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September 21, 2000What is new with RAMA?1 Naomi Avigdor.

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Presentation on theme: "September 21, 2000What is new with RAMA?1 Naomi Avigdor."— Presentation transcript:

1 September 21, 2000What is new with RAMA?1 Naomi Avigdor

2 September 21, 2000What is new with RAMA?2 Review RAMA is a network file system Disks are divided into fixed size lines (rings) File blocks are hashed to a specific disk and a specific line on that disk No need to search inodes for file location Files have parity protection for fault tolerance (optional)

3 September 21, 2000What is new with RAMA?3 Overflow When a line is full, the entire disk is full, regardless of data in other lines What to do? Reserve overflow area on each disk Simulation: How full is the disk before overflow happens? How much overflow area is needed? Smaller / Larger systems? Any effect on overflow?

4 September 21, 2000What is new with RAMA?4 Overflow Math C = capacity of each line N = number of disks * number of lines per disk T = total blks at ov point fullness = total blks at ov point / total space = T / N*C

5 September 21, 2000What is new with RAMA?5 Explanation of Results

6 September 21, 2000What is new with RAMA?6 Conclusions Hash function is very good 1 overflow line is sufficient

7 September 21, 2000What is new with RAMA?7 What is parity? (single) Fault tolerance by creating a sum (mod2) of the data and saving it on another disk. Ex: d1 =3, d2=5, d3 = 2, d4 = 1 p = 3 xor 5 xor 2 xor 1 = 5 D1, d2, d3, d4, p, each reside on a different disk (why?) D1, d2, d3, d4, p, together are called a stripe If d3 is lost: restore d3 by D3 = p xor d1 xor d2 xor d4 = 5 xor 3 xor 5 xor 1 = 2 If d3 is changed: new p = old p xor old d3 xor new d3 or: new p = old d1 xor old d2 xor new d3 xor old d4

8 September 21, 2000What is new with RAMA?8 What happens on a write? If d3 is changed: Traditional: new p = old p xor old d3 xor new d3 Cost = read old p, read old d3, write new p, write new d3 = 2 reads + 2 writes RAMA: New p = read old d1, d2, d4, write new p, write new d3 = 3 reads + 2 writes

9 September 21, 2000What is new with RAMA?9 What happens on a write (Cont.)? If d3, d4 are changed: Traditional: new p = repeat as before, two times Cost = 4 reads + 4 writes RAMA: New p = read old d1, d2, write new p, write new d3, write new d4 Cost = 2 reads, 3 write

10 September 21, 2000What is new with RAMA?10 What happens on a write (Cont.)? If d1, d3, d4 are changed: Traditional: new p = repeat as before, three times Cost = 6 reads + 6 writes RAMA: New p = read old d2, write new p, write new d1, d3, d4 Cost = 1 reads, 4 write

11 September 21, 2000What is new with RAMA?11 Conclusion – RAMA parity Parity updates can be delayed without the data writes being delayed for more efficiency (since old data is not needed) For applications where minimum changes are made to each stripe, RAMA may be slower.


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