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

Total distance for RTT = 4 Km

Transfer rate =2 *10^8   m/s  

Time to transfer =(4*10^3)/(2*10^8)=2*10^-5 sec

Data rate  =  10^7 bps

Packet size  = 2*10^-5*10^7 bits = 200 bits

so its option D ....

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bps means bits per second not Bytes per second

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(doubt) yes the packet size ahould be 25 bytes, but question says what could be minimum size .

a packet should be equal to or more than 25bytes so can't we say 50 bytes(option a) is fulfilling the question criteria?

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@skyby

50B is valid but it is not minimum.

but if it is asking minimum from the options then 50B is correct.

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well that's quite dependent on how we interpret question in one go during Exam, anyway thanks for clearing it

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@Arjun   @Bikram  

Sir,Why we are not adding jamming signal 32 bits long  when collision occurs .i.e

Tx of frame = 2*propogation time+ Tx of Jamming Signal  in this question it is not mention about jamming signal but whenever collision occurs jammming signal is generated  

Similar type of question is asked in Gate 2005 https://gateoverflow.in/1397/gate2005-74 but in that jamming signal was mentioned in the question

 

 

 

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@skyby

In general, we have to choose the correct answer from the options. If none of them are correct, then we choose the best possible one.

Here “none of these” is already given as an option so it has to be the answer (because 25B is not present). But say, the options were 50B, 100B, 200B, 250B...then you could select 50B as the answer

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 @Abhrajyoti00 @Kabir5454

Tt >= 2*Tp

or     Tt > 2*Tp

which is correct

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@JAINchiNMay It is $Tt >=2*Tp$.

In order to detect the collision, in worst case keep in transmitting the data→ Until collision is detected (got back). Now suppose Stations $A$ and $B$ are at extreme  ends, and $A$ start transmitting it’s packet at $t=0$. Assume exactly at $Tp-1 s$ (say units are in sec), B also starts, then at $t = Tp$ there will be collision. So it will take one more $Tp$ for the collision to reaach at $A$. So $A$ must transmit atleast for $2*Tp$.

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thanks :)

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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