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GATE CSE 2015 | Set 3 | Question: 6

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47 47 votes

Consider a CSMA/CD network that transmits data at a rate of $100\;\textsf{Mbps}\; (10^8\;\text{bits}$ per second) over a $1\;\textsf{km}$ (kilometre) cable with no repeaters. If the minimum frame size required for this network is $1250\;\text{bytes},$ What is the signal speed $\textsf{(km/sec)}$ in the cable?

  1. $8000$
  2. $10000$
  3. $16000$
  4. $20000$

11 Answers

Best answer
77 77 votes
For collision to be detected, the frame size should be such that the transmission time of the frame should be greater than twice the propagation delay (So, before the frame is completely sent, any possible collision will be discovered).

So, $\dfrac{1250\times 8}{10^8} \geq\dfrac{2\times 1}{ x}$

$\implies x = 2 \times 10^4 = 20000$

Correct Answer: $D$
• edited by
25 25 votes
@arjun sir is fully correct now I am going to discuss about why frame transmission time must be >= 2 * prop. Time


In csma/cd the sender doesn't holds a copy after it has sent a frame so the sender sends a frame now if a collision is detected it resends after  a random amount of time. In the worst case the collision report can take 2*prop time to reach the sender. If the sender is still transmitting then only it can resend it else not.  That's reason why frame transmission time must be > 2 prop. Time
4 4 votes
Given

Length of Frame L=1250 Bytes

Data Transmission Rate B=100 mbps

Distance d= 1km

Let Signal Speed (km/sec) in cable is V.

In CSMA/CD, L=2*dB/V

By putting all the given value we get  Signal speed in cable V=20,000 km/sec
4 4 votes
Answer D.

Since its a CSMA/CD protocol, the formula TT=2*PT applies. TT is transmission time , PT is propagation time

TT= data/bandwidth

PT=distance/ velocity.

 putting values we get 20000
3 3 votes

It is correct

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