CS469/569 - Linux and Unix Administration and Networking Spring 2022

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Drop deadline - Monday
- just keep participating in the class.
- if you're not going to take quizzes, do HWs, take the exam then you will fail.

h7 - questions?  late credit once you get things done.

h8
- Assigned, due 4/6.

Attendance

q8
- Due by midnight 4/7.
- Over lessons 9-12, things you have done on h7, h8
- Timed, 30 minutes

RAID
- basic idea - combine many physical drives into one logical drive
- basic idea - performance and/or redundancy
- note - abstraction
- levels: 0, 1, 5
- level 0, picture from https://cs.indstate.edu/cs469/lesson.php?lesson=lesson12
  /dev/sdc1, /dev/sdd1, /dev/sde1 => one RAID drive /dev/md0
  read sector 1 from /dev/md0 => sector 1 from /dev/sdc1
  read sector 2 from /dev/md0 => sector 1 from /dev/sdd1
  read sector 3 from /dev/md0 => sector 1 from /dev/sde1
  read sector 4 form /dev/md0 => sector 2 from /dev/sdc1
  ...
  read sector k from /dev/md0 => sector (k-1)/3+1 from the k%3 drive in the RAID
- level 0 - throughput is higher, hopefully hopefully k times as high
            extremely important with spinning hard disk drives
	  - probability non-failure for /dev/md0 something like to the kth power
	    e.g., 99% of non-failure => /dev/md0 (.99)**3 = 97% chance of non-failure
- level 1, picture from ...
  /dev/sdc2, /dev/sdd2, /dev/sde2 if we made this RAID1, say /dev/md2
  read sector 5 from /dev/md2 => read sector 5 from any of /dev/sdc2, /dev/sdd2, /dev/sde2
  write sector 5              => write to all of them
- level 1 - throughput - read throughput could be about k times faster,
                       - write throughput maybe about the same, but likely slower
          - probability of non-failure: 1 - (1 drive failure rate)**k, 1 - .01**3 = .999999
	  - amount of useable storage is just one drive (others are duplicates)
- level 5 - see picture from lesson
          - suppose Drive 3 goes down.
	    we knew that: Ap = A1 XOR A2 XOR A3 (for each bit in those sectors)
	    note that:    Ap XOR A1 XOR A2
	               (A1 XOR A2 XOR A3) XOR A1 XOR A2
		       A1 XOR A1 XOR A2 XOR A2 XOR A3 =
		          0      XOR     0     XOR A3
			         0             XOR A3 = A3
		       (recall that 0 xor 0 is 0, 1 xor 1 is 0, 0 xor 1 is 1, 1 xor 0 is 1)
	  - throughput - reads are about k times as fast
	               - writes are something like k-1 times faster, or something...
	  - probability of non-failure - ok with 1 drive failing, not with 2
	  - amount of useable storage - k-1 drives