Huawei H35-211_V2.5 Exam - Topic 2 Question 51 Discussion

Actual exam question for Huawei's H35-211_V2.5 exam

Question #: 51
Topic #: 2

(Single) A total of 100 multicast users on OLT are watching 20 multicast programs, assuming that each multicast program occupies 2M bandwidth, then how much multicast traffic is currently on the OLT upstream port

A200M

B100M

C40M

D50M

Show Suggested Answer

Suggested Answer: B

by Olive at Jul 14, 2024, 12:29 AM

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Billye

3 months ago

Wow, I didn't realize multicast could add up like that!

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Margery

3 months ago

I'm not so sure about that, seems high for 100 users.

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Freeman

3 months ago

Wait, isn't it 100M? Only counting unique streams?

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Jennie

4 months ago

Totally agree, 200M makes sense!

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Afton

4 months ago

It's 200M, right? 20 programs x 2M each.

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Albert

4 months ago

I feel like I might have miscalculated in practice. If each program is 2M and there are 20 programs, I thought it would be 200M, but that seems too high for just the upstream.

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Luis

4 months ago

I'm a bit confused because I thought the number of users would impact the total traffic, but maybe it only matters for the downstream?

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Margart

4 months ago

I remember a similar question where we calculated total bandwidth based on active streams. If each program is 2M and there are 20 programs, that sounds like 40M total.

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Monroe

5 months ago

I think the total multicast traffic should be the number of programs multiplied by the bandwidth per program, but I'm not sure how the number of users affects that.

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Millie

5 months ago

The key is to multiply the number of users by the number of programs and the bandwidth per program. That should give us the total multicast traffic. I'm pretty confident in this approach.

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Flo

5 months ago

Wait, I'm a bit confused. How do we calculate the total multicast traffic on the upstream port? I want to make sure I understand this properly.

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Marguerita

5 months ago

Hmm, let me think this through step-by-step. I want to make sure I don't miss anything.

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Elouise

5 months ago

This looks like a straightforward calculation. I think I can handle this one.

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Carey

5 months ago

Okay, so we have 100 multicast users and 20 multicast programs, each using 2M of bandwidth. I think the answer is A, 200M.

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Maurine

5 months ago

Okay, I think I've got this. The key is to look at the total capacity increase needed - 60 cores and 640 GB of memory. Option B seems like the best fit, adding three c4.metal hosts to the cluster.

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Maryanne

5 months ago

Hmm, I'm not entirely sure about this one. I know Multi-Org Access Control is related to operating units, but I'm not confident which profile option is required to enable it. I'll have to think this through carefully.

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Shawnda

10 months ago

Hmm, let me think this through. If each program is 2M and there are 20 programs, that's 40M. But with 100 users, it should be 100M, so I'll go with B. Definitely not A, that would be like a multicast party on the OLT!

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Samira

9 months ago

User 3: Yeah, 100M seems to be the correct answer

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Brett

9 months ago

User 2: I agree, 100M sounds right

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Georgene

10 months ago

User 1: I think it's 100M

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Gerardo

10 months ago

Okay, I got this. If each program takes 2M, and there are 20 programs, that's 40M. But wait, there are 100 users, so it must be 100M. I'll choose B.

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Annamae

9 months ago

User 4: Yeah, that makes sense. Option B) 100M it is.

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Kristeen

9 months ago

User 3: I agree, I'll go with option B) 100M.

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Odette

9 months ago

User 2: But there are 100 users, so it must be 100M.

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Wilda

10 months ago

User 1: I think it's 40M because each program takes 2M.

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Annabelle

10 months ago

Woah, this one's tricky! I'm thinking the total multicast traffic would be the 20 programs multiplied by the 2M each, which is 40M. But then you have to factor in the 100 users, so I'm going with C.

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Sommer

10 months ago

User 2: Yeah, but don't forget about the 100 users. So it's actually 200M.

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Cassie

10 months ago

User 1: I think it's 40M because each program is 2M and there are 20 programs.

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Krissy

10 months ago

Hmm, let's see. If each multicast program takes up 2M bandwidth, and there are 20 programs, that's 40M. But with 100 users, wouldn't that make it 100M? I'll go with B.

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