One Surprising Way Calculus Can Help You Have A Better Vacation – Predict How Long The Line Will Be.
You find yourself running a ski resort in Lake Tahoe, which is not the worst thing to have happened to your life. Frequently, you find long lines at the chairlift building up around midday which cause people to stay away from the mountain. In order to boost attendance, you're going to try and make the time spent waiting for the ski lift more entertaining. To be able to plan for this you need an estimate of how many people will be on the line.
For simplicity, your resort has just one run and one chairlift (maybe that is the real reason people stay home?). The ski run has a length of l r u n = 4 0 0 0 m , and all skiers have an average speed of v s k i = 8 m/s during their descent. The ski lift has a speed v l i f t = 1 m/s , has chairs spaced every l s p a c i n g = 4 0 m , and each chair can hold g = 6 people. Finally, there is a total of N T = 1 0 0 0 people on the mountain.
Question : At steady state (when the average length of the line becomes constant), how many people do you expect to be waiting on the line?
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Details and assumptions
- The length of the chairlift is the same as the length of the run, i.e. l l i f t = l r u n .
- As soon as people get off the lift at the top they begin skiing straight down the mountain.
- As soon as people get to the bottom of the run they get on the line for the chairlift.
- The effective length of the ski run is always l r u n , no matter how long the line becomes.
- People wait for the chairlift in one single file line.
The answer is 325.
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3 solutions
I got the answer of 322 because it says as long as people get to the top, they will ski straight down. I assume therefore this means that all 6 people in each chair reaching the top can all ski down at the same time. This gives the answer of 1000 - 600 - 13*6 = 322. I seriously think this question is flawed in its phrasing.
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All 6 people in a chair will ski down at the same time. The question is correct in its language though it may seem a bit vague. The thing is that at any point of time the no. Of people in the Queue will be either equal to 322 or 328. NOTE THAT IT IS ASKING FOR AVERAGE LENGTH OF THE LINE.
Let's say in the starting the no. of people in the line is 1000. Now the chairlift starts and people start entering it. Now consider the FIRST BATCH OF 6, it will take them 4000/1= 4000 seconds to reach the top and another 4000/8 =500 seconds to reach the bottom.
So it takes the first batch a total of 4500 seconds to reach the bottom.
By that time (the time when the first batch reaches the bottom) there will be 12*6=72 people on the ski run. Also note that 1-by this time exactly 100 chairs will be filled (since the chair nearest to the top will be 20 metres away from the top and same at the bottom) so the number of chairs filled=[(4000-40)/40]+1=100 2-it will take another 20 seconds for the next chair to arrive at the queue. So by this time the number of people in the line = 1000 - 72 -600= 328 Now the next batch of 6 will reach the bottom(queue) after 40 seconds. And the next chair will arrive to the queue in 20 seconds.
Sooo after 20 seconds the length of the line will reduce to 322. And the time it will take for the next 6 to arrive to the bottom(queue) will be 20 seconds. After 20 seconds those 6 will reach the queue and the length will become 328!!
After another 20 seconds the length will again reduce to 322 and again 20 seconds after that moment the line will increase to 328 since the next batch of 6 will arrive.Thus
The no. Of people in the line oscillate between 322 and 328, giving an average of 325.
It says expected number of people. That means there will be either 3 2 2 people at one time, or 3 2 8 people at one time, with either outcome being equally likely. The question isn't flawed.
According to "Details and assumptions" "As soon as people get off the lift at the top they begin skiing straight down the mountain. " Therefore people should always be moving in blocks of 6! Now we can consider the descent as a continuation of the lift that instead of 4000m is only 4000/8=500 meters long. As 4500/40=112.5 We have that there are 113 full chairs every time. 113*6=678 1000-678=322
This is totally understandable but I don't understand why the consideration that lift would come back empty and not with people on it is taken!?
sorry but we must have 101 chairs not only 100 so the no. of people will be 606 ??
Very nice explanation Sir! You must be a good math teacher. Truth be told, I didn't answer this problem because I got bored when I read the question. Hahaha...
Consider that the ski resort has just opened and all 1 0 0 0 people are on line. The first 6 people go on the ski lift, and it takes them 1 4 0 0 0 + 8 4 0 0 0 = 4 5 0 0 seconds to come back. Every 4 0 seconds, 6 more people leave the line. By the time the first 6 people come back, this would complete a cycle and we would be in the steady state. Thus, the expected number of people on line at any time would be 1 0 0 0 − 4 0 6 ( 4 5 0 0 ) = 3 2 5 people.
Notice that the rate at which people come and people leave is the same, 6 people every 40 seconds. So the length of the line is just 1 0 0 0 minus the amount of people on the ski lift and the people on the slope. We thus get: l l i n e = 1 0 0 0 − 6 × 4 0 4 0 0 0 − 6 × 8 × 4 0 / 1 4 0 0 0 = 1 0 0 0 − 6 0 0 − 7 5 = 3 2 5
Since the lift is 4000m long, and there is a chair every 40m, there are 100 chairs, each carrying 6 people. So at any given time, there are 600 people on the lift.
The lift is primarily what regulates the rate. Since its rate is 1m/sec, it empties a chair of people at the top every 40 seconds. So the rate of "people-flow" throughout the cycle is 6/40 or 0.15 people per second.
How long does a trip down the mountain take? If it's 4000m long, and the average rate of skiing is 8m/s, that puts it at 500 seconds. Taking the 0.15 people/s rate from earlier, that tells us that every 500 seconds, 75 people begin their run, and 75 people end their run. So at any given time, 75 people are on the slope.
If there are 600 on the lift, and 75 people on the slope, that leaves 1 0 0 0 − 6 0 0 − 7 5 = 3 2 5 people waiting in line.