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.
Related
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
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.
I have a client which issues parallel requests for data from a server. Each request uses a separate TCP connection. I would like to estimate the available throughput (bandwidth) based on the received data.
I know that for one connection TCP connection I can do so by dividing the amount of data the has been download by the duration of time it took to download the data. But given that there are multiple concurrent connections, would it be correct to sum up all the data that has been downloaded by the connections and divide the sum by the duration between sending the first request and the arrival time of the last byte (i.e., the last byte of the download that finishes last)? Or am I overlooking something here?
[This is a rewrite of my previous answer, which was getting too messy]
There are two components that we want to measure in order to calculate throughput: the total number of bytes transferred, and the total amount of time it took to transfer those bytes. Once we have those two figures, we just divide the byte-count by the duration to get the throughput (in bytes-per-second).
Calculating the number of bytes transferred is trivial; just have each TCP connection tally the number of bytes it transferred, and at the end of the sequence, we add up all of the tallies into a single sum.
Calculating the amount of time it takes for a single TCP connection to do its transfer is likewise trivial: just record the time (t0) at which the TCP connection received its first byte, and the time (t1) at which it received its last byte, and that connection's duration is (t1-t0).
Calculating the amount of time it takes for the aggregate process to complete, OTOH, is not so obvious, because there is no guarantee that all of the TCP connections will start and stop at the same time, or even that their download-periods will intersect at all. For example, imagine a scenario where there are five TCP connections, and the first four of them start immediately and finish within one second, while the final TCP connection drops some packets during its handshake, and so it doesn't start downloading until 5 seconds later, and it also finishes one second after it starts. In that scenario, do we say that the aggregate download process's duration was 6 seconds, or 2 seconds, or ???
If we're willing to count the "dead time" where no downloads were active (i.e. the time between t=1 and t=5 above) as part of the aggregate-duration, then calculating the aggregate-duration is easy: Just subtract the smallest t0 value from the largest t1 value. (this would yield an aggregate duration of 6 seconds in the example above). This may not be what we want though, because a single delayed download could drastically reduce the reported bandwidth estimate.
A possibly more accurate way to do it would be say that the aggregate duration should only include time periods when at least one TCP download was active; that way the result does not include any dead time, and is thus perhaps a better reflection of the actual bandwidth of the network path.
To do that, we need to capture the start-times (t0s) and end-times (t1s) of all TCP downloads as a list of time-intervals, and then merge any overlapping time-intervals as shown in the sketch below. We can then add up the durations of the merged time-intervals to get the aggregate duration.
You need to do a weighted average. Let B(n) be the bytes processed for connection 'n' and T(n) be the time required to process those bytes. The total throughput is:
double throughput=0;
for (int n=0; n<Nmax; ++n)
{
throughput += B(n) / T(n);
}
throughtput /= Nmax;
I have 2 questions:
While calculating RTT, should we take transmission time into consideration or not?
The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometers is t seconds. Let T bits/sec be the channel capacity. Assuming that processing delay is negligible, what is the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used?
1) RTT is the measured time from segment transmission until ACK received. So transmission time will be considered obviosly. One can ignore the retransmission time.
RTT is calculated by considering the values of propagation delay, queing delay, transmission delay and processing delay..
Rtt = 2*[propagation delay + N(queing delay+transmission delay+processing delay)]
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.).