Is there a maximum in specifications,do they start to interfer if many try to connect at the same time?
what are the modes of communication is there a secured mode or something else ?
Maximum packet size ?
can I send an image or a sound using ble ?
There is no limit in the specification. In reality at least around a millisecond must be allocated to serve each connection event. So if you use a 7.5 ms connection interval you could not expect more than at maximum 10 connections without getting dropped packets (and therefore larger latency). Connection setup/scanning will also miss a large amount of packets if the radio is busy handling current connections.
The maximum packet length is 31 bytes for advertisements (up to Bluetooth 4.2). While connected the longest packet length is 27 bytes. Bluetooth 4.2 defines a packet length extension allowing larger packets but far from all implementations support that.
The security that BLE offers is the bonding procedure. After bonding the devices have established a shared secret key which is then used to encrypt and sign all data being sent.
Sending normal-sized images or sounds will take several seconds since the throughput is quite low.
I think you should really read the Bluetooth specification or some summary to get the answer to your questions.
I have read some articles online and I got a pretty good idea about the TCP and UDP in general. However, I still have some doubts which I am sure not completely clear to me.
What is the proper way to calculate throughput ?
(Can't we just divide Total number of bytes received by total time taken ?)
What is that key feature in TCP that makes it have much much higher
throughput than UDP ?
UPDATE:
I understood that TCP uses windows which is nothing but that much segments can be sent before actually waiting for Acknowledgements. But my doubt is that in UDP segments are continuously sent without even bothering about Acknowledgements. So there is no extra overheads in UDP. Then, why the throughput of TCP is much much higher than that of UDP ?
Lastly,
Is this true ?
TCP throughput = (TCP Window Size / RTT) = BDP / RTT = (Link Speed in Bytes/sec * RTT)/RTT = Link Speed in Bytes/sec
If so then TCP throughput is always equals to the Know Link speed. And since the RTTs cancels out each other, the TCP throughput does not even depends on RTT.
I have seen in some network analysis tools like iperf, passmark performance test etc. that the TCP/UDP Throughput changes with Block size.
How is throughput dependent on Block size ?
Is Block size equals TCP window or UDP datagram size ?
What is the proper way to calculate throughput?
There are multiple ways, depending on what exactly you want to measure. They all boil down to dividing some number of bits (or bytes) to some duration, as you mention; what varies is which bits you are counting or (more rarely) which moments of time you are considering for measuring the duration.
The factors you need to take into account are:
At which layer in the network stack are you measuring throughput?
If you measure at the application layer, all that matters is what useful data you transmit to the other endpoint. For example, if you are transferring a file of 6 kB, the amount of data you count when measuring throughput is 6 kB (that is 6,000 bytes, not bits, and note the multiplier of 1000, not 1024; these conventions are common in networking).
This is usually called goodput and it may be different from what is actually sent at the transport layer (as in TCP or UDP), for two reasons:
1. Overhead due to headers
Each layer in the network adds a header to the data that introduces some overhead due to its transmission time. Moreover, the transport layer breaks your data into segments; this is because the network layer (as in IPv4 or IPv6) has a maximum packet size called MTU, typically 1,500 B in Ethernet networks. This value includes the network layer header size (e.g. the IPv4 header, which is variable in length but usually 20 B long) and the transport layer header (for TCP, it is also variable in length but usually 40 B long). This leads to a maximum segment size MSS (number of data bytes, without headers, in one segment) of 1500 - 40 - 20 = 1440 bytes.
Thus if we want to send 6 kB of application-layer data, we must break it into 6 segments, 4 of 1440 bytes each and one of 240 bytes. However at the network layer we end up sending 6 packets, 4 of 1500 bytes each and one of 300 bytes, for a total of 6.3 kB.
Here I have not considered the fact that the link layer (as in Ethernet) adds its own header and possibly also a suffix, which increases the overhead further. For Ethernet this is 14 bytes for the Ethernet header, optionally 4 bytes for VLAN tag, then a CRC of 4 bytes and a gap of 12 bytes, for a total of 36 bytes per packet.
If you consider a fixed-rate link, say of 10 Mb/s, depending on what you measure you will get a different throughput. Normally you want one of these:
The goodput, i.e. application layer throughput, if what you want to measure is application performance. For this example, you divide 6 kB by the transfer duration.
The link-layer throughput, if what you want to measure is network performance. For this example, you divide 6 kB + TCP overhead + IP overhead + Ethernet overhead = 6.3 kB + 5 * 36 B = 6516 B by the transfer duration.
Retransmission overheads
The Internet is a best-effort network, meaning that the packets will be delivered if possible, but may also be dropped. Packet drops are corrected by the transport layer, in case of TCP; for UDP, there is no such mechanism, which means that either the application does not care if some parts of the data do not get delivered, or the application implements retransmission itself on top of UDP.
Retransmission reduce goodput for two reasons:
a. Some data needs to be sent again, which takes time. This introduces a delay which is inversely proportional to the rate of the slowest link in the network between the sender and the receiver (a.k.a the bottleneck link).
b. Detecting that some data was not delivered needs feedback from the receiver to the sender. Due to propagation delays (sometimes called latency; caused by the finite speed of light in the cable), feedback can only be received by the sender with some latency, which slows down the transmission even more. In most practical cases, this is the most significant contribution to the extra delay caused by the retransmission.
Clearly, if you use UDP instead of TCP and you do not care about packet loss, you will of course get better performance. But for many applications, data loss cannot be tolerated, so such a measurement is meaningless.
There are some applications that do use UDP for transferring data. One is BitTorrent, which may use either TCP or a protocol they designed called uTP, which emulates TCP on top of UDP, but aims at being more efficient with many parallel connections. Another transport protocol implemented over UDP is QUIC, which also emulates TCP and offers multiplexing multiple parallel transfers over a single connection, and forward error correction to reduce retransmissions.
I will discuss forward error correction a little since it is related to your question about throughput. A naive way of implementing it is by sending every packet twice; in case one gets lost, the other still has a chance of being received. This reduces the amount of retransmissions to half, but also halves your goodput since you send redundant data (note that the network or link layer throughput remains the same!). In some cases this is fine; especially if the latency is very large, such as on intercontinental or satellite links. Moreover, some mathematical methods exist where you don't have to send a full copy of the data; for instance for every n packets you send, you send another reduntant one which is the XOR (or some other arithmetic operation) of them; if the redundant one gets lost, it doesn't matter; if one of the n packets gets lost, you can reconstruct it based on the redundant one and the other n-1. You can thus configure the overhead introduced by forward error correction to whatever amount of bandwidth you can spare.
How you are measuring the transfer time
Is the transfer completed when the sender finished sending the last bit over the wire, or does it also include the time it takes for the last bit to travel to the receiver? Additionally, does it include the time it takes to get a confirmation from the receiver, stating that all data has been received successfully and no retransmission is neede?
It really depends on what you want to measure. Note that for large transfers, one extra round-trip-time is insignificant in most cases (unless you are communicating, for instance, with a probe on Mars).
What is that key feature in TCP that makes it have much much higher throughput than UDP?
This is not true, although a common misconception.
In addition to retransmitting data when needed, TCP will also adjust its sending rate so that it will not cause packet drops by congesting the network. The adjustment algorithm has been perfected over decades, and usually converges quickly to the maximum rate supported by the network (actually, the bottleneck link). For this reason it is usually difficult to beat TCP in throughput.
With UDP, there is no rate limiting at the sender. UDP lets the application send as much as it wants. But if you try to send more than the network can handle, some of the data will be dropped, lowering your throughput, and also making the admin of the network you are congesting very angry. This means that sending UDP traffic at high rates is impractical (unless the goal is to DoS a network).
Some media applications are using UDP but rate-limiting the transfer at the sender at a very small rate. This is typically used in VoIP applications or Internet Radio, where you require very little throughput but low latency. I suppose this is one of the reasons for the misconception that UDP is slower than TCP; that is not the case, UDP can be as fast as the network allows.
As I said before, there are protocols such as uTP or QUIC, implemented over UDP, which achieve performance similar to TCP.
Is this true ?
TCP throughput = (TCP Window Size / RTT)
Without packet loss (and retransmissions), this is correct.
TCP throughput = BDP / RTT = (Link Speed in Bytes/sec * RTT)/RTT = Link Speed in Bytes/sec
This is correct only if the window size is configured to the optimal value. BDP/RTT is the optimal (maximum possible) transfer rate in the network. Most modern operating systems should be able to auto-configure it optimally.
How is throughput dependent on Block size ? Is Block size equals TCP window or UDP datagram size?
I don't see any block size in the iperf documentation.
If you refer to the TCP window size, if it is smaller than BDP, then your throughput will be suboptimal (because you waste time waiting for ACKs instead of sending more data; if needed I can explain further). If it is equal or higher to the BDP, then you achieve optimal throughput.
It depends on how you define "Throughput". It usually can be one of the followings.
Number of bytes (or bits) sent in a fixed period of time;
Number of bytes (or bits) sent and received on the receiver end in a fixed period of time;
You can apply these definition to every layer when people talking about throughput. In application layer, 2nd definition means the bytes have really been received by the receiver end of the application. Some people refer to it as "goodput". In Transport layer, say TCP, 2nd definition means the corresponding TCP ACKs are received. To me, most of people should be only interested in the bytes are really received by the receiver end. So, 2nd definition is usually what people mean by "Throughput".
Now, once we have a clear definition of throughput (2nd definition). We can discuss how to measure the throughput correctly.
Usually, people either use TCP or UDP to measure the network throughput.
TCP: People usually measure TCP throughput only on the sender end. As for packets successfully received by the receiver end, ACKs will be sending back. So, sender itself will know how many bytes are sent and received on the receiver end. Divided this number by the measuring time, we will know the throughput.
But, there are two things need to be noticed during TCP throughput measurement:
Is sender side always full buffer during the measurement? i.e. During the measurement period, sender should always has packets to send. It is important for correct throughput measurement. e.g. if I set my measuring time to be 60 seconds, but my file has been finished transmission in 40 seconds. Then there are 20 seconds the network is actually idle. I will under-estimate the throughput.
TCP rate is regulated by its congestion window size, slow-start duration, sender window (and receiver window) size. Sub-optimal configuration of these parameters will result in under-estimated TCP throughput. Although most of the modern TCP implementation should have a quite good configuration of all of these, it is hard for a tester to 100% sure all these configurations are optimal.
Due to these limitations/risks of TCP in network throughput estimation, quite a number of researchers will use UDP for measuring network throughput.
UDP: As UDP has no ACK sending back once the packets are successfully received, people has to measure the throughput in the receiver end. Or, if the receiver end is not easily accessed, people can compare the logs on both sender and receiver sides to determine the throughput. But, this inconvenience is mitigated by some throughput measuring tools. For example, iperf has embedded sequence numbers in its customized payload, so that it can detect any loss. Also, a receiver's report will be sent to the sender to show the throughput.
As UDP by nature is just sending whatever it has to the network and not waiting for the feedback. Its throughput (remember the 2nd definition) once measured will be the actual capacity (or bandwidth) of the network.
So, usually, the throughput measured by UDP should be higher than that from TCP although the difference should be small (~5%-10%).
One biggest drawback of UDP throughput measuring is that, when using UDP one should also make sure that sender-side buffer must be full. (Otherwise, it results in under-estimated throughput as TCP). This step will be little tricky. In iperf, one can specify the sending rate by -b option. Increase -b value in different rounds of testing will converge the throughput measured. For example, in my gigabit ethernet, I first use -b 100k in the test. The throughput measured is 100Kbps. Then I perform the following iterations to converge the maximum throughput which is the capacity of my ethernet.
-b 1m --> throughput: 1Mbps
-b 10m --> throughput: 10Mbps
-b 100m --> throughput: 100Mbps
-b 200m --> throughput: 170Mbps
-b 180m --> throughput: 175Mbps (this should be quite close to the actual capacity)
When transferring, say, 1GB worth of data over the internet this data is split into packets, each packet containing a small piece of data and aach of these packets are part of a frame.
Eg. Windows reports that you are transferring the file in 100kb/s over a TCP connection, but this appears to be the amount of data from the file being transferred per second, and does not seem to include the ip or tcp header, or ethernet frame.
What is the actual amount of traffic on the network needed to transfer at this speed? Or is that data actually already included in the transfer speed, but just small enough that it makes no significant difference?
Also, IP supports up to 1500 byte / packet (I think?), but what is the common size of data packets when loading, say, an HD image on reddit?
Sorry for the rather basic questions I probably should have figured out myself by now...
It depends on where you look at the transfer rate:
Task Manager will report all the transferred bytes (i.e. the sum of all the packets including their headers).
A file transfer program will report the transmitted payload.
Task Manager
If you look at Task Manger / Network, you can see the transmitted bytes together with the number of transmitted packets (unicast or non-unicast).
That data comes from the network driver (or at least something close to it), so it makes sense to report the total amount of data here (otherwise each packet would need to be inspected to calculate the payload).
There is also a graph showing the transfer rate. Those numbers could easily be compared with the reported numbers in file transfer software.
File transfer program
A file transfer program on the other hand, does not know the details about the packets being created in the lower layers (those could be any size). So the only option here is to report the amount of transmitted payload data / part of the file, which also makes more sense to the user.
Network packets
On normal networks (there could also be jumbo frames), a TCP-packet (full ethernet frame) is around 1500 bytes when fully loaded (on my system (IPv4) the packets are 1514 bytes with a total header size of 54 bytes -- 14 for the Ethernet header, 20 for the IP header and 20 for the TCP header). Those could be split in smaller packets along the way in the network, but in most cases they won't.
Transfer rate
When transferring a file (or other large datastream), on average 2 full packets (1514 bytes) will be sent each time, and 1 small packet (54 bytes) is received (the [ACK] packet). In this optimal case we have 2 x 1460 payload, and 2 x 54 bytes of overhead on the sending side + 54 bytes on the receiving side. When comparing to the maximum transfer rate of the internet connection, we also have to consider some latency.
Not all transmissions are optimal:
There could be packets that never arrived or where the checksum was wrong, so a retransmit would be needed.
In some cases data could be sent in smaller parts, causing a higher overhead/payload ratio (but with small chunks Nagle's algorithm could take care of that).
Certain software could be reading the file contents into small buffers (say 4096 bytes). Those could then be split in 2 x 1460 and 1 x 1176, introducing some extra overhead.
Conclusion
It's hard to tell or calculate the exact ratio transferred_bytes/payload. It depends on the quality of the internet connection (lost packets, retransmits), the software or API calls used to transfer the data, and even the underlying network (small frames vs jumbo frames for example).
A typical full-size TCP/IPv4 packet on the Internet is of size 1500B (maximum transmission unit (MTU)), of which (minimum) 20B are of TCP header and (minimum) 20B are of IPv4. This MTU was chosen to be compatible with Ethernet. Furthermore, there are application headers (e.g., HTTP for web, SIP/RTP/RTCP for voice call, etc.) included in this packet. The minimum MTU is 576B for IPv4 and 1280B for IPv6. One can see MTU on Linux with ifconfig command.
The best way to figure these values is by using a pcap tool/network analyzer such as Wireshark. Also refer to wiki pages or a good networking book for headers and fields of the protocols.
I'm pretty sure that the reported transer rate doen't include all the headers and overhead of the different layers in the protocal stack, since the reported thruput usually comes from some user-space application which would only get the actual data from the network stream object. It would need to do additional work to find out about all the headers and frames and other overhead that occurred in the different layers and affected the actual physical transmission.
I know that the definition of an iBeacon is a fixed specification of the advertising packet that it is transmitting:
9 bytes iBeacon prefix
16 byte UUID
2 bytes Major
2 bytes Minor
1 byte TX power
That being said, is there anything that would prevent a beacon from both sending out advertising iBeacon packets to wake up a phone's app and also transmit actual data content as part of a BLE packet? Would there be a lot of handshaking required in order to send / transmit additional data?
Is there some other way for a beacon to transmit data? One of my large concerns is spoofing of my beacons to falsify the data I am attempting to collect. I was hoping that being able to transmit some data along with an iBeacon packet would allow me to limit the spoofing.
Is something like that even feasible?
A few possibilities:
You can tack on one extra data byte to the end of the iBeacon transmission before it reaches its max advertisement length. This byte cannot be read by iOS devices, though, because Apple blocks reading raw data of iBeacon adverts. It would work on Android/Mac/Linux.
You can interleave a second advertisenent with mostly data fields and line the two up with a common identifier like the minor. The more bytes you allocate to lining up the advertisements, the fewer you have to use for data. You can't use the mac tobline them up, because that is unreadable in iOS for the iBeacon transmission.
You can make the beacon connectable via GATT, and read data fields with GATT attributes. The beacon will stop advertising, though, when connected. This limits throughput and reliability.
All of these options require you to build a custom BLE beacon that does multiple advertisements. It is not a trivial undertaking.
I'm currently building an application that is intended to run on an embedded system hooked up to a cellular data card. I've been made aware of some low-data plans from several carriers, and our application only generates about 5 bytes/second, lending itself to such plans.
However, I can't seem to figure out if the TCP/IP header overhead (about 40 bytes, give or take) is included in the calculation for data usage. Since I need real-time data, I've disabled Nagle's algorithm. This means for each 5 byte burst I send out, I'm sending out a new header. If TCP/IP headers are factored into the data usage pricing, it will dwarf the amount of data I'm sending.
I can't answer definitively, but I would assume they must. Otherwise this could be exploited by adding extra data to the headers. With TCP you send a 40 byte packet and then you receive a 40 byte acknowledgement packet. You could try using UDP instead of TCP so that you don't have to waste data with the acknowledgement packets.
According to an email from Sprint network engineering, "Any data that goes through our network, including network Header [sic.] would be billed or count towards your plan."