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3. Given the following characteristics for a disk pack with

Customer Question

3. Given the following characteristics for a disk pack with 10 platters yielding 18 recordable surfaces: rotational speed = 10 ms transfer rate = 0.1 ms/track density per track = 19,000 bytes number of records to be stored = 200,000 records size of each record = 160 bytes block size = 10 logical records number of tracks per surface = 500

Find the following:
a. Number of blocks per track
b. Waste per track
c. Number of tracks required to store the entire file
d. Total waste to store the entire file
e. Time to write all of the blocks (Use rotational speed; ignore the time it takes to move to the next track.)
f. Time to write all of the records if they're not blocked (Use rotational speed; ignore the time it takes to move to the next track.)
g. Optimal blocking factor to minimize waste
h. What would be the answer to (e) is the time it takes to move to the next track were 5 ms?
i. What would be the answer to (f) is the time it takes to move to the next track were 5 ms?

Read more: http://www.justanswer.com/questions/2vsuh-3-given-the-following-characteristics-for-a-disk-pack-with#ixzz0fR1oOjYl
Submitted: 4 years ago.
Category: Computer
Expert:  Molinari replied 4 years ago.

HiCustomer Thank you for asking your question on JustAnswer.

 

 

 

a. Number of blocks per track:

Size of block = 10*160 bytes = 1600 bytes

1 track can hold = 19000 bytes

so no. of blocks per track will be = 19000/1600 = 11.875(approx 11)

b. Waste per track:

Data that can be stored on 1 track is = 19000 Bytes
But block size is fixed = 1600 bytes
So only 11 blocks can be saved = 1600*11 = 17600 bytes
remaining all gets wasted
so Waste per track will be = 19000 - 17600 = 1400 bytes

c. Number of tracks required to store the entire file:

Total number of records to be stored = 200000
1 record size = 160 bytes
so total required space will be = 200000*160 = 32000000 bytes

Track size is = 19000 bytes

So, No of tracks required will be = 32000000/19000 = 1684.21 (1685 approx)

d. Total waste to store the entire file:

waste per track was = 1400 bytes.
waste per 1685 track will be = 1400*1685 =(NNN) NNN-NNNNbytes

e. Time to write all of the blocks:

Total no. of blocks are = 1685 * 11 = 18535
will be written at the rate of .1ms/track
and total no of track are 1685 so time taken will be = 1685*0.1 = 168.5 ms

h.
answer will be = 1685*0.2 = 337 ms

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