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 What are these? Drift rate  Clock skew  Resynchronization interval R Max drift rate  implies: (1- ) ≤ dC/dt < (1+ ) Challenges (Drift is unavoidable) Accounting for propagation delay Accounting for processing delay Faulty clocks

Internal synchronization 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. Berkeley Algorithm A simple averaging algorithm that guarantees mutual consistency |c(i) - c(j)| < 

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? Bad clock A faulty clocks exhibits 2-faced or byzantine behavior

Internal synchronization Lamport & Melliar-Smith’s algorithm (continued) The maximum difference between the averages computed by two non-faulty nodes is (3td / n) To keep the clocks synchronized, 3td / n < d So, 3t < n k B a d c l o c k s

Cristian’s method External Synchronization Client pulls data from a time server every R unit of time, where R <  / 2. 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) Time server

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 Time server Level 0 Level 1 Level 1 Level 1 Level 2 Level 2 Level 2 The tree can reconfigure itself if some node fails.

P2P mode of NTP  = (T2 -T4 -T1 +T3) / 2 - (TPQ - TQP) / 2 Let Q’s time be ahead of P’s time by . Then T2 = T1 + TPQ +  T4 = T3 + TQP -  y = TPQ + TQP = T2 +T4 -T1 -T3 (RTT)  = (T2 -T4 -T1 +T3) / 2 - (TPQ - TQP) / 2 T2 T3 Q P T1 T4 x Between y/2 and -y/2 So, x- y/2 ≤  ≤ x+ y/2 Ping several times, and obtain the smallest value of y. Use it to calculate 

Problems with Clock adjustment 1. What problems can occur when a clock value is Advanced from 171 to 174? 2. What problems can occur when a clock value is Moved back from 180 to 175? 1.What happened to the instant 172 and 173? 2. The instants 175 -180 appear twice

Mutual Exclusion CS p0 CS p1 p2 CS CS p3

Why mutual exclusion? Some applications are: Resource sharing Avoiding concurrent update on shared data Controlling the grain of atomicity Medium Access Control in Ethernet 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 measures a 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 Client do true  send request; reply received  enter CS; send release; <other work> od server busy: boolean queue release req Server do 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 queue od reply clients

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