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Two hosts are connected via a packet switch with $10^7$ bits per second links. Each link has a propagation delay of $20$ microseconds. The switch begins forwarding a packet $35$ microseconds after it receives the same. If $10000$ bits of data are to be transmitted between the two hosts using a packet size of $5000$ bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is ______.
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Best answer
111 votes
111 votes

No. of packets sent $=\dfrac{10000}{5000}= 2$.

Time for the first packet to reach switch $=\text{Transmission time + Propagation delay}$ 
$= \left(\dfrac{5000}{10^7}\right)\times {10^6}\mu s+20\mu s$

$= 520\mu s.$

$($Another $520\mu s$ is required for the same packet to reach the destination from the switch and in between there is a forwarding delay of $35\mu s.$ So, first packet is received at destination at $2\times 520+ 35=1075 \mu s. )$

After $520 \mu s,$ the switch can start receiving the second packet and at $520 + 500 = 1020 \mu s,$ second frame is completely received by the switch $($we don't need to add propagation time here as packet $2$ can just follow packet $1 )$.

So, at $1055\mu s$ from the start the switch starts sending the second packet and this will be received at destination after another $520 \mu s=1575 \mu s.$ Since we added transmission time, this ensures that the last bit of data is received at the sender. 


EDIT:-

(Alternate solution)

We can think the same question in terms of last packet, argument here is: The moment last packet reaches to destination, all other packets are already reached.

Total time $=\text{Transmission time of all packets + Propagation time for first link}$
$\text{+ Switch Delay + Transmission time of last packet for Switch}$
$\text{+ propagation time for 2nd link}.$

$=\left (\dfrac{10^4}{10^7}\;\text{sec} = 1\;\text{ms} = 1000\mu s \right) 1000 + 20 + 35 + 500 + 20=1575\mu s.$

edited by
404 votes
404 votes

​​​​​$\text{number of packets send }=\frac{10000}{5000}=2$

$\text{Transmission Time }=\frac{5000}{10^{7}}=500 \mu s$

$\text{Propagation Time }=20 \mu s$

packet 1

At t=0
   packet starts from source 
At t=500
   
packet 1 is fully transmitted by source. Now it is available in the link

At t=520
     packet 1 is available at switch

At t=555
    packet transmission starts from switch after waiting for 35μs
  
At t= 1055
    packet 1 is fully transmitted by switch . Now it is available in the link
   
At t=1075
  
packet 1 reaches receiver.

  
packet 2

At t=500
   packet starts from source

At t=1000
  packet 2 is now available in the Link. Fully transmitted from source .packet 2 Now, begins to propagate to Switch

At t=1020
 
packet 2 now reaches switch

At t= 1055
 
packet transmission starts from switch after waiting for 35μs

At t=1555
 
packet 2 is fully transmitted by switch . Now it is available in the link

At t=1575
 
packet 2 Reaches destination
 

edited by
9 votes
9 votes

Transmission time for a packet $\frac{5000}{10^{7}}=500microsec$

Propagation time for a packet $20microsec$

Swithching delay $35microsec$

So, overall transmission time of 1st packet =$2\times 500+2\times 20+35=1075 microsec$...................i

Second packet needs only last transmission $500microsec$..................................................................ii

So, total transmission time $1075+500=1575 microsec$

6 votes
6 votes

Transmission time, Tt = ( 5000 / 10) *106 μs  = 500 μs
Tp= 20 μs, switch delay= 35 μs
Total time to reach destination (For packet1) = (500+20+35+500+20) = 1075 μs

Now, I am going to make it easy for you. Always think like this.
There is a train.
Packet2 -> Packet1->
(Transmission time is time for the train to go to track from the station, the propagation time is to cover the track.)

Just tell me in a train, if 1st Bagi reaches in time x, when will second bagi reach destination? Is it that hard? 2nd bagi (Packet2) is 500 μs apart from Packet1. Because Tt from sender was 500 μs.
So it will reach 500 μs after it. Very simple.

But there is something to think about in this question.
In the way, they stop the train, take a delay of 35 μs. We have included it for bagi1(Packet1).
So do we need to add it to bagi2 too? Will the two bagis get separated due to this?
No. Why?
Before checking, you think yourself.
.
.
.
.
Because see the time for bagi1 again.  
500+20+35+500+20
At time 500, bagi2 started. at 520 bagi1 is at that stoppage.
At time 555 bagi1 left station, it should come to track from the railway station in 1055, For bagi2 transmission time is not over yet. 
at time 1000, bagi2 is at the stoppage. For bagi1 55time left to leave the stoppage station.
During this bagi2 took 35 delay, Bagi 1 has 20more time to leave stoppage station. So Bagi2 is ready too. So after 20 bagi2 will just be connected to bagi1. So there will be no gap in between them. So bagi2 will reach just after 500 μs of bagi1.

So I prefer this way of doing.
By the way, it is not always going to happen that they will be connected always and you will just add Tt, that is the extra time for packet2.
They may get separated in the way. So you can calculate that separation between them, you add it at last.
Here bagi2 covered it's delay when bagi1 was still transmitting. So they did not get separated. If the delay would be more than 55μs then they would get separated and you would have to add that distance too.

Anyway answer is 1075 μs + 500 μs = 1575 μs

Answer:

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