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Optimal Inter-Object Correlation When Replicating for Availability. Haifeng Yu National University of Singapore Phillip B. Gibbons Intel Research Pittsburgh. Multi-object Operations. Data replication for better availability - PowerPoint PPT Presentation
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Optimal Inter-Object CorrelationOptimal Inter-Object CorrelationWhen Replicating for AvailabilityWhen Replicating for Availability
Haifeng YuNational University of Singapore
Phillip B. GibbonsIntel Research Pittsburgh
Haifeng Yu (National University of Singapore) 2
Multi-object OperationsMulti-object Operations
Data replication for better availability Traditional research focuses on availability of
individual data objects
E.g., individual files or database objects
User-level tasks may access multiple data objects / files: Multi-Object Operations Compile a project / Latex a paper
Aggregation queries for databases
Haifeng Yu (National University of Singapore) 3
Availability of Multi-Object OperationsAvailability of Multi-Object Operations
Availability of single object is not the same as the availability for multi-object operations: An operation requesting 1,000 objects may observe
nearly 1,000 times higher failure probability
But there’s more..... Our recent experimental study shows that the
assignment from object replicas to machines has critical effects on such availability:
“Availability of Multi-Object Operations” [NSDI’06]
Haifeng Yu (National University of Singapore) 4
Object Assignment ExampleObject Assignment Example
Compile a project need all 4 files
4 files: A B C D
4 machines:
A B C DA B C DA B C DA B C D
Computing average may tolerate one
missing object
still OKnot OK
which is better?
A B
A B C D
C Dbetter
A B
A C
C D
B D
better
Haifeng Yu (National University of Singapore) 5
Important Observations from the ExampleImportant Observations from the Example
For individual objects, it does not matter which machines it is assigned Distinguishes from replica placement problems
Same “concentration” / “spread”
Difference is in inter-object correlation: Obj A is “fully correlated” with one obj (B) VERSUS
Obj A is “partially correlated” with two objs (B and C)
which is better?A B
A B C D
C D A B
A C
C D
B D
Haifeng Yu (National University of Singapore) 6
Practical ImportancePractical Importance
Applicable to almost all replication systems CAN, CFS, Chord, Coda, FARSITE, GFS, GHT,
Glacier, Pastry, R-CHash, RIO, …
Failure probability of TPC-H varies by 4 orders of magnitude under different assignments All with the same storage overhead
Haifeng Yu (National University of Singapore) 7
Formal ModelFormal Model N objects each with k replicas
Each machine holds l objects (total Nk / l machines)
Machines fail / crash i.i.d.
Object unavailable if all k machines holding it fail
Assignment: Mapping from objects to machines No machine holds multiple replicas of the same obj
Two specific assignments: PTN: Partition objects into N/l groups of size l, and map
each group to k machines
RAND: A uniformly random assignment
Haifeng Yu (National University of Singapore) 8
Formal Model (continued)Formal Model (continued)
A multi-object operation requests n specific objs This talk assumes n = N; See paper for n < N.
Operation is successful if ≥ t objects are available
t depends on application semantics
We will consider a single (multi-object) operation See paper for more discussion on this…
Availability of the assignment defined as Prob[the operation succeeds]
Haifeng Yu (National University of Singapore) 9
Previous Experimental ResultsPrevious Experimental Results
0 .9 8 5 0 .9 9 0 .9 9 5 10.985 0.99 0.995 1
1.0
0.1
0.01
0.001 t/n
Several other assignments experimented
(including Chord) fall between
PTN and RAND
RAND
PTN
failure probability (i.e., 1 – availability)
Haifeng Yu (National University of Singapore) 10
Our GoalOur Goal Limitations of experimental method:
Can only study a small number of assignments
Results only for specific parameter values (e.g., specific machine failure probabilities)
Our goal: Find assignment with best/worst availability among all
possible assignments
This is the first theoretical study on this subject…
Haifeng Yu (National University of Singapore) 11
Summary of Our ResultsSummary of Our Results
Impossible to remain optimal for all t values
Calculating the availability of an assignment is #P-hard
Best Worst
t = n (general case)
PTN(within small constants)
RAND(within small constants)
t = n (restricted case)
PTN n/3 rings of size 3
t = l + 1 Trivial construction (also RAND
within small constants)
PTN
This talk focuses on t = n only
Haifeng Yu (National University of Singapore) 12
The General CaseThe General Case
Upper and lower bounds Leveraging Janson’s inequality – a tail approximation
for sum of dependent Bernoulli trialsNot to be confused with Jensen’s inequality
Approaching upper and lower bounds by PTN and RAND Leverage Janson’s inequality a second time
See paper for details…
Haifeng Yu (National University of Singapore) 13
The Restricted CaseThe Restricted Case
Results for general case Have constants
RAND is a distribution – what is the “structure” of the worst assignment?
We consider significantly restricted scenarios Each object has two replicas
Each machine holds two objects (l = 2)
Corresponds to our example earlier
Haifeng Yu (National University of Singapore) 14
Using Rings to Represent AssignmentsUsing Rings to Represent Assignments Each assignment corresponds to a set of rings
Ring size from 2 to n
Sum of sizes of all rings is n
Availability uniquely determined by ring sizes
A B
A B C D
C D A B
A C
C D
B D= =
A B
C D
A B
DC
2 rings of size 2 1 ring of size 4
Haifeng Yu (National University of Singapore) 15
Hill ClimbingHill Climbing
Adjustment step: merge two rings into one or split one ring into two Can always transform an assignment to another
within a finite number of steps
Crux: How does availability change when merging two rings of size x and y (x y) into a ring of size (x+y) ? Theorem: Availability improves iff y is odd.
Haifeng Yu (National University of Singapore) 16
The Restricted Case: ResultsThe Restricted Case: Results
Theorem: n/2 rings of size 2 is the best. This corresponds to PTN
Theorem: n/3 rings of size 3 is the worst.
What about a single ring of size n?
What about rings of other sizes? See paper for answers…
Parity of ring sizes matter a lot…
Haifeng Yu (National University of Singapore) 17
Impossibility ResultImpossibility Result
0 .9 8 5 0 .9 9 0 .9 9 5 10.985 0.99 0.995 1
1.0
0.1
0.01
0.001
RAND
PTN
Area bounded by the curve is constant
Impossible to remain optimal under all t values
Haifeng Yu (National University of Singapore) 18
ConclusionsConclusions Availability of multi-object operation critically
affected by inter-object correlation
First theoretical study of object assignment Best/worst assignments for t = n and t = l + 1
Impossible to remain optimal under all t values
See paper for full results…
Open questions: Other t values?
Erasure coding?