3k views

A hard disk has $63$ sectors per track, $10$ platters each with $2$ recording surfaces and $1000$ cylinders. The address of a sector is given as a triple $\langle c, h, s \rangle$, where $c$ is the cylinder number, $h$ is the surface number and $s$ is the sector number. Thus, the 0$^{th}$ sector is addresses as $\langle 0, 0, 0 \rangle$, the 1$^{st}$ sector as $\langle 0, 0, 1 \rangle$, and so on

The address of the 1039$^{th}$ sector is

1. $\langle 0, 15, 31 \rangle$
2. $\langle 0, 16, 30 \rangle$
3. $\langle 0, 16, 31 \rangle$
4. $\langle 0, 17, 31 \rangle$

edited | 3k views
+3
In this question we can go for options elimination:

A. $15*63+31 = 976$ //15 is taken because we need to cross 0 to 14 platters (15 platters) each with 63 sectors.
B. $16*63+30=1038$
C.$16*63+31=1039$
D. $17*63+31=1102$

Hence, (C) is the correct option!
0
The Q is illuding enough bcs it considers the data fills the first cylinder from top to bottom then starts filling next cylinder, indeed option elimination looks appealing.
0

15 is taken because we need to cross 0 to 14 platters (15 platters) each with 63 sectors.

can u explain this line bit more. i m not getting how 1 platter has 63 sectors. There are 63 sectors per track and there may be many tracks per surface.

$1039^{th}$ sector will be stored in track number $(1039 + 1)/63 = 16.5$ (as counting starts from $0$ as given in question) and each track has $63$ sectors. So, we need to go to $17^{th}$ track which will be numbered $16$ and each cylinder has $20$ tracks $(10$ platters $\times 2$ recording surface each) . Number of extra sectors needed $= 1040-16 \times 63 = 32$ and hence the sector number will be $31$. So, option (C).

by Boss (22.8k points)
edited by
+4
what is the significance in adding 1 to 1039 ?
couting starts from 0 and 0th sector is counted as 0 itself right ?
So 1039th sectors  is actually numbered 1039 itself right ?
+19

check option C:

$\langle 0,16,31\rangle$
Number of sectors to be crossed $= 0+ (16 \times 63) + 31 \\ = 1039$

0
correct..

point to be noted here that secor 31 means that it is 32th sector since numbering starts from 0.

16*63 =1008 sectors have been crossed and we need to read further 32 sectors to get 1039th sector(sector number 1040)
0
The answer will not change either we take the 0th sector as 1st sector or 1st sector as 1st sector.

a sector address is given like this $\langle c, h, s \rangle$ so it means that, we traverse in this way:

this all is happening in a single cylinder:
means first move sector wise(covering a track on a surface),
when all sector are over change the surface (also means a single track is over)

then move in all surfaces(a cylinder has $2\times 10 = 20$ total surfaces) as done previously,
when all surface are over then at last move into next cylinder

for a single track, it has 63 sectors
for a single surface, it has 1 track = 63 sectors
this means that for a single cylinder, it has 20 surfaces = 20 tracks = 20 $\times$ 63 sectors

on checking option C:
$\langle0,16,31\rangle$
Number of sectors to be crossed $= 0+ (16 \times 63) + 31 \\ = 1039$

@cse23

by Boss (30.8k points)
edited
+1
@amarVashishth sir,
How could one  interpret that data on disk is stored cross sectionally ? and not latteraly ?
0

Thus, the 0th sector is addressed as ⟨0,0,0⟩, the 1st sector as ⟨0,0,1⟩, and so on.

If we proceed in a similar fashion we have the address <0,1,0> after all the sectors in a track are addressed. Which means the addresses are numbered in a cross sectional fashion.

0

" for a single surface, it has 1 track = 63 sectors "

can anyone explain me how this interpretation comes. How we are considering that in one surface there is only one track. ?

It may be silly question but i m not getting this visually nor conceptually.

0

@Satbir bhai mujhe bhi ye doubt hai jo kuljeet bhai ne pucha. could u explain ?

0
They have given $<c,h,s >$ to reach a sector and on a surface we will have $1$ track in which there will be $63$ sectors. If there would had been multiple tracks per surface then they would have either mentioned it or given address format like $<c,h,t,s>$ where $t$ means track number.

On the contrary , if you assume more than $1$ track per surface then how will you address a sector using $<c,h,s>$ only ?
track no= $\frac{1039}{63}$ =$16.492063492063492063492063492063 ....$ so track no = $16 ..$

now i am considering the fractional part ... sector no=$0.492063492063492063492063492063* 63 = 31$

so the address is $\langle0,16,31\rangle$....

correct me if i am wrong ...
by Boss (12.1k points)
edited
0
Can u explain what's the logic behind multiplying the fractional part by no of sectors to get the sector number
0
think of it like this way, 1 track has 63 sectors, therefore to find the location of 1039 sector we need to divide it  by 63.
Now after dividing we get 16.4920634920634920  , this means that our sector lies somewhere between 16 and 17 track that means 16 track for sure.
now we reached 16 track , now we need to go to sector on 16th track. So multiplying this fractional part should give us the sector number. This is why we multiply it.
0
better do like this to get the remainder : 1039 - 16*63 = 31