Raft consensus Landon Cox February 15, 2017

Paxos ACM TOCS: Submitted: 1990. Accepted: 1998 Introduced: Transactions on Computer Systems Submitted: 1990. Accepted: 1998 Introduced:

Butler W. Lampson http://research.microsoft.com/en-us/um/people/blampson/ Butler Lampson is a Technical Fellow at Microsoft Corporation and an Adjunct Professor at MIT…..He was one of the designers of the SDS 940 time-sharing system, the Alto personal distributed computing system, the Xerox 9700 laser printer, two-phase commit protocols, the Autonet LAN, the SPKI system for network security, the Microsoft Tablet PC software, the Microsoft Palladium high-assurance stack, and several programming languages. He received the ACM Software Systems Award in 1984 for his work on the Alto, the IEEE Computer Pioneer award in 1996 and von Neumann Medal in 2001, the Turing Award in 1992, and the NAE’s Draper Prize in 2004.

Barbara Liskov MIT professor 2008 Turing Award “View-stamped replication” PODC ’88 Very similar to Raft Different election process

At any moment, machine exists in a “state” State machines At any moment, machine exists in a “state” What is a state? Should think of as a set of named variables and their values

State machines What is your state? 4 5 3 2 6 1 Client My state is “2” Clients can ask a machine about its current state. What is your state? 4 5 3 2 6 1 Client My state is “2”

“actions” change the machine’s state State machines “actions” change the machine’s state What is an action? Command that updates named variables’ values

“actions” change the machine’s state State machines “actions” change the machine’s state Is an action’s effect deterministic? For our purposes, yes. Given a state and an action, we can determine next state w/ 100% certainty.

“actions” change the machine’s state State machines “actions” change the machine’s state Is the effect of a sequence of actions deterministic? Yes, given a state and a sequence of actions, can be 100% certain of end state

Replicated state machines Each state machine should compute same state, even if some fail. Client What is the state? What is the state? Client What is the state? Client

Replicated state machines What has to be true of the actions that clients submit? Applied in same order Client Apply action c. Apply action a. Client Apply action b. Client

State machines How should a machine make sure it applies action in same order across reboots? Store them in a log! Action …

Replicated state machines Can reduce problem of consistent, replicated states to consistent, replicated logs … … … …

Replicated state machines How to make sure that logs are consistent? Two-phase commit? … … … … …

Replicated state machines What is the heart of the matter? Have to agree on the leader, outside of the logs. Leader=L Leader=L Apply action a. … … Client Leader=L Leader=L … …

Key elements of consensus Leader election Who is in charge? Log replication What are the actions and what is their order? Safety What is true for all states, in all executions (including failures)? e.g., either we haven’t agreed or we all agree on the same value

Key elements of consensus Leader election Who is in charge? Log replication What are the actions and what is their order? Safety What is true for all states, in all executions (including failures? e.g., either we haven’t agreed or we all agree on the same value

Server states Three states: leader, follower, candidate F L C

Server state: follower Passive state: respond to candidates and leaders F L C

Server state: leader Server handles all client requests F L What should happen if a client sends a request to a follower? Follower forwards request to leader.

Server state: candidate An intermediate state, used during elections F L C

Time divided into terms Election Normal operation Election Normal operation Leader unknown Leader known What happened here? Election failed

Terms as a logical clock All servers maintain the current term Terms increase monotonically Maintained as a logical clock Terms are exchanged whenever servers communicate What if A’s term is bigger than B’s? B updates its term to A’s What if A’s term is bigger and B thinks of itself as the leader? B reverts to a follower state What if A’s term is bigger, and it receives a request from B? A rejects B’s request B must be up to date to issue requests

Server state: follower C Current term = 0 S1 F L C Current term = 0 S3 F L C Current term = 0 All servers start as followers. All servers have local timers. Note: no bootstrapping problem!

Server state: follower C Current term = 0 S1 F L C Current term = 0 S3 F L C Current term = 0 Server remains follower as long as it receives periodic valid messages from a leader or candidate. Called a “heartbeat” message.

Server state: follower C Current term = 0 S1 F L C Current term = 0 S3 F L C Current term = 0 What should server assume if no heartbeat? Assume no viable leader, start election.

Server state: follower C Current term = 0 S1 F L C Current term = 0 S3 F L C Current term = 0 Who should the server nominate? How about itself? At least it knows that it’s running.

Server state: candidate F L C Current term = 0 S1 F L C Current term = 1 S3 F L C Current term = 0 To start an election: Increment current term and set state to candidate

Server state: candidate F L C Current term = 0 S1 F L C Current term = 1 S3 F L C Current term = 0 Need to collect votes. For whom should the server vote? Itself, of course! Major qualification: It’s running.

Server state: candidate F L C Current term = 0 S1 Vote in term 1 F L C Current term = 1 S3 Votes S1=S1 S2=? S3=? Vote in term 1 F L C Current term = 0 How should S2, S3 respond to vote request?

Server state: candidate F L C Current term = 1 S1 S1 in term 1 F L C Current term = 1 S3 Votes S1=S1 S2=? S3=? S1 in term 1 F L C Current term = 1 How should S2, S3 respond to vote request? Increment term, vote for S1 … why vote for S1? Our goal is consensus, and we know the collector voted for itself.

Server state: candidate F L C Current term = 1 S1 F L C Current term = 1 S3 Votes S1=S1 S2=S1 S3=S1 F L C Current term = 1 What should S1 do next? Count votes (majority wins) Make itself the leader, start sending heartbeats.

Server state: candidate F L C Current term = 1 Leader = S1 S1 S1 is leader F L C Current term = 1 S3 Leader = S1 S1 is leader F L C Current term = 1 Leader = S1

Server state: candidate F L C Current term = 1 Leader = S1 S1 F L C Current term = 1 S3 Leader = S1 F L C Current term = 1 Leader = S1 How many faults can we tolerate? One. Need two/three to vote for same new leader

Server state: candidate F L C Current term = 1 S1 Vote in term 1 F L C Current term = 1 S3 Votes S1=S1 S2=? S3=? Vote in term 1 F L C Current term = 1 C Votes S1=? S2=? S3=S3 F Who votes for whom if S1 and S3 both call elections?

Server state: candidate F L C Current term = 1 S1 Vote in term 1 F L C Current term = 1 S3 Votes S1=S1 S2=S3 S3=S3 Vote in term 1 F L C Current term = 1 C Votes S1=S1 S2=S3 S3=S3 F Who votes for whom if S1 and S3 both call elections? S2 votes for the server that asked first

Server state: candidate F L C Current term = 1 S1 F L C Current term = 1 S3 Votes S1=S1 S2=S3 S3=S3 F L C Current term = 1 C Votes S1=S1 S2=S3 S3=S3 F What does S1 do if it loses the election?

Server state: candidate F L C Current term = 1 S1 F L C Current term = 1 S3 Votes S1=S1 S2=S3 S3=S3 F L C Current term = 1 C Votes S1=S1 S2=S3 S3=S3 F What does S1 do if it loses the election? Moves back to a follower state.

Server state: candidate F L C Current term = 1 S3 is leader Leader = S3 S1 L Current term = 1 C S3 Leader = S3 F F L C Current term = 1 C Leader = S3 F

Server state: follower C Current term = 0 S1 F L C Current term = 0 S3 F L C Current term = 0 Can all three servers nominate themselves?

Server state: candidate F L C Current term = 1 Votes S1=S1 S2=S2 S3=S3 S1 F L C Current term = 1 S3 Votes S1=S1 S2=S2 S3=S3 F L C Current term = 1 Votes S1=S1 S2=S2 S3=S3 Can all three servers nominate themselves? Sure!

Server state: candidate F L C Current term = 1 Votes S1=S1 S2=S2 S3=S3 S1 F L C Current term = 1 S3 Votes S1=S1 S2=S2 S3=S3 F L C Current term = 1 Votes S1=S1 S2=S2 S3=S3 Could this happen indefinitely?

Server state: candidate F L C Current term = 1 Votes S1=S1 S2=S2 S3=S3 S1 F L C Current term = 1 S3 Votes S1=S1 S2=S2 S3=S3 F L C Current term = 1 Votes S1=S1 S2=S2 S3=S3 Could this happen indefinitely? Yes, there is not way to prevent this from happening. The worst possible thing is still possible.

Server state: candidate F L C Current term = 1 Votes S1=S1 S2=S2 S3=S3 S1 F L C Current term = 1 S3 Votes S1=S1 S2=S2 S3=S3 F L C Current term = 1 Votes S1=S1 S2=S2 S3=S3 How do we make this less likely to occur?

Server state: candidate F L C Current term = 1 Votes S1=S1 S2=S2 S3=S3 S1 F L C Current term = 1 S3 Votes S1=S1S2=S2 S3=S3 F L C Current term = 1 Votes S1=S1 S2=S2 S3=S3 How do we make this less likely to occur? Randomize election timeouts i.e., wait a random period before new election

Server state: candidate F L C Current term = 1 S1 F L C Current term = 1 S3 F L C Current term = 1 Can we ever have two leaders?

Server state: candidate F L C Current term = 1 S1 F L C Current term = 1 S3 F L C Current term = 1 Can we ever have two leaders? No, votes either split or converge w/ three nodes.

Key elements of consensus Leader election Who is in charge? Log replication What are the actions and what is their order? Safety What is true for all states, in all executions (including failures)? e.g., either we haven’t agreed or we all agree on the same value

Each node maintains an action log. Leader=L Each node maintains an action log. Each entry contains an action and a term. The term indicates when the leader received the action. Leader=L 1 x  3 1 y  1 Leader=L 1 x  3 1 y  1 1 x  3 1 y  1

When a request comes in, leader appends action to its log. Leader=L When a request comes in, leader appends action to its log. Client y  5 Leader=L 1 x  3 1 y  1 Leader=L 1 x  3 1 y  1 1 y  5 1 x  3 1 y  1

Next, the leader tells other servers to append action. Leader=L Next, the leader tells other servers to append action. Client y  5 1 y  5 Leader=L 1 x  3 1 y  1 Leader=L 1 y  5 1 x  3 1 y  1 1 y  5 1 x  3 1 y  1

The leader waits for confirmation that the entry was appended. Leader=L The leader waits for confirmation that the entry was appended. Client y  5 ack Leader=L 1 x  3 1 y  1 1 y  5 Leader=L 1 x  3 1 y  1 1 y  5 1 x  3 1 y  1 1 y  5

Leader=L y  5 Client 3/3 … commit! Leader=L Leader=L As with leader election, majority rules. Entry is committed once the leader that received it has replicated the entry on a majority of servers (for now). Leader=L Client y  5 3/3 … commit! Leader=L 1 x  3 1 y  1 1 y  5 Leader=L 1 x  3 1 y  1 1 y  5 1 x  3 1 y  1 1 y  5

If action commits, leader updates state machine. Leader=L Client y  5 Leader=L 1 x  3 1 y  1 1 y  5 Leader=L 1 x  3 1 y  1 1 y  5 1 x  3 1 y  1 1 y  5

Leader reports success to client and other servers if action commits Leader=L Client y  5 C C Leader=L 1 x  3 1 y  1 1 y  5 Leader=L 1 x  3 1 y  1 1 y  5 1 x  3 1 y  1 1 y  5

What happens if one follower fails? Action will still commit, since 2/3 are alive. Leader=L Leader=L 1 x  3 1 y  1 Leader=L 1 x  3 1 y  1 1 y  5 1 x  3 1 y  1 1 y  5

Leader=L Leader=L Leader=L What happens if both followers fail: (1) after adding entry to logs, and (2) before acking? Leader=L The leader will keep trying to append. If one server comes back, the action will eventually commit. Leader=L 1 x  3 1 y  1 1 y  5 Leader=L 1 x  3 1 y  1 1 y  5 1 x  3 1 y  1 1 y  5

Leader=L Leader=L Leader=L How long might a client have to wait in this case? Leader=L A long time, i.e., until two machines append the action Leader=L 1 x  3 1 y  1 1 y  5 Leader=L 1 x  3 1 y  1 1 y  5 1 x  3 1 y  1 1 y  5

Leader=L Leader=L Leader=L What happens if leader fails: (1) after adding entry to log, and (2) before contacting followers? Leader=L Action will not commit. New entries will occur under next term. Leader=L 1 x  3 1 y  1 Leader=L 1 x  3 1 y  1 1 y  5 1 x  3 1 y  1

As with leader election, majority rules for commit. For now: entry is committed once the leader that received it has replicated the entry on a majority of servers 1 2 3 4 5 6 7 8 Log index 1 x  3 y  1 y  9 2 x  2 3 x  0 y  7 x  5 x  4 Leader Followers

In this example, which committed entry has the highest index? Entry 7 1 2 3 4 5 6 7 8 Log index 1 x  3 y  1 y  9 2 x  2 3 x  0 y  7 x  5 x  4 Leader Followers

<sigh>This picture is far too orderly and easy to understand. Committing a log entry also commits all entries in leader’s log with lower index. Note: the current leader’s index is the one that counts! 1 2 3 4 5 6 7 8 Log index 1 x  3 y  1 y  9 2 x  2 3 x  0 y  7 x  5 x  4 Leader Followers <sigh>This picture is far too orderly and easy to understand. No guarantee the world will look like this.</sigh>

This can be the state of the logs when a leader comes to power. Each server has assigned each entry (1) a term, and (2) an index. Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader (a) 1 1 1 4 4 5 5 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 6 Followers (d) 1 1 1 4 4 5 5 6 6 6 7 7 (e) 1 1 1 4 4 4 4 (f) 1 1 1 2 2 2 3 3 3 3 3

What is the current term? We are in term 8. Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader (a) 1 1 1 4 4 5 5 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 6 Followers (d) 1 1 1 4 4 5 5 6 6 6 7 7 (e) 1 1 1 4 4 4 4 (f) 1 1 1 2 2 2 3 3 3 3 3

Why aren’t there any entries for term 8? Because the leader hasn’t received any requests yet. Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader (a) 1 1 1 4 4 5 5 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 6 Followers (d) 1 1 1 4 4 5 5 6 6 6 7 7 (e) 1 1 1 4 4 4 4 (f) 1 1 1 2 2 2 3 3 3 3 3

This can be the state of the logs when the leader comes to power. Each server has assigned each entry (1) a term, and (2) an index. Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader (a) 1 1 1 4 4 5 5 6 6 (b) 1 1 1 4 What’s wrong with the logs of (a) and (b)? They are missing log entries.

How might this have happened? This can be the state of the logs when the leader comes to power. Each server has assigned each entry (1) a term, and (2) an index. Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader (a) 1 1 1 4 4 5 5 6 6 (b) 1 1 1 4 How might this have happened? They could have gone offline and come back; (a) during term 6, (b) during term 4

This can be the state of the logs when the leader comes to power. Each server has assigned each entry (1) a term, and (2) an index. Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader What’s wrong with the logs of (c) and (d)? They have extra log entries. (c) 1 1 1 4 4 5 5 6 6 6 6 (d) 1 1 1 4 4 5 5 6 6 6 7 7

How might this have happened? This can be the state of the logs when the leader comes to power. Each server has assigned each entry (1) a term, and (2) an index. Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader How might this have happened? (c) was leader for term 6, added entry and crashed; (d) was leader for 7, added entries and crashed (c) 1 1 1 4 4 5 5 6 6 6 6 (d) 1 1 1 4 4 5 5 6 6 6 7 7

This can be the state of the logs when the leader comes to power. Each server has assigned each entry (1) a term, and (2) an index. Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader What’s wrong with the logs of (e) and (f)? They have extra log entries and missing log entries. (e) 1 1 1 4 4 4 4 (f) 1 1 1 2 2 2 3 3 3 3 3

This can be the state of the logs when the leader comes to power. Each server has assigned each entry (1) a term, and (2) an index. Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader (f) was leader for term 2, added several entries and crashed. (f) quickly restarted and became leader for term 3, added more entries and crashed before any entries from terms 2 or 3 could commit How could this have happened to (f)? (e) 1 1 1 4 4 4 4 (f) 1 1 1 2 2 2 3 3 3 3 3

Goal: (eventually) converge on a sane state from logs like this. Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader (a) 1 1 1 4 4 5 5 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 6 Followers (d) 1 1 1 4 4 5 5 6 6 6 7 7 (e) 1 1 1 4 4 4 4 (f) 1 1 1 2 2 2 3 3 3 3 3

Servers have to keep track of committed log entry with highest index Servers have to keep track of committed log entry with highest index. What are those here? Leader=L 2 for all three servers. Client y  5 Leader=L 1 x  3 1 y  1 1 2 Leader=L 1 x  3 1 y  1 1 y  5 1 2 3 1 x  3 1 y  1 1 2

Leader=L Client y  5 Leader=L 1 2 Leader=L 1 2 3 1 2 Leader reports its highest index of committed action when forwarding request to followers. This is how followers update their state machines. Leader=L Client y  5 Term=1 y  5 High=2 Leader=L 1 x  3 1 y  1 1 2 Leader=L Term=1 y  5 High=2 1 x  3 1 y  1 1 y  5 1 2 3 1 x  3 1 y  1 1 2

Leader also reports highest index immediately preceding current append request. Leader=L Client y  5 Term=1 y  5 High=2 Pred=2 Leader=L 1 x  3 1 y  1 1 2 Leader=L Term=1 y  5 High=2 Pred=2 1 x  3 1 y  1 1 y  5 1 2 3 1 x  3 1 y  1 1 2

Could this happen? Yes, if follower failed before it received action with index 2 from term 1 Leader=L Client y  5 Term=1 y  5 High=2 Pred=2 Leader=L 1 x  3 1 Leader=L Term=1 y  5 High=2 Pred=2 1 x  3 1 y  1 1 y  5 1 2 3 1 x  3 1 y  1 1 2

Should the recovered follower append the new action? If it did, then it would have a different action in index 2 during term 1. Better to reject new action and append missing committed actions first. Leader=L Client y  5 Term=1 y  5 High=2 Pred=2 Leader=L 1 x  3 1 Leader=L Term=1 y  5 High=2 Pred=2 1 x  3 1 y  1 1 y  5 1 2 3 1 x  3 1 y  1 1 2

Yes, since we can still achieve majority w/o it. Can new action commit while recovered server receives actions it missed while down? Leader=L Yes, since we can still achieve majority w/o it. Client y  5 Term=1 y  5 High=2 Pred=2 Leader=L 1 x  3 1 Leader=L Term=1 y  5 High=2 Pred=2 1 x  3 1 y  1 1 y  5 1 2 3 1 x  3 1 y  1 1 2

This gives us a very important system property: Two entries in different logs w/ same index and term always have the same action. Leader=L Client y  5 Term=1 y  5 High=2 Pred=2 Leader=L 1 x  3 1 Leader=L Term=1 y  5 High=2 Pred=2 1 x  3 1 y  1 1 y  5 1 2 3 1 x  3 1 y  1 1 2

Leader=L Client y  5 Leader=L 1 Leader=L 1 2 3 1 2 Two entries in different logs w/ same index and term always have the same action. Why? Leader=L In a term, leader creates at most one entry with a given index. Followers catch up first when behind. Client y  5 Term=1 y  5 High=2 Pred=2 Leader=L 1 x  3 1 Leader=L Term=1 y  5 High=2 Pred=2 1 x  3 1 y  1 1 y  5 1 2 3 1 x  3 1 y  1 1 2

Log Matching Property Two entries in different logs with same index and term  The entries store the same action The logs are identical in all preceding entries Does the matching property hold below? Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader 1 1 1 4 4 5 5 6 6 6 7 7 1 1 1 4 4 4 4 Followers 1 1 1 2 2 2 3 3 3 3 3

Log Matching Property Two entries in different logs with same index and term  The entries store the same action The logs are identical in all preceding entries Why is the first part always true? One leader/term If leader fails, term changes Leader inserts a new entry once at a given index NOTE: this is not enough to ensure the second part

Log Matching Property Two entries in different logs with same index and term  The entries store the same action The logs are identical in all preceding entries Ensuring the second part requires extra check by follower Leader sends followers append(term, last_index, action) Follower must check that last entry has same term and last_index If not, the follower refuses the new entry Otherwise, follower may append new entry containing action

Log Matching Property Two entries in different logs with same index and term  The entries store the same action The logs are identical in all preceding entries Do logs always agree on indexes, terms? No, indexes at logs may have different actions and terms Goal of repairing logs is to increase entries w/ same index, term 1 1 1 4 4 5 5 6 6 6 7 7 1 1 1 4 4 4 4 1 1 1 2 2 2 3 3 3 3 3

How do we make the logs look more like one another? Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader (a) 1 1 1 4 4 5 5 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 6 Followers (d) 1 1 1 4 4 5 5 6 6 6 7 7 (e) 1 1 1 4 4 4 4 (f) 1 1 1 2 2 2 3 3 3 3 3

How do we make the logs look more like one another? Raft: How do we make the logs look like the leader’s? Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader (a) 1 1 1 4 4 5 5 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 6 Followers (d) 1 1 1 4 4 5 5 6 6 6 7 7 (e) 1 1 1 4 4 4 4 (f) 1 1 1 2 2 2 3 3 3 3 3

Repairing logs Leader Append-only Property Leader’s log is treated as ground truth Leader never overwrites or deletes log entries Leader only appends to its log Basic idea: make the leader’s life simple How would a leader know that follower is missing entries? Follower refuses an append request (based on last_index) Leader must monitor each follower’s progress (next_index) Send log entries until followers are caught up

Leader initially assumes followers’ logs look like hers next_index[a]=11 next_index[b]=11 next_index[c]=11 Leader initially assumes followers’ logs look like hers Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 Leader (a) 1 1 1 4 4 5 5 6 6 Followers (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 6

next_index[a]=11 next_index[b]=11 next_index[c]=11 Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 8 Leader (a) 1 1 1 4 4 5 5 6 6 8 Prev={term 6, index 10} (b) 1 1 1 4 8 Prev={term 6, index 10} (c) 1 1 1 4 4 5 5 6 6 6 6 8 Prev={term 6, index 10}

next_index[a]=10 next_index[b]=10 next_index[c]=10 Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 8 Leader ✗ (a) 1 1 1 4 4 5 5 6 6 8 Prev={term 6, index 10} ✗ (b) 1 1 1 4 8 Prev={term 6, index 10} ✗ (c) 1 1 1 4 4 5 5 6 6 6 6 8 Prev={term 6, index 10}

next_index[a]=10 next_index[b]=10 next_index[c]=10 Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 8 Leader (a) 1 1 1 4 4 5 5 6 6 6 Prev={term 6, index 9} (b) 1 1 1 4 6 Prev={term 6, index 9} (c) 1 1 1 4 4 5 5 6 6 6 6 Prev={term 6, index 9} 6

What should happen to this log entry? next_index[a]=10 next_index[b]=10 next_index[c]=10 Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 8 Leader ✓ (a) 1 1 1 4 4 5 5 6 6 6 Prev={term 6, index 9} ✗ (b) 1 1 1 4 6 Prev={term 6, index 9} ✓ (c) 1 1 1 4 4 5 5 6 6 6 Prev={term 6, index 9} 6 6 What should happen to this log entry? It should be deleted

What entry does the leader send to (a) next? next_index[a]=11 next_index[b]=9 next_index[c]=11 Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 8 Leader (a) 1 1 1 4 4 5 5 6 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 What entry does the leader send to (a) next? Entry at index 11 (term 8); this succeeds.

What entry does the leader send to (b) next? next_index[a]=11 next_index[b]=9 next_index[c]=11 Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 8 Leader (a) 1 1 1 4 4 5 5 6 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 What entry does the leader send to (b) next? Entry at index 9 (term 6); this fails and next_index[b]  8. Eventually, (b) will accept entry from term 4 at index 5, and it will catch up with everyone else.

What entry does the leader send to (c) next? next_index[a]=11 next_index[b]=9 next_index[c]=11 Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 8 Leader (a) 1 1 1 4 4 5 5 6 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 What entry does the leader send to (c) next? Entry at index 11 (term 8); this succeeds.

next_index[a]=11 next_index[b]=9 next_index[c]=11 Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 8 Leader (a) 1 1 1 4 4 5 5 6 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 As described, this should make you uncomfortable. Why? We could choose a bad leader (e.g., one w/ an empty log). This could delete committed entries!

How do we prevent this from happening? next_index[a]=11 next_index[b]=9 next_index[c]=11 Log index 1 2 3 4 5 6 7 8 9 10 11 12 1 1 1 4 4 5 5 6 6 6 8 Leader (a) 1 1 1 4 4 5 5 6 6 6 (b) 1 1 1 4 (c) 1 1 1 4 4 5 5 6 6 6 How do we prevent this from happening? We have to be smarter about electing an leader…

Refined leader election Goal Leader must have all committed actions How actions are committed Leader accepts action from client Leader forwards action to followers Majority append must occur before next term If new term starts before majority append, no commit Commit cannot occur until later action commits

Refined leader election Initial election criterion Any candidate can be elected Normally the candidate who starts the election wins Problem: leader can force stale log on followers What must be true of an elected leader? Leader must have all prior committed actions If leader is missing a committed action, it may be lost

Refined leader election New election criterion Vote for candidate with most “up to date” log If voter’s log is more recent than candidate’s, vote self Which log is more up to date? (a) because its last entry is from term 4 (> term 3) If last log entry is from later term, log is more up to date (a) 1 1 1 4 4 4 4 (b) 1 1 1 2 2 2 3 3 3 3 3

Refined leader election New election criterion Vote for candidate with most “up to date” log If voter’s log is more recent than candidate’s, vote self Which log is more up to date? (b) because its last term-4 entry has higher index When last log entries are from same term, higher index is more up to date (a) 1 1 1 4 4 4 4 (b) 1 1 1 4 4 4 4 4

Refined leader election New election criterion Vote for candidate with most “up to date” log If voter’s log is more recent than candidate’s, vote self How to decide which log is more up to date If last log entry is from later term, log is more up to date When last log entries are from same term, higher index is more up to date

Refined leader election Goal Leader must have all committed actions Recall how actions had been committed Leader accepts action from client Leader forwards action to followers If a majority of followers append action, it’s committed If action commits, leader returns to client If leader fails, future leaders can commit Unfortunately, there’s a problem …

Committing actions Log index 1 2 (a) 1 2 (b) 1 2 (c) 1 (d) 1 (e) 1

How was (e) elected if (b) has a more up-to-date log? Committing actions Log index 1 2 (a) 1 2 How was (e) elected if (b) has a more up-to-date log? (b) 1 2 (c) 1 Votes from (c) and (d) (d) 1 (e) 1

Committing actions Log index 1 2 (a) 1 2 (b) 1 2 (c) 1 (d) 1 (e) 1 3

Committing actions Log index 1 2 (a) 1 2 (b) 1 2 (c) 1 (d) 1 (e) 1 3

What is the first thing that (a) will do? Committing actions Log index 1 2 3 (a) 1 2 4 What is the first thing that (a) will do? (b) 1 2 (c) 1 Forward entry at index 3 to followers, but this will fail at (c), because (c) does not have entry at index 2 yet (d) 1 (e) 1 3

Which entry will (a) send to (c) first? Committing actions Log index 1 2 3 (a) 1 2 4 Which entry will (a) send to (c) first? (b) 1 2 (c) 1 2 Entry at index 2 (d) 1 (e) 1 3

Has the entry at index 2 been committed? Committing actions Log index 1 2 3 (a) 1 2 4 Has the entry at index 2 been committed? (b) 1 2 (c) 1 2 By our previous definition, yes. But note that it reached a majority of nodes in term 4, not term 2 (i.e., after it was created) (d) 1 (e) 1 3

Committing actions 1 2 4 1 2 1 2 1 1 3 Log index 1 2 3 (a) (b) (c) (d)

How could (e) have been elected? Committing actions Log index 1 2 3 (a) 1 2 4 How could (e) have been elected? (b) 1 2 (c) 1 2 Votes from (b),(c), and (d) (d) 1 (e) 1 3

What is the first thing that (e) will do? Committing actions Log index 1 2 3 (a) 1 2 4 What is the first thing that (e) will do? (b) 1 2 (c) 1 2 Forward entry at index 2 to followers (d) 1 (e) 1 3

What will happen to other logs’ entries at index 2? Committing actions Log index 1 2 3 (a) 1 2 4 What will happen to other logs’ entries at index 2? (b) 1 2 3 (c) 1 3 2 They will be replaced by the leader’s entry from term 3 (d) 1 3 (e) 1 3

What will happen to (a)’s entry at index 3? Committing actions Log index 1 2 3 (a) 1 3 2 4 What will happen to (a)’s entry at index 3? (b) 1 2 3 (c) 1 3 2 It will be deleted (d) 1 3 (e) 1 3

Committing actions So, are actions stored on a majority really committed? No, only if they reach a majority in the term they’re created. Log index 1 2 3 (a) 1 3 2 (b) 1 2 3 (c) 1 2 3 (d) 1 3 (e) 1 3

Can (e) win the election? Committing actions Log index 1 2 (a) 1 2 Can (e) win the election? (b) 1 2 (c) 1 2 No, only (d) will vote for it (d) 1 (e) 1

Can (b) or (c) win the election? Committing actions Log index 1 2 (a) 1 2 Can (b) or (c) win the election? (b) 1 2 (c) 1 2 Yes, either can win. (d) 1 (e) 1

What will (b) send to (d) and (e) first? Committing actions Log index 1 2 3 (a) 1 2 What will (b) send to (d) and (e) first? (b) 1 2 3 STOPPED (c) 1 2 3 Entry at index 2 (d) 1 2 3 (e) 1 2 3

Committing actions (say 2 reached majority in term 4) 1 2 4 1 2 1 2 1 Log index 1 2 3 (a) 1 2 4 Can entry 2 ever commit? (b) 1 2 (c) 1 2 Yes, if a subsequent entry (e.g., index 3, term 4) commits, all prior entries are also committed (d) 1 (e) 1 3

Entry 3 is now committed, so entry 2 is as well Committing actions (say 2 reached majority in term 4) Log index 1 2 3 (a) 1 2 4 Entry 3 is now committed, so entry 2 is as well (b) 1 2 4 (c) 1 2 4 (d) 1 (e) 1 3

Intertwining elections and log repair Log index 1 2 3 4 5 6 7 8 9 10 11 12 (a) 1 1 1 4 4 5 5 6 6 6 (b) 1 1 1 4 4 4 4 (c) 1 1 1 2 2 2 3 3 3 7 Is it possible for servers’ logs to reach this state? Why or why not?

Intertwining elections and log repair Log index 1 2 3 4 5 6 7 8 9 10 11 12 (a) 1 1 1 5 5 6 6 6 6 (b) 1 1 1 5 5 5 5 (c) 1 1 1 2 4 Is it possible for servers’ logs to reach this state? Why or why not?

Intertwining elections and log repair Log index 1 2 3 4 5 6 7 8 9 10 11 12 (a) 1 1 1 2 6 (b) 1 1 1 2 4 (c) 1 1 1 2 5 Is it possible for servers’ logs to reach this state? Why or why not?

Key elements of consensus Leader election Who is in charge? Log replication What are the actions and what is their order? Safety What is true for all states, in all executions? e.g., either we haven’t agreed or we all agree on the same value

Review: Log Matching Property Two entries in different logs with same index and term  The entries store the same action The logs are identical in all preceding entries Critical for Safety proof that follows

Log Matching Property Two entries in different logs with same index and term  The entries store the same action The logs are identical in all preceding entries Why is the first part always true? One leader/term If leader fails, term changes Leader inserts a new entry once at a given index  every action is assigned a unique term and index NOTE: this is not enough to ensure the second part

Log Matching Property Two entries in different logs with same index and term  The entries store the same action The logs are identical in all preceding entries Ensuring the second part requires extra check by follower Leader sends followers append(term, last_index, action) Follower must check that last entry has same term and last_index If not, the follower refuses the new entry Otherwise, follower may append new entry containing action

Log Matching Property Two entries in different logs with same index and term  The entries store the same action The logs are identical in all preceding entries Do logs always agree on indexes, terms? No, indexes at logs may have different actions and terms Goal of repairing logs is to increase entries w/ same index, term 1 1 1 4 4 5 5 6 6 6 7 7 1 1 1 4 4 4 4 1 1 1 2 2 2 3 3 3 3 3

Safety Leader Completeness Property Proof setup If a log entry is committed in a given term entry will be in leaders’ logs in all future terms Proof setup Majorities required to commit and elect What must be true of any two majorities? They must overlap This will be the linchpin of the proof Question: is a node with committed entries to be elected?

Proof by contradiction Assume that node missing a committed entry e is elected Entry e was committed in term T Leader missing committed entry e is elected in term U U is earliest term after T whose leader is missing e

Proof by contradiction Assume that node missing a committed entry e is elected Entry e was committed in term T Leader missing committed entry e is elected in term U U is earliest term after T whose leader is missing e Majorities are needed to commit and elect At least one node accepted e in term T and voted for leader in U Call this node “the voter”

Proof by contradiction Assume that node missing a committed entry e is elected Entry e was committed in term T Leader missing committed entry e is elected in term U U is earliest term after T whose leader is missing e Majorities are needed to commit and elect At least one node accepted e in term T and voted for leader in U Call this node “the voter” Possible that the voter obtained entry e after voting in U? If so, the voter would have rejected e from the leader in term T And we know that the voter accepted e in term T So, we know that the voter obtained e before voting in term U

Proof by contradiction Assume that node missing a committed entry e is elected Entry e was committed in term T Leader missing committed entry e is elected in term U U is earliest term after T whose leader is missing e Majorities are needed to commit and elect At least one node accepted e in term T and voted for leader in U Call this node “the voter” Voter obtained e before voting for leader in U We know that the voter voted for leader in U  Leader in U’s log must have been as up to date as the voter’s log

Proof by contradiction Assume that node missing a committed entry e is elected Entry e was committed in term T Leader missing committed entry e is elected in term U U is earliest term after T whose leader is missing e Majorities are needed to commit and elect At least one node accepted e in term T and voted for leader in U Call this node “the voter” Voter obtained e before voting for leader in U Leader in U’s log must have been as up to date as the voter’s log This will give us our contradictions

Proof by contradiction Leader in U’s log must have been as up to date as the voter’s log If leader’s log was the same as the voter’s, then what must be true? Leader’s log must have contained committed entry e This contradicts our assumption, so leader’s log must ≠ voter’s

Proof by contradiction Leader in U’s log must have been as up to date as the voter’s log If leader’s log was the same as the voter’s, then what must be true? Leader’s log must have contained committed entry e This contradicts our assumption, so leader’s log must ≠ voter’s What if leader’s log has same last term as voter’s? Then leader’s log must have been longer than the voter’s If longer with same last term, then there must be an entry from last term in common Log Matching Property: entries with same index and term  all preceding are same Thus, the leader’s log must contain entry e This contradicts our assumption, so leader’s last log term > voter’s last log term

Proof by contradiction Leader in U’s log must have been as up to date as the voter’s log If leader’s log was the same as the voter’s, then what must be true? Leader’s log must have contained committed entry e This contradicts our assumption, so leader’s log must ≠ voter’s What if leader’s log has same last term as voter’s? Then leader’s log must have been longer than the voter’s If longer with same last term, then there must be an entry from last term in common Log Matching Property: entries with same index and term  all preceding are same Thus, the leader’s log must contain entry e This contradicts our assumption, so leader’s last log term > voter’s last log term What if leader’s last log term is later than voter’s last log term?

Log index 1 2 3 4 5 6 7 8 9 10 11 12 L Leader(U) Voter T Leader (L) Last term in Leader(U)’s log is L, and L > the last term in Voter’s log What does this tell us about the relationship of L and T? L > T, since Voter has entry e from T Entry e: term T at index 6 in Voter’s log

Log index 1 2 3 4 5 6 7 8 9 10 11 12 L Leader(U) Voter T Leader (L) T Last term in Leader(U)’s log is L, and L > the last term in Voter’s log Do we know whether the leader for L, Leader(L), has e? Yes, it must, since we assumed that Leader(U) was first w/o e Entry e: term T at index 6 in Voter’s log

Log index 1 2 3 4 5 6 7 8 9 10 11 12 L Leader(U) Voter T Leader (L) T L Last term in Leader(U)’s log is L, and L > the last term in Voter’s log Do we know whether Leader(L) has the last entry in Leader(U)’s log? Yes, it must, since updates only flow from leaders to followers Entry e: term T at index 6 in Voter’s and Leader(L)’s log

Log index 1 2 3 4 5 6 7 8 9 10 11 12 T L Leader(U) Voter T Leader (L) T L Last term in Leader(U)’s log is L, and L > the last term in Voter’s log Finally, apply the Log Matching property to infer that Leader(U) has e. Since Leader(L) and Leader(U) both have the last entry in Leader(U)’s log at the same index and term, all preceding entries are identical. And since Leader(L) has e and (T < L), Leader(U) must have e. Entry e: term T at index 6 in Voter’s and Leader(L)’s log

Log index 1 2 3 4 5 6 7 8 9 10 11 12 T L Leader(U) Voter T Leader (L) T L Last term in Leader(U)’s log is L, and L > the last term in Voter’s log Finally, apply the Log Matching property to infer that Leader(U) has e. Since Leader(L) and Leader(U) both have the last entry in Leader(U)’s log at the same index and term, all preceding entries are identical. And since Leader(L) has e and (T < L), Leader(U) must have e. ✓ Entry e: term T at index 6 in Voter’s, Leader(L)’s, and Leader(U)’s log

Proof by contradiction Leader in U’s log must have been as up to date as the voter’s log If leader’s log was the same as the voter’s, then what must be true? Leader’s log must have contained committed entry e This contradicts our assumption, so leader’s log must ≠ voter’s What if leader’s log has same last term as voter’s? Then leader’s log must have been longer than the voter’s If longer with same last term, then there must be an entry from last term in common Log Matching Property: entries with same index and term  all preceding are same Thus, the leader’s log must contain entry e This contradicts our assumption, so leader’s last log term > voter’s last log term What if leader’s last log term is later than voter’s last log term? Last term in leader’s log is L, and L > T since voter’s log contains entry e Leader in L must have had e, since leader in U is first to not have it By Log Matching Property, leader in U’s log must also contain e

Proof by contradiction Assume that node missing a committed entry e is elected Entry e was committed in term T Leader missing committed entry e is elected in term U U is earliest term after T whose leader is missing e Majorities are needed to commit and elect At least one node accepted e in term T and voted for leader in U Call this node “the voter” Voter obtained e before voting for leader in U Leader in U’s log must have been as up to date as the voter’s log Nodes missing committed entries cannot be elected Committed entries are always preserved across terms/elections

One final issue Cluster membership changes Membership is part of the cluster’s configuration What is the easy way to handle this? Take the cluster down Change configuration on all machines Restart the cluster Why isn’t this ideal? Would rather not take the availability hit Manually editing configuration can be dangerous

Should never have two leaders Problem: cannot guarantee when change happens What makes this period dangerous? Disjoint majorities (a) old new (b) old new (c) old new (d) new (e) new Time

Should never have two leaders Problem: cannot guarantee when change happens How can you have disjoint majorities? Old (a,b) New (c,d,e) (a) old new (b) old new (c) old new (d) new (e) new Time

Raft’s approach Change configuration in two phases First, “joint consensus” Second, transition to new configuration Joint consensus rules Log entries replicated to all servers in both configs Any server, from either config, may serve as leader Agreement requires majorities from both configs

Configuration change Leaders initiate configuration change from old to new They send out a special log entry, Cold,new, describing both As soon as a server logs Cold,new, it plays by joint consensus rules For Cold,new to commit, it must reach a majority of Cold and Cnew If the leader fails, the new leader could be in Cold or Cold,new Important thing is that membership of Cnew cannot act unilaterally In other words, majority of Cold still required for commit, elections Eventually, Cold,new will commit (we may have to try several times) If Cold,new commits, what must be true of future leaders? By Leader Completeness, they must be running Cold,new config

Configuration change Once, Cold,new commits, we’re in good shape Leader can now try to commit Cnew Leader logs and uses Cnew immediately i.e., only requires a majority of Cnew to commit entries Nasty corner case: what if the leader isn’t in Cnew? (i.e., managing a cluster of which they are not a member) It’s fine! Majorities don’t have to include the leader The leader continues to count votes and push updates When and how should the leader exit? Wait until Cnew commits, then shut down Cnew will elect a new leader

Midterm exam Logistics Topics Format Wednesday, February 22nd during lecture In-class, closed-note Ali will proctor Topics Concurrency and virtual memory Single-host: THE, Multics, and UNIX Distributed: RPC/MAUI, Grapevine, VAXClusters, and Raft Anything from reading/lecture is fair game Format Multi-part, short answer