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if a disk system has an average seek time of 30ns and rotation rate of 360RPM. each track of the disk has 512 sectors each of size 512 Bytes. what is the time taken to read 4 successive sectors,also compute the effective data transfer rate

a) 0.0843 sec , 1536 kbps

b) 0.123 sec , 1436 kbps

c) 0.156 sec ,1326 kbps

d) 0.135 sec , 1252 kbps
asked in CO & Architecture by Loyal (5k points) 1 3 9 | 74 views
Dear user question from another test series or class are not allowed please change title of this question else it will get closed

1 Answer

+3 votes
Given Data,

Average seek time = 30ns
Rotation rate = 360 RPM
Number of sector = 512
Sector capacity = 512 Bytes

Here we have to find data transfer rate and disk access time of 4 sector i.e 4 * 512B = 2048 B

a) Lets find data transfer rate first

Its given disk is doing 360 rotation in 60 second
so to complete 1 rotation we need $\frac{1}{6}$ seconds
In 1 rotation we read 1 track i.e 512 sectors i.e 512*512 Bytes i.e 256 KB
To read 256 KB we need $\frac{1}{6}$ seconds
so in 1 second we read 256 * 6 KB i.e 1536 KB
Therefore data transfer rate is 1536 KB/sec

b) Lets calculate disk access time for 4 sector i.e 4 * 512B = 2048 B

We know data transfer rate is 1536KB/sec
i.e 1 B data can be transferred in $\frac{1}{1536 \times 10^{3}}$ sec
i.e 2048 B can be transferred in $\frac{2048}{1536 \times 10^{3}} sec = 0.00133 sec$

Now 1 rotation is done in $\frac{1}{6}$ sec, so Avg rotation latency is $\frac{1}{12}$ sec i.e 0.0833 sec

As we know Disk access time = Avg seek time + Avg rotation latency + Data transfer rate
DAT = 30ns + 0.0833 sec + 0.00133 sec
we can neglect avg seek time as its too small
Therefore Disk Access time = 0.0846 secs

So option A is correct.

Also there is mistake in options given in question, change every b to B
answered by Boss (6.4k points) 3 11 32

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