I need to calculate Round Trip Time in an Ethernet segment 100 Mbit. The calculation is made via UDP protocol with Record TimeStamp option, the block of data in UDP is 64byte.
Also I need to draw the structure of data transported in the channel.
to calculate RTT it would be proper to use tcp syn sent segment compared with syn received .
calculating boths timestamps would give you req o/p.. IN order to structure udp streams us wireshark's FOLLW UDP STREAM OPTIN.. AND convert it into ascii or raw format....
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Host A is sending data to host B over a full duplex
link. A and B are using the sliding window
protocol for flow control. The send and receive
window sizes are 5 packets each. Data packets
(sent only from A to B) are all 1000 bytes long
and the transmission time for such a packet is
50 ps. Acknowledgment packets (sent only from
B to A) are very small and require negligible
transmission time. The propagation delay over
the link is 200 trrs. What is the maximum
achievable throughput in this communication?
This question was asked in gate my question. I have calculated it, but what is the meaning of word 'maximum'? The calculation was just for throughput. How would one calculate minimum throughput?
I think maximum means assuming no packets loss and therefore no retries. Also, no additional transmission time above the 50ms. Basically, given the above transmission time and propagation delay, how many bytes can be sent and acknowledged per sec?
My intuition is to figure out how long it takes to send 5 packets to fill up the window with the propagation delay added. Then add the time for the acknowledgement for the first packet to arrive at the sender. That's your basic window send and acknowledgement time because as soon as the acknowledgement arrives the window will slide forward by one packet.
Since the window is 5 packets and each packet is 1,000 bytes then the maximum throughput should be 5,000 bytes / the time you calculated for the above cycle.
MTU (Maximum transmission unit) is the maximum frame size that can be transported.
When we talk about MTU, it's generally a cap at the hardware level and is for the lower level layers - DataLink and Physical layer.
Now, considering the OSI layer, it does not matter how efficient are the upper layers or what kind of magic-sauce they are applying, data-link layer will always construct frames of size < 1500 bytes (or whatever is the MTU) and anything in the "internet" will always be transmitted at that frame size.
Does the internet's transmission rate really capped at 1500 bytes. Now-a-days, we see speeds in 10-100 Mbps and even Gbps. I wonder for such speeds, does the frames still get transmitted at 1500 bytes, which would mean lots and lots and lots of fragmentation and re-assembly at the receiver. At this scale, how does the upper layer achieve efficiency ?!
[EDIT]
Based on below comments, I re-frame my question:
If data-layer transmits at 1500 byte frames, I want to know how is upper layer at the receiver able to handle such huge incoming data-frames.
For ex: If internet speed in 100 Mbps, upper layers will have to process 104857600 bytes/second or 104857600/1500 = 69905 frames/second. Network layer also need to re-assemble these frames. How network layer is able to handle at such scale.
If data-layer transmits at 1500 byte frames, I want to know how is
upper layer at the receiver able to handle such huge incoming
data-frames.
1500 octets is a reasonable MTU (Maximum Transmission Unit), which is the size of the data-link protocol payload. Remember that not all frames are that size, it is just the maximum size of the frame payload. There are many, many things with much smaller payloads. For example, VoIP has very small payloads, often smaller than the overhead of the various protocols.
Frames and packets get lost or dropped all the time, often on purpose (see RED, Random Early Detection). The larger the data unit, the more data that is lost when a frame or packet is lost, and with reliable protocols, such as TCP, the more data must be resent.
Also, having a reasonable limit on frame or packet size keeps one host from monopolizing the network. Hosts must take turns.
For ex: If internet speed in 100 Mbps, upper layers will have to
process 104857600 bytes/second or 104857600/1500 = 69905
frames/second. Network layer also need to re-assemble these frames.
How network layer is able to handle at such scale.
Your statement has several problems.
First, 100 Mbps is 12,500,000 bytes per second. To calculate the number of frames per second, you must take into account the data-link overhead. For ethernet, you have 7 octet Preabmle, a 1 octet SoF, a 14 octet frame header, the payload (46 to 1500 octets), a four octet CRC, then a 12 octet Inter-Packet Gap. The ethernet overhead is 38 octets, not counting the payload. To now how many frames per second, you would need to know the payload size of each frame, but you seem to wrongly assume every frame payload is the maximum 1500 octets, and that is not true. You get just over 8,000 frames per second for the maximum frame size.
Next, the network layer does not reassemble frame payloads. The payload of the frame is one network-layer packet. The payload of the network packet is the transport-layer data unit (TCP segment, UDP datagram, etc.). The payload of the transport protocol is application data (remember that the OSI model is just a model, and OSes do not implement separate session and presentation layers; only the application layer). The payload of the transport protocol is presented to the application process, and it may be application data or an application-layer protocol, e.g. HTTP.
The bandwidth, 100 Mbps in your example, is how fast a host can serialize the bits onto the wire. That is a function of the NIC hardware and the physical/data-link protocol it uses.
which would mean lots and lots and lots of fragmentation and
re-assembly at the receiver.
Packet fragmentation is basically obsolete. It is still part of IPv4, but fragmentation in the path has been eliminated in IPv6, and smart businesses, do not allow IPv4 packet fragments due to fragmentation attacks. IPv4 packets may be fragmented if the DF bit is not set in the packet header, and the MTU in the path shrinks smaller than the original MTU. For example, a tunnel will have a smaller MTU because of the tunnel overhead. If the DF bit is set, then a packet too large for the MTU on the next link, the packet is dropped. Packet fragmentation is very resource intensive on a router, and there is a set of steps that must be performed to fragment a packet.
You may be confusing IPv4 packet fragmentation and reassembly with TCP segmentation, which is something completely different.
I have been learning wireshark lately. While inspecting TCP segments, I saw a strange situation, at least for me. There was a mismatching SEQ,ACK numbers. Then I realized that difference between two ACK's same as 1 and half package size. However, as far as I know, ACKs are only growing by the full packet size. So what happened here?
SEQ ACK
1 1
2897 8689
5793 13033 <--
8689 14481
11585
14481
Well, there's not much to go by here, but I'm going to guess that you are capturing packets on the machine sending the large packets, i.e., the packets with 2896 bytes of TCP payload. As such, you are seeing the packets before they are IP-fragmented and actually transmitted out on the wire.
But you can't just send 2896 bytes of data onto the wire; Ethernet links typically impose a 1500 byte MTU (Maximum Transmission Unit), and when you account for IP and TCP header overhead, you usually end up with 1460 bytes of available payload or MSS (Maximum Segment Size). In your case it looks like you're only getting 1448 bytes of available MSS, most likely due to the addition of one or more IP and/or TCP header options.
In any case, the 2896 bytes of payload are going to be fragmented over 2 IP fragments, each containing 1448 bytes of TCP payload. I'm pretty sure that what you're seeing is an ACK from the receiver after having received 1 full segment plus 1 IP fragment from the next segment.
The previous ACK number was 8689, and 8689 + 2896 = 11585. Now add 1/2 of the data segment (2896 / 2 = 1448) and you get 11585 + 1448 = 13033. That's the ACK number you're seeing. Now add the other 1/2 and you get 13033 + 1448 = 14481, which is the ACK number of the next packet.
I hope that makes sense?
For an in depth look at the drawbacks of local packet captures, I direct you to a well-written blog by Jasper Bongertz titled, "The drawbacks of local packet captures".
Since I do not have reputation to comment the answer of Christopher Maynard, I comment it here.
As Christopher stated correctly the MTU (Maximum Transmission Unit) is standard Ethernet networks is 1500Bytes. However, crafting 1460Bytes (1500Bytes - 20Bytes IP Header - 20Bytes TCP Header, ignoring possible TCP options) TCP segments imposes a high load on the network stack since every segment must be handled separately.
As an example, if I want to transmit 1Mb (equals to 1024*1024Bytes = 1048576Bytes) of data, it would result in 1Mb/1460Bytes = 719 segments. In contrast to 1Mb/65525Bytes (theoretical max TCP segment size) = 16 segments. Since the overhead in the Kernel is mostly independent of the segment/packet size, small segments require much more processing.
To counteract this fact, TCP Segmentation Offloading (TSO) was developed. TSO allows the Kernel to craft TCP segments with maximum size and the NIC (Network Interface Card) uses hardware acceleration to split these segments into the separate TCP segments of 1460Bytes. Therefore, no IP fragmentation is required.
This question is on a test review and I'm not really sure of the answer.
TCP packets are being sent from a client to a server. The MMS is equal to 1460 bytes, and each TCP packet is sent with the maximum capacity. How many TCP packets can be sent before the sequence number field in the TCP header will wrap around?
How much time in seconds will this take on a 1 Mbit/s link?
How much time in seconds will this take on a 1Gbit/s link?
Is there some sort of formula used to figure this out?
Thanks!
Each TCP segments contains 1460 bytes, and sequence number in TCP header is 4 bytes=32 bits so there need to be send 2^32 bytes (because sequence number measure bytes and not bits) in order to sequence number field to wrap around.
In order to calculate the delay you need to consider:
Transmission time - time it takes to push the packet's bits onto the link.
Propagation time - time for a signal to reach its destination.
Processing delay - time routers take to process the packets header.
Queuing delay - time the packet spends in routing queues.
In your questions the transmission time is 1 Mbit/s and 1Gbit/s, and I assume the other delays are 0; so the time it will take to send 2^32 bytes= 8*2^32 bits on:
1 Mbit/s link:
8*2^32 / 10^6 = 34359 seconds
1Gbit/s link:
8*2^32 / 10^9 = 34 seconds
Hope this help you
For the TCP stream, i want to get the Network Jitter values as shown for the RTP packets under RTP statistics as Max Jitter and Mean Jitter in Wireshark.
Is there any option in Wireshark which gives the Jitter values for a TCP stream.
I have only found graphs related to RTT, Throughput, Seq. no. and Windows scaling but nothing related to Jitter is there for TCP streams.
Anyone has idea how can i get the Jitter values in Wireshark for TCP stream?
Regards
Nitin
I don't know of a way to calculate TCP jitter in Wireshark (or any other tool) but I assume that you could do this by comparing the deltas between the TCP timestamp fields and comparing to the actual arrival times.
You can export received timestamp values by adding a custom column with the expression tcp.options.timestamp.tsval then exporting as CSV or XML