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GATE CSE 2014 | Set 2 | Question: 26

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83 83 votes

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 $T3$ be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?

  1. $T1<T2<T3$
  2. $T1>T2>T3$
  3. $T2=T3, T3<T1$
  4. $T1=T3, T3> T2$

8 Answers

Best answer
128 128 votes

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

• edited by
42 42 votes

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
Header Size = 100 bytes
Transmission Speed of all  links = 10^6 bytes/sec

Ist Case: 
 Transmission time for one link 
                = 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
2 2 votes
1100 μs

              1100 μs

                            1100 μs

 

 

200μs

           200μs

                      200μs                        9 more packets (200μs)

 

 

150μs

           150μs

                      150μs                         19 more packets(150μs)

 

 

 

 

If u calculate this time, you will get to know that T1=T3 >T2
2 2 votes
Using formula which can be derived easily,

Transmission time = ((P-1) + H)*L/B , where P is number of packets, H is number of Hops(router + destination host),               L = packet size, B = bandwidth

in this question,

a) P = 1 H=3, T = 3*1100 ms/1000 = 3.3ms

b) P = 10 H=3, T = (9 + 3)*200 ms/1000 = 2.4ms

c) P = 20 H=3, T = (19 + 3)*150 ms/1000 = 3.3ms

Correct option D.
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