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Airplanes enter a rectangular land as shown in the following figure. The sector length is $50 \text{ nm}$. The spacing between airplane as they enter the land is $20 \text{ nm}$ plus an exponentially distributed random variable with a mean of $1 \text{ nm}$. Suppose that airplane travels at $300 \text{ nm}$ per hour. What is the average number of airplanes in a land?

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My thought: Even though I spent many hours thinking of the way to use Little's Law to link the relationship of the distances between airplanes and the sector length, I don't really see the relationship between those two. Furthermore, why do we need the information about the width of the land $= 20 \text{ nm}$? In short, I'm completely stumbled upon this problem. If anyone could give some hints on how to approach this problem, I would really appreciate it.

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First let us suppose the spacing is $20$ nm, without the added random variable. If we measure distance from the left edge of the box we will have three planes in the box when the leftmost is between $0$ and $10$ nm and two planes in the box when the leftmost is between $10$ and $20$ nm. This leads to an average of $2.5$ airplanes in the box. The width of the box is immaterial.

Adding in the random variable to the spacing makes it possible that there are less planes in the box. When the leftmost plane is at $0$, it will be the only one in the box if the random variable of the spacing in front of it exceeds $30$, which is rather unlikely but possible. There will be only two if the sum of two random variables exceeds $10$, still unlikely. For each position from $0$ to $20$ for the leftmost plane you need to compute the chance that there is one or two more planes in the box. It is now possible that the leftmost plane is at a position greater than $20$, so you need to compute the chance of that and the expected number of planes in front of it.

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    Thank you very much for your help. But I have a question: for the last case, there would be at most 2 planes (including the leftmost) in the land if the random variable is less than 10 and the leftmost plane is between 20 and 30 (inclusive). And it would be exactly 1 plane if the leftmost > 30nm. Is this correct? But even if this is correct, how could we compute the chance of the leftmost for each of these 4 cases? – 2017-01-29
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    Yes, your maxima are correct. To have the leftmost plane at $25$ you need the random variable to its left to be at least $5$ as otherwise the plane to its left would be in the box – 2017-01-29
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    thanks for your quick answer. But what is the probability for each of the leftmost case? For example, what is the probability that the leftmost plane is less than 10nm? And I don't think my argument above is correct for the case if the leftmost is > 30nm (There should be 2 or 3 airplanes in that case). Btw, does the speed have anything to do with this problem though? – 2017-01-29
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    I don't think the speed has anything to do with it – 2017-01-29
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    could you show me how you determined the probability that the leftmost airplane is at most 5nm away from the left border? I think the Little's Law could be used given the information on the speed (it's like the arrival rate, but it's not clear to me if this interpretation for the speed is correct). Btw, could you please give this problem a try as well? http://math.stackexchange.com/questions/2116115/average-time-spent-in-the-system-for-a-customer – 2017-01-29
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    my apology for asking a lot of questions above. But could you please be patient to give your thought to them? – 2017-01-29
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    I think you are right you can use Little's law. The speed does then matter as it gives you the residence time. The arrival times vary as 300/(20+ the random variable). You need to convert that to an average arrival rate. The residence time is 10 minutes. – 2017-01-30
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    thank you very much for your help. Could you explain why arrival time = speed/distance between 2 aircrafts? If that's the case, is the average arrival rate equal to $15 - 15*\int_{0}^{\infty} \frac{xe^(-x)}{20+x}\approx 0.0456 = 14.3155$ airplanes per hour, since $x$ follows $exp(1)$? Finally, the average time in the system would be $50/(300/60) = 10$ minutes, so by using Little's Law with $\lambda = 14.3155$, and $W = 10$ mins we would get: $L = 14.3155/60*10 = 2.4$ airplanes, so approximately $\fbox{2}$ airplanes. – 2017-01-30
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    The time for the next airplane to arrive is the distance between divided by the speed. That gives the first relation. I agree with the rest, but do not believe you should round $2.4$ down to $2$. If you remove the random variable the average number present is $2.5$, so you are calculating that decrease of $0.1$. I think I would report the $2.4$ to more places, but that is between you and the teacher. – 2017-01-30
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    thank you so much for your patience! It's all clear now. Could you please help me with another (a bit tricky) problem here: http://math.stackexchange.com/questions/2116115/average-time-spent-in-the-system-for-a-customer. Really appreciate if you could give me some thoughts on my current solution. – 2017-01-30
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    The unit apparently doesn't match. The formula for average arrival rate would have unit in $hours$ ($\frac{\text{nm/hours}}{\text{nm}}$), whereas the average time in the system has unit in $hours$. Multiplying these two does not make any sense in terms of unit, so the formula for average arrival rate is somehow wrong?! – 2017-01-30