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

edited | 6.1k views
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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.
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Here Average latency  = R/2 = 12.5 ms

Rotational latency by 2.

@naha What value do you get for Q?
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P = Average Rotational Latency = 12.5 ms

Q will depend upon how many bytes to be transferred. But how much is that?
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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
+1
Track capacity is given directly rt? Why divide by 200?
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sir i thought total capacity is given..
thats why m dividing 62500 by 200 to find single track capacity.
+3
These kind of words are usual in GATE :)
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After the syllabus change in 2016 is this topic currently within the syllabus?
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$\\ 1.\ Avg \ lateny:2400R\rightarrow 60sec\\ \\ 1R\rightarrow \dfrac{1}{40}sec\\ \\ \dfrac{1}{2}R\rightarrow \dfrac{1}{80}sec=12.5ms\\ \\ \\$

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

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.
by Veteran (434k points)
edited by
+5
sir, Disk rotation speed = 200 rpm or 2400 rpm ... in question ,need correction ...
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Thanks :) Corrected..
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Arjun sir why we are not multiplying 200 in 62500 bits ?

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Here Track storage capacity = 62,500 bits for 1 track only

not for 200 tracks
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When asked for disk transfer rate does it mean track transfer rate? What is significance of 200 tracks per surface?
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Because at a time we would be accessing any one track so size of only one track is considered not all,As Data transfer rate is bits accessed per sec by Read/Write Header

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

by Active (1.9k points)
edited by
Average Rotational Latency = 12.5 ms
Transfer rate = 2.5Mbps
by Veteran (61k points)
edited
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
by (435 points)