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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. $15$ $\text{and}$ $22$

edited | 3.6k views
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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.

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-time (that is transmitting).

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

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

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 large no. of stations.

For that we'll calculate expected value of bandwidth usage (if more data comes we store in buffer and due to expectation, the buffer will be emptied soon):

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

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

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

by Active (5k points)
edited
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+2
@Aspi here 1/3 is on time of the perticular station

As here is a large buffer we have to search for how much time it is actually working
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What I did not get, In the first case : since sources are transmitting at low speed, if we multiplex among sources (allow one source at a time to transmit ) there would not be 100% utilization of 100Mbps. If we overlap 10 stations , and each transmitting @10Mbps then, net link traffic becomes 100Mbps. Thats ok , because we need 100% utilization. But if we allow simultaneous transmission of 10 stations, multiplexing does not make any sense ? Correct me if wrong some where.
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for that we'll calculate expected value of bandwidth usage (if more data comes we store in buffer and due to expectation, the buffer will be emptied soon):

Can someon explain this line?
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Whta dow we mean by data loss here and why are we not taking on -off part for first case?
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How the sources are multiplexed ?

as according to this ans 10 sources asynchronously sending the data at 10 Mbps
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For the first part even if 10 hosts transfer at 10 Mbps  still it is not possible to get 100 Mbps because hosta are off for 2/3 of the time
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can someone tell from which topic this question is?
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@rahul

minimum number of sources that can be multiplexed on the link so that link capacity is not wasted

so if link capacity should be fully utilised we have to assume it to always have the traffic on and then what happens

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If we have buffer then why there is upper bound on number of hosts.

If 30 are max, hosts with buffer ,if all transfer at same time ,then some data goto buffer.

Now same can hold for 40 hosts also?
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@rahul now i am also confused ....will think about it

may be this part of explanation will give the answer

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 large no. of stations..

for that we'll calculate expected value of bandwidth usage (if more data comes we store in buffer and due to expectation, the buffer will be emptied soon):

but am not fully cleared with the above

+15

I think the second part involves the idea of Time Division method of channel access.

Since, our duty cycle is 1:2, a source S will transmit for one time unit and keep quiet for 2 time units.

So, if my source S1, say transmits at t=1, it will transmit again at t=4 and keep quiet at t=2,3.

Since the question says

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.

So, I can synchronize my sources such that I group them into three groups say G1,G2 and G3.

G1 transmits in slot1, G2 in Slot2 and G3 in slot3.

Now the maximum number of stations that I can allow in each group will be

$N*10=100 , N=10$.

10 stations in each group. Any data that could not be sent would be buffered. So, total stations for S2

3*10=30

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did u get this part??
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rahul sharma

the doubt is that when we don't have any limit on buffer size, how can we impose any restriction on minimum no of hosts?
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@sushmita From @Ayush Upadhyaya solution which is correct, you can directly conclude there won't be any synchronization among the stations if they happen to be $>30$ provided we dont want data loss.  And question tells stations are synchronised.

Assuming that all sources are synchronized

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Can't we overap such that in the 2 units of off-time of a station, other 2 stations transmit i.e 2 stations utilize the off time when a particular station has completed transmission in its on-time and is currently in off-time.

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.

by Active (3.3k points)
edited by
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@akb

you have just assumed 3 slots right ? any reason behind it ? if not

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

then if u assume 4 slots answer will be 40

if 5 then 50 and so on

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"The duty cycle, which is the ratio of on-time to off-time, is 1: 2"
Because of this I have assumed 3 slots(1in which it will be on and in another two it will be off)
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@akb

if on off duty part was not ther in the question. Just assuming that always traffic is 10Mbps

then the minimum would be 10 only....right ?

what would the max be.....i think any number .......correct me