Cs603 clock synchronization
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CS603 Clock Synchronization. February 4, 2002. What is the best we can do? Lundelius and Lynch ‘84. Assumptions: No failures No drift Fully connected network of n nodes Uncertainty of ε in message delivery time Best guarantee: ε (1 – 1/ n ) This is a tight lower bound.

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CS603 Clock Synchronization

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Cs603 clock synchronization

CS603Clock Synchronization

February 4, 2002


What is the best we can do lundelius and lynch 84

What is the best we can do?Lundelius and Lynch ‘84

  • Assumptions:

    • No failures

    • No drift

    • Fully connected network of n nodes

    • Uncertainty of ε in message delivery time

  • Best guarantee:

    • ε(1 – 1/n)

    • This is a tight lower bound


Lower bound proof

Lower bound proof

  • Idea: Based on view of each node

    • Views indistinguishable even if real time not the same

    • Shift execution of a node relative to real time

  • Shift of global view and local view equivalent if message delays changed

    • Can always shift by at least ε(1 – 1/n) without changing local views


Proof induction

Proof: Induction

  • Clocks synchronized to within γ

  • Assume messages one way take time μ, return takes time μ+ε (e1)

  • Induction: Assume node i-1 sends with delay μ, receives with delay μ+ε

    • Shift processes < i by ε

  • Let V1,…,Vn be local times at termination of e1.

    • In e1, Vn ≤ V1 + γ

    • In ei, Vi-1 ≤ Vi + y – ε

  • ∑ Vi ≤ ∑ Vi+nγ – (n-1) ε

    • (n-1) nγ

    • γ ≥ ε(1-1/n)


Synchronization with faulty clocks dolev halpern strong 84

Synchronization with Faulty Clocks(Dolev, Halpern, Strong ‘84)

  • Problem: What if some sites are really bad?

    • Bad clocks

    • Don’t follow protocol

  • Notation

    • C: Logical clock

    • D: Physical clock

    • TAR: Time Adjustment Register

      • C = D + TAR

    • Δ: Uncertainty in message delay

    • C(t), D(t) – value of clock at REAL time t


Assumptions

Assumptions

  • Fully connected, but not necessarily complete

  • Recipient knows source of message

  • Given nodes p,q; H(p,q) and L(p,q) are upper/lower bounds on transmission time

    • ρ is min(H/L)

  • A real time frame (not directly observable)

  • Correct physical clock has bounded drift rate: R such that time u>v, (1/R)(u-v) ≤ D(u)-D(v) ≤ R(U-v)

  • Correct processor has correct clock, implements algorithm

  • No assumptions on behavior of faulty processor

    • Don’t care if faulty processor knows correct time

  • All processors start within time B (can easily show B ≤ R(n-1)H)


Weak synchronization

Weak Synchronization

  • Weak Clock Synchronization Condition: Constants PER, DMAX, ADJ such that:

    • TAR changes only at times that are multiples of PER by amount less than ADJ

    • Difference between clocks bounded by DMAX

  • Theorem: There is an algorithm that achieves WCSC, independent of faults, for which C(t) is unbounded

  • Proof: Set TAR(t’) = logPER(D(t))-D(t)


Real clock synchronization

Real clock synchronization

  • Clock Synchronization Condition: Add

    • PER > ADJ

    • Changes occur only first time C reads iPER

      • If change when C(t)=iPER, then C(t’) ≠ iPER  t’<t

  • Gives Linear Envelope Synchronization:

    • at+b < C(t) < ct+d, a>0

  • Theorem:Linear Envelope Synchronization impossible if  1/3 processors faulty


Proof sketch

Proof Sketch

  • Construct algorithm that forces a correct processor to run at rate greater than aρn

  • Idea: faulty processor p uses one algorithm for processor q, other for others

    • Two-faced behavior

    • Can’t tell which is two-faced

    • Correct processor caught in the middle – follow fast clock or slow clock?


Three processor case p q r

Three-processor case (p, q, r)

  • Assume algorithm A synchronizes in time N and tolerates one fault

  • F0 = A

  • Fm+1: p pretends its clock runs at ρ times q’s rate

  • p pretends r sends messages soCp(t) > aρmDp(t)+b-mDMAX

    • Fm gives these messages

  • q cannot distinguish from case where p’s clock is fast, r is sending p messages according to Fm

  • Cq(t) > Cp(t) – DMAX> aρmDp(t) + b – (m+1) DMAX= aρm+1Dq(t)+b-(m+1) DMAX (since Dp(t) = ρDq(t)


Possibility fischer lynch merritt

Possibility(Fischer, Lynch, Merritt)

  • If no uncertainty in message delay, f faulty, can do with 2f+1 processors

    • Send messages to all neighbors

    • Send all messages back

    • Round trip gives time

    • Faulty processor will be detected if it tries to be worse than round-trip time

      • Messages out of order


Possibility dolev halpern simons strong

Possibility(Dolev Halpern Simons Strong)

  • We CAN do better

    • Requires authentication

  • Assumptions:

    • Messages will be received with bounded delay

    • Bounded drift

    • Digital signature

    • If p has set of messages M at time t with more than f distinct signers, one signer was correct at time signed

    • 2ρ(f+1) < 1

  • Key: Synchronization time known in advance

    • At time, send signed “time is now”

    • If receive f+1 messages saying “time is now” before getting to that time, update local time


Recruiting bulletin

Recruiting Bulletin

  • Harris Corporation is in the CS lobby until 3pm today


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