Physical clock synchronization

Question 1.

Why is physical clock synchronization important?

Question 2.

With the price of atomic clocks or GPS coming down,

should we care about physical clock synchronization?

Classification

Types of Synchronization

External Synchronization Internal Synchronization Phase Synchronization

Types of clocks

Unbounded 0, 1, 2, 3, . . .

Bounded 0,1, 2, . . . M-1, 0, 1, . . .

Unbounded clocks are not realistic, but are easier to

deal with in the design of algorithms. Real clocks are

always bounded.

Terminologies

R R

Newtonian time

c

l

o

c

k

t

i

m

e

clock 1

clock 2

≤ δ

= drift rate ρ

What are these?Drift rate ρClock skew δResynchronization interval R

Max drift rate ρ implies: (1- ρ) ≤ dC/dt < (1+ ρ)

Challenges(Drift is unavoidable)Accounting for propagation delayAccounting for processing delay

Faulty clocks

Internal synchronization

Berkeley Algorithm

A simple averaging algorithm

that guarantees mutual

consistency |c(i) - c(j)| < δ

Step 1. Read every clock in the system.Step 2. Discard outliers and substitute

them by the value of the local clock. Step 3. Update the clock using the

average of these values.

Resynchronization interval will depend on the drift rate.

Internal synchronization

Lamport and Melliar-Smith’s

averaging algorithm handles

byzantine clocks too

Assume n clocks, at most t are faulty

Step 1. Read every clock in the system.Step 2. Discard outliers and substitute them by the

value of the local clock. Step 3. Update the clock using the average of

these values.

Synchronization is maintained if n > 3t

Why?

i j

k

c

c+ δ

-c δ

-2c δ

A faulty clocks exhibits 2-faced or byzantine behavior

Bad clock

Internal synchronization

Lamport & Melliar-Smith’s algorithm (continued) The maximum difference between

the averages computed by two

non-faulty nodes is (3tδ / n)

To keep the clocks synchronized,

3tδ / n < δ

So, 3t < n

i j

k

c

c+ δ

-c δ

-2c δ

B a d c l o c k s

k

Cristian’s method

Client pulls data from a time server

every R unit of time, where R < δ / 2ρ.

(why?)

For accuracy, clients must compute

the round trip time (RTT), and

compensate for this delay

while adjusting their own clocks.

(Too large RTT’s are rejected)

Timeserver

External Synchronization

Network Time Protocol (NTP)

Tiered architecture Broadcast mode

- least accurate

Procedure call

- medium accuracy

Peer-to-peer mode

- upper level servers use this for max accuracy

Timeserver

The tree can reconfigure itself if some node fails.

Level 1Level 1

Level 1Level 0

Level 2Level 2

Level 2

P2P mode of NTPLet Q’s time be ahead of P’s time by δ. Then

T2 = T1 + TPQ + δT4 = T3 + TQP - δ

y = TPQ + TQP = T2 +T4 -T1 -T3 (RTT)

δ = (T2 -T4 -T1 +T3) / 2 - (TPQ - TQP) / 2

So, x- y/2 ≤ δ ≤ x+ y/2

T2

T1 T4

T3Q

P

Ping several times, and obtain the smallest value of y. Use it to calculate δ

x Between y/2 and -y/2

Problems with Clock adjustment

1. What problems can occur when a clock value isAdvanced from 171 to 174?

2. What problems can occur when a clock value is Moved back from 180 to 175?

Mutual Exclusion

CS

CS

CS

CSp0

p1

p2

p3

Why mutual exclusion?

Some applications are:

1. Resource sharing

2. Avoiding concurrent update on shared data

3. Controlling the grain of atomicity

4. Medium Access Control in Ethernet

5. Collision avoidance in wireless broadcasts

Specifications

ME1. At most one process in the CS. (Safety property)ME2. No deadlock. (Safety property)ME3. Every process trying to enter its CS must eventually succeed.

This is called progress. (Liveness property)

Progress is quantified by the criterion of bounded waiting. It measuresa form of fairness by answering the question: Between two consecutive CS trips by one process, how many times other processes can enter the CS?

There are many solutions, both on the shared memory model and the message-passing model

Message passing solution:Centralized decision making

clients

Clientdo true

send request;wait until a reply is received;enter critical section (CS)send release;<non-CS activities>

od

Serverdo request received and not busy send reply; busy:= true request received and busy enqueue sender release received and queue is empty busy:= false release received and queue not empty send reply

to the head of the queueod

busy: boolean

server

queue

req replyrelease

Comments

- Centralized solution is simple.

- But the server is a single point of failure. This is BAD.

- ME1-ME3 is satisfied, but FIFO fairness is not guaranteed. Why?

Can we do better? Yes!

Decentralized solution 1:Lamport’s algorithm

{Life of each process}

1. Broadcast a timestamped request to all.

2. Request received enqueue sender in local Q;.

Not in CS send ack

In CS postpone sending ack (until

exit from CS).

3. Enter CS, when

(i) You are at the head of your own local Q

(ii) You have received ack from all processes

4. To exit from the CS,

(i) Delete the request from Q, and

(ii) Broadcast a timestamped release

5. Release received remove sender from local Q.

0 1

2 3

Q0 Q1

Q2 Q3

Completely connected topology

Can you show that it satisfies all the properties (i.e. ME1, ME2, ME3) of a correct solution?