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+15 votes
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A computer on a $10$ $Mbps$ network is regulated by a token bucket. The token bucket is filled at a rate of $2$ $Mbps$. It is initially filled to capacity with $16$ $Megabits$. What is the maximum duration for which the computer can transmit at the full $10$ $Mbps$?

  1. $1.6$ seconds
  2. $2$ seconds
  3. $5$ seconds
  4. $8$ seconds
asked in Computer Networks by Veteran (59.5k points)
edited by | 3.8k views
0
Is token bucket in gate syllabus 2018 since token ring is not in syllabus ?
+3
Its not related to token ring . It is a congestion control algorithm in network layer (falls under traffic shaping).

5 Answers

+26 votes
Best answer

New tokens are added at the rate of $r$ bits/sec which is
$2$ $Mbps$ in the given question.

Capacity of the token bucket (b) = $16$ $Mbits$
Maximum possible transmission rate (M) = $10$ $Mbps$
So, the maximum burst time = b/(M-r) = $16$/($10$-$2$) = $2$ $seconds$
Here is the animation for token bucket hope this will help us to understand the concept.

answered by Boss (13.6k points)
edited by
+1
Can you please explain how the formula is derived?
+9
say till t sec it operate at 10mbps,

t*2+16=10*t

8t=16

so  t=2
+1

I tried this easiest method! Is my approach correct, yes or no? plz let me know!

+4
That animation link is not working. Those who are interested to see that: http://webmuseum.mi.fh-offenburg.de/index.php%3Fview=exh&src=8.html
+13 votes
TIME TOKEN IN BUCKET TOKEN SEND LEFT IN BUCKET
First sec 16Mb+ 2Mb =18Mb 10Mb 8Mb
next sec 8Mb+2Mb 10Mb 0Mb

Hence for 2 seconds we can send the tokens at 10 Mbps.

answered by Active (3.3k points)
+9 votes
It's 'kind of' aptitude question:

packets leaving bucket at 10Mbps   &&   packets entering bucket at 2Mbps.

So "actual rate" of bucket being empty is 10-2=8Mbps.

so (capacity/transmission rate) = 16/8 =2seconds.
answered by (325 points)
0 votes

Answer is 2 sec...

answered by (321 points)
0 votes
Simple way :

Token rate into bucket = 2 Mbps

Output rate of bucket = 10 Mbps

Effective rate = 10-2 = 8 Mbps

Bucket capacity is full, 16 Mb

So, at effective rate bucket will drain in 16/8 = 2s

In these 2s, output capacity of bucket will be 10Mbps, after bucket gets empty, it's capacity will reduce to effective rate of 8Mbps.
answered ago by (227 points)


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