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+22 votes

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 $P$ ms and the data transfer rate is $Q$ bits/sec. Write the values of $P$ and $Q$.

Number of tracks/recording surface $= 200$

Disk rotation speed $= 2400$ rpm

Track storage capacity $= 62,500$ bits

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

0

i didn't have clear idea about the average latency ,average latency is time needed to traverse all the track or is that same as rotational latency.

0

P = Average Rotational Latency = 12.5 ms

Q will depend upon how many bytes to be transferred. But how much is that?

Q will depend upon how many bytes to be transferred. But how much is that?

0

200 tracks having capacity 62500 //single head so recording surface =1

no of bits/track = 62500/200 bit

cycle time = 25ms

transfer rate = 62500/200*25*10^(-3)

= 12500bps

= 12.5 kbps

no of bits/track = 62500/200 bit

cycle time = 25ms

transfer rate = 62500/200*25*10^(-3)

= 12500bps

= 12.5 kbps

+1

sir i thought total capacity is given..

thats why m dividing 62500 by 200 to find single track capacity.

thats why m dividing 62500 by 200 to find single track capacity.

+3

$\\ 1.\ Avg \ lateny:2400R\rightarrow 60sec\\ \\ 1R\rightarrow \dfrac{1}{40}sec\\ \\ \dfrac{1}{2}R\rightarrow \dfrac{1}{80}sec=12.5ms\\ \\ \\ $

$2.\ Data \ transfer\ rate: \\ 1R\rightarrow \dfrac{1}{40}sec\\ \\ \dfrac{1}{40}sec\rightarrow 62,500\ bits\\ \\ 1sec\rightarrow 40 \times 62,500 bits \\ \\ means\ 2500000\ bits/sec\\$

$2.\ Data \ transfer\ rate: \\ 1R\rightarrow \dfrac{1}{40}sec\\ \\ \dfrac{1}{40}sec\rightarrow 62,500\ bits\\ \\ 1sec\rightarrow 40 \times 62,500 bits \\ \\ means\ 2500000\ bits/sec\\$

+52 votes

Best answer

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 \ s = 2.5 \times 10^6 $ bps.

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 \ s = 2.5 \times 10^6 $ bps.

+21 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 *10^{3 }ms

therefore 1 rotation = x ms, where,

x= 60 *10^{3} / 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 *10^{3 }ms

therefore 1 rotation = x ms, where,

x= 60 *10^{3} / 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

0 votes

1.

In 60 seconds, the disk rotates 2400 times. So it takes 25ms to rotate it one time. There are 200 tracks. On average, 100 tracks need to be traveresed to reach a specific track.So 25ms*100 tracks= 2.5s time is taken on average to reach a particular track on avg.

2.

In 25 ms, one rotation is completed and all the 62500 bits are transferred. So in 1ms, 2500 bits are transferred. Accordingly, in 1s, 2.5Mb are transferred. Thats why the speed is 2.5Mb/s

In 60 seconds, the disk rotates 2400 times. So it takes 25ms to rotate it one time. There are 200 tracks. On average, 100 tracks need to be traveresed to reach a specific track.So 25ms*100 tracks= 2.5s time is taken on average to reach a particular track on avg.

2.

In 25 ms, one rotation is completed and all the 62500 bits are transferred. So in 1ms, 2500 bits are transferred. Accordingly, in 1s, 2.5Mb are transferred. Thats why the speed is 2.5Mb/s

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