Dell EMC Exam D-VXR-OE-23 Topic 1 Question 14 Discussion

Actual exam question for Dell EMC's D-VXR-OE-23 exam

Question #: 14
Topic #: 1

An environment has two VxRail Clusters with the following configuration

* Cluster-A is using FTT = 2 and FTM = Erasure Coding

* Cluster-B is using FTT = 2 and FTM = Mirroring

A 100 GB VM is planned to be vMotioned from Cluster-A to Cluster-B. How much more vSAN capacity is consumed in Cluster-B than in Cluster-A?

A100 GB

B167 GB

C150 GB

D0 GB

Show Suggested Answer

Suggested Answer: B

To calculate the additional vSAN capacity consumed when a 100 GB VM is vMotioned from Cluster-A to Cluster-B, consider the following:

Cluster-A: FTT = 2 and FTM = Erasure Coding (RAID 6), which uses approximately 1.5 times the data size, thus 100 GB VM uses about 150 GB of vSAN capacity.

Cluster-B: FTT = 2 and FTM = Mirroring (RAID 1), which uses 3 times the data size, thus 100 GB VM uses 300 GB of vSAN capacity.

The difference in vSAN capacity consumed between Cluster-B and Cluster-A is: 300 GB (Cluster-B) - 150 GB (Cluster-A) = 150 GB

So, Cluster-B consumes 150 GB more vSAN capacity than Cluster-A.

VMware vSAN Storage Policies: vSAN Storage Policies

by Audrie at Sep 14, 2024, 09:58 PM

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Macy

7 months ago

Hmm, I wonder if the answer is hidden in a secret compartment, like a treasure chest. Or maybe it's written in invisible ink on the back of the question. Either way, I'm sure the answer is here somewhere!

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Isadora

7 months ago

This is a no-brainer! If Cluster-A is using erasure coding and Cluster-B is using mirroring, then Cluster-B is going to consume more capacity. The answer is clearly B) 167 GB.

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Anglea

7 months ago

Exactly, that's why Cluster-B will consume 167 GB more.

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Stephaine

7 months ago

Cluster-A with erasure coding is more efficient in terms of capacity.

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Chau

7 months ago

So, the answer is B) 167 GB.

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Mollie

7 months ago

Cluster-B will consume more capacity because it is using mirroring.

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Wynell

7 months ago

I've got it! Cluster-B is using mirroring, so it needs to store two copies of the 100 GB VM, which is 200 GB. Cluster-A only needs to store the original data plus the parity bits, which is less. The answer must be B) 167 GB.

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Rebeca

7 months ago

Okay, let's think this through. Cluster-A is using erasure coding, which means it only needs to store the original data plus the parity bits, so it should consume less capacity than Cluster-B's mirroring.

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Leonida

7 months ago

Hmm, this is a tricky one. Since Cluster-A uses erasure coding, the capacity consumed would be less than Cluster-B's mirroring, but I'm not sure of the exact numbers.

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Lovetta

8 months ago

Haha, this question is like a game of 'Guess the Storage Overhead'! I'll go with C) 150 GB, just to keep things interesting.

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Stevie

6 months ago

The correct answer is D) 0 GB.

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Annette

7 months ago

I believe it's C) 150 GB.

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Verona

7 months ago

I'm going with A) 100 GB.

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Iraida

7 months ago

I think the answer is B) 167 GB.

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Alica

8 months ago

I'm not so sure about this one. Isn't Mirroring supposed to be more efficient than Erasure Coding when it comes to storage consumption? I'm going with D) 0 GB.

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Ernest

8 months ago

I agree with Loren. Mirroring always consumes twice the storage compared to Erasure Coding, so the answer has to be B) 167 GB.

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Thurman

7 months ago

So, the vSAN capacity consumed in Cluster-B is 167 GB more than in Cluster-A.

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Leontine

7 months ago

Yes, you are correct. Mirroring consumes more storage than Erasure Coding.

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Pansy

7 months ago

I think the answer is B) 167 GB

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Page

8 months ago

Why do you think it's 150 GB?

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Whitney

9 months ago

I disagree, I believe the answer is C) 150 GB.

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Loren

9 months ago

Hmm, I think the answer is B) 167 GB. If Cluster-A is using Erasure Coding, it will only consume 100 GB, but if Cluster-B is using Mirroring, it will consume 200 GB, so the difference is 167 GB.

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