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Home/ Questions/Q 3277762
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T19:24:20+00:00 2026-05-17T19:24:20+00:00

I am solving some examples from Data Networks by Gallager and i didnt quite

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I am solving some examples from Data Networks by Gallager and i didnt quite understand this particular question. The question goes as follows:

Suppose the expected frame length on a link is 1000 bits and the standard deviation is 500 bits
Find the expected time and standard deviation of the time required to transfer a million frames on a 9600bps link.
The follow up question is: The rate at which frames can be transmitted is generally defined as the reciprocal of the expected transmission time. Using the result u find in the previous problem, discuss whether this definition is reasonable.

My attempt is as follows:

Time to send one frame is : 1000/9600 = 0.104 seconds
Hence, time to send million frames is = 104, 000 seconds

I did not understand the second part, where it asks to find standard deviation and the follow up question. I also didnt understand what does it mean when we say standard deviation of 500bits, does that mean error loss, which over here is 50%?

This is not a homework problem. I have a midterm in a few days, and im solving these to improve my grip on the subject.

Any help/hints will be appreciated

Thanks,
Chander

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  1. Editorial Team
    Editorial Team
    2026-05-17T19:24:20+00:00Added an answer on May 17, 2026 at 7:24 pm

    Assuming the distributions are normal then you have a sum of normally distributed variables. In this case both the expectation and the variance are easy to compute, you can just add them.

    1 frame ~ N(1000,500)
    1 mil frams ~ (1E9,5E8)
    

    to get the times you just divide both by 9600.

    E[time] ~= 104167 seconds
    std[time] ~= 52083 seconds
    

    One detail is that it is stupid for them to use a normal distribution since there is some non-zero probability a frame has negative bits. And consequently there is some non-zero chance you will send all of your (negative) bits in negative time… Since they don’t specify a distribution I don’t see what else they could have meant though.

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