Math is Wonderful! O_o

There are a lot of things I like about planning Project Destiny each year. One of them is working alongside people who are very different from me. Here's an unexpected e-mail conversation among the team that took off into outer space before landing quite nicely. It was too good to keep to ourselves. Enjoy at your own risk.

Mathematical Formula for Creating Classes
20 messages

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Steve Liu	Tue, Nov 17, 2009 at 1:25 PM
To: Vernon Cheung, Kam Ho Lau, Emily Wu
Maybe this is really simple and I am overcomplicating it... but, is there a mathematical formula to figure out how many creating classes we need? Here are some parameters: x campers 10 campers/ class session y class sessions/ day 15 days z classes How do you solve for z? Am I looking at this problem the wrong way? Steve

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Vernon Cheung	Tue, Nov 17, 2009 at 1:49 PM
To: Steve Liu Cc: Kam Ho Lau, Emily Wu
z = (x campers /10) classes/cycle * 3 cycles AKA: z = 3x/10

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Steve Liu	Tue, Nov 17, 2009 at 1:55 PM
To: Vernon Cheung Cc: Kam Ho Lau, Emily Wu
What if the number of cycles is unknown? (i.e. what if we go back to 5 cycles with 3 creating classes each?) Is it possible to factor in that variable using only x? Steve

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Vernon Cheung	Tue, Nov 17, 2009 at 2:04 PM
To: Steve Liu Cc: Kam Ho Lau, Emily Wu
It would still be: z = x * c / r where z = classes, x = campers, c = cycles, and r = max campers per class Actually, to be technical, it would be z = c * roundup(x / r) if you wanted r to be the max.

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Steve Liu	Tue, Nov 17, 2009 at 2:37 PM
To: Vernon Cheung Cc: Kam Ho Lau, Emily Wu
I'm not sure if this factors in the limits of having only 15 days of classes and having to fit it all in that time span. For example, if I have 50 campers and want to do 5 cycles, I end up with 25 classes, meaning there would have to be 5 classes per cycle, still. This would take us 25 creating days to do it (5 creating classes per week). But we only have 15 days (3 per week). Is there a way to factor in the fact that we only have 15 days? I think this means that C has to be relative to X and a constant of 15 or something like that. Thanks, Steve

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Vernon Cheung	Tue, Nov 17, 2009 at 3:22 PM
To: Steve Liu
It still works. Referring to my previous equation: z = c * ceiling(x / r)... I assumed that the number of days of classes (in this case, d = 15 days) are factored into determining c, the number of cycles you want. In order to decide on c, you must assume a set d, since c must be a factor of d. And without first determining c, you can't determine z. So I suppose you could add a phrase to the equation: z = c * ceiling(x / r) where z = classes, x = campers, c = a factor of 15, and r = max campers per class In your example with x = 50 campers and c = 5 cycles, you didn't include r, the max campers per class. If you choose r = 10, you would get: z = c * ceiling(x / r) = 5 * ceiling(50 / 10) = 5 * ceiling(5) = 5 * 5 = 25 classes which, with c = 5 cycles, means there would be 5 classes per cycle.

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Vernon Cheung	Tue, Nov 17, 2009 at 3:25 PM
To: Steve Liu Cc: Kam Ho Lau, Emily Wu
Oops, forgot to send to everyone. And oops, I think you did use r = 10, but just didn't mention it. I think the confusion came in the final number, 25, being total classes, not creating days.

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Emily Wu	Tue, Nov 17, 2009 at 3:31 PM
To: Vernon Cheung Cc: Steve Liu, Kam Ho Lau
I'm glad you like math, Vernon! (whoa -- I first typed 'Bernon'!!) I'm sure Kam Ho would've been happy to jump into this conversation too, but he's probably been at school (HS) all day. math makes my head swim..... =(

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Vernon Cheung	Tue, Nov 17, 2009 at 4:00 PM
To: Emily Wu Cc: Steve Liu, Kam Ho Lau
That's my should-be name!

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Steve Liu	Tue, Nov 17, 2009 at 4:27 PM
To: Vernon Cheung, Emily Wu, Kam Ho Lau
But there can't be 5 classes in a cycle if there are 5 cycles-- in actual weeks, that would mean having a creating class every day for 5 weeks! How can we limit the results to include the reality? From a correct result, we need limits on what C can possibly be, since you can't have 5 cycles of 5 classes (giving you 25 days of class!) nor can you have 5 cycles of 3 classes (giving you 15 days of class, but more than 10 kids per class!) My assumption here has been that C cannot be arbitrarily set. Everything has to somehow be able to be determined through X and the other constants only. Maybe we need a related equation for C. Steve

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Steve Liu	Tue, Nov 17, 2009 at 4:33 PM
To: Vernon Cheung, Emily Wu, Kam Ho Lau
and k= z/c....so. OK, now I'm really stuck.

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Kam Ho Lau	Tue, Nov 17, 2009 at 7:51 PM
To: Steve Liu, Vernon Cheung, Emily Wu
Wait, the question that started this thing:  is there a mathematical formula to figure out how many creating classes we need? The Number of Creating classes are determined by cycles and class per cycle. These are independent variables. This independent variable will determine the dependent variable such as however many students will be in the class (r).  So back to the how many creating classes we need,  It depends on how many cycles we want and how many classes per cycle.  So let K=number of class per cycle and let C=number of cycle, and Z= Number of Creating class  Z=KC. And we want Z<15, which will ended up by setting a restraint on the side. So let Z=15, Then 15=KC. For Creating Classes should be whole numbers, the only factors of 15 are 1, 3, 5, 15. So from that, we found that possible patterns are 1 cycle of 15 creating classes. (which is impossible, since that mean 15 classes have to go on at the same time) 3 cycle of 5 creating classes (this is what we are doing).  5 cycle of 3 creating classes (a cycle a week) 15 cycle of 1 creating classes (this is not what we want, since it means that there will be just 1 class).  Now that we analyzed the data,  We can plug in to the formula. Let R=campers in each class in the cycle, X=total number of campers at camp, we will have the formula that X/K=R For each case, we can put C=number of cycle and K=number of classes.  Let X=50  For C=1, K=15, 50/15=3.333 students per class For C=3, K=5, 50/5=10 students per class For C=5, K=3, 50/3=16.667 students per class For C=15, K=1, 50/1=50 students per class The equation that Vernon had, Z=X*C/R is right, but it is just that because you have given too much independent variable. There actually arent that much independent variables, so you cant actually plug in numbers as you did earlier.  Z is a number less than 15, so thats a constraint With that, C and K are suppose to be whole numbers which is another constraint.  The number of students in a class is determined by the K (the number of classes), so you cant just plug a number in for R.  Basically, Only X (Number of campers) and Z (number of creating class) are independent variable. So with your formula in looking for an independent variable, it does not work like this, since mathematics, you plug in an independent variable to obtain the dependent variable.  I set a spreadsheet, further similate the process. You can change number in the green columns (A and D) and other numbers will appear.  http://spreadsheets.google.com/ccc?key=0AnVBxZYXnTWndGRoVlhnWXJfNGVaNHNNWUdON1NfNGc&hl=en Remember only whole numbers work, since you cant have part of a class or part of a student (thats why we found factors earlier).  Let me know if you have nay questions.  Kam Ho

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Vernon Cheung	Tue, Nov 17, 2009 at 8:46 PM
To: Kam Ho Lau Cc: Steve Liu, Emily Wu
kh, Your equation works, but not for what Steve asked for. Steve wanted to find Z, but you're setting Z and X to begin with. Steve wants Z, given the rest of the variables. So the question remains unsolved.

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Kam Ho Lau	Tue, Nov 17, 2009 at 8:55 PM
To: Vernon Cheung Cc: Steve Liu, Emily Wu
Given the constraint that steve gave, Steve had turn Z into an independent variable. Thus you cant solve for just one equation when Z is pre-determined, which turns it more a dependent. In an equation there should exist some independent and some variable that are dependent. If both C Cycle and K Classes are independently selected, then Z can be anything. In order to set a constraint, Both K and C are to be a factor. Steve wants Z given the rest of the variable. It will not be possible, since the other variable depends on Z. because steve has set Z<15. So basically plugging in other numbers only give us a number with no factor that Z<15. This is actually something that is harder to limit its function behavior. So what steve actually needs to do is to have systems of equations for which we can solve for the feasible region and putting it back into the Z equation to find which is more efficient. Kam Ho

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Steve Liu	Tue, Nov 17, 2009 at 9:23 PM
To: Kam Ho Lau Cc: Vernon Cheung, Emily Wu
Thank you all for the math lesson. This has now officially gone off the deep end of my understanding. I have a sense that Kam Ho's last sentence is what I was thinking all along, but didn't know how to articulate in math terms. But I didn't know that there would be no way to put this into even a very complicated formula. Can't multi-variable calculus help us? Steve

Vernon Cheung Tue, Nov 17, 2009 at 9:49 PM To: Kam Ho Lau Cc: Steve Liu, Emily Wu Take a look at Sheet 2 of the spreadsheet. Given the number of campers = 50, we choose a max number of campers we want in a class in each cycle. Let's say 11. So we need 4.54545... classes per cycle, which we would have to round up to 5 (each class would have 10 campers, which still satisfies the max allowable). Then we choose a number of cycles, which we agree must be a factor of 15 since there are 15 days of creating classes. 1 and 15 are unrealistic, so we choose either 3 or 5 cycles for the summer. Column E are the total number of classes we need for the entire summer if we have 3 cycles. Column F are the total classes if have 5 cycles. So for 50 campers, 10 campers per class, 5 cycles, you need 25 classes (as stated several emails ago). Of course, with 25 classes and 5 cycles, that means we have 5 classes over the course of 3 days. Which means not every camper will experience every creating class. To solve this dilemma, we realize that with 5 cycles, each class may only meet 3 times, which means there should only be 3 classes per cycle. Similarly, with 3 cycles, we see that each class may only meet 5 times, which means there should only be 5 classes per cycle. In either case, K * C must be 15. In other words, K * C = Z = 15. Z must be 15. So we should only look at the pink boxes with Z = 15 (I bolded them). I believe this solves the case! So it turns out that Z is constant, and K must either be 3 or 5. All you have to input is the number of students, and what you get in the end is R, the max number of campers you can allow in a class. So this equation actually solves for how many campers we can let into a class, since Z will always be 15 given the way our camp days are set up. So to solve the equation for R, I've simplified everything in Sheet 3. v
Steve Liu Tue, Nov 17, 2009 at 9:58 PM To: Vernon Cheung Cc: Kam Ho Lau, Emily Wu Props! We have solved it (mostly Kam Ho and Vernon... no credit to Emily except for moral support). Sheet three is actually pretty useful. Steve Steve Liu Tue, Nov 17, 2009 at 10:05 PM To: Vernon Cheung Cc: Kam Ho Lau, Emily Wu I'm really impressed! Math is wonderful! I'm so close to understanding how this all came together... but it does exactly what I wanted it to AND it clarified which variable we were looking for (not Z but R!!!) I never would have guessed that. Steve Vernon Cheung Tue, Nov 17, 2009 at 10:11 PM To: Steve Liu Cc: Kam Ho Lau, Emily Wu I guess, to put it simply, if each camper were to go to a creating class each of the 15 days and participate in all creating classes by the end of summer, it makes sense that we will always need 15 classes. Z shouldn't have been a variable, but a constant! In the end, the equation comes out to be: R = roundup(X / K), where K = Z / C, with R is the max campers per class per cycle, X is the # campers, K is the number of classes per cycle, Z = 15 is the number of classes, and C is the number of cycles, which would realistically be 3 or 5. Kam Ho Lau Tue, Nov 17, 2009 at 10:11 PM To: Steve Liu Cc: Vernon Cheung, Emily Wu wow impressive. ... Yes, Math is wonderful! And this is one way how we can apply 4 years of Math Education into Ministry! =P Kam Ho