Operating Systems

Operating Systems Practical Session 11Distributed synchronization

Motivation • Interest in distributed computation models is rapidly growing (Grids, Cloud computation, internet relay, etc) • A good model for computation is difficult to come up with: • Concurrent computation • No global time and no global state • Hard to capture effects of possible failures

The model • Each instance is executed on a different processor • Assume no shared memory, communication is handled with messages of the following format: <destination, action; parameters> • Sending is non-blocking and reliable • A processor waits for events (messages) – a timeout mechanism is possible but we will not discuss it here

Global states and causality • It is impossible to determine the global state of a distributed system: • Noninstantaneous communication (delays, lost messages, etc) • Can’t synchronize with a timer mechanisms (drift, initial synchronization) • Local interruptions (can’t trust simultaneous reactions) • Thus, we must find global system properties which we can depend on – causal order of events

Happened before • We would like to define some order over system events – a “happened before” relation (denoted <H): • If (e1 <p e2) then e1 <H e2 • If (e1 <m e2) then e1 <H e2 • If (e1 <H e2 && e2 <H e3) then e1 <H e3 • Defines a partial order • Can be defined as a DAG Same processor event Send – receive event Transitivity of <H

Global time – Lamport’s timestamps • Defines a global (and total) order on events • Order is consistent with <H • Created on the fly • Will assume that each event has a timestamp attached to it • An ID is appended to the timestamp and allows for tie breaking • Lamport’s algorithm: If e1<He2 then e1.TS < e2.TS 1 Initially my_TS=0 2 Upon event e, 3 if e is the receipt of message m 4 my_TS=max(m.TS, my_TS) 5 my_TS++ 6 e.TS=my_TS 7 If e is the sending of message m 8 m.TS=my_TS

Causality violation and vector timestamps • Lamport’s algorithm does not guarantee that if e.TS < e’.TS then e <H e’ • This make it difficult to detect causality violation • A causality violation occurs if a message mis sent to a remote processor p before another message m’ is sent, but p receives m’ before mWritten as: m<cm’ and r(m’)<pr(m) • We will use a vector of timestamps to overcome this problem

Global time – vector timestamps 1 Initially my_VT=[0,…,0] 2 Upon event e, 3 if e is the receipt of message m • for i=1 to M • my_VT[i] = max(m.VT[i],my_VT[i]) • My_VT[self]++ • e.VT=my_VT • if e is the sending of message m • m.VT=my_VT • Vector timestamp: • e.VT ≤ve’.VTiffe.VT[i] ≤ e’.VT[i], 1 ≤ i ≤ M • e.VT <ve’.VTiffe.VT ≤ve’.VT and e.VT≠e’.VT • Can be used to detect causality violations • VT algorithm: e1<He2iffe1 <VT e2

Question 1 Consider the following interaction between four processors: P1 P2 P3 P4 Time e1 e8 e14 e20 e2 e9 e10 e15 e21 e3 e4 e11 e22 e23 e16 e12 e5 e17 e18 e24 e6 e13 e7 e19

Question 1 • What is the largest Lamport’s timestamp value? (hint: you can answer without calculating all time stamps) • List the Lamport timestamp of each event. • List the vector timestamp of each event. • Is there a potential causality violation? What can indicate this violation?

Question 1 The Lamport timestamp mechanism calculates the longest chain of event occurring within a system. Thus, the largest timestamp value would be the number of vertices included in the longest path of events of the underlying DAG.In this case the answer is 12. P1 P2 P3 P4 e1 e8 e14 e20 e2 e9 e10 e15 e21 e3 e4 e11 e22 e23 e16 e12 e5 e17 e18 e24 e6 e13 e7 e19

Question 1 Time P1 P2 P3 P4 e1 e8 e14 e20 e2 e9 e10 e15 e21 e3 e4 e11 e22 e23 e16 e12 e5 e17 e18 e24 e6 e13 e7 e19

Question 1 A possible causality violation exist. Note that the send event e3 (a message from p1 to p3) happens before e23 (a message from p4 to p3) but is received afterward. When using VT, e3.VT=(3,1,0,0) but right before the reception of this message (e17) the clock’s VT is e16.VT=(4,3,3,4). Thus, when P3 receives this message it knows that e23 arrived “too soon”. P1 P2 P3 P4 e1 e8 e14 e20 e2 e9 e10 e15 e21 e3 e4 e11 e22 e23 e16 e12 e5 e17 e18 e24 e6 e13 e7 e19

The Ricart - Agrawala algorithm • An algorithm for handling distributed mutual exclusion • Uses Lamport’s timestamps • Each process only uses the following set of variables / data structures: • Timestampcurrent_time • Timestampmy_timestamp • integerreply_count • booleanisRequesting • booleanreply_deferred[M]

The Ricart - Agrawala algorithm The following code is used to enter the critical section: Request_CS: • my_timstamp:= current_time • isRequesting:= TRUE • Reply_count:= M-1 • for every processor j≠i • send(REMOTE_REQUEST; my_timestamp) • wait until reply_count= 0

The Ricart - Agrawala algorithm A listener thread is used so that the node responds to requests from others: CS_monitoring: Wait until a REMOTE_REUQUEST or REPLY message is received REMOTE_REQUEST(sender; request_time) • Let j be the sender of the REMOTE_REQUEST message • if (not is_requesting or my_timestamp > request_time) • send(j, REPLY) • else • reply_deferred[j] = TRUE REPLY 6. reply_count := reply_count-1 Ties are broken with processor IDs

The Ricart - Agrawala algorithm Releasing the CS: Release_CS_monitoring: • is_requesting:= false • For j=1 through M (other than this processor's ID) • if (reply_deferred[i]=TRUE) • send(j, REPLY) • reply_deferred[j]=FALSE

Question 2 Assume that N processors are handling mutual exclusion with the aid of the Ricart-Agrawala’s mutual exclusion algorithm. • How many messages will be passed in the system whenever a processor wishes to enter the critical section? Are there scenarios where this number is lower/greater? • Why is this algorithm deadlock free? • What can happen if a single message is lost?

Question 2 • To enter the CS a processor must request permission from all other processors – i.e. N-1 messages are sent.Only after the processor received a REPLY response from all other processors may it enter the CS. Note, that these may be deferred for a while…That is, entering the CS will require a total of 2(N-1) messages passed. One means to reduce this network load is by keeping several requests deferred for a while. This will allow agents to enter the CS more than once without having to send messages to all N-1 nodes in the system [Roucairol & Carvalho].

Question 2 2. The algorithm relies on the fact that each timestamp is unique (based on the Lamport’s time and processor ID). Thus, a total order over request can be easily deduced, and CS access is handled through this order. 3. The algorithm assumes that the system is failure free and its correctness heavily relies on this condition. It is easy to see that if a single message is lost a deadlock can easily occur.

Raymond’s algorithm • Solve the mutual exclusion problem via a token (only the token holder may enter CS) • Communication is based on an underlying tree structure of all nodes • The tree is always oriented towards token holder • Uses a FIFO queue to prevent starvation • Good performance (number of messages per CS entry decreases as the load increases!) • Uses O(log n) messages only

Raymond’s algorithm • Each process only uses the following set of variables / data structures: • Booleantoken_holder • BooleaninCS • Processorcurrent_dir • Queuerequests_queue

Raymond’s algorithm Request_CS: • If not token_holder • if requests_queue.isEmpty( ) • send(current_dir, REQUEST) • requests_queue.enqueue(self) • wait until token_holder is true • inCS:= true Release_CS: • inCS:= false • If not requests_queue.isEmpty( ) • current_dir:= requests_queue.dequeue( ) • send(current_dir, TOKEN) • token_holder:= false • if not requests_queue.isEmpty( ) • send(current_dir, REQUEST)

Raymond’s algorithm Monitor_CS: • while (true) • wait for a REQUEST or a TOKEN message REQUEST • if token_holder • if inCS • requests_queue.enqueue(sender) • else • current_dir:= sender • send(current_dir, TOKEN) • token_holder:= false • else • if requests_queue.isEmpty() • send(current_dir,REQUEST) • requests_queue.enqueue(sender)

Raymond’s algorithm (cont.) TOKEN • current_dir:= requests_queue.dequeue( ) • if current_dir = self • token_holder:= true • else • send(current_dir, TOKEN) • if not requests_queue.isEmpty( ) • send(current_dir, REQUEST)

Question 3, Moed B 2006 The following 8 processor network is using Raymond’s algorithm to solve the mutual exclusion problem. In the initial state, the token is with processor A at the root of the tree (and wants to enter the critical section), and no requests for the CS are recorded.

Question 3, Moed B 2006 Directed edges correspond to the current_directionvar A B F C D E G H

Question 3, Moed B 2006 To allow for a convenient representation we define agent steps as the invocation of a procedure or an action. Use the sketch above to describe the result of applying Raymond’s algorithm if nodes C,D,F and G request the token. Provide a detailed description of all concurrent steps (in which a single step is taken by all relevant nodes) by sketching the system’s state after each one and up until three of the four agents receive the token.Note: assume that ties are broken based on ID.

Question 3, Moed B 2006 F G D C A REQUEST B F REQUEST REQUEST REQUEST C D E G H

Question 3, Moed B 2006 F G D C C F A G D REQUEST B F C D E G H

Question 3, Moed B 2006 G D C C F F A G D B B F C D E G H

Question 3, Moed B 2006 F G D C C B A G D REQUEST B F C D E G H

Question 3, Moed B 2006 G D C C B G A A D B F C D E G H

Question 3, Moed B 2006 G D C A B C A D B F REQUEST C D E G H

Question 3, Moed B 2006 B C C D F A A D B F C D E G H

Question 3, Moed B 2006 B C C D A A D B F C D E G H

Question 3, Moed B 2006 B C C D A D B F C D E G H

Question 3, Moed B 2006 C C D A D B F C D E G H

Question 3, Moed B 2006 D C D A B F REQUEST C D E G H

Question 3, Moed B 2006 D B D A B F C D E G H

Operating Systems