GATE IT 2006 | Question: 67

Question

A link of capacity $100$ $\text{Mbps}$ is carrying traffic from a number of sources. Each source generates an on-off traffic stream; when the source is on, the rate of traffic is $10$ $\text{Mbps}$, and when the source is off, the rate of traffic is zero. The duty cycle, which is the ratio of on-time to off-time, is $1: 2$. When there is no buffer at the link, the minimum number of sources that can be multiplexed on the link so that link capacity is not wasted and no data loss occurs is $S1$. Assuming that all sources are synchronized and that the link is provided with a large buffer, the maximum number of sources that can be multiplexed so that no data loss occurs is $S2$. The values of $S1$ and $S2$ are, respectively,

• A. $10$ $\text{and}$ $30$
• B. $12$ $\text{and}$ $25$
• C. $5$ $\text{and}$ $33$
• D. $15$ $\text{and}$ $22$

Comments

what is meant by  saying large buffer size?? I mean if we have very very large buffer we can accumulate any amount of data and can send it at 10Mbps?? I didnt how is the max no of stations computed?

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Not able to understand. Please someone give a more clear and lucid explanation

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@Tuhin Dutta ji what is not clear to you @Vicky Bajoria ji has already given very good explanation.

Answer 1

Bandwidth is given as = 100 Mbps.

 

for calculating S1 : We have no buffer. Buffer role is to have all the bits that are currently in the queue and are needed to be sent but because of the bandwidth utilization is complete the bits can’t be send as of right now.

 Since no buffer is there and each source have a data transfer rate = 10 Mbps, a total of 10 sources can send the data without any hassle.

 

for calculating S2 : For this case we are given we will need a buffer to store outstanding bits in the queue.

But how many such bits can be stored ?

We can calculate it by using the duty cycle ratio i.e 1:2

meaning if 1 source is on than 2 sources are off (or buffered up)

 

so, from S1 we had 10 sources at max we can send. But now since there is a buffer available we can queue up sources.

And from duty cycle,

for 1 source On → 2 sources need to be off

therefore for 10 sources on → 20 sources need to be off

and the total equals 10 on and 20 off = 30 sources.

 

therefore a minimum of 10 sources and a max of 30 sources can send data is the answer.

Answer 2

Bandwidth is given as = 100 Mbps.

for calculating S1 : We have no buffer. Buffer role is to have all the bits that are currently in the queue and are needed to be sent but because of the bandwidth utilization is complete the bits can’t be send as of right now.

 Since no buffer is there and each source have a data transfer rate = 10 Mbps, a total of 10 sources can send the data without any hassle.

 

for calculating S2 : For this case we are given we will need a buffer to store outstanding bits in the queue.

But how many such bits can be stored ?

We can calculate it by using the duty cycle ratio i.e 1:2

meaning if 1 source is on than 2 sources are off (or buffered up)

 

so, from S1 we had 10 sources at max we can send. But now since there is a buffer available we can queue up sources.

And from duty cycle,

for 1 source On → 2 sources need to be off

therefore for 10 sources on → 20 sources need to be off

and the total equals 10 on and 20 off = 30 sources.

 

therefore a minimum of 10 sources and a max of 30 sources can send data is the answer.

Answer 3

It's literally an aptitude question. Just word play.. And, that is what GATE is known for!

If anyone feel stuck.. just read question 2-3 times, even if you can't get.. take a sip of water and try one more time.

 

If still feel stuck, feel free to drop the query :)