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Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is $10^6$ bytes / sec. A user on host $A$ sends a file of size $10^3$ bytes to host $B$ through routers $R1$ and $R2$ in three different ways. In the first case a single packet containing the complete file is transmitted from $A$ to $B$. In the second case, the file is split into $10$ equal parts, and these packets are transmitted from $A$ to $B$. In the third case, the file is split into $20$ equal parts and these packets are sent from $A$ to $B$. Each packet contains $100$ bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let $T1$, $T2$ and $T34 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT? 1.$T<T2<T3$2.$T1>T2>T3$3.$T2=T3, T3<T1$4.$T1=T3, T3> T2$asked edited | 3.6k views +10 T1 =$\frac{3*(1100)}{10^6} = 3.3 ms$T2 =$\frac{3*(200)}{10^6} + \frac{9*(200)}{10^6} = 2.4ms$T3 =$\frac{3*(150)}{10^6} + \frac{19*(150)}{10^6} = 3.3 ms$Hence, (D) is correct option! 0 @Manu Thakur, Bandwidth is given in Byte/sec and data is also given in Byte. So no need to convert bytes into bits. 0 thanks @Hemant, i updated my comment +1 ## 2 Answers +65 votes Best answer In this question we have used the concept of pipelining. In second and Third case, First packet will take$3\times T_t$time and all subsequent packets will be delivered in one$T_t$​​​​​​ time.$T_1=3\times T_t=3\times \dfrac{(1000+ 100)}{B}T_t=\dfrac{\text{(data + header)}}{\text{Bandwidth}}$data$= 1000\text{ Bytes};$header$=100\text{ Bytes}T_1 =\dfrac{3300}{B}\text{ seconds}T_2 = 3\times T^{'}_{t} +9\times T^{'}_{t}=12\times T^{'}_{t} $​$T^{'}_{t}=\dfrac{\text{(data + header)}}{\text{Bandwidth}}T_2=\dfrac{12\times (100 + 100)}{B}=\dfrac{2400}{B}\text{ seconds}T_3 =3\times T^{"}_{t}+ 19\times T^{"}_{t}=22\times T^{"}_{t}T^{"}_{t}=\dfrac{(50 + 100)}{B}T_3 =\dfrac{22\times 150}{B}=\dfrac{3300}{B}$So$T_1 = T_3$and$T_3 > T_2;\$

option D

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plz explain what are thes 3,9,19.........
+14

you can think of this figure as stages in the pipeline system.

so to process n instructions (here packets) with number of stages (link between hosts/router) = k (3)

Time taken is  = n - 1 + k

So for 1st case put n (number of packets) = 1

time taken is = 1 - 1 + 3 = 3 units

for 2nd case n = 10

time taken = 10 - 1 + 3 = 9 + 3 = 12 units

for 3rd case n = 20

time taken  = 20 - 1 + 3 = 22 units

In this figure when the 1st packet reach the destination, 2nd packet will be at R2 and 3rd packet is at R1 and so on..

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great explanation sir!!
thank you
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Is this necessary

In this figure when the 1st packet reach the destination, 2nd packet will be at R2 and 3rd packet is at R1 and so on..

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Great explanation .. thank u sir
+3

Refer this video link......nicely explained here.....

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but this is store and forward method. does not it mean that all packets 1st should be collected at node and than forward to next link?
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hi sheshnag,

Did u get any solution for ur doubt?
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I think store forward means we will not fragment the packet and we will only transmit the packet once it is completely received . Correct me if I am wrong
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@Sheshang packet switiching is actually store and forward approach, because here when a packet is recieved completely then routing protocol decide its path and then forward it. So, it does'nt mean that all packets should be collected first and then forwarded.

The important thing to note here is in first case, the whole packet is being transmitted, so no piplelining of packet happens. In second and third case, we have advantage of pipelining (While packet 'i' is being transmitted from R1 to R2, packet 'i-1' is being transmitted from A to R1 at the same time). Following are complete calculations.

File Size = 1000 bytes
Transmission Speed of all  links = 10^6 bytes/sec

Ist Case:
= packetsize/bandwidth
= (1000 + 100)/10^6
= 1100 micros
Total time = 3*1100
= 3300 microsec.

Second case:
Transmission time for one link and one part
= (100 + 100)/10^6
=  200 microsec

[Note the pipe-lining in packets.  While
packet 'i' is being transmitted from R1 to R2,
packet 'i-1' is being transmitted from A to R1
at the same time]
Total time = 3*200 + 9*200
= 2400 micro sec

Third Case:
Transmission time for one link and one part
= (50+100)/10^6
= 150microsec
Total time = 3*150+19*150
= 3300 microsec 

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