Help With McK Mock Question
Just out of interest, I recently found myself watching some MBB Mock interviews on Youtube (the particular one you can find below). I was trying to construct answers as I watched to compare them to the model answer given in the video to see how well I could do.
I came across a quantitative question (14m30s timestamp) which was loosely as follows:
"Our client, the local govt, is looking to introduce free wi-fi with no deadspots across the whole city. They have been quoted $30m to carry out this project by a third party. Using the below data, figure out if you would recommend the government do it themselves or if they should go with the 3rd party. For reference, we will use general units of measurement (just named "units").
- The city is 40 x 25 units.
- Each Hotspot covers circular area with radius of 5 units.
- Each wifi hotspot costs 1 million dollars.
- There should be no deadspots i.e. no areas without wifi coverage."
So, I got a wrong answer clearly - I worked out the area of the city (40*25=1,000), divided by the area of each hotspot circle (5^2 * pi) & got around 13 - therefore a cost of 13 million.
The correct answer is in the region of 27 million. I'm still not sure on how this works; completely loses me in the video. Could someone help explain this to me? I would be extremely grateful.
Best,
(skip to: 14m 30 secs)
I did not watch the video, but based on your data, you cannot simply divide the total area by the area of the circle.
That method assumes each circle can be broken down into a continuous area. Imagine trying to stack circular coins side by side on a square table. You will see there will be empty spots forming between 4 coins after every coin and in the corners of the table - hence dead spots.
This would not occur if we had square/rectangles as they would not have empty areas when stacking side by side.
So you need the largest concentric squares that each circle can fit (square inside the circle) to be stacked side by side. Meaning the circles need to be overlapping. It will easily be more than 13M.
it’s 2 am, haven’t slept much this week, and on a phone browser, so too tired plus I can’t go back to see the numbers on your post while typing this.
But for a radius r, a square of area 2r^2 should be formed (cause half-diagonal of the square should be equal to radius and length will be Sqrt(r^2 + r^2) = r.sqrt(2). ) [Pythagorean theorem].
Dividing the total rectangular area by the area of these individual squares might be the answer of how many routers or whatever it was in the question.
But again, you will have to factor in the edges as these squares may be half stacked at the end [say you have 10.2 squares needed along the length but you might need to add whole 11 to ensure no dead spots].
Meaning you will need a few extra routers around the edges of the city as exact lengths of the square might not fit perfectly. So you will account for routers needed along the length and along the breadth of the city limits separately.
Again, too tired to do this on paper. I hope this was helpful enough for you on how to calculate.
Yes, very helpful & makes a lot more sense now. Thank you.
If anyone wants a vid I found this, too:
https://study.com/skill/learn/how-to-inscribe-a-square-in-a-circle-expl…
This still isn’t technically correct. The most efficient way to cover a rectangle isn’t with a square lattice but a hexagonal one. Doubt they would’ve expected you to do that math in an interview but just fyi
Thanks for sharing in a non condescending way; I wouldn't have thought of that. Out of interest, would you be able to elaborate on how that then works for filling out the entire 40 by 25 with hexagonal lattices?
Haha yeah happy to help. The answer turns out to be… very non-trivial (this is an open problem in mathematics, see https://math.stackexchange.com/questions/216211/what-is-the-optimal-sol…). I think what I would have done in an interview is show the square-packing solution to establish an upper bound (since hexagonal packing is weakly better) and then say since the upper bound is lower than the 3rd party cost, the city should do the installation themselves, with the caveat that there might be even more efficient layouts than what I demonstrated. Lmk if that helps
That's helpful. Thank you
Carl above is correct in the approach indeed. Tried doing it while browsing just now, and I'd say there are two tricky parts:
I'll write it out in full below for completeness but 90% of the thinking is done above by carl.sagan
The right maths answer (but wrong practically)
The right practical answer (but boring mathematically)
I'll be honest and say that I only realised that my initial answer I did while reading it (ie 20) was wrong after starting to type it out here, and noticing that you can't make fractional units of most items (assuming you're not talking liquids or gas). And this, kids, is why you don't charge with your answer without thinking first :D
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