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I read Token ring protocol from a book Forouzon.
According to book,
Let N be the number of stations in the ring, THT the token holding time, Tt be the transmission time of packet, Tp be the propagation time of packet on Channel/ Link.
Then Cycle Time = n * THT + Tp (this is cycle time for token)
and efficiency = (useful time)/(Cycle Time)
Here useful time is stated as N * Tt. (justified as transmission time at each station in single cycle of token passing)
And thus proven efficiency = (N * Tt)/(n*THT + Tp)
My question is: why not this (N*Tt) is added in Cycle time?
so the efficiency could become efficiency = (N * Tt)/(n*THT + Tp +N * Tt)
Yes. But it has already been included.
Token Holding time is THT = Transmission Time + Ring Latency time (for single round of packet transmission)
As (THT = Tt + Tp) .
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A lot of ISP's sell their products saying: 100Mbit/s Speed.
However, compare the internet to a packet service, UPS for example.
The ammount of packages you can send every second(bandwith) is something different then the time it takes to arrive(speed).
I know there are multiple meanings of the term 'bandwith' so is it wrong to advertise with speed?
Wikipedia( http://en.wikipedia.org/wiki/Bandwidth_(computing) )
In computer networking and computer science, bandwidth,[1] network
bandwidth,[2] data bandwidth,[3] or digital bandwidth[4][5] is a
measurement of bit-rate of available or consumed data communication
resources expressed in bits per second or multiples of it (bit/s,
kbit/s, Mbit/s, Gbit/s, etc.).> In computer networking and computer
science, bandwidth,[1] network
bandwidth,[2] data bandwidth,[3] or digital bandwidth[4][5] is a
measurement of bit-rate
This part tells me that bandwith is measured in Mbit/s, Gbit/s.
So does this mean the majority of ISP's are advertising wrongly while they should advertise with 'bandwith' instead of speed?
Short answer: Yes.
Long answer: There are several aspects on data transfer that can be measured on an amount-per-time basis; Amount of data per second is one of them, but perhaps misleading if not properly explained.
From the network performance point of view, these are the important factors (quoting Wikipedia here):
Bandwidth - maximum rate that information can be transferred
Throughput - the actual rate that information is transferred
Latency - the delay between the sender and the receiver decoding it
Jitter - variation in the time of arrival at the receiver of the information
Error rate - corrupted data expressed as a percentage or fraction of the total sent
So you may have a 10Mb connection, but if 50% of the sent packages are corrupted, your final throughput is actually just 5Mb. (Even less, if you consider that a substantial part of the data may be control structures instead of data payload.
Latency may be affected by mechanisms such as Nagle's algorythm and ISP-side buffering:
As specified in RFC 1149, An ISP could sell you a IPoAC package with 9G bits/s, and still be true to its words, if they sent to you 16 pigeons with 32GB SD cards attached to them, average air time around 1 hour - or ~3,600,000 ms latency.
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Maybe it's a stupid question!
Assume a P2P network in which peers independently try to find and connect to good nodes.
good nodes are those that are closer(in term of RTT) and has higher bandwidth.
in this scenario RTT is more important (for example RTT has 90% weight).
I want to obtain the liner combination of RTT and bandwidth in a meaningful way.
but it's obvious that the nature of these two metric is inconsistent.
How can I combine these two metrics ?
One of the key unicast routing protocols (it is a point to point and not P2P), EIGRP, uses several metrics to come up with a single metric. EIGRP is a very popular routing protocol. Delay (aka RTT) and bandwidth are typically the two key params for EIGRP. So, your questions is a relevant one!
This should help:
http://www.cisco.com/en/US/prod/collateral/iosswrel/ps6537/ps6554/ps6599/ps6630/whitepaper_C11-720525.html
Given the nature of bandwidth and delay, you could have multiple peers scoring the same final metric. For example, you could have one peer, P1, with values B and D. and another peer, P2, with values 2B and 2D. Assuming the bandwidth represents the capacity of the pipe, you would be able to transport a given amount of data with both of these peers in the same time!
You should certainly use a default value of weights, let us say, 0.50/0.5, for these two params. But, you might find that for a given goal (faster routing convergence, say), a different value of weights might be better.
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Recall that propagation delay d/s is the time to transmit one bit over a link
And transmission delay is the time to transmit a whole packet over a link
Then, why isn't packet length * propagation delay = transmission delay?
Because they're measuring different things.
Propagation delay is how long it takes one bit to travel from one end of the "wire" to the other (it's proportional to the length of the wire, crudely).
Transmission delay is how long it takes to get all the bits into the wire in the first place (it's packet_length/data_rate).
The transmission delay is the amount of time required for
the router to push out the packet.
The propagation delay, is the time it takes a bit to propagate from one router to the next.
the transmission and propagation delay are completely different! if denote the length of the packet by L bits, and denote the transmission rate of the link from first router to second router by R bits/sec. then transmission delay will be L/R. and this is depended to transmission rate of link and the length of packet.
then if denote the distance between two routers d and denote the propagation speed s, the propagation delay will be d/s. it is a function of the Distance between the two routers, but
has no dependence to the packet's length or the transmission rate of the link.
Obviously , packet length * propagation delay = trasmission delay is wrong.
Let us assume that you have a packet which has 4 bits 1010.You have to send it from A to B.
For this scenario,Transmission delay is the time taken by the sender to place the packet on the link(Transmission medium).Because the bits(1010) has to be converted in to signals.So it takes some time.Note that here only the packet is placed.It is not moving to receiver.
Propagation delay is the time taken by a bit(Mostly MSB ,Here 1) to reach from sender(A) to receiver(B).
Transmission Delay : Amount of time required to pump all bits/packets into the electric wire/optic fibre.
Propagation delay : It's the amount of time needed for a packet to reach the destination.
If propagation delay is very high than transmission delay the chance of losing the packet is high.
Transmission Delay:
This is the amount of time required to transmit all of the packet's bits into the link. Transmission delays are typically on the order of microseconds or less in practice.
L: packet length (bits)
R: link bandwidth (bps)
so transmission delay is = L/R
Propagation Delay:
Is the time it takes a bit to propagate over the transmission medium from the source router to the destination router; it is a function of the distance between the two routers, but has nothing to do with the packet's length or the transmission rate of the link.
d: length of physical link
S: propagation speed in medium (~2x108m/sec, for copper wires & ~3x108m/sec, for wireless media)
so propagation delay is = d/s
The transmission delay is the amount of time required for the router to push out the packet, it has nothing to do with the distance between the two routers.
The propagation delay is the time taken by a bit to to propagate form one router to the next
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I'm reading one of the Distance vector protocol RIP and come to know maximum hop count it uses is 15 hops but My doubt is why 15 is used as maximum Hop count why not some other number 10,12 or may be 8 ?
My guess is that 15 is 16 - 1, that is 2^4 - 1 or put it otherwise: the biggest unsigned value that holds in 4 bits of information.
However, the metric field is 4 bytes long. And the value 16 denotes infinity.
I can only guess, but I would say that it allows fast checks with a simple bit mask operation to determine whether the metric is infinity or not.
Now the real question might be: "Why is the metric field 4 bytes long when apparently, only five bits are used ?" and for that, I have no answer.
Protocols often make arbitrary decision. RIP is a very basic (and rather old protocol). You should keep that in mind when reading about it. As said above, the max hop count will be a 4 byte field, where 16 is equivalent to infinity. 10 is not a power of 2 number. 8 was probably deemed too small to reach all the routers.
The rationale behind keeping the maximum hop count low is the count to infinity problems. Higher max hop counts would lead to a higher convergence time. (I'll leave you to wikipedia count to infinity problem). Certain versions of RIP use split horizon, which addresses this issue).
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Why does TCP wait for three duplicate ACK before fast retransmit?
RFC 2001 says
Since TCP does not know whether a duplicate ACK is caused by a lost
segment or just a reordering of segments, it waits for a small number
of duplicate ACKs to be received. It is assumed that if there is
just a reordering of the segments, there will be only one or two
duplicate ACKs before the reordered segment is processed, which will
then generate a new ACK. If three or more duplicate ACKs are
received in a row, it is a strong indication that a segment has been
lost. TCP then performs a retransmission of what appears to be the
missing segment, without waiting for a retransmission timer to
expire.
The reasoning for not doing the retransmit until the third duplicate seems to be that until that point it's more likely to just be out-of-order delivery and the retransmit isn't really needed.