# Why K8S HA chooses 3 instead of 5..6..7 as the size of masters?

**Theory**: In order to take any action, there has to be n/2 + 1 members reach the consensus. 

**Algorithm**: RAFT

| Cluster size | Majority | Fault Tolerance |
| ------------ | -------- | --------------- |
| 1            | 1        | 0               |
| 2            | 2        | 0               |
| **3**        | **2**    | **1**           |
| **4**        | **3**    | **1**           |
| 5            | 3        | 2               |
| 6            | 4        | 2               |
| 7            | 4        | 3               |
| 8            | 5        | 3               |
| 9            | 5        | 4               |

{% hint style="info" %}
cluster\_size / 2 + 1 = majority

fault\_tolerance  = cluster\_size - majority  - In order to let the system work properly on failures , the number of majority has to be reached. So the number of fault\_tolerance is the cluster size minus the required number of majority.
{% endhint %}

**Why choosing 3:**

If all other factors are the same, the larger cluster size, the more expensive. Choosing 4 does not increase the number of fault tolerance. However, if choosing 9, one needs to consider the cost of the cluster.

**Sometime people choose 5 as well in production:**

You have 2 fault tolerances. That means you could have one fails, one is down for maintainance.