0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">This is from theidi=
scussion of Raid 3 Stallings Text which the slide

come from.=C2=A0 It is related to the problem you asked about the parity bi=
t

calculation for the Hamming Code



Not the methodology for logical calculation, it is straight math



Note that X is XOR for whatever UFT reason





REDUNDANCY In the event of a drive failure, the parity drive is accessed>
and data is reconstructed from the remaining devices. Once the failed

drive is replaced, the missing data can be restored on the new drive and>
operation resumed.



Data reconstruction is simple. Consider an array of five drives in which>
X0 through X3 contain data and X4 is the parity disk.The parity for the

ith bit is calculated as follows:



X4(i) =3D X3(i) { X2(i) { X1(i) { X0(i)



where { is exclusive-OR function.



Suppose that drive X1 has failed. If we add X4(i) { X1(i) to both sides

of the preceding equation, we get



X1(i) =3D X4(i) { X3(i) { X2(i) { X0(i)





file:///home/ruben/william stallings - computer organization and

architecture designing for performance (8th edition).pdf



Page 202







ank">http://www.nylxs.com/images/raid3_parity_math.png



Ruben