Test by Bikram | Mock GATE | Test 3 | Question: 37

Question

In a college, the $CSMA / CD$ protocol has a bandwidth of $512$ $Mbps$ and distance of $2$ $km/s$.

If the signal speed is $2,00,000$ $km$, then the minimum frame size in order to detect a collision is _________ $bytes$.

Comments

Almost same previous year question :

1. https://gateoverflow.in/8400/gate2015-3_6
2. https://gateoverflow.in/3834/gate2005-it-71

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@Bikram sir,

What value should we consider for translating bandwidth if it is not mentioned in question?

If we consider bandwidth 512Mbps = 2^29 bits per second instead of 512 * 10^6 bps , answer will vary.

F = (2 * 2 * 512 * 2^20) / (200000 * 8) B

therefore, F = 1342.177B 

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change the UNIT of distance and speed. both are interchanged.

Answer 1

Let the minimum frame size is F .

F ≥  2 × bandwidth × t_d ....(i)

where t_d  = propagation delay = distance / link speed

now putting value into (i) we get F  ≥  2 x 512 x 10⁶ x ( 2 / 200, 000 ) 

F  ≥  1024000000 * 2 / 200, 000

F ≥   2048000000  / 200, 000

F ≥  10240  bits 

 8 bits = 1 Byte

10240 bits =  (10240 / 8 ) bits  = 1280 Bytes