Data on link layer propagation correlation time calculation

Generally this kind of topic will make me feel very suffering, do not know how to calculate. Finally get through this kind of topic, summed up here.

First look at the common expression of this type of topic: as shown in the figure, the router in the image of the storage-forwarding way, all the link propagation rate is 100Mbps (can be other numbers), the packet size is 1000B (also can be other numbers), where the packet head size (some can be ignored), group disassembly time, Propagation delay, etc. can be ignored. This article will talk about not ignoring or affecting the macro data transfer time, only in the last group slightly affected. Based on the topology map, what is the time to transfer the size of 10MB?

First of all, here to explain the difference m, in 10MB data here, is the storage value, with 1mb=220bits 1MB = 2^{20}bits, in the transmission rate, m=106 M = 10^6.

This is a very important distinction.

We know that sending a piece of data from one host to another needs to go through a computer network. This process needs to be elaborated again to clarify how to calculate the data transmission time consumption problem.

Let's assume that host A to host B.
A large chunk of data, whether it is a direct transmission or divided into multiple groups of transmissions, but first asked, where the time has gone .

Host a think, the task came, have to pass it to B, I don't care about the b,c,d or what, I only know to let it on the road.

Right, on the road, on what road, on the link.
A actually like a port, the goal is to load the cargo into the container and then let the container go by boat. The packing process is time-consuming, and we cannot think that this is a short time and there is no clear concept.

So, how long does it take to get the data on the road? The right is the transfer rate, also known as the transmit rate.

The process of sending is clear, depending on the rate of the link.

OK, now the data is on the link, in fact, it is the electrical signal running on the link, at the speed of light running. Most of the time, the speed of transmission on the link is not counted, as if an instant can reach the router connected to a. In fact, you can think so.

We go from a to a connected router after a very short period of time.

Here, if I ask, what is the time between A and b directly connected through the link, a to B?
is not the blink shift.
No, it's time-consuming. A to send the data to the link.

To understand this direct-connect model, you don't have to worry about introducing other routers into the data transfer problem. Because, in essence, the router is also a computer host, and a transmission of things no difference, since it is the storage-forwarding mechanism, it means that the data came in the router, and then go out, also have to go through a boxing process. As if the goods from a port to a transit station, the first to write down the goods, so that the ship also rest, and so on (the rest of the time can be likened to queuing, can also be considered to rest time), and to be boxed. The router is also the case, it will be re-boxed, up and down a section of the link, specifically to calculate the transmission time (send time).

In this way, each router is counted, arriving at B, finally do not have to calculate the storage and forwarding time B.

In the process described above, we only see how a piece of data is transmitted.
Here I'll start with a macro concept: The whole transmission time is much larger than the packet transfer time.
Why this, intuitively, is that a large group is stored every time – forwards, and computes N times. When divided into multiple groups, although the process is the same, but considering the parallelism, in fact, we only consider a large group of data transmission time, the details of the time only to consider the last group.

In fact, the above paragraph is all about the point.

And after drawing, calculation, thinking, as long as the packet continuous transmission of the same interval, regardless of the test without considering the queue delay, propagation delay, etc., the total formula is:
When grouping:
(Total data Volume ÷ data rate) + the last packet from the first link to the end time (total data volume \div transfer rate) + the last packet from the first link to the end time

It is necessary to say the last packet transfer time, we consider: a point the last packet has been sent to the first link, which is exactly the total amount of data divided by the link transfer rate of time.

So we just need to take the last piece of data and calculate the time from the first link to the end. Let's say that the last youngest was taken care of by a and gave it a ride.

If there is no grouping, you can also divide the total data by the link transfer rate, and then analyze the transmission of this packet. Or start analysis from a send.

Recognizing that there are basically no problems with such topics.

Light says do not practice false bashi.

Let's talk about it in real hands.

The data transfer rate for all links in a packet-switched network using "store-and-forward", as shown in the following figure, is 980 Mbps, and the packet size is $ B, where the packet head size is B. If the host H1 to the host H2 send a size of 000 b in the case of packet disassembly time and propagation delay, from H1 Send to H2 received, the time required is at least (C)

A. 80.08 Ms B. 80.16 Ms C. 80.24 MS
Answer: Group size is 1000B, with x x groups

Then: 1000x=980,000+20x 1000x = 980,000 + 20x
Get x=1000 x = 1000 groupings.
is simply a mathematical calculation, not by instinct to judge 1000 groups.

Well, the 1000 groupings, according to the conclusion we formed, are:
Total data Volume ÷ link rate = (1000 packet ⋅1000b) ÷100mbps= (106⋅8bit) ÷ (100⋅106BIT/S) =80ms total data \div link rate = {(1000 packet \cdot1000b)}\ div100mbps = (10^6\cdot8 bit) \div (100\cdot 10^6bit/s) = 80ms

Consider the last grouping from the first link to the H2 time:
It is clear that there are two links left, two routes (that is, left-to-right lines).
A grouping is 1000B.

A group sends the time on a link:
1000b÷