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

Actual exam question for Huawei's H35-211_V2.5 exam
Question #: 51
Topic #: 2
[All H35-211_V2.5 Questions]

(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

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Suggested Answer: B

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

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Shawnda
1 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
4 days ago
User 3: Yeah, 100M seems to be the correct answer
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Brett
21 days ago
User 2: I agree, 100M sounds right
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Georgene
25 days ago
User 1: I think it's 100M
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Gerardo
2 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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Kristeen
3 days ago
User 3: I agree, I'll go with option B) 100M.
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Odette
22 days ago
User 2: But there are 100 users, so it must be 100M.
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Wilda
30 days ago
User 1: I think it's 40M because each program takes 2M.
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Annabelle
2 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
1 months ago
User 2: Yeah, but don't forget about the 100 users. So it's actually 200M.
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Cassie
1 months ago
User 1: I think it's 40M because each program is 2M and there are 20 programs.
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Krissy
2 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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Melodie
2 months ago
But each multicast program occupies 2M bandwidth, so with 20 programs, it should be 40M, which is option C).
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Royce
2 months ago
I disagree, I believe the answer is B) 100M.
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Melodie
2 months ago
I think the answer is A) 200M.
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