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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

  1. $5$
  2. $4.45$
  3. $3.45$
  4. $0$
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90 votes
90 votes

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 $4.45$.

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12 votes
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
9 votes
9 votes

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

 

Answer:

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