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Nokia 4A0-107 Exam - Topic 3 Question 62 Discussion

Assume all transmitted frames on a SONET port are 1500 bytes and that the average frame overhead has been configured as 10%. If the total bandwidth of all queues, excluding overhead, needs to be guaranteed at 55 Mbps, how much bandwidth should be configured for the egress port scheduler applied on that port?
D) 55 Mbps * 1.1 = 60.5 Mbps
A) 55 Mbps / 1.1 = 50 Mbps
B) 55 Mbps
C) 55 Mbps + 55 Mbps * 0.1 / 1.1 = 60 Mbps

Nokia 4A0-107 Exam - Topic 3 Question 62 Discussion

Actual exam question for Nokia's 4A0-107 exam
Question #: 62
Topic #: 3
[All 4A0-107 Questions]

Assume all transmitted frames on a SONET port are 1500 bytes and that the average frame overhead has been configured as 10%. If the total bandwidth of all queues, excluding overhead, needs to be guaranteed at 55 Mbps, how much bandwidth should be configured for the egress port scheduler applied on that port?

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Suggested Answer: D

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Dylan
6 months ago
Gotta go with option C, seems right to me!
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Cyril
6 months ago
Isn't it just 55 Mbps? Why add more?
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Roosevelt
7 months ago
Wait, how does 55 Mbps plus overhead equal 60 Mbps?
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Tequila
7 months ago
I think option A makes the most sense here.
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Deangelo
7 months ago
The overhead is 10%, so we need to account for that.
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Noe
7 months ago
I vaguely remember that we need to account for the 10% overhead, so it seems like we should be looking at something like 60 Mbps, but I'm not confident.
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Beckie
7 months ago
I feel like the answer should be higher than 55 Mbps because of the overhead, but I can't recall the exact formula we used in class.
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Ariel
8 months ago
I think we practiced a similar question where we had to account for overhead percentages. Maybe it's about dividing the guaranteed bandwidth by the effective rate?
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Jamal
8 months ago
I remember something about calculating bandwidth with overhead, but I'm not sure if I should just add the overhead directly or adjust it somehow.
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Karon
8 months ago
This seems straightforward. The question states the total bandwidth, excluding overhead, needs to be 55 Mbps. So we just need to configure the port scheduler for 55 Mbps.
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Mi
8 months ago
I think the answer is C. We need to add the overhead to the 55 Mbps, so 55 Mbps + (55 Mbps * 0.1) = 60 Mbps.
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Chandra
8 months ago
Okay, let's think this through step-by-step. We need to account for the 10% overhead, so we'll need to configure the port scheduler for a higher bandwidth than the 55 Mbps we need to guarantee.
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Dorathy
8 months ago
Hmm, I'm a bit confused. Do we need to add the overhead to the 55 Mbps, or divide the 55 Mbps by 1.1 to get the actual bandwidth we need to configure?
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Cruz
8 months ago
I'm pretty confident I know this one. The Oracle instance is a combination of the database files, parameter file, and background processes.
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Lizbeth
8 months ago
I think enterprise networking products mostly include WAN and LAN, right? They seem essential for connectivity.
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Miles
8 months ago
I've got it! The file will inherit the destination folder's permissions, so the correct answer is C.
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Deandrea
1 year ago
This question is like a bad joke. It's asking me to configure the bandwidth higher than the actual requirement. I bet the test writer is having a good laugh at our expense.
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Gerald
1 year ago
Yeah, it's all about making sure you understand the requirements and not falling for the trick answers.
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Marnie
1 year ago
I think the key is to remember to account for the overhead in the calculation.
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Tonette
1 year ago
I know, it's frustrating when the question seems tricky like that.
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Lorrine
1 year ago
Okay, let me think this through. If I factor in the 10% overhead, that's gotta be the answer. *scratches head* Wait, which one was it again?
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Gerardo
1 year ago
C) 55 Mbps + 55 Mbps * 0.1 / 1.1 = 60 Mbps
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Nobuko
1 year ago
Yes, that's correct. The answer is C) 55 Mbps + 55 Mbps * 0.1 / 1.1 = 60 Mbps
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Gilberto
1 year ago
I think it's C) 55 Mbps + 55 Mbps * 0.1 / 1.1 = 60 Mbps
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Ciara
1 year ago
C) 55 Mbps + 55 Mbps * 0.1 / 1.1 = 60 Mbps
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Glory
1 year ago
B) 55 Mbps
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Meghann
1 year ago
A) 55 Mbps / 1.1 = 50 Mbps
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Noah
1 year ago
A) 55 Mbps / 1.1 = 50 Mbps
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Daren
1 year ago
Haha, classic exam trickery. They're trying to make us overthink this, but I'm not falling for it. I know the answer is C, 60 Mbps. Bring it on, certification gods!
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Matt
1 year ago
Wait, what? Did they really just throw in a 10% overhead factor? That's like trying to fit 55 Mbps into a 50 Mbps pipe. Good luck with that!
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Tayna
1 year ago
But if we consider the frame overhead, the answer should be C) 60 Mbps.
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Tawanna
1 year ago
Ah, the classic SONET port bandwidth conundrum. I bet the correct answer is hiding right under our noses.
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Cheryl
1 year ago
D) 55 Mbps * 1.1 = 60.5 Mbps
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Hyman
1 year ago
C) 55 Mbps + 55 Mbps * 0.1 / 1.1 = 60 Mbps
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Delisa
1 year ago
B) 55 Mbps
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Hoa
1 year ago
A) 55 Mbps / 1.1 = 50 Mbps
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Johanna
1 year ago
I disagree, I believe the correct answer is D) 60.5 Mbps.
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Tayna
1 year ago
I think the answer is A) 50 Mbps.
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