Minimum number of servers required to run the Zookeeper is called Quorum.
Zookeeper replicates whole data tree to all the quorum servers. This number is also the minimum number of servers required to store a client’s data before telling the client it is safely stored.
Quorum size should be calculated by Majority Rule

Majority rule: QN = (N + 1) / 2

QN: Minimum number of servers in quorum
N: Total number of servers. (should be an odd number)
So, if we have 5 servers, then Quorum should be minimum of 3 servers.

Now, we will see why majority rule is good for eventual consistency:

Points to remember about Zookeeper before moving further:
1. The Zookeeper client has all the Zookeeper server IP list. If Zookeeper client disconnects from any server, then Zookeeper client tries to connect with another IP(Zookeeper server).
2. All the write requests are processed via Leader. In the following example, S1 is the Leader.

Case 1 - Less Than Majority Rule (Less than 3 servers):

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In this case, Qn = 2 (we are not following the Majority rule). If 2 servers are available in Quorum, then Zookeeper will process the request in the following way:
1) Initially, we have 5 Zookeeper servers in quorum Q (as shown in above image a). A client wants to create a node N on the Zookeeper server. So, the client will get connected to any of the servers. The server will transfer the request to the Leader of the quorum Q(S1).
2) At the same time, due to network partition, quorum Q is left with only 2 servers(S1 and S2).
So, only S1 and S2 servers will have the data (Node N) as server S3, S4 and S5 got disconnected. (as shown in above image b).
3)
After some time, the client makes a request to get data from the same node N but server S1, S2 got disconnected from the client due to network partition(as shown above image c).
So now the client got connected with any of the servers S3, S4, S5. As we know, none of these servers have the data (Node N).
This will create the Split Brain problem.

Case 2 - With Majority Rule (Minimum 3 servers):

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In this case Qn = 3. If 3 servers are available in Quorum, then Zookeeper will process the request in the following way:
1. The client connects to the S1(Leader) server to create a node.
2. To process the client request or to create a quorum, a minimum of 3 servers are required.
So say as, at the time client requests, Server S1, S2 got disconnected from servers S3, S4, S5. Hence the request will not be processed and the client has to handle this disconnection error.
3. Let’s suppose, S1, S2, and S3 connected and created a quorum Q. Server S4 and S5 are still disconnected.
Hence, node request will be processed only on three servers S1, S2, S3.
4. Later on, the client requests to get data for the same node N from any of the servers S1, S2, S3.
Now suppose, the client got disconnected from servers(S1, S2) and after a while, S3, S4, and S5 make a quorum after they got connected.
In this case, S3 is the leader as it has the requested data but before processing the request, S3 will be syncing with S4 and S5 to have the same data in both the nodes.
After sync only, client request will be processed.

Case 3 - More than Majority rule(More than 3 servers):
In this case Qn = 4, If 4 servers are available in Quorum, then Zookeeper will process the request. But quorum servers more than Majority rule has one disadvantage. Before return success result to the client, we have to write on one extra server. This will increase latency.
If the total number of servers (N) is even, Then the above case also happens. Let's suppose N = 6. According to Majority rule Qn = 4. This extra server will increase the latency to perform write operation.

Majority Wins Rule Proof:

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Majority wins

As we know, we required 3 servers for successful write,
W: Write successfully.
N: Write failed, due to a network partition.
When we say we write on the majority of servers and Quorum create with the majority of servers then one of a server(that have the latest data) intersect with both quorum. you can see in the above case1 in majority wins rule proof. S3 intersect with both left (S1, S2)and right quorum (S4, S5).

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