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27 votes
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A certain moving arm disk storage, with one head, has the following specifications:

  • Number of tracks/recording surface $= 200$
  • Disk rotation speed $= 2400$ rpm
  • Track storage capacity $= 62,500$ bits

The average latency of this device is $\text{P}$ ms and the data transfer rate is $\text{Q}$ bits/sec. Write the values of $\text{P}$ and $\text{Q}$.

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Best answer
66 votes
66 votes
RPM $= 2400$

So, in $60$ s, the disk rotates $2400$ times.

Average latency is the time for half a rotation $= 0.5 \times 60 / 2400 \ s = 3/240 \ s = 12.5$  ms.

In one full rotation, entire data in a track can be transferred. Track storage capacity $= 62500$ bits.

So, disk transfer rate $= 62500 \times 2400 / 60 \ \text{s}  = 2.5 \times 10^6 $ bps.
edited by
22 votes
22 votes

 

Disk rotation speed= 2400 rpm

Track storage capacity = 62,500

Average Latency = p 

Data Transfer rate = q 

Data transfer rate is nothing but how much data can be read in one second 

WE have 2400 rotation in 1 minute

2400 rotation = 60 *10ms

therefore 1 rotation = x ms, where,

x= 60 *103 / 2400

therefore we get x=25 ms

Hence 25 mill sec is required to read one complete track (1 rotation )

In 25 m sec, we  read track capacity = 62,500 bits

therefore in 1 sec we can read (62,500 bits )*1000/25 = 2.5 Mbits

i.e., Data transfer rate, Q = 2.5 Mbits/ sec

Now come The average Latency

Average latency is nothing but 0.5 * rotation delay 

We have 2400 rotations in 1 minute

2400 rotation = 60 *10ms

therefore 1 rotation =  x ms, where,

x= 60 *103 / 2400

therefore we get x = 25 ms

And Rotation delay = (0.5)*25= 12.5 ms 

Hence P =12.5 ms

and Q= 2.5 M bits / s

edited by
1 votes
1 votes
RPM = 2400
So, in DOS, the disk rotates 2400 times.
Average latency is the time for half a rotation
= 0.5×60/2400 s
= 12.5 ms
In one full rotation, entire data in a track can be transferred. Track storage capacity = 62500 bits
So, disk transfer rate
= 62500 × 2400/60
= 2.5 × 106 bps
So,
P = 12.5, Q = 2.5×106

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