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Consider the following two networks 
Network 1 : (Ethernet LAN)
(i) Transmission speed is 10 Mbps
(ii) Signal travelling speed 2/3 of velocity of light
(iii) Length of cable = 3 km
Network 2 : (CSMA / CD Network)
(i) Transmission speed is 200 Mbps
(ii) Signal travelling speed 1/3 of velocity of light
(iii) Length of cable = 6 km
The minimum frame size required for network is x bits and for Network 2 is y bits. So what is value of x + y = ______.

Your Answer:

15000

Correct Answer: 24300    Status: incorrect
 

asked in Computer Networks by Active (2.6k points) | 166 views

1 Answer

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In Ethernet the Transmission Time should be greater than equal to twice that of the propagation time.(Ethernet uses CSMA/CD).

For network 1: L(length of the data)=x bits                                          For network 2:L(length of the data)=y bits

                        Bandwidth=10Mbps                                                                             Bandwidth=200Mbps

                        distance i.e d=3 km                                                                               distance i.e d=6 km

                        v=2/3 * velocity of light                                                                          v=1/3*velocity of light

                       L/B>=2*d/v

                  so putting these values in the above equation we obtain x=300 bits and y=24000bits

               So the answer i.e. x+y=24000+300=24300 bits.
answered by (137 points)

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