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The network consists of 4 hosts distributed as shown below: Assume this network uses CSMA/CD . And signal travels at $3$ $X$  $10^{5} \text{km/sec}$

If sender sends at $1 \text{Mbps}$. What could be the minimum size of packet?

(a).$600$ $\text{bits}$

(b). $400$ $\text{bits}$

(c). $6000$ $\text{bits}$

(d). $1500$ $\text{bits}$

5 Answers

Best answer
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Cannot solve this question without knowing sender and receiver.  Question needs clarity.  Let us assume A is a sender and D is receiver. Algorithm to calculate distance should be specified in GATE question Use the following equation to solve answer once you know the distance

 

 

 

 

L>= B*2*d/v
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The minimum size of packet in CSMA/CD is
TT>=2*TP.
Length of packet=2*(distance/Speed of signal)*Bandwidth.

Here all terms are common b/w all nodes except distance. So minimum the distance, minimum will be length of packet.
Here minimum distance=40km
L=2*(40/3*105)*106=800/3=266.66 bits

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in CSMA/CD-

L/B  = 2(D/V)

SO "L " IS DIRECTLY PROPORTIONAL TO "D"

=>MINIMUM SIZE=MINIMUM DISTANCE,,

AND WITH MINIMUM DISTANCE OF 90KM WE WILL GET 600BITS ..WITH DISTANCE OF 40,51,54,57 WE WILL GET <400 BITS....

SO 90 KM SATISFIES
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Suppose u get a value of the frame size by considering 40 km. There u will get frame size to be 266.67 bits. Now if u consider the frame size to be 266.67 bits and if a collision takes place between A and D, it will not suffice with the value of frame size as 266.67 bits. Because when u put the value in Tt ( transmission time ) and compare it with propagation time ( Tp ), then it will not satisfy the collision detection equation in CSMA / CD that is Tt > = 2*Tp. That is why you have to consider the longest path of 90 km to get the minimum frame size which is coming out to be 600 bits.

Option A.

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