how many bits fit in a cable

Question

A 100 Km long cable runs at 154 Mbps data rate. The prop speed in the cable is 2/3 of light. how many bits fit in the cable ?

1) 772
2) 700
3) 782
4) 770

Comments

No of bits send in 1 RTT= RTT* bw ...
Is it correct or not ????

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i thing some thing wrong in this question ....

Answer 1

t_p = $\frac{d}{v} = \frac{10^{5}}{2\times 10^{8}} = 0.5 \times 10^{-3}$

No of bits = t_p x bandwidth = $0.5 \times 10^{-3} \times 154 \times 10 ^{6} = 77000$

Comments

but if channel is considered  as half duplex then number of bits fitted in the cable will be of what count?

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I have a doubt shouldn't it be Total bits = RTT * Bandwidth = Bandwidth delay product ?? Please explain here :(

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Capacity of half duplex link=$T_{p}*Bandwidth$

Capacity of full duplex link=$2*T_{p}*Bandwidth$

$T_{p}=0.5$ ms

if the link is half duplex then capacity=0.5 ms*154 Mbits=77000 bits

None of the options matches

Answer 2

Transmission time > 2 * Propagation time
We have to use this only concept ?

Comments

This concept is used in small ethernets with Time Division Multiplexing.

Answer 3

tp=d/v= 1/(2*10^5) sec no. of bit =tp*bandwidth =1/(2*10^5)*154*10^6 =770 bits

Comments

tp= 10⁵ / 2*10⁸ = 0.5 *10³ 

??

Answer 4

Consider reading d link given below. The ques is exactly similar. I hope it solves ur query. :)

https://gateoverflow.in/9562/how-many-bits-fit-in-a-cable