Raft consensus Landon Cox March 4 2016 Paxos

Raft consensus Landon Cox March 4, 2016

Paxos • ACM TOCS: • 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

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

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

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

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? Client

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

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

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 Leader=Llogs. Leader=L Apply action a. Leader=L … … Client 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 C F L

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

Server state: leader • Server handles all client requests C 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 C F L

Time divided into terms Term 1 Electio n Leader unknown Term 2 Normal operation Leader known Electio n T 3 Term 4 Normal operation 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 S 2 C S 1 F C F L Current term = 0 S 3 L C Current term = 0 All servers start as followers. All servers have local timers. Note: no bootstrapping problem! F L Current term = 0

Server state: follower S 2 C S 1 F C F L Current term = 0 S 3 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. F L Current term = 0

Server state: follower S 2 C S 1 F C F L Current term = 0 S 3 L C Current term = 0 What should server assume if no heartbeat? Assume no viable leader, start election. F L Current term = 0

Server state: follower S 2 C S 1 F C F L Current term = 0 S 3 L C Current term = 0 Who should the server nominate? How about itself? At least it knows that it’s running. F L Current term = 0

Server state: candidate S 2 C S 1 F C F L Current term = 0 S 3 L C Current term = 1 To start an election: Increment current term and set state to candidate F L Current term = 0

Server state: candidate S 2 C S 1 F C F L Current term = 0 S 3 L C Current term = 1 Need to collect votes. For whom should the server vote? Itself, of course! Major qualification: It’s running. F L Current term = 0

Server state: candidate S 2 C S 1 F L t Vo C Votes S 1=S 1 S 2=? S 3=? nt i e erm L Vo te Current term = 0 in ter S 3 m 1 C Current term = 1 How should S 2, S 3 respond to vote request? F L Current term = 0

Server state: candidate S 2 C S 1 S C Votes S 1=S 1 S 2=? S 3=? F L n 1 i S 1 in erm 1 t F L Current term = 1 ter m S 3 1 C Current term = 1 How should S 2, S 3 respond to vote request? Increment term, vote for S 1 … why vote for S 1? Our goal is consensus, and we know the collector voted for itself. F L Current term = 1

Server state: candidate S 2 C S 1 F C Votes S 1=S 1 S 2=S 1 S 3=S 1 F L Current term = 1 S 3 L C Current term = 1 What should S 1 do next? Count votes (majority wins) Make itself the leader, start sending heartbeats. F L Current term = 1

Server state: candidate S 2 C S 1 er s i 1 S C Leade r = S 1 F L S 1 is d lea F L Leade r = S 1 Current term = 1 S 3 lea de r C Current term = 1 F L Current term = 1 Leade r = S 1

Server state: candidate S 2 C S 1 F C Leade r = S 1 F L Leade r = S 1 Current term = 1 S 3 L C Current term = 1 How many faults can we tolerate? One. Need two/three to vote for same new leader F L Current term = 1 Leade r = S 1

Server state: candidate S 2 C S 1 F L t Vo C Votes S 1=S 1 S 2=? S 3=? nt i e erm L Vo te Current term = 1 in ter S 3 m 1 C Current term = 1 Who votes for whom if S 1 and S 3 both call elections? F L Current term = 1 Votes S 1=? S 2=? S 3=S 3

Server state: candidate S 2 C S 1 F L t Vo C Votes S 1=S 1 S 2=S 3 S 3=S 3 nt i e erm L Vo te Current term = 1 in ter S 3 m 1 C Current term = 1 Who votes for whom if S 1 and S 3 both call elections? S 2 votes for the server that asked first F L Current term = 1 Votes S 1=S 1 S 2=S 3 S 3=S 3

Server state: candidate S 2 C S 1 F C Votes S 1=S 1 S 2=S 3 S 3=S 3 F L Current term = 1 S 3 L C Current term = 1 What does S 1 do if it loses the election? F L Current term = 1 Votes S 1=S 1 S 2=S 3 S 3=S 3

Server state: candidate S 2 C S 1 F C Votes S 1=S 1 S 2=S 3 S 3=S 3 F L Current term = 1 S 3 L C Current term = 1 What does S 1 do if it loses the election? Moves back to a follower state. F L Current term = 1 Votes S 1=S 1 S 2=S 3 S 3=S 3

Server state: candidate S 2 C S 1 L ea d er F F L Current term = 1 S 3 Leade r = S 3 is l C S 3 is Leade r = S 3 Current term = 1 S 3 C lea de r F L Current term = 1 Leade r = S 3

Server state: follower S 2 C S 1 F C F L Current term = 0 S 3 L C Current term = 0 Can all three servers nominate themselves? F L Current term = 0

Server state: candidate S 2 C S 1 F C Votes S 1=S 1 S 2=S 2 S 3=S 3 F L Current term = 1 S 3 L C Current term = 1 Can all three servers nominate themselves? Sure! Votes S 1=S 1 S 2=S 2 S 3=S 3 F L Current term = 1 Votes S 1=S 1 S 2=S 2 S 3=S 3

Server state: candidate S 2 C S 1 F C Votes S 1=S 1 S 2=S 2 S 3=S 3 F L Current term = 1 S 3 L C Current term = 1 Could this happen indefinitely? Votes S 1=S 1 S 2=S 2 S 3=S 3 F L Current term = 1 Votes S 1=S 1 S 2=S 2 S 3=S 3

Server state: candidate S 2 C S 1 F C Votes S 1=S 1 S 2=S 2 S 3=S 3 F L Current term = 1 Votes S 1=S 1 S 2=S 2 S 3=S 3 C F L Could this happen indefinitely? Yes, there is not way to prevent this from happening. The worst possible thing is still possible. Current term = 1 Votes S 1=S 1 S 2=S 2 S 3=S 3

Server state: candidate S 2 C S 1 F C Votes S 1=S 1 S 2=S 2 S 3=S 3 F L Current term = 1 S 3 L C Current term = 1 How do we make this less likely to occur? Votes S 1=S 1 S 2=S 2 S 3=S 3 F L Current term = 1 Votes S 1=S 1 S 2=S 2 S 3=S 3

Server state: candidate S 2 C S 1 F C Votes S 1=S 1 S 2=S 2 S 3=S 3 F L Current term = 1 S 3 L C Current term = 1 How do we make this less likely to occur? Randomize election timeouts i. e. , wait a random period before new election Votes S 1=S 1 S 2=S 2 S 3=S 3 F L Current term = 1 Votes S 1=S 1 S 2=S 2 S 3=S 3

Server state: candidate S 2 C S 1 F C F L Current term = 1 S 3 L C Current term = 1 Can we ever have two leaders? F L Current term = 1

Server state: candidate S 2 C S 1 F C F L Current term = 1 S 3 L C Current term = 1 F Can we ever have two leaders? No, votes either split or converge w/ three nodes. L Current term = 1

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. 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

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

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

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 y 5 Client Leader=L 3/3 … commit ! 1 x 3 1 y 1 1 y 5 Leader=L 1 x 3 1 y 1 1 y 5

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

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

What happens if both followers fail: (1) after adding entry to logs, and (2) before acking? 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

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

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 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 Log index 1 2 3 4 5 6 7 8 1 x 3 1 y 1 1 y 9 2 x 2 3 3 x 0 y 7 3 x 5 3 x 4 1 x 3 1 y 1 1 y 9 2 x 2 3 x 0 1 x 3 1 y 1 1 y 9 2 3 3 x 2 x 0 y 7 3 x 5 3 x 4 1 x 3 1 y 1 1 y 9 2 x 2 3 x 5 3 3 x 0 y 7 Leader Followers

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

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! Log index 1 2 3 4 5 6 7 8 1 x 3 1 y 1 1 y 9 2 x 2 3 3 x 0 y 7 3 x 5 3 x 4 1 x 3 1 y 1 1 y 9 2 x 2 3 x 0 1 x 3 1 y 1 1 y 9 2 3 3 x 2 x 0 y 7 3 x 5 3 x 4 1 x 3 1 y 1 1 y 9 2 x 2 3 x 5 3 3 x 0 y 7 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. 1 2 3 4 5 6 7 8 9 10 1 1 1 4 4 5 5 6 6 6 (a ) (b ) 1 1 1 4 4 5 5 6 6 1 1 1 4 (c ) 1 1 1 4 4 5 5 6 6 6 (d ) (e ) 1 1 1 4 4 5 5 6 6 6 7 1 1 1 4 4 (f) 1 1 1 2 2 2 3 3 Log index 11 12 Leader 6 3 Followers 7

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

Why aren’t there any entries for term 8? Because the leader hasn’t received any requests yet. 1 2 3 4 5 6 7 8 9 10 1 1 1 4 4 5 5 6 6 6 (a ) (b ) 1 1 1 4 4 5 5 6 6 1 1 1 4 (c ) 1 1 1 4 4 5 5 6 6 6 (d ) (e ) 1 1 1 4 4 5 5 6 6 6 7 1 1 1 4 4 (f) 1 1 1 2 2 2 3 3 Log index 11 12 Leader 6 3 Followers 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 (a ) (b ) 1 2 3 4 5 6 7 8 9 10 1 1 1 4 4 5 5 6 6 6 1 1 1 4 4 5 5 6 6 1 1 1 4 What’s wrong with the logs of (a) and (b)? They are missing log entries. 11 12 Leader

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 (a ) (b ) 1 2 3 4 5 6 7 8 9 10 1 1 1 4 4 5 5 6 6 6 1 1 1 4 4 5 5 6 6 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 11 12 Leader

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 1 1 1 4 4 5 5 6 6 6 What’s wrong with the logs of (c) and (d)? 11 12 Leader They have extra log entries. (c ) 1 1 1 4 4 5 5 6 6 6 (d ) 1 1 1 4 4 5 5 6 6 6 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 1 1 1 4 4 5 5 6 6 6 11 12 Leader (c) was leader for term 6, added entry and crashed; (d) was leader for 7, added entries and crashed How might this have happened? (c ) 1 1 1 4 4 5 5 6 6 6 (d ) 1 1 1 4 4 5 5 6 6 6 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 1 1 1 4 4 5 5 6 6 6 What’s wrong with the logs of (e) and (f)? (e ) 1 1 1 4 4 (f) 1 1 1 2 2 2 3 11 12 Leader They have extra log entries and missing log entries. 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 1 1 1 4 4 5 5 6 6 6 1 1 1 4 4 (f) 1 1 1 2 2 2 3 12 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 ) 11 3 3

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

Servers have to keep track of committed log entry with highest index. What are those here? Leader=L 2 for all three servers. y 5 Client 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 reports its highest index of committed action when forwarding request to followers. Leader=L This is how followers update their state machines. y 5 Client Leader=L Term=1 y 5 High=2 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 y 5 Client Leader=L Term=1 y 5 High=2 Pred=2 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 y 5 Client 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 y 5 Client 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

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. y 5 Client 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 y 5 Client 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

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. y 5 Client 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 1 1 1 4 4 5 5 6 6 6 1 1 1 4 4 1 1 1 2 2 2 3 11 12 Leader 7 7 Followers 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 1 1 1 4 4 1 1 1 2 2 2 3 6 6 6 7 3 3 7

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

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

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

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

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

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

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

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

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

next_index[a]=11 next_index[b]=9 next_index[c]=11 1 2 3 4 5 6 7 8 9 10 11 1 4 4 5 5 6 6 6 8 (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 Log index 12 Leader 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.

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

next_index[a]=11 next_index[b]=9 next_index[c]=11 1 2 3 4 5 6 7 8 9 10 11 1 4 4 5 5 6 6 6 8 (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 Log index 12 Leader 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!

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

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? Votes from (c) and (d)

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

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

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

Committing actions Log index 1 2 3 (a) 1 2 4 (b) 1 2 (c) 1 2 (d) 1 (e) 1 Has the entry at index 2 been committed? 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) 3

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

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

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

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

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

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 (a) 1 3 2 (b) 1 2 3 (c) 1 2 3 (d) 1 3 (e) 1 3 3

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

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

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

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

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

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 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 1 1 1 4 4 1 1 1 2 2 2 3 6 6 6 7 3 3 7

Safety • Leader Completeness Property 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 L Voter 12 Leader(U) 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 Voter T Leader (L) T Leader(U) 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 L Voter T Leader (L) T 12 Leader(U) 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

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

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