Question

I submitted this question twice and i got two different answers, i'm so confused CAN YOU PLEASE tell what answer is the correct one and WHY.

Given the following characteristics of magnetic tape, and using linear recording:

- Density=2000 bpi

- Speed= 1500 inches/second

- Size=3000 feet

- Number of records to be stored: 200,000 records

- Size of each records=100 bytes

- Block size=10 records

- IBG=.5 inches

Calculate the following:

1- Number of blocks needed

2- Size of the block in bytes

3- Time required to read one block (explain this part.why do we add both data time and IBG time)

4- Time required to read all blocks

5- Amount of tape used for data only in inches

6- Total amount of tape used (data + IBG) in inches

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++

first answer----

-------

1)number of blocks needed=no of records stored/block size

=200,000/10

no of blocks needed      =20,000 blocks

2) size of block in Bytes=block size * record size

=10 * 100

=1,000 bytes/block

3)time required to read one block=(block size /density)/speed

=(1,000 bytes/block/2000 bpi) / 1500 in/sec

=0.5/1500

=0.000333 sec/block

4) Time required to read all blocks= no of blocks * time to read one block

=20,000 * 0.000333

=6.66 ms

5)Amount of tape used for data only in inches=no of blocks needed*(block size/density)

=20,000 blocks *(0.5in/block)

= 10,000 in

6) Total amount of Tape used(data + IBG) in inches=Amount of tape used+(no of blocks needed-1)*IBG

=10,000 in +(20000-1)*0.5

=10,000+ 9999.5

=19999.5 in

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Second answer--

-------

1. Blocks needed = Total number of records / number of record per block = 200,000 / 10 = 20,000 blocks

2. Size of block in bytes = Size of record * number of records per block =100 bytes/rec * 10 rec/block = 1000 bytes/block

3. Time to read one block:

• data time = 1000 bytes/block / 2000 bytes/in / 1500 in/sec = 0.00033 sec/block
• ibg time = 0.5 in/block / 1500 in/sec = 0.00033 sec/block
• total = 0.00033 + 0.00033 = 0.00066 sec/block

4. Time to read all blocks = time to read 1 block * total number of blocks = 200,000 blocks * 0.00066 sec/block = 133.33 sec

5. Data length = 200,000 blocks * 0.00033 sec/block * 1500 in/sec = 99,000 in

6. Total amount of tape used = 133.33 sec * 1500 in/sec = 199995 in

Answer 1

why do we add both data time and IBG time?

Actually we need to add both the ibg time and data time to get the time to read one block completely, because for each data process like read or write happens, ibg time will also affect the total time.

ibg - Inter block gap.

- It separates one block from another, so depending the number of blocks and the gap, the total time required to read one block will vary.

Looking at above 2 answers, second answer calculation is correct.