Questions about RAID 5?
Hello to all professionals,
I'm still self-taught in Linux server administration and am currently on the topic of RAID levels. According to my textbook, RAID level 5 states that the data is distributed across at least 3-5 disks (other sources say something like 3+, so it could be more than 5, I'm guessing).
According to my textbook, the data is striped across the hard drives, and if one fails, the others continue to run, including the data on the failed one.
But how can that be?
If I understand the procedure correctly, the data is distributed across n hard drives. If one hard drive fails and needs to be replaced, the data from the failed drive will be missing.
Or is the principle of RAID level 5 the same as RAID level 1, where the data is mirrored to the other hard drives?
Furthermore, through research, I found out that there is a RAID alternative called Multi-Copy Mirroring, which is supposed to have the advantage that the data is copied multiple times and stored on different hosts in the network.
I looked into whether dedicated server hosts offered this option because I wanted to find out about the prices.
Unfortunately, I didn't find a provider (maybe I entered the wrong keywords in Google?).
Do you happen to know any?
And is this MCM recommended if you want to use multiple programs with a potentially higher data stream/volume on a dedicated server?
Thank you in advance for your answers
GLG
Tichu player
(2 votes)
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You need to know how this works with the parity bit.
RAID 3 and 5 have a hard drive with parity. At RAID 3, it is a separate one, and at RAID 5, the parity is distributed to all hard drives.
If now a hard drive fails a few data. But thanks to the parity, you can count back what data the missing plate has included.
This basically runs in such a way that at the same points on the different plates the Mits (0 and 1) are counted together. Now the question is whether you have straight or odd parity. The parity bit basically only gives what you need to get straight from the individual bits.
Example:
Straight parity with 5 plates
0 1 Patity bit 0
1 1 1 0 Patity bit 1
You can count back on the bit after a failure
1 – 0 1 parity bit 0 It was a 1
1 1 1 – Paritybit 1 It was a 0
Basically, you can make a RAID 5 from 3 records. Open End.
Hello,
thank you for your answer.
Unfortunately, the term “parity bit” is not in my textbook (maybe it still comes, but where I’m just stuck, it is not listed. In other words, I am learning something new again :-).
The one with the 3 hard drives to open end… I conclude that the five is only a recommendation/direction you don’t need to hold.
Right. The RAID says just how it is built. But in the end, you can find endless hard drives.
Okay. Thanks for your help 🙂
Raid 5, just like 4, is easily explained.
If you have an equation with only known values, you can calculate it from the other values if one is missing.
Example:
1+2+3+4=10
1 is now on drive 1, 2 on drive 2, 3 on Lauwerk 3, 4 on drive 4 and 10 is your parity bit on drive 5.
If drive 3 fails, you have:
1+2+x+4=10
So the lost value is 10-1-2-4=x x=3
To MCM
A dedicated server cannot offer this logically, because it contradicts the idea. At MCM, the idea is the same as at Raid1, only that you don’t have the data on a computer’s drives, but you have many. Ergo needs at least 2 servers. This keeps the system alive when a server fails.
And because we are already there, Erasure codes.
Here the idea is the same as with Raid5, only that you don’t distribute the data to the drives from a server, but you have many servers.
Hello flauski,
Thank you very much for your answer, which makes me slightly confused. Maybe you like to briefly dissolve my question marks:
If LW 1 = 1, LW 2 = 2 LW 3 = 3, LW 4 = 4, why is the parity value on LW 5 = 10 and not 15?
That’s what the MCM is.
And Erasure Code, unfortunately, I will not be able to use: my financial resources are only enough for a dedicated server :-/
But with the term Erasure Code, I have added something new. Also a big thank you 🙂
These are numerical values. The numerical value 1 is stored to 1, 2 to 2, 3, to 3, 4 to 4 and 10 to 5. And 1+2+3+4 is 10 and not 15, so much should be paid in school. I could have taken 2 to 1, 5 to 2, 6 to 3, 1 to 4 and 14. But I’m lazy.
And then it would have been in Example 2+5+6+1=14.
Right, Raid 1 is just mirroring.
Ah! That’s the hook. Now I understand it:-) Thank you for giving me the help. And I can confirm that the RAID 1 means 50% loss of capacity, which is the 1:1 data. So if I have 2 LW a 500 GB and data on LW1 250 GB, then I have 250 GB of data on LW 2 and consequently instead of 1 TB “Usable Area” indirectly only 500 GB
Yes, you pay the capacity from a drive at Raid4 and 5. But what is significantly more efficient than Raid 1, there is 50%.
This is true, but would that mean that only the parity value of LW 1 – 4 is stored on LW 5?
Hi.
that with parity etc – can sound quite confusing. I’m doing a little different.
Example with three plates:
Plate 1 has 1 1 1 0 0 0 0
Plate 2 has 1 0 0 0 0 0 0 0 0
Parity is 0 1 0 0 0 0 because 1 XOR 1 = 0, 0 XOR 0 = 1, 1 XOR 0 = 1, 0 XOR 1 = 1
If plate 2 falls away, then you simply calculate
Plate 1 has 1 1 1 0 0 0 0
Parity is 0 1 0 0 0 0 0
gives the original content
Plate 2 has 1 0 0 0 0 0 0 0 0
To practice, we do this on a note or with Excel.
This bill works from three plates (all the same size), but on any number. From the entire capacity of your RAID, a plate for the parity / checksum always goes on. Whether the parity info is stored on a separate plate or distributed over all plates is not necessary for the calculation.
Hello norbertk62,
thank you for your answer. I have to confess, at the moment it seems to me like Bohemian villages, but I’m getting this together.
I’m wondering once more here…;-)
Don’t lose the courage. If there’s one more thing to do, log in by PN. Everybody’s got it.
Thank you. With this you make me courage:-)
Raid 5 needs minimum three data carriers
Source for reading Storage Basics: RAID at a glance – RAID 0 to 7 – TecChannel Workshop
Hello,
also thank you for your answer.
The page to which your posted link leads, I look at, but I had written that in my textbook, it must be at least 3 hard drives.
Addendum: Puh, on the page are many information. I print out and work through them alone
I read, I just wanted to confirm
More are, of course, better, Raid5 I have never built up among five data carriers
yes, Lvl0, Lvl1, Lvl5 and Lvl10 are the tangent lists and 10 is a combination of 0 and 1
Ah, okay.
I have now printed out the pages (unfortunately my Kyocera printed everything to me in a small letter: /) and worked out the information in peace. I find it interesting that there are far more than the RAID systems listed in my book. In my book, “only” stands as levels 0.1,5 and 10, where the book explains that the 10 is a combination of level 0 and level 1.
On the website I find much more levels.
Hei, through you I will be really smart 🙂