24 (msecs)
Total time for accessing the file in sequence
= 57 x 300 + 263.24
= 17100 + 263.24
= 17363.24(msec)
» 17.4(secs)
4.
Transfer rate/sector = 12/63 » 0.19 (msecs/sector)
Average time to read one sector = Average seek time + Average rotational delay + Transfer time
= 12 + 6 + 0.19 = 18.19(msecs)
Average total time to access the entire file randomly = Number of records x Average time to read
one sector
= 350000 x 18.19
= 6366500(msecs) = 6366.5(secs)
= 106.1(mins)
» 1.77 (hours)
Question
2.2.
1.
Suppose blocking factor = a
b = length of data block(in inches) = (a x 80)/1600 = 0.05a (inches)
n = 350000/a
s = 350000/a x (0.05a + 0.5)
The minimum blocking factor should satisfy the following requirement:
s <= (2400 x 3 x 12) = 86400 (inches)
i.e. (350000/a x (0.05a + 0.5)) <= 86400
3500 x (0.05a + 0.5) <= 864a
(175a + 1750) <= 864a
689a >= 1750
a >= 2.53
a = 3
The maximum blocking factor should satisfy the following requirement:
s > (2400 x 2 x 12) = 57600 (inches)
i.e (350000/a x (0.05a + 0.5)) > 57600
3500 x (0.05a + 0.5) > 576 a
(175a + 1750) > 576a
401a < 1750
a < 4.4
a = 4
2.
Blocking factor = 50
b = length of data block(in inches) = (50 x 80)/1600 = 2.5 (inches)
Number of bytes per block = 50 x 80 = 4000 (bytes)
Effective Recording Density = Number of bytes per block/number of inches to store a block
= 4000/(2.5 + 0.5) = 4000/3
» 1333.3 (bpi)
3.
Effective Transmission Rate = Effective Recording Density x tape speed
= 4000/3 x 150 =200,000(bytes/sec)
= 200 KB/sec