Question from a networking class:
"In a csma/cd lan of 2 km running at 100 megabits per second, what would be the minimum frame size to hear all collisions?"
Looked all over and can't find info anywhere on how to do this. Is there a formula for this problem? Thanks for any help.
bandwidth delay product is the amount of data on transit.
propagation delay is the amount of time it takes for the signal to propagate over network.
propagation delay=(lenght of wire/speed of signal).
assuming copper wire i.e speed =2/3* speed of light
propagation delay =(2000/(2*3*10^8/3))
=10us
round trip time is the time taken for message to travel from sender to receiver and back from receiver to sender.
round trip time =2*propagation delay =20us
minimum frame size =bandwidth *delay (rtt)
frame size = bandwidth *rtt
=100Mbps*20us=2000bits
Related
Q: Why the vulnerable time in ALOHA depends on the frame
transmission time (Tfr), but in CSMA it depends on frame propagation
time (Tp)
-->I've understood that Vulnerable time is the time when there is a possibility for collision
There is no proper explanation about Transmission time and propagation time anywhere.
Please help
This is because of the difference in the magnitude of Transmission time and Propogation time. Most propogation is done through EM waves and thus for most distances the propogation delay is very negligible as compared to transmission time which is limited by the Bandwidth of a channel.
In the case of ALOHA, both these times contribute to the vulnerable time and thus the dominant factor is retained , however incase of CSMA the transmission time factor is avoided and thus only the propogation delay in considered although its very small , it must be taken into account to validate the efficiencies of CSMA protocols.
Please help me out understanding a concept in computer networks. I have understood what propagation delay and transmission time means. I can't understand how propagation time can be greater than transmission time? Please explain with one such example.
Transmission time = time it takes to send (dispatch) the data.
Propagation delay = time it takes for the data to reach the other side.
Source
For example, suppose you're sending 10 bits of data from the North Pole to the South Pole. If, for example, your computer (which is very old) has a 1bit/ms bit rate for sending, the transmission time would be 10ms. However, due to the length and the great number of hops required between the north and south poles, it could take 30ms for the first bit (and assuming everything goes well, for the other bits as well) to reach the south pole.
Thus, propagation time > transmission time.
According to Wikipedia Jitter is the undesired deviation from true periodicity of an assumed periodic signal, according to a papper on QoS that I am reading jitter is reffered to as delay variation. Are there any definition of the jitter in the context of real time applications? Are there applications that are sensitive to jitter but not sensitive to delay? If for example a streaming application use some kind of buffer to store packets before show them to the user, is it possible that this application is not sensitive to delay but is sensitive to jitter?
Delay: Is the amount of time data(signal) takes to reach the destination. Now a higher delay generally means congestion of some sort of breaking of the communication link.
Jitter: Is the variation of delay time. This happens when a system is not in deterministic state eg. Video Streaming suffers from jitter a lot because the size of data transferred is quite large and hence no way of saying how long it might take to transfer.
If your application is sensitive to jitter it is definitely sensitive to delay.
In Real-time Protocol (RTP, RFC3550), a header contains a timestamp field. The value of it usually comes from a monotonically incremented counter and the frequency of the increment is the clock-rate. This clock-rate must be the same all over the participant wants something with the timestamp field. The counters have different base offsets, because the start time may different or they contains it because of security reason, etc... All in all we say the clocks are not syncronized.
To show it in an example consider if we refer to snd_timestamp and rcv_timestamp the most recent packet sender timestamp from the RTP header field and receiver timestamp generated by the receiver using the same clock-rate.
The wrong conclusion is that
delay_in_timestamp_unit = rcv_timestamp - snd_timestamp
If the receiver and sender clock-rate has different base offset (and they have), this not gives you the delay, also it doesn't consider the wrap around the 32bit unsigned integer.
But monitoring the time for delivering packets is somehow necessary if we want a proper playout adaption algorithm or if we want to detect and avoid congestions.
Also note that if we have syncronized clocks delay_in_timestamp_unit might be not punctually represent the pure network delay, because of components at the sender or at the receiver side retaining these packets after and/or before the timestamp added and/or exemined. So if you calculate a 2seconds delay between the participant, but you know your network delay is around 100ms, then your packets suffer additional delays at the sender or/and at the receiver side. But that additional delay is somehow (or at least you hope that it is) constant, so the only delay changes in time is - hopefully - the network delay. So you should not say that if packet delay > 500ms then we have a congestion, because you have no idea what is the actual network delay if you use only one packet sender and receiver timestamp information.
But the difference between the delays of two consecutive packets might gives you some information about weather something wrong in the network or not.
diff_delay = delay_t0 - delay_t1
if diff_delay equals to 0 the delay is the same, if it greater than 0 the newly arrived packets needed more time then the previous one, and if it smaller than 0 it needed less time.
And from that relative information based on two consecutive delays you could say something.
How you determine the difference between two delay if the clocks are not syncronized?
Consider you stored the last timestamps in rcv_timestamp_t1 and snd_timestamp_t1
diff_delay = (rcv_timestamp_t0 - snd_timestamp_t0) - (rcv_timestamp_t1 - snd_timestamp_t1)
but that would be problem without maintaining the base offsets of the sender and the receiver, so reordering it:
diff_delay = (rcv_timestamp_t0 - rcv_timestamp_t1) - (snd_timestamp_t0 - snd_timestamp_t1)
and here you can subtract rcv timestamps from each other and it eliminates the offset rcv and snd contain, and then you can extract the rcv_diff from snd_diff and it gives you the information about the difference of the delays of two consecutive packets in the unit of the clock-rate.
Now, according to RFC3550 jitter is "An estimate of the statistical variance of the RTP data packet interarrival time".
In order to finally get to the point your question is
"What is the difference between the delay and the jitter in the context of real time applications?"
Tiny note, but real-time applications usually refer to systems processing data in a range of nanoseconds, so I think you refer to end-to-end systems.
Also despite of several altered definition of jitter, it all uses the difference of the delays of arrived packets and thus provide you information about the relative changes of the network delay, meanwhile delay itself is an absolute value of the time of delivery.
Its given that the bandwidth-delay product defines the number of bits that can fill the link.
The sender should send a burst of data of (2*bandwidth*delay) bits.
I am not getting why the term bandwidth*delay multiplied by 2.Please Explain the reason???
It depends what you mean by "delay". If delay is the round trip time (RTT) then you wouldn't multiply it by two. Presumably, in the formula you are looking at, the delay is the unidirectional transmiasion time, so you multiply it by 2 to estimate the RTT.
One RTT is the earliest time you could get an acknowledgement back for the first bit you transmitted, so that's why your window should be that big in order to fill the pipe.
Delay in your case is the propagation delay which is the time taken by the signal(message) to propagate from sender to receiver.
It is multiplied by 2 because the link is bidirectional i.e sender and receiver can both send the data at the same time i.e in order to completely fill the link you need to multiply the propagation delay by 2 and this term is known as round trip time(RTT).
bandwidth-delay product = RTT * bandwidth
bandwidth-delay product = 2 * propagation delay * bandwidth
where
RTT = 2 * propagation delay
i guess this product only work to tcp/ip, not udp/ip. as only tcp need confirmation about the sending data.
My understanding is that Bandwidth delay product refers to the maximum amount of data "in-transit" at any point in time, between two endpoints.
The thing that I don't get is, why multiply bandwidth by RTT. Bandwidth is a function of underlying medium, such as copper wire, fire optics etc and RTT is function of how busy intermediate nodes are, any scheduling applied at the intermediate nodes, distance etc. RTT can change, but bandwidth for practical purposes can be considered as fixed. So how does multiplying a constant value (capacity aka bandwidth) by fluctuating value (RTT) represents total amount of data in transit?
Based on this, will a really really slow have very large capacity? Chances are the "Causes" of RTT will start dropping.
Look at the units:
[bandwidth] = bytes / second
[round trip time] = seconds
[data volume] = bytes
[data volume] = [bandwidth] * [round trip time].
Unit-wise, it is correct. Semantically,
What is bandwidth * round trip time? It's the amount of data that left the sender before the first acknowledgement was received by the sender. That is, bandwidth * round trip time = the desired window size under perfect conditions.
If the round trip time is measured from the last packet and the sender's outbound bandwidth is perfectly stable and fully used, then the measured window size exactly calculates the number of packets (data and ACKs together) in transit. If you want only one direction, divide the quantity by two.
Since the round trip time is a measured quantity, it naturally fluctuates (and gets smoothed out). The measured bandwidth could fluctuate as well, and thus the estimated total volume of data in transit fluctuates as well.
Note that the amount of data in transit can vary with the data transfer rate. If the bottleneck is wire delay, then RTT can be considered constant, and the amount of data in transit will be proportional to the speed with which it's sent to the network.
Of course, if a round trip time suddenly rises dramatically, the estimated max. amount of data in transit rises as well, but that is correct. If there is no accompanying packet loss, the sliding window needs to expand. If there is packet loss, you need to reconsider the bandwidth estimate (and the bandwidth delay product drops accordingly).
To add to Jan Dvorak's answer, you can think of the 'big fat pipe' as a garden hose. We are interested in how much water is in the pipe. So, we take its 'bandwidth' i.e. how fast it can deliver water, which for a hose is determined by its cross-sectional area, and multiply by its length, which corresponds to the RTT, i.e. how 'long' a drop of water takes to get from one end to the other. The result is the volume of the hose, the volume of the pipe, the amount of data 'in the pipe'.
First, BDP is a calculated value used in performance tuning to determine the upper bounds of data which could be outstanding/unacknowledged. This, almost always, does not represent the quantity of "in-transit" data, but a target which tuning parameters are applied. If it represented "in-transit" data, always, there would be no room for performance tuning.
RTT does in fact fluctuate. This is why the expected worse case RTT is used in calculations. By tuning to the worse case, throughput efficiency will be at maximum when RTT is poorest. If RTT improves, we get outstanding Acks sooner, the pipe remains full and maximum throughput (efficiency) is maintained.
"Full pipe" is a misnomer. The goal is to keep the Tx side full, as the Rx contains Ack packets which are typically smaller than the transmitted packets.
RTT also aggregated asymmetrical upstream and downstream bandwidths (ADSL, satellite modem, cable modem, etc.).