One frame how many bytes

Specifically for TCP, sequence numbers are used so the receiving host knows how to reassemble the data. So the size of a PDU at a particular layer is determined by the layer above? I understand the negatives of fragmentation; ideally one packet would fit into one frame. Do routers care about Layer 2 MTUs - are packets sent regardless of how many bytes can be transmitted at a time on the local link?

Where does TCP segment reassembly fit into all this? I guess if a packet is missing at the destination host TCP knows a segment is missing and so can't pass Application data up a layer. Hi Matt partially correct. I think the confusion is with the terms frame segment, packet and frame.

The application sends data as a stream of bytes down to TCP. IP adds a header onto the TCP segment to define the final source and destination. Finally the IP packet is passed down to the Ethernet Protocol who adds an Ethernet header and trailer into the IP packet making a frame. So in conclusion TCP decides how much data to send at one time. It will then add a TCP header onto the data to define the sending application and receiving application. The packet is then transited.

It is important to understand that once we have our data that we want to send all we are doing with encapsulation is adding additional information onto the data so that it gets from point A to point B.

For example imagine we are sending bytes of data at once. TCP sends data as a stream. I can think of two ways in which this might be done. TCP reassembles the stream on the receiving end by use of sequence numbers. Very true Sdavids. TCP determines how much data we are going to send by packing the stream of bytes it has received into a large chunk of bytes.

Once we have that data we are going to send and we move down the OSI model from layers 4 - 2 all we are doing is adding additional information onto that data in the form of headers and trailers so that there is enough information to move that data from point A to point B. When IP says that it can send 65, bytes at once that is a theoretical max on how much data a single IP packet can carry.

However TCP defines how much data we are sending as we move down the OSI model we are adding additional information onto that data which increases the overall size. Here we see how many fragments are required to send this data note, for some reason each fragment produces two log entries so you have to divide the number by 2 :.

R1 show log inc. R1 show log count. From above you can see that the packet is bytes long. The last fragment was:. Like in Sdavids example it is usually another protocol defining how much data we are sending. Thanks for the explanations and examples, it makes much more sense to me now.

I don't know if it is good etiquette on this forum to ask more than one question in a thread but I have another query, about packet transmission. Talking about the LSA example from above will those 45 packets be sent one after another till they are all sent and received or could they be intermingled with other packets? Then the receiving router sends the packets on to their correct destination?

Yes, if the router has a reason to use the link over which the LSA is being transmitted. The router will also route other packets while carving up the LSA so on the wire you might see dozens and dozens of packets, destined for dozens and dozens of different locations, intermingled with the LSA fragments.

The router will prioritize the LSA fragments over other traffic during times of congestion because the routing protocol packets are critical to the operation of the network. I'm wondering now about end to end speed between host devices. I'm looking at a games console game update as an example. Then there is the path to the server online and back to my LAN.

The Ethernet adapter will send the frame bits for some X uSec and pause the transmission to check for a collision on the medium and then continue. When it is checking like this, the 1st bit signal of current own packet hit the other end of network and echoed back in the opposite direction. Our device is still busy in sending the remaining bytes and checking for collision. And if it does like this, it will never be able to successfully transmit a complete packet.

So it should ignore this echo and continue its transmission. But there is no special technique to differentiate the echo and collisions, then how to ignore the echo collision? The trick is timing, by the time the own frame echo is reaching back, all the other systems in the network detected a collision and stopped their transmission. So that the device can safely ignore this echo. Now the next question is, after how much time, it should start ignoring the collisions? For everyone to notice the collision, we should keep transmitting the bits for the RTD Time or more.

So the minimum frame size must be greater then the number of bits that can be transmitted with-in the RTD time. The recent advancement in the Ethernet speed leads to the increase in the minimum frame size. But to maintain the backward compatibility the same minimum frame size retained. And instead of increasing the frame size, the length of the high speed Ethernet network is reduced. The Maximum frame size or for VLAN support was decided to prevent a particular node from hogging the network for long time, efficient buffer handling, retransmission, error recovery, QoS, etc.

Hi Bharathi,It was a nice explanation. I have a general about MTU. When i send a ping with a size of say bytes, we can see 2 fragmented packets followed by a ping request. Similarly for Ping reply, we can see 2 fragmented packets followed by a ping reply.

This is the actual working scenario. Ethernet header size is 20 bytes, and IP header size is 20 bytes. Even though this Bytes moves out from the NIC, but in First fragmentes packet, we can see Fragment offset as 0 , in second Fragmented packet we can see as ,in third Fragmented packet Ping request Packet we can see fragment offset as Please Clarify.

Thanks in Advance,Satish. MAC and Type. There is no Data field and Length field. Only in How the upper layer calculates packet length field, when i use Ethernet2 since there is no Lenght field??

If I understood your Q properly: 1. MTU is Maximum frame size, which is controlled at L2. L3 will fragment the packet based L2 MTU value.

L3 or L4 has to know and record the data size in the packet, that they are sending to lower layer. At L2, start and end of the frame is identified by the Start and End of frame bit patterns. This one is widely applied. Now we have pretty much get the Bit Depth clear.

And in the following content related to Bit Depth, we will take the 8-bit per channel as an example to help us. And at this time, we can also calculate how much storage one frame, or one image will take up.

The formula, which we have mentioned above, is,. Like we have said, we take the 4K resolution x and 8-bit per channel as examples. Well, bits are less likely to see out there. It would be much better to convert into megabytes MB. That is to say, for 4K video file with 8-bit Bit Depth, the size of one frame is 8. Now we can answer the question of how to calculate video file size, and here is the video file size formula,.

Please let me explain to you the meaning of each item in this formula. Time refers to how long your video is; Frame per Second, or called FPS, means how many frames will be played per one second for this video; Pixel per Frame, or the Resolution, and Bit Depth have been talked about above.

This is kind of scary. Probably many of you even would not believe this number, right? But it is true. If we want to transfer such a 4K video to one of our friends, and the uploading internet speed is 10M, it would take us about 25 hours.

Seriously, this is very inconvenient for information sharing. So the engineers came up with the solution - encoding the video files. The main purpose of video encoding is reducing its size.

There are many ways to encode a video file. If you are interested in this, you can refer to this guide from Wikipedia. Each encoding method comes with its unique compression algorithm and ratio. Besides, we also need to know that even if different devices, for example iPhone X and Android Pixel, use the same encoding way to compress a same video file, the final size probably would be not same. So considering its complexity, we will not continue with how to calculate encoded video file.

But this online tool on this page can work it out easily.