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

(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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Shawnda

2 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

12 days ago

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

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Brett

29 days ago

User 2: I agree, 100M sounds right

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Georgene

1 months 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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Annamae

8 days ago

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

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Kristeen

11 days ago

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

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Odette

1 months ago

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

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Wilda

1 months 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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