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So, here's the problem:

An access point is alternately sending L-sized packets to 2 stations, A and B.

The access point is transmitting to A at 10 MBits /s and to B at 2 MBit/s.

What's the average rate the access point is transmitting at?

The answer is not 10 + 2 / 2 = 6, but instead it's 3.33 Mbit/s.

How?

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    Ask your instructor. (hint: there's more to his answer than is in your question)
    – Ricky
    Jan 19 '16 at 20:28
  • Unfortunattely that will not be possible, since classes are over. Do you know any kind of webpages where I can read and try to figure out the question?
    – Luís Pedro Borges
    Jan 19 '16 at 20:34
  • Resource recommendations are off-topic.
    – Ron Maupin ♦
    Jan 19 '16 at 20:41
  • Ok then, so any clue on how to think or what to search?
    – Luís Pedro Borges
    Jan 19 '16 at 20:45
  • You'd have to ask whoever screwed up the math. If it sent 10Mb in the 1st second, and 2Mb in the 2nd second, then over 2sec the average rate is 6Mb.
    – Ricky
    Jan 19 '16 at 21:44
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First, an example:

Suppose the frames are 1500 bytes long, it is the same as 12000 bits.

The transmission of the frame to A at 10 Mb/s takes 1.2 milliseconds.

The transmission of the same size frame to B takes 6 milliseconds.

In total, to transmit 24000 bits took 7.2 milliseconds. It gives an speed of 3.33 Mb/s.

The result isn't the arithmetic mean, is the Harmonic Mean.

The harmonic mean can be expressed as the reciprocal of the arithmetic mean of the reciprocals.

For our case:

=1/(((1/10)+(1/2))/2) =1/((6/10)/2) = (2/1)/(6/10) = 20/6 = 3.33

In general, for two speeds s1 and s2 will be:

=(2*s1*s2)/(s1+s2)

In this case =(2*10*2)/(10+2) =(40/12) =3,33

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