CN22 doubt

Question

Answer 1

My approach:

1. Given the distance between each pair, find dmax = maximum distance b/w any 2 nodes ( source !=destination).

=> dmax = 300 km

2. Tt >= 2*Tp (CSMA/CD condition)

=> L/B >= 2*dmax/v

=> L >= 2*(dmax/v)*B

L(min) = 2*(dmax/v)*B  = 1500 bits

Here, B = bandwidth = 10^6 bits/sec (given), v = velocity = 2*10^8 m/s (given)

Comments

suppose we are sending packet between nodes at a distance $120$, then packet will surely be transmitted it no other transmits, that is $1200$ bit is needed.

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As mentioned before, that will not cover the worst case. If you take L=1200 bits,  Tt = L/B= 1.2 ms. The worst case collision signal takes time = 2*Tp = 2*(300km/v) = 1.5 ms. Q will not be able to detect collision. You need to take care that all nodes are able to detect every possible collision. So you have to take distance as maximum.

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it will be 3000 bits