first we will see how the data recovery takes place in grade level 1 so in raid level 1 the redundant data is going to be duplicated completely so let us consider the example of 4-bit data so here I am considering data 1 let us say D 1 is 0 1 0 1 B 2 is 1 0 1 1 D 3 is 0 triple 1 d 4 is 1 0 0 1 something like this and this is the example what I will be considering for all the raid levels data recovery right and now in data disk 1 now let us say this is a raid level 1 so here this is redundant data so what happens like this is original data and this is redundant data in the case of ride level one we are having the mirroring methodology so what happens in the case here is this D 1 this is G 2 this is D 3 and this is deep or so because it is mirroring so everything will be replaced everything will be mirrored and as a duplicate copy now what happens here is like suppose let us say that D 2 D 2 and D 3 are lost then easily we can make a copy of them using redundant data so this is how the data recovery takes place in the case of raid level 1 it’s very simple now coming to raid level 2 this is a bit level stripping and Hamming code is used for maintaining the redundant data now let us see the Hamming code process so if yum data bits are there k check deaths will be occurred so the relation between m and K is 2 power K minus 1 greater than or equal to M plus K okay now we are having four data bits that is M is equal to 4 so now what happens to power k minus 4 1 greater than or equal to 4 plus K if K is equal let us say for k equal to 2 2 power 2 4 minus 1 3 is greater than four plus two which is not correct let us say for k equal to three so 8 minus 1 7 is greater than or equal to 4 plus 3 so this is satisfying the inequality so K is equal to 3 that is 3 check box will be required now how the data is maintained so first for radar is 4 because we are having 4 bits of information in each data and 3 Hamming code check bits so 3 dispar Hamming go so this is depending upon the number of data bits in each data and this depends upon the number of check bits that is the value of K so what happens because the value of D 1 is 0 1 0 1 here 0 comes here or one comes here 0 comes and here are 1 comes now D 2 is 1 0 1 1 so 1 comes young 0 consumed one comes here and 1 comes here this is nothing but bit level interleaving we say the clevon stripping we can say now D 3 is 0 triple 1 will have like this and d 4 is 1 w 0 1 so 1 0 0 1 like this will be having the bit level stripping now we are supposed to calculate the Hamming code values now when you are calculating the Hamming code values we are supposed to have some Czech bit formulas these are the formulas for calculating the chine bits and you have to store the calculator check bits in these 3 discs so you are we are going to see how the check bits are calculated so these bits did put according to the bit positions we are considering 0 1 0 1 so d1 he this one D 2 D 3 d 4 okay so D 1 is 1x or D 2 is 0 X or D 4 is 0 so it is 1 D 1 is 1 B 3 is 1 B 4 is 0 so this is 0 so here D 2 is 0 XR d 3 is 1 XR d 4 is 0 so this is 1 so here we have to store 1 0 & 1 this is C 1 C 2 C 4 like this we can maintain and Jimmy let me calculate the other check bits and also store them now we can see that all the check bits are calculated unfilled in the corresponding disk if you want to know which disk is in error these are all the data rest which disk is in error we are supposed to know all the like C 1 C 2 C 4 values now if you see the formulas yum we can say that disk D 1 is in error if both C 1 and Z 2 are in error and if C 1 and C 4 check bits are in error then we can say this D 2 is in error and similarly if C 2 and C 4 are in error we are getting the wrong data then D 3 is in error similarly C 1 D 4 depends on on C 1 C 2 C 4 Utley and that is in order to identify the error or disk which is damaged we are supposed to know all the check bits data but if we want to recover the particular failed disk one of the check bits is enough now let us consider that data does d3 is an error so which can be recovered using C two formulas C 2 is equal to D 1 X or D 3 XR d 4 so this formula and changing like this the 3 is equal to C 2 XR d 1 XR d for now just substitute the values of C 2 D 1 and D for and we are getting 1 0 1 0 that is nothing but D 3 value this is how the data recovery takes place in the case of grade level 2 now let us see how the data recovery takes place in Dre level 3 according to the raid level 3 similar to raid level 2 it is bit level stripping so the same copy I am showing your d1 d2 d3 d4 and parity is used in order to maintain the redundant disk so parity is considered to be even parity here so if you see your 0 1 0 1 here in number of buns are already there so we will get 0 here and 1 0 1 1 3 ones are these so one comes you and 0 1 1 1 3 ones are the 8 so one comes here 1 0 0 1 2 ones are there so 0 comes here this is how the redundant data is maintained in the case of raid level 3 now let us say that some d2 disk is damaged or lost how we can recover this d2 data disk using B 4 D 3 D 1 and parity disk so we don’t know about this value so how you are killing D D 2 value so now again like same for a same concept even parity should be maintained using d 4 D 3 D 1 and parity so 0 1 1 0 here the number of one’s is there so 0 1 0 1 and 1 so 3 odd number of ones is there so one should come here 0 1 1 so 0 1 one odd number of ones so again 1 1 0 1 0 so 0 comes here so this is how the data the two will be recovered in the case of a level 3 it is very simple in the gears over a level 3 to recover the data now let us see how the data recovery takes place in raid level 4 in the raid level 4 it is actually byte level stripping takes place and parity will be used in order to maintain the redundant data the difference between the bit level and byte level let us see in the case of raid level 3 if they did this is D 1 D 2 D 3 and E 4 we are storing the first bit of each and every data in one particular disk but if you see in the case of third level for complete data here that is like D 1 will be stored in one this D 2 will be stored in wonders D 3 will be stored in one disk and E 4 will be stored in one disk actually this is nothing but the block level stripping actually will be taking in the case of a level 4 here we consider a block to be equal to one byte and because we are having only four bits not equal to bite also we are storing complete one data into one this clear ok so the advantage of reading a data in the level 4 suppose if I want to read data 2 then this disc a low and we can read but if it is if it is the case of raid level 3 if I want to read level read data to all the second bits of all the disk need to be read so this is the advantage when compared to the Ray level 3 in the case of raid level 4 now let us go for the data recovery process so how it is like 0 1 0 1 even parity so it comes 0 here 1 0 1 0 0 here and again like 0 1 1 0 one one one one zero again so this is a redundant parity now let us say that we are losing d1 completely so how to recover that d1 now one zero one zero even parity is 4 0 0 1 1 sorry 0 1 0 0 so on here 1 1 1 0 0 0 1 1 1 0 so 1 comes here so you can see that the 1 is matching now so this is how the raid level for data recovery will be happening now let us consider how the data will be recovered in the case of raid level 5 here it is a block level stripping and we are considering the block is of each 4 bits and the parity will be distributed here distributed parity in the sense the parity calculated from these data will be stored here and the calculated parity from these particular data will be stored here that if this is one disk and this is another disk and this is another disk this is another disk and this is another disk here we are constrained like this and now let us say that we are losing one particular disk so let us say that this disk is lost now how we are recovering that particular desk we can easily identify that so 0 0 1 0 we are getting 1 here and then 1 1 0 0 we are getting 0 here and then 0 1 0 0 obviously will get 1 and when 1 1 1 and 0 will get 1 again similarly here we can get using the other bits and we can get this value so this is a how the data recovery takes place in the case of raid level high you

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