Media diagnostics · Primary-source explanation; no UK IPTV implementation or account test. Instructions are based on publisher guidance; a UK IPTV account was not tested on this platform.
Two equal files, very different waits
Consider an invented test with two consecutive files, each containing exactly 10,000,000 bytes of counted payload. The first takes one second; the second takes nine seconds. Their individual rates are 80 Mbit/s and approximately 8.89 Mbit/s. Taking the arithmetic mean gives about 44.44 Mbit/s. However, the viewer transferred 160,000,000 bits in ten seconds, so the aggregate payload rate is 16 Mbit/s. The slow transfer occupied most of the time. Giving each file an equal vote in the rate average hides that fact.
Decide what belongs in the denominator
This example excludes gaps between the files. If there were another five seconds of waiting that the reported experience is intended to include, the elapsed interval becomes fifteen seconds and the result changes again. State whether the timer covers active transfer, the entire test, or the period from selecting a programme to receiving a first picture. Those are useful measurements of different things. They should not be exchanged merely because each can be displayed as a speed or a duration.
Use the calculation for the right question
RFC 6349 describes a TCP throughput testing framework for managed networks; this arithmetic does not claim to run that framework. It helps check whether a summary follows its stated byte and time counters. Keep the payload definition, test endpoint, units and interval with the number. A high result from these two files still does not establish decoder support, account access, or uninterrupted delivery during a later live session.