I don't accept the argument that the numbers add up correct because I can go to and get a solid 12MB/s test which defies the logic about what is presented here. 1.8MB/s is absolutely abhorrent in today's day and age. I just got off the phone with AT&T support and just read through this post. Thus, under perfectly ideal conditions, an 18 Mbps Internet service should download a file at 2.06 MB/sec as displayed in Windows. Now convert this to the Windows download value in megabytes per second (Windows uses powers of 2 prefixes for Mega- and Giga-, so 1024 multipliers are used here):Ģ,158,344 Bytes/second * 1 KB/1024 Bytes * 1 MB/1024 KB = 2.06 MB/sec. Of the 1500 byte payload, you have 20 bytes of overhead for the IP header, and another 20 bytes of overhead for the TCP header, leaving your file transfer data at only 1460 bytes per packet.Īdding all this together, you get 18,000,000 bits per second to carry 1522-byte frames, of which only 1460 bytes of each frame is counted towards the Windows download speed that is displayed in the download window.ġ8,000,000 bps * 1 Byte/8 bits = 2,250,000 Bytes/sec * 1460 bytes file / 1522 bytes frame = 2,158,344 Bytes of the file per second That's 22 bytes of overhead for a 1500 byte payload (payload size matches the MTU defined in the modem). It's 6 bytes for source MAC, 6 bytes for destination MAC, 4 bytes for 802.1Q QOS/VLAN, 2 bytes for type, and 4 bytes at the end for FCS. VDSL framing encapsulation is similar to Ethernet. This bits per second figure is for ALL traffic that crosses the VDSL line, including framing encapsulation. Your download speed that you're getting is a combination of a bits-to-bytes conversion as well as allowances for protocol overhead.ġ8 Mbps = 18,000,000 bits per second (AT&T and other internet providers use Mega- or Giga- as SI units, which are powers of 10, thus all multipliers are 1000). Then you have to take into account that the data you are requesting may be coming from a server more than 200 miles away with multiple connection points along the way each adding their own set of added overhead as well. There's handshaking, confirmation, and other various other forms of traffic that's traveling along with the raw data of your download. Bytes are bigger than bits.Īs for why you're not getting the full 2.25 MBps that you think you should be getting is that TCP/IP which is the protocol that the whole Internet is based upon has a lot of overhead. ![]() "b" stands for bits, "B" stands for bytes. Notice that the last number has a big "B" and not a small "b" like the first number has. Now, we take 2,359,296 and divide it by 1,048,576 (1 Megabyte) which then comes out to be 2.25 Megabytes. Now, since 8 bits are in a byte, we have to take the 18,874,368 bits and divide it by 8, that comes out to be 2,359,296 Bytes. ![]() 18 Mbps is equal to 18,874,368 bits per second.
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