1.Given the following characteristics for a magnetic disk pack with 10 platters

Question

1.Given the following characteristics for a magnetic disk pack with 10 platters yielding 18 recordable surfaces (not using the top and bottom surfaces):

Rotational speed = 13 ms              

Transfer rate = 0.15 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.)

Optimal blocking factor to minimize wasted space

Explanation / Answer

Answer:-

Given:

Rotational speed = 13 ms
Transfer rate = 0.15 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

a)Number of blocks per track

Since there are 10 platters and sizeof each record is 160 bytes
Then Size of block = 10*160 bytes = 1600 bytes

Density per track = 19000 bytes

So no. of blocks per track will be = 19000/1600 = 11.875 Ans.

b) Waste per track:

1 track can hold = 19000 Bytes

But block size is = 1600 bytes
and no. of blocks per track =11
So only 11 blocks can be saved = 1600*11 = 17600 bytes

All the remaining gets wasted.

so Waste per track will be = 19000 - 17600 = 1400 bytes Ans.

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

The Number of records to be stored = 200000 records
Size of 1 recorde = 160 bytes
so total space required will be = 200000*160 = 32000000 bytes

1 track can hold = 19000 bytes

Therefore No of tracks required will be = 32000000/19000 = 1684 Ans.

d). Total waste to store the entire file:

waste per track generated = 1400 bytes.
so waste per 1685 track will be = 1400*1684 =2,357,600 bytes

e). Time to write all of the blocks:

Total no. of blocks are = no. of tracks * blocks per track = 1684 * 11 = 18524
Transfer rate = 0.15 ms/track
and total no of track are 1685 so time taken will be = (13+((.15*11)*10))*18524 =546458 ms Ans.

f)Time to write all of the records if they’re not blocked.

=(13+(19,000*.15))*200,000
=(13+2850)*200,000
=2863*200,000
=57,26,00,000 MS

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