Rating Ski SlopesDate: 02/07/99 at 17:52:38 From: Chris Conte Subject: How to rate the best ski slope in New England I am doing a science project rating the best ski slope value for kids on a weekend. I selected 20 slopes - 5 each in Massachusetts, New Hampshire, Maine, and Vermont - and collected information on each slope: Jr. Price lift ticket, number of trails, vertical drop, number of lifts, longest trail, and cost per trail. I gave a rating to each variable so the lowest price ticket got a 1 and the next a 2. I did that for each variable, and now have a total score for each mountain obtained by adding the ranking for each variable for each mountain. Killington has the best score of 24, and Ward Hill in Massachusetts the worst with a score of 64 (a perfect score is 6 and the worst score is 90). How can I give more weight to one variable over another and come up with scores that are more accurate and that are based on 100? Date: 02/09/99 at 10:19:04 From: Doctor Nick Subject: Re: How to rate the best ski slope in New England Hi Chris - One thing you might do is reverse your scoring system so that the worst score in each variable gets 1, the second worst gets 2, and so on, so that the best gets the most points. Then figure the maximum score that a slope could get, and re-scale it by an appropriate factor. For instance, if 90 is the best possible score, multiply all the scores by 100/90 and then they'll all be based on 100. As for giving weight to one variable over another, the easiest thing to do is to multiply each variable by a weighting factor, say a number between 0 and 1, with close to zero values for unimportant factors and values close to 1 for important ones. This changes the maximum possible score, so you'll have to determine what the maximum score is with your weighting, and re-scale to get the scores to be based on 100. For instance, if a slope had rankings of 3, 10, 7, 9, 11, and 2 for the six variables, and there are 15 slopes, then the scores for this slope would be 13, 6, 9, 7, 5, 14 for a total score of 54 which on a basis of 100 is 54*(100/90) = 60. Now suppose you don't want give the first variables as much weight as the others. You could multiply that one by 1/2 (for example) and the rest by 1 (so they stay the same). This gives the slope a score of 13/2 + 6 + 9 + 7 + 5 + 14 = 47.5 Now a score of 90 is no longer possible. The best a slope could do is 15/2 + 15 + 15 + 15 + 15 + 15 = 82.5. This slope's scaled score is now 47.5*(100/82.5) = 57.5757, or about 58. Notice that it makes sense that this slope's score went down, since its first score was one of its best. A difficulty you'll have to consider in this is that while weighting the scores is a good idea, there is no one way to weight them. Perhaps the number of lifts is important to you, but it might not be to me. Each person would give a different weight to a different variable. You should experiment and try a bunch of weightings to see how they affect the final scores. Is one slope generally the best, even with quite different weightings? Can you make each slope the best by choosing a weighting that emphasizes the good aspects of that slope? This is a very important area of applied mathematics. Have fun with it, and know that it's not easy. Try a lot of things to get a feel for how different methods might work. Here's something to consider: what happens if you multiply the scores, instead of adding? - Doctor Nick, The Math Forum http://mathforum.org/dr.math/ |
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