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A and C are the only two stations on an Ethernet with data rate is 75 Mbps, the distance between station A and C is 4 km, and the propagation speed is 2××108 m/s. Station A start sending a long frame at time t1 = 1; station C starts sending a long frame at time t2 = 4.5 µs. The size of the frame is long enough to guarantee the detection of a collision by both stations. Find the time in microseconds when station A hears the collision(t3) and when station C hears the collision(t4).
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T3( A will hear) after  = 11.75 microsecond ?
the ans is t3=24.5 and t4=21
Yup correct
Let after x km detects collision

Let packet length = L bits

Time taken by packet from A to reach x km = (L/(75*10^6)) + (x*1000)/2*10^8

Time taken by packet to reach from C to x = (L/(75*10^6)) +( (4-x)*1000)/2*10^8

by condition ,

(L/(75*10^6)) + (x*1000)/2*10^8  - (L/(75*10^6)) +( (4-x)*1000)/2*10^8 = 3.75*10^-6   [as C start to tansmit after 3.5microsecond after A)

so, solving this x = 11.75*10^-6

now, time taken a packet to come and retuen to x = 2 * (11.75*10^-6)/2*10^8 = 23.5*10^-6

so A will hear collison at (1+23.5)*10^-6 [ as A started at 1 micro second}

    = 24.5microsecond

@Manas Mishra thanks!

welcome !!!!

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