GATE CSE 2003 | Question: 83

Question

A $2$ $km$ long broadcast LAN has $10^7$ bps bandwidth and uses CSMA/CD. The signal travels along the wire at $2 \times 10^8$ m/s. What is the minimum packet size that can be used on this network?

• A. $50$ $\text{bytes}$
• B. $100$ $\text{bytes}$
• C. $200$ $\text{bytes}$
• D. None of the above

Answer 1

In CSMA/CD, to detect a collision the transmission time (which depends on the
packet size) must be greater than twice the propagation delay.

Propagation delay here = $\dfrac{2 km}{  2 \times 10^8 m/s}=10 \text{ microseconds}$

Now, transmission time for $x$ bytes = $\dfrac{x \times 8}{10^7} = 0.8x\text{ microseconds}$

So, $0.8x > 2 \times 10 \implies x > 25 \text{ bytes}$

So, None of these.

Correct Answer: $D$

Comments

Can you explain why Tt >= 2*Tp ? this expression can be valid for acquiring the channel, in that the minimum time should be 2Tp. But in collision, how are you using this, It should be Tt<=2Tp na??? please clear my doubt

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I didn't exactly got your doubt ,but who said there is collision?

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Though I cleared this doubt afterwards. But can you confirm that Tt>=2*Tp is for acquiring the channel here.

Answer 2

For CSMA/CD Protocol formula of minimum packet size 

L >= 2*T_p*B 

where B is the bandwidth and T_p is the propagation delay which is D/V D is the distance and V is the velocity that signal travels through the wire so by putting all these values we get that L>= 200 bits which is L>= 25 bytes 

Ans D

Answer 3

In CSMA/CD, the transmitting node is listening for collisions while it transmits it's frame. Once it has finished transmitting the final bit without hearing a collision, it assumes that the transmission was successful. In this worst-case collision scenario, the time that it takes for a Node to detect that its frame has been collided with is twice the propagation delay. Hence to confirm that the collision has not occurred the condition for the minimum size of the packet is:

 

RTT = 2 * Propagation Time

Transmission Time = Length of packet / Bandwidth

RTT = 2 (d/v) = 2(2000/2×10⁸) 

Therefore to find minimum size of the packet, 

RTT <= Length of packet / Bandwidth

Length of packet >= RTT x Bandwidth

                         = 2(2000/2×10⁸) x 10⁷ = 200bits = 25bytes

 

Therefore, minimum size of the packet = 25bytes

Comments

RTT = 2* Propagation Time

Answer 4

L/B=2D/V ( we are using 2 because we want minimum packet size)

L=(2*B*D)/V

2*(10^7)/8 * 2000/2*10^8

=>25 bytes

so the answer is D