Dell EMC Exam D-ISM-FN-23 Topic 2 Question 11 Discussion

Actual exam question for Dell EMC's D-ISM-FN-23 exam

Question #: 11
Topic #: 2

If the erasure coding technique used divides data into 12 data segments and 5 coding segments, what is the maximum number of disk drive failures against which data is protected by the configuration?

D17

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Suggested Answer: A

by Ellsworth at Jul 15, 2024, 05:42 PM

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Alexis

3 months ago

I remember learning that the answer is actually C) 3.

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Jonelle

3 months ago

Wait, if I lose 17 disk drives, I think my data is the least of my problems. I'll just go buy a new server and start over.

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Ryan

3 months ago

17 disk drive failures? What is this, a Raid setup for the Death Star? I'm going with 5, keep it simple.

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Leonora

3 months ago

Psh, you both are way off. It's clearly 3. Why else would they give us 17 as an option? Gotta be a trick question.

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Amira

3 months ago

I'm not sure, but I think the correct answer is A) 5.

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Dong

4 months ago

I disagree, I believe the answer is D) 17.

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Starr

4 months ago

Hold on, I think the answer is 7. Isn't that how erasure coding works? You can lose up to half the segments and still recover the data.

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Leigha

2 months ago

Exactly, erasure coding is designed to provide fault tolerance by allowing for multiple disk failures without data loss.

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Kenny

2 months ago

So the maximum number of disk drive failures against which data is protected in this configuration is 7.

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Ashton

2 months ago

Yes, with 12 data segments and 5 coding segments, losing up to 7 segments would still allow data recovery.

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Kimberely

2 months ago

I think you're right, the answer is 7. That's how erasure coding provides data protection.

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Trinidad

4 months ago

With 12 data segments and 5 coding segments, the maximum number of disk drive failures it can protect against should be 5. After all, that's the number of coding segments, right? Simple math.

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