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Consider a disk with $c$ cylinders,  $t$ tracks per cylinder, $s$ sectors per track and a sector length $s_l$. A logical file $d_l$ with fixed record length $r_l$ is stored continuously on this disk starting at location $(c_L, t_L, s_L)$, where $c_L, t_L$ and $S_L$ are the cylinder, track and sector numbers, respectively. Derive the formula to calculate the disk address (i.e. cylinder, track and sector) of a logical record n assuming that $r_l=s_l$.
in Operating System by Veteran (416k points)
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3 Answers

+1 vote
disk address= cL(C)+tL(t)+SL(s)

we must know the cylinder number and track number in advance then only we can calculate the address of of record
by Loyal (10k points)
+1 vote

Here logical file size d1

Now , d1 has n records of size r1

So, d1= n r1

No of sectors for the file say (sec)=n r1 % s [Now, r1=s1, then sec=0]

No of tracks say (track) =sec % t

No. of cylinder say (cyl)= sec / t

Now, disk address of the logical record is <cL ,tL , sL>+  <cyl, track,sec>

by Veteran (113k points)
+1 vote

GIVEN:Consider a disk with c cylinders,  t tracks per cylinder, s sectors per track

from this we can conclude that 1 cylinders contain = t*s sectors

and one track contain =s sectors

now we have to drive formula of logical address n

so the cylinder no is $\left \lfloor \left ( \frac{n}{ts} \right ) \right \rfloor$

and track no will be floor of ( (n%ts)/s)

and sector no will be n%s

by Loyal (9.9k points)
0
any video resource to study?

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