Sequential and Concurrent events

1. Sequential and Concurrent events Sequential = Totally ordered in time. Total ordering is feasible in a single process that has only one clock. This is not true in a distributed system. Two issues are important here: • How to synchronize physical clocks ? (We already discussed this) • Can we define sequential and concurrent events without using physical clocks, since physical clocks cannot be perfectly synchronized?

2. Causality Causality helps identify sequential and concurrent events without using physical clocks. Joke  Re: joke ( implies causally ordered before or happened before) Message sent  message received Local ordering: a  b  c (based on the local clock)

3. Defining causal relationship Rule 1. If a, b are two events in a single process P, and the time of a is less than the time of b then a b. Rule 2. If a = sending a message, and b = receipt of that message, then a  b. Rule 3.a  bb  c a  c

4. a  d since (a  bb  cc  d) e  d since (e  ff  d) (Note that  defines a PARTIAL order). Is g f or f g? NO.They are concurrent. . Example of causality Concurrency = absence of causal order

5. LC is a counter. Its value respects causal ordering as follows a  b  LC(a) < LC(b) Note that LC(a) < LC(b) does NOT imply a  b. Each process maintains its logical clock as follows: LC1. Each time a local event takes place, increment LC. LC2. Append the value of LC to outgoing messages. LC3. When receiving a message, set LC to 1 + max (local LC, message LC) Logical clocks

6. Total order is important for some applications like scheduling (first-come first served). But total order does not exist! What can we do? Strengthen the causal order to define a total order (<<) among events. Use LC to define total order (in case two LC’s are equal, process id’s will be used to break the tie). Let a, b be events in processes i and j respectively. Then a << b iff -- LC(a) < LC(b) OR -- LC(a) = LC(b) and i < j a  b  a << b, but the converse is not true. Total order in a distributed system The value of LC of an event is called its timestamp.

7. Causality detection can be an important issue in applications like group communication. Logical clocks do not detect causal ordering. Vector clocksdo. a  b  VC(a) < VC(b) Vector clock C may receive Re:joke before joke, which is bad!

8. {Sender process i} 1. Increment VC[i]. 2. Append the local VCto every outgoing message. {Receiver process j} 3. When a message with a vector timestamp Tarrives from i, first increment the jth component VC[j] of the local vector clock, and then update the local vector clock as follows: k: 0 ≤ k ≤N-1:: VC[k] := max (T[k], VC[k]). Implementing VC jth component of VC

9. Example [3, 3, 4, 5, 3, 2, 1, 4] < [3, 3, 4, 5, 3, 2, 2, 5] But, [3, 3, 4, 5, 3, 2, 1, 4] and [3, 3, 4, 5, 3, 2, 2, 3] are not comparable Vector clocks Let a, b be two events. Define. VC(a) < VC(b) iff i : 0 ≤ i ≤ N-1 : VC(a)[i] ≤ VC(b)[i], and  j : 0 ≤ j ≤ N-1 : VC(a)[j] < VC(b)[j], VC(a) < VC(b)  a  b Causality detection

10. Mutual Exclusion CS p0 CS p1 CS p2 CS p3

11. 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

12. 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

13. Message passing solution:Centralized decision making Client do true  send request; wait until a reply is received; enter critical section (CS) send release; <non-CS activities> od server busy: boolean queue release Server dorequest received and not busy  send reply; busy:= true request received and busy  enqueue sender release received andqueue is empty  busy:= false release received and queue not empty  send reply to the head of the queue od req reply clients

14. 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!

15. {Life of each process} 1. Broadcast a timestampedrequest 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. Decentralized solution 1:Lamport’s algorithm Completely connected topology • Can you show that it satisfies • all the properties (i.e. ME1, ME2, • ME3) of a correct solution?