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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

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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.

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.

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