Solved

Calculating the peak source data rate

Posted on 2013-05-27
5
121 Views
Last Modified: 2016-01-22
For an exam I am soon to sit there is a question I have no idea how to answer, it is in past papers but there is no explanation to work it out in our lecture slides. The question is:

A company requires high-quality video conferencing service. The holding times are between 1000 to 10000 seconds. The Burstiness is 3 and the average data rate is between 1.5 - 15 Mb/s. Calculate the peak source data rate.

I've looked around on the net for information relating to this but I can't find anything. Can anyone figure out the formula necessary for this question?

I would really appreciate it.

Thanks
0
Comment
Question by:pbzzs
  • 2
5 Comments
 
LVL 37

Expert Comment

by:TommySzalapski
ID: 39199116
That is a very odd academic question. How can an "average" be a range like that?
Anyway, if you are looking for a "peak" data rate, then you need to look at the max so we can use the 15 Mb/s as the "average" in the peak case.

How are you defining "holding time"? 10000 seconds is close to 3 hours. You can't do anything in that time with video conferencing. Is that really supposed to microseconds or nanoseconds? Is it how long to buffer a packet before dropping it? Holding time shouldn't affect data rate from the source.

What are the units in "burstiness"? What does 3 burstiness mean?

The most simplistic definition would say that burstiness of 3 means it bursts up to 3 times the data rate, so the peak data rate would be 45 Mb/s, but that looks too easy?
0
 

Author Comment

by:pbzzs
ID: 39199210
Yes I completely agree with this being an odd question, that's why I posted it - our lecturer is terrible, this question has been in the exam the last 2 years however his lecture notes do not in any way explain how to work this out. The closest I have got is:

peak source data rate = burstiness x average data rate

We know that burtsiness is 3 but average data rate we have to calculate. Assignment says that it is between 1.5 - 15 Mb/s so I will obtain mean average:

1.5 + 15 = 16.5 / 2 = 8.25.

Which gives:

peak source data rate = 3 x 8.25 Mb/s = 24.75 Mb/s

But this completely ignores the holding time.
0
 
LVL 37

Accepted Solution

by:
TommySzalapski earned 500 total points
ID: 39199271
Holding time should not affect peak source rate. That data is just in there to trick you. For peak rate, you should use the worst case average of 15, so 45 would be the peak.

Now that you defined things a bit better, I am confident that 45 is the expected answer. It's the peak, so it means the max.
0

Featured Post

Master Your Team's Linux and Cloud Stack

Come see why top tech companies like Mailchimp and Media Temple use Linux Academy to build their employee training programs.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
Simplify expression 3 99
IP Calculator 10 75
Why do I get "media disconnected" when I run ipconfig? 2 46
Math Equation 23 88
Complex Numbers are funny things.  Many people have a basic understanding of them, some a more advanced.  The confusion usually arises when that pesky i (or j for Electrical Engineers) appears and understanding the meaning of a square root of a nega…
When we purchase storage, we typically are advertised storage of 500GB, 1TB, 2TB and so on. However, when you actually install it into your computer, your 500GB HDD will actually show up as 465GB. Why? It has to do with the way people and computers…
This is a video describing the growing solar energy use in Utah. This is a topic that greatly interests me and so I decided to produce a video about it.
Finds all prime numbers in a range requested and places them in a public primes() array. I've demostrated a template size of 30 (2 * 3 * 5) but larger templates can be built such 210  (2 * 3 * 5 * 7) or 2310  (2 * 3 * 5 * 7 * 11). The larger templa…

823 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question