Log In
23 votes

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
in Computer Networks
edited by
Is token bucket in gate syllabus 2018 since token ring is not in syllabus ?
Its not related to token ring . It is a congestion control algorithm in network layer (falls under traffic shaping).

Confusion in the analysis of the problem.

at $t = 0$ Bucket occupancy $= 16 \ Mb$
between $t = 0 \text{ and } t = 1$

$10 \ Mb$ worth of token (and packet data) is drained out.

$2 \ Mb$ worth of token is added to the bucket.

at the end of $t = 1$ Bucket occupancy $= (16 - 10 + 2) \ Mb = 8 \ Mb$
between $t = 1 \text{ and } t = 2$ What will happen here?

My understanding: $2$ possibilities

  1. $2 \ Mb$ worth of token is first added to the bucket. And then, $10 \ Mb$ worth of token is removed. This leaves bucket occupancy $= 0 \ Mb$. This gives a full rate transmission time of $2$ seconds.
  2. $8 \ Mb$ worth of token is first drained out of the bucket, and then, $2 \ Mb$ worth of token is added. This leaves bucket occupancy $= 2 \ Mb$. This gives a full rate transmission time of $1$ second.

As per the discussions here, it is evident that the former approach is preferred over the latter. But, intuitively, I do not see any reason for doing this.

  • Why would the transmitter wait for the bucket to fill again with $2 \ Mb$ worth of token (possibility $1$) when it already has $8 \ Mb$ worth of token in the bucket and can use it to transmit $8 \ Mb$ data?

I would appreciate an intuitive explanation, rather than using the formula $T_{trans} = \frac{16 \ Mb}{(10 - 2) \ Mbps}$

Is this there in the syllabus?

congestion control is there is syllabus
Ok, Thanks.

$\\ Max\ rate=\dfrac{c+rt}{t}\\ \\ 10Mbps=\dfrac{16Mb+2Mbps\times t}{t}\\ \\ 10Mbps\times t-2Mbps\times t=16Mb\\ \\ 8Mbps\times t=16Mb\\ \\ t=2sec$

Ans: 2sec

6 Answers

39 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.

Correct Answer: $B$

edited by
Can you please explain how the formula is derived?
say till t sec it operate at 10mbps,



so  t=2

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

That animation link is not working. Those who are interested to see that:
@partner it is also not working

Yes, it is not working now, unfortunately!

25 votes
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.

15 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.
3 votes
  • I/P rate of tokens=2 Mega bps. 2 Megabits tokens come 1 sec
  • O/P rate of tokens=10 Mega bps. 10 Megabits tokens comes out in 1 sec
  • Capacity of the bucket=16 megabits
  • At t=0, bucket=16 Megabits
  • At t=1, 10 Megabits tokens comes out(as it is filled so first tokens have to pulled out then it will be filled), 2 Megabits tokens come in, (16-10+2)=8 Megabits tokens. At t=1 bucket has 8 Megabits tokens
  • At t=2, bucket has 8 Megabits tokens(it is not full no first tokens will come in then pulled out), 2 Megabits tokens comes in, bucket=8+2=10 Megabits. 10 Megabits tokens will come out. Now bucket has 0 Megabits tokens
  • At t=3, bucket has 0 tokens, 2 Megabits tokens comes in and 2 Megabits tokens goes out. So here tokens are send at the rate of 2 Mega bps and this will continue
  • In t=1 sec 10 Megabits tokens comes out
  • At t=2, 10 Megabits tokens come out
  • So till 2 sec we're able to send tokens at 10 Mega bps

edited by


Good answer.

Just correct the last second line. You missed to write 10 there .


@aditi19 @`JEET

I think rate of adding tokens to bucket and rate of consuming tokens from bucket happens according to the capacity of Bucket (for example, it takes 1 sec to consume a token and 2 sec to add a token, then for every 2 sec - 2 tokens are used and 1 token is added). 

Correct me, if i am wrong.


2 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.
Its capacity will reduce to effective rate of 2 Mbps, not 8 Mbps.
0 votes

Answer is 2 sec...


Related questions

34 votes
3 answers
A client process P needs to make a TCP connection to a server process S. Consider the following situation: the server process S executes a $\text{socket()}$, a $\text{bind()}$ and a $\text{listen()}$ system call in that order, following which it is preempted. ... $\text{connect()}$ system call returns an error $\text{connect()}$ system call results in a core dump
asked Sep 12, 2014 in Computer Networks Kathleen 6.5k views
19 votes
2 answers
If a class $B$ network on the Internet has a subnet mask of $$, what is the maximum number of hosts per subnet? $1022$ $1023$ $2046$ $2047$
asked Sep 12, 2014 in Computer Networks Kathleen 4.3k views
21 votes
3 answers
In the slow start phase of the TCP congestion algorithm, the size of the congestion window: does not increase increase linearly increases quadratically increases exponentially
asked Sep 12, 2014 in Computer Networks Kathleen 3.5k views
24 votes
4 answers
Which of the following system calls results in the sending of SYN packets? socket bind listen connect
asked Sep 12, 2014 in Computer Networks Kathleen 5.2k views