HOTSPOTYou have four Hyper-V hosts named Server1, Server2, Server3 and Server4 that run Windows Server 2016.
The hosts are nodes in a failover cluster.
The failover cluster is configured to balance virtual machines when a node is more than 80 percent loaded.
You need to configure the failover cluster to balance the virtual machines when the load of a node is more than
five percent higher than the average for the nodes.
Which command should you run? To answer, select the appropriate options in the answer area.
Hot Area:
Answer: 
Explanation:
References: https://www.vembu.com/blog/hyper-v-virtual-machine-load-balancing/
Answer is OK. But cited site is higgledy-piggledy, if not simply wrong.
There is nice, correct description: https://blogs.msdn.microsoft.com/clustering/2016/04/29/failover-cluster-node-fairness-in-windows-server-2016/
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Answer seems correct
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Pic: http://freecram.com/uploads/70-740/2b20488fa2c47d5e992c117e89d93a7e.jpg
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The command Get-Cluster AutobalanceLevel = 3 is according to the referenced article the most aggressive level which behavior is “When the host is more than 60% loaded it is moved”.
The question states that “You need to configure the failover cluster to balance the virtual machines when the load of a node is more than five percent higher than the average for the nodes.” However no average load number for the nodes is given.
Can anybody add some knowledge to this question? For now I’m assuming that “Get-Cluster AutobalanceLevel = 3” is the correct answer since the other options don’t seem to match.
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Get-cluster autobalancelevel=3
Average nodes and move when host is more than 5% above average
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(Get-Cluster).AutoBalancerLevel =
AutoBalancerLevel Aggressiveness Behavior
1 (default) Low Move when host is more than 80% loaded
2 Medium Move when host is more than 70% loaded
3 High Average nodes and move when host is more than 5% above average
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I don’t see that in the book!?
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That’s right. You also need experience. 😛
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Confirmed what Chris said, on MSDN. Answer is correct.
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