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Determine the maximum length of the cable (in km) for transmitting data at a rate of $500$ Mbps in an Ethernet LAN with frames of size $10,000$ bits. Assume the signal speed in the cable to be $2,00,000$ km/s.

1. $1$
2. $2$
3. $2.5$
4. $5$
edited | 4k views

$\text{transmission time} \geq \text{round trip time of 1 bit}$

$\text{transmission time} \geq 2 \times \text{propagation time}$

$\dfrac{10,000\ bits}{500\ Mbps} \geq 2 \times \dfrac{d}{2\times 10^{5}\text{ km per sec}}$

$2 \text{ km} \geq d$

Option B is correct.

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Why are you considering the case of "small packet"? In sliding window packet size is fixed, and hence we have to consider that packet size. The below one is not an authentic link but this is the only one I could find online. You can check your network book - but see the use of RTT in sliding window section only.

http://stackoverflow.com/questions/14662365/round-trip-time-sliding-window

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RTT definition is fine for LAN. But it will confuse for sliding window case.

http://robotics.eecs.berkeley.edu/~wlr/Tutorials/nets_files/Page921.htm

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Let me put a perspective :

That link you gave has a very authentic nature, that guy EJP

asked a very nice question. i.e. RTT of what?

the RTT is the answer to "The time to get ACK after sending the packet".

Suppose we ask this question to the Sliding Window Protocol. It measures its RTT by checking after how much time packet is getting received once send. asking this protocol was transmission time included in your measurement of RTT is meaningless.

transmission was involved as well as not. that protocol never did the transmission(i.e. the act of placing bits on channel) coz it was done by some different protocol, so it includes that time BUT when it transfers the collected data (in form of fixed sized packets) to the next layer, there was no time(just few machine cycles which made it happen) involved.

Similarly when this question is asked to TCP then it does this(link). It determines its RTT during connection establishment phase, using SYN and ACK packets as shown in the post. Here, small sized packets were used.

Small packet case is considered because RTT is defined like that only, see my last comment.

saying $\text{transmission time} \geq \text{round trip time}$ in the above question is ok coz For Ethernet we check if One single bit is involved in the collision, anyhow. That bit's RTT.

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Saying here is correct but it is because "for one bit" case or the "small" packet assumption. If you use this in exam, you will get full mark. But using that here will confuse most people when they do sliding window problem- because they don't really care about "assumptions" and won't ever change the definition as per the situation. You can see the Berkeley link or any genuine source for sliding window protocol problems- transmission time is added as part of RTT- simply because packet size is significant there, unless propagation delay is too high..
+2

for sliding window protocol problems- transmission time is added as part of RTT- simply because packet size is significant there, unless propagation delay is too high..

I'm in support of this.

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you guys mean that the transmission time of Acknowledgement should be added to RTT?
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There is no doubt in that. But that is usually "0" or negligible.
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the above discussion was about that only, right? or something else?
+1
@Arjun sir: Is the discussion between amarVashishth and you about transmission time of the frame/packet being sent? I also have a doubt regarding that, but I used the following reasoning:

In sliding window, we send one frame/packet but receiver sends acknowledgement only when it receives the complete packet (i.e. the last bit of the packet), so RTT would consist of transmission time of the packet being sent + Propagation delay of packet + Transmission delay of Ack +Propagation delay of Ack

But in the case of Ethernet, we're considering only cases of collisions, and the worst case of collision can occur when the first bit is just about to reach the destination. So, this bit will have to travel one propagation time (to reach the destination) and collision signal will again take one propagation time (to reach the sender), so in total RTT (needed to detect collision) is 2*Propagation time.

Is this reasoning correct?
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@Amar, Why you've multiplied 2 with the propagation time??
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@arjun sir please clear this doubt the reason we considering that transmission time ≥ 2* propagation delay because in ethernet protocol such as CSMA there is no ack mechanism and to detect if there is any collision the sender should have packet length of 2*propagation delay so thst if there is any collision it could be detected.

Please clear pratyush doubt as well he also have same question
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@devshree coz propagation delay is one way latency to get round trip time we need to multiply it by 2 and add queuing delay + processing delay + transmit time of ack is involved at receiver or else it is cinsidcons negligible

In Ethernet LAN, there is a relation that transmission time >= 2*propagation time (To detect collision)
=> 10000/(500*1000000) >= 2*length/200000
=> length = 2km
so,
Answer is B i.e 2 Km

ans b)

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