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