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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,

  1. $10$ $\text{and}$ $30$
  2. $12$ $\text{and}$ $25$
  3. $5$ $\text{and}$ $33$
  4. $15$ $\text{and}$ $22$
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7 Answers

Best answer
62 votes
62 votes

Since there is no buffer and constraint given is there should not be any data lost, and no wastage of capacity as well.

Since data should not be lost, we calculate for the extreme case when all sources are on (that is transmitting).

$10\ Mbps\times \text{n-station} \leqslant 100 Mbps$

$\implies \text{n-station}= 10.$

In the next part of the question it is given that the link is provided with large buffer and we are asked to find out the maximum number of stations.

For this we'll calculate the expected value of bandwidth usage (if more data comes we store in buffer and due to expectation, the buffer will be emptied soon or in other words buffer space will never run out):

$E =\dfrac{1}{3}\times 10 +\dfrac{1}{3}\times 10+\ldots \text{n-station times}\leqslant 100\ Mbps$
[ total time is $(1+2)=3$ then on time is $1$ so $\dfrac{1}{3}$ of BW]

$\Rightarrow \dfrac{1}{3}\times 10\times \text{n-station}\leqslant 100\ Mbps$

$\Rightarrow \text{n-station} =30$

So, option (A) is answer.

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26 votes

Part 1(for S1)

Suppose there are three slot 1,2 and 3

Every station can send traffic only during one slot and in remaining two slots it will not send traffic(bcz on off duty cycle is 1:2)

Here we have to find minimum number of station that can generate traffic so that we have maximum utilization with no loss of data.

Any station can generate traffic during any slot But there is no buffer it means the traffic generated by stations during slot 1 will also be on link during slot2 and slot 3 also traffic generated by stations during slot 2 will also be on link during slot 3

But we have to find out minimum number of stations which will be possible when all stations generate traffic during same slot.

Traffic generated by n stations= Bandwidth

10*n=100

So n=10 stations

Alternate method for part 1(for S1)

Let n1,n2 and n3 be the no of stations that generate traffic during slot 1, slot 2 and slot 3 respectively.

Any station can generate traffic during any slot But there is no buffer it means the traffic generated by stations during slot 1 will also be on link during slot 2 and slot 3 also traffic generated by stations during slot 2 will also be on link during slot 3.

Traffic generated during slot 1 =n1*10

Traffic generated during slot 2 =n2*10

Total Traffic during slot 3 =n3*10 + n1*10 + n2*10.

Total traffic=100 Mbps

n3*10 + n1*10 + n2*10 =100

n3 + n1 + n=10

Part 2(For S2)

Suppose there are three slot 1,2 and 3

Every station can send traffic only during one slot and in remaining two slots it will not send traffic(bcz on off duty cycle is 1:2)

The stations who generate traffic during slot 1 their data during slot 2 and and during slot 3 can be stored in buffer.

The stations who generate traffic during slot 2 their data during slot 3 can be stored in buffer.

So lets's find how many maximum   stations can send data during slot 1 so that we will have maximum utilization with no loss of data.

Let n station generate traffic during slot 1 as each station generate 10 Mbps.

So n station generate 10*n Mbps it must be equal to.100Mbps (for maximum utilization and no loss of data)

10n=100

So n=10 during slot 1.

It is also true for other slots also.

So Total Stations=10(slot1)+10(slot2)+10(slot3)=30.

Correct me if you think my approach is wrong.

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7 votes
7 votes

I will try to explain the S2 part only in the simplest way possible!

Given that Duty Cycle =ontime/offtime =½


=> on/off=1/2
=>  on=1/2 times of off


means if a station is off for 2 times then it will be on for half of (2 times)= 1 time

So we can say that in a time duration of 3 slots for one slot station will be on and the remaining two slots the station will be off!

groupd=s


We have been told that we have a Buffer of infinite Capacity!

Also we know that link has a max capacity of 100Mbps!

In computer science we always assume the worst case!

Since our link has a max capacity of 100Mbps so at max during a particular time instant(say t=0) all the 10 stations of group1 got on and now all started transmitting at same time at there max capacity of 10Mbps each,So since our link capacity is 100Mbps so Group1 can contain at max 10 stations.

Now assume the data which group1 stations produced got stored in the buffer!

Now at the end of slot1 ,Group1 will become silent for slo2,slot3{they will not produce any data till end of Slot3} Since as i mentioned above bcoz of that duty cycle.

Now Suppose Group2 started transmitting ,similarly group2 will also contain at max 10 stations as i already mentioned the reason above.Now in the slot2 when group2 stations were transmitting ,assume again the data of group2 got stored in the buffer again.
Now group 2 will remain silent for the next two time slots.{slot3,4}

Now Suppose Group3 started transmitting ,similarly group3 will also contain at max 10 stations as i already mentioned the reason above.Now in the slot3 when group3 stations were transmitting ,assume again the data of group3 got stored in the buffer again.
Now group3 will remain silent for the next two time slots.{slot4,5}

So till now we got 30 stations,all the data has been stored onto the buffer till this point.

Now at this point{end of slot3} if you assume (just a hypothetical scenario) one more station (say of group4),now this would create a problem bcoz now at the end of slot3 the stations of group1 will become active and since group1 already contains 10 members so no more extra stations can be afforded,bcoz then we will be exceeding the limit of 10stations{10*10Mbps=100Mbps(the max cap of out link)}
As the moment group1 stations gets activated they will again try to send the data{and in worst case all the 10 Stations of group1 can try to send the data at 10Mbps/station)

So even if we have a infinite size buffer, we can afford a max of 30 Stations in whatever worstcase we assume,as stated in the question that no data loss should occur and also capacity must not exceed!

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First, we have to find a minimum no of Sources so, We have 100mbs link capacity, and for having 10mbps for each source, no of sources should be 10 if there is no buffer provided so no data can be stored,

In the second situation, we have buffers, so suppose a source is transmitting and the link is full, then it can be stored in a buffer and can be transmitted later. As the duty cycle ratio is 1:2 and for not wasting any link’s bandwidth, the no of sources should be 30. according to duty cycle if we consider 3 sources per 10mbps then , one of the source will always be transmitting in any situation.
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