edited by
25,231 views
78 votes
78 votes

Consider a source computer $(S)$ transmitting a file of size $10^{6}$ bits to a destination computer $(D)$ over a network of two routers $(R_{1}\text{ and }R_{2})$ and three links $(L_{1},L_{2},\text{ and } L_{3})$. $L_{1}$ connects $S$ to $R_{1}$; $L_{2}$ connects $R_{1}$ to $R_{2}$; and $L_{3}$ connects $R_{2}$ to $D$. Let each link be of length $\text{100 km}$. Assume signals travel over each link at a speed of $10^{8}$ meters per second. Assume that the link bandwidth on each link is $\text{1 Mbps}$. Let the file be broken down into $1000$ packets each of size $1000$ bits. Find the total sum of transmission and propagation delays in transmitting the file from $S$ to $D$?

  1. $\text{1005 ms}$
  2. $\text{1010 ms}$
  3. $\text{3000 ms}$
  4. $\text{3003 ms}$
edited by

8 Answers

Best answer
130 votes
130 votes
Routers are store and forward devices.

Propagation time $=\dfrac{100\;\text{km}}{10^8\;\text{m/s}}=1\;\text{milli second}$

Transmission time for a packet $=\dfrac{1000 \text{ bits}}{10^6 \text{ bits/sec}}=1\;\text{milli second}$

Packets will be forwarded in a pipelined manner after the first packet reaches the receiver, in every $1\;\text{ms}$ a new one arrives.

Now time taken by packet no $1$ to reach destination is :

$\text{1 ms ($T_x$ at sender) + 1 ms ($Tp$ from sender to R1) + 1ms  ($T_x$ at R1)}$

$\text{+ 1 ms($T_p$ from R1 to R2) + 1ms ($T_x$ at R2) + 1 ms ( $Tp$ from R2 to destination) = 6ms}$

So, time for packet, $1000=6\;\text{ms} + 999\;\text{ms} =1005\;\text{ms}$

Correct Answer: $A$
edited by
91 votes
91 votes
First all data needs to be transmitted from source and after all packets transmission from source, just focus on last packet jorney, and u will get it.

Transmission time for all packets from Source:= $\frac{10^6}{1\times 10^6}$ = 1 sec = $1000$ms.
(Now the last packet is our main focus, bcoz the moment last packet reaches, all previous are already reached to Destination)

Last packet time = $3\times \text{propagation time of link} + 2 \times \text{tranmsion time of router} $ (Transmission time for source is already included in above $1000$ msec)
= $(3 \times 1) + (2 \times 1)$ = 5ms

Total time= $1000+5 = 1005 \text{ms}$
15 votes
15 votes

We have been given following info:

Signal Speed = 10^8 m/s

Bandwidth, BW = 1 Mbps = 10^6 bps

Propagation time, Tp = Link Distance/ Link Speed = (100x10^3)/(10^8) = 10^-3 sec = 1 ms.

[Propagation time is the time taken by packet to travel through the link]

Transmission time for 1 packet, Tt = Packet Size / BW = 1000 / 10^6 = 10^-3 sec = 1 ms.

[Transmission time is the time taken by source or router to process or put the data on Link]

Link filling time can be calculated as:

Pipeline Filling time
Time L1 R1 L2 R2 L3 D
t=0            
t=1 P1          
t=2 P2 P1        
t=3 P3 P2 P1      
t=4 P4 P3 P2 P1    
t=5 P5 P4 P3 P2 P1  
t=6 P6 P5 P4 P3 P2 P1

This way time taken by packet P1 to reach Destination (D) = 6 ms

Now, as you can see, at every ms, 1 packet will reach D.

Therefore, time taken for remaining 999 packets = 999 ms.

So, total time taken to transmit 1000 packets = 999 + 6 = 1005 ms.

Hence Option (A) is correct answer.

 

3 votes
3 votes
S------R1--------R2--------D        here we will use the concept of packet switching first of all ,transmission time = L/B which is equal to 1mili second from S to D  total  Tt= s to R1(1milisec) +R1to R2(1milisec) +R2 to D(1milisec)=3milisec  ,now for remaining 999 packets it will be 999*1milisec=999milisec   (We send 1 packet and now we are sending remaining 999 packets)

so total Tt=3+999=1002

now the propogation delay Tp=D/V , here it is mentioned that (Let each link be of length 100 km) Total distance from S To D is  300km    so Tp=300*10^3/10^8=3milisec.

now the sum of Tt + Tp=1002+3=1005 milisecond. ANSWER IS A.
Answer:

Related questions

0 votes
0 votes
3 answers
3
1 votes
1 votes
1 answer
4
go_editor asked Jul 7, 2016
1,331 views
Match the following with link quality measurement and handoff initiation :$\begin{array}{clcl} \text{(a)} & \text{Networked-Controlled}& \text{(i)} & \text{MS connect to...