GATE IT 2007 | Question: 62

Question

Let us consider a statistical time division multiplexing of packets. The number of sources is $10$. In a time unit, a source transmits a packet of $1000$ bits. The number of sources sending data for the first $20$ time units is $6, 9, 3, 7, 2, 2, 2, 3, 4, 6, 1, 10, 7, 5, 8, 3, 6, 2, 9, 5$ respectively. The output capacity of multiplexer is $5000$ bits per time unit. Then the average number of backlogged of packets per time unit during the given period is

• A. $5$
• B. $4.45$
• C. $3.45$
• D. $0$

Comments

Is this in syllabus?

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Yes! it is a part of MAC sublayer

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TDMA is a  channel-access scheme is based on a multiplexing method, that allows several data streams or signals to share the same communication channel or physical medium.

 Multiplexing is provided by the physical layer .

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some necessary information before solving this question:

. In case of tdm we create a combine frame which will contain time slot from each data source and we send this combined frame to the destination but here the problem is that if some source is not sending the data then also we have to reserve slots for it. Due to which bandwidth required is very high we can resolve this issue in statistical tdm here any data can be send in the slot time and there is no empty space also it may be required to send some data in next frame so backlog It in next frame.

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The capacity of single communication line that is used to carry the various transmission should be greater than the total speed of input lines in synchronous time division multiplexing. Can someone explain why??

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Sir this topic is in syllabus or not?

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Aboveallplayer

this is not in syllabus

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Can you tell us what all things are not in the syllabus of CN?

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Is this in syllabus of 2021 ?

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Is this a tricky/difficult question?

Answer 1

Answer is B.

Here, we can send at max $5$ packets per Time unit $\dfrac{5000}{1000}$.

So, whatever which is not sent is backlog.

So,

First Time Unit $\Rightarrow 6,$

Backlog in First time unit $\Rightarrow 6-5\Rightarrow 1$ This one gets added to next Time units load

Second time unit $\Rightarrow 9 + 1$ (One from Previous Time Unit)

Backlog in Second time Unit $=10-5\Rightarrow 5$ (This one gets added to next Time Units load.)

Total Backlog this way  $=1+5+3+5+2+0+0+0+0+1+0+5
+7+7+10+8+9+6+10+10=89$

Avg Backlog $=\dfrac{89}{20}=4.45$

The average number of backlogged of packets per time unit during the given period is $4.45$.

Comments

@Arjun @Bikram @srestha @Dixith Reddy @Akash Kanase @Milicevic3306 @Ishrat Jahan @Radha mohan

How exactly TDM works for Access Control? I have read that, it is similar to Round-Robin method where every Station gets a slot to transmit a data packet. Is it really, we do multiplexing (i.e. take data packets from each station (which is ready to transmit) into a single frame and sent over the channel in a single time slot).

Correct me, if i am wrong.

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https://www.youtube.com/watch?v=5R0DGCV7hvQ&list=PLC36xJgs4dxHT-TxTy3U1slr5RaBJGaLd&index=39

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Should we consider the 5 backlogged packets in the 21^st time unit. Then the answer comes to 94/21 = 4.476 approx.

Answer 2

Answer: C

In statistical TDM, the bandwidth is divided into slots each for a source if the source requires. There is no dedicated slot for each source in the bandwidth.
STDM does not reserve a time slot for each terminal, rather it assigns a slot when the terminal is requiring data to be sent or received.

Multiplexer bandwidth = 5000bits = 5 packets can be send in a time unit.

If x number of source want to transmit in a particular time unit then MBW will be divided among x sources. Therefore,

if [x <= 5] {
    all x sources can successfully transmit their 1000bits because [x*1000]<=[5000]
} else {
    all sources cannot transmit their complete 1000 bits and there will be backlog of x (incompletely transmitted) packets.
}

So,

6 [Backlog], 9 [Backlog], 3 [-], 7 [Backlog], 2 [-], 2 [-], 2 [-], 3 [-], 4 [-], 6 [Backlog], 1 [-], 10 [Backlog], 7 [Backlog], 5 [-], 8 [Backlog], 3 [-], 6 [Backlog], 2 [-], 9 [Backlog], 5 [-]

Average number of backlogged packets per time unit during the given period is = [6+9+7+6+10+7+8+6+9]/20 = 3.4

Comments

Thanks a lot @ Bikram sir for your efforts :) yes sir u are correct .....we should not make any assumptions about this and follow standard book.

sir do we know the official answer for this question ?

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@gari

This is Gate 2007 paper and that time no official answer key posted .  And this question also match with that Stallings link, just check yourself the value given in question and value given in that snap , answer is B and it is correct.

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Thanks sir :)

Answer 3

The capacity of multiplexer is 5000 bits per time unit. This means there are 5 packets per unit time since each source transmits a packet of 1000 bits in a unit time.
If the no. of packets transmitted is larger than 5 then the extra packets are backlogged. This means gets added to the next number and further backlog is calculated.

 

Average no. of backlogged packets = 89/20 = 4.45

 

Comments

Thanks Utkarsh. All the best for GATE 2021

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This is part of which section??