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I am currently enrolled in an Intro to Networking course and I have been studying for an upcoming exam by doing practice problems in the course textbook. I came across this question that stumped me. It deals with probability, which I have always had a hard time understanding and never really studied in depth.

Suppose an IP packet is fragmented into 10 fragments, each with a 1% (independent) probability of loss. To a reasonable approximation, this means there is a 10% chance of losing the whole packet due to loss of a fragment. What is the probability of net loss of the whole packet if the packet is transmitted twice

A) Assuming all fragments received must have been part of the same transmission?

B) Assuming any given fragment may have been part of either transmission?

I assume for part A there is a 20% chance of losing the whole packet, because since we send two transmissions the probability doubles. What I don't understand is how the probability is affected in part B. How does the probability change if the two packets are mixed within each transmission? Can someone explain this to someone who is not a math major? Thanks in advance.

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3 Answers

With all respect, never ask a mathematician to solve an engineering problem; the lack of precision inherent in the real world gives them heart palpitations:-)

As the question states, " to a reasonable approximation ... 10% chance of losing the whole packet" (mathematicians can't read either:-)). There is also an inherent assumption that the loss of fragments is independent; in a real network this is unlikely to be the case.

So the chance of losing both is $.1\times.1=.01$. So the chance of getting at least 1 is $1-0.01=0.99$.

For part B, the chance of losing any single fragment is $0.01\times0.01=0.0001$. Now, if any of the 10 fragments is lost twice the whole packet is lost. So we must get 10 good fragments, this happens with a probability of $(1-0.0001)^{10}\approx0.999$ which gives a loss probability of $0.001$.

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For part A, you need to flip back and forth between the probability of success and failure to get the correct result.

Prob of loss of the first fragment is 0.01; thus the probability of the first fragment getting through is (flip!) 0.99.

For a good transmission, all ten packets must get through, independently. The probability of this is $(0.99)^{10}$ or $0.9043820...$

So the probability of the first transmission failing is (flip!) $0.0956179...$

So the probability of both transmissions failing is $(0.0956179...)^2$ or $0.00914278...$

So, finally, the probability that at least one transmission will succeed is (flip!) $0.9908572$

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The approximation of $10\%$ given in the exercise is not quite exact. For the exact value: Each fragment has a probability of $0.99$ to arrive. Assuming these probabilities are independent, we get a chance of $0.99^{10}$ of all fragments to arrive, which makes the probability of a packet loss $$1 - 0.99^{10} \approx 9.562\%.$$

For part A, with a probability of $$(1 - 0.99^{10})^2 \approx 0.914\%$$ this happens twice in a row.

For part B, a single fragment is lost if it is lost in both transmissions, with occures with a probability of $0.01^2$. So it is transmitted correctly with a probability of $1 - 0.01^2$. So all $10$ fragments are transmitted correctly with a probability of $(1 - 0.01^2)^{10}$, which yields the probability of a packet loss as $$1 - (1 - 0.01^2)^{10} \approx 0.100\%.$$

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you are out by an order of magnitude in your final calculation –  Dale M Mar 25 '13 at 0:33
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