With all the drinking I've been doing this weekend, I waited until today to start on a group project that's due tomorrow. Well, I don't have a group, and I'm stuck on the problem.
So, do any of you have experience with linear programming and Excel Solver? Basically, I'm trying to find a way to link a rating variable to my objective function.
I can post the problem if needed. I'm fu[i][/i]cking stumped.
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Jophiel wrote:
I managed to be both retarded and entertaining.
I'd help but I'm only in level 20 maths. Good luck though.
Excel solver as in microsoft excel? or somethign different
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Yes, Microsoft Excel has a neat little add-on that can be used to solve linear systems by maximizing or minimizing some objective fuction bounded by user-defined constraints. The problem that I'm facing is how to set up a constraint that lets me link an average rating value to the value of a variable in my objective function.
Jophiel wrote:
I managed to be both retarded and entertaining.
I'm going to go ahead and ask you a question dem (or anyone else who is familiar with minimum and maximum values).
"One positive interger is 3 greater than 4 times another positive interger. If the product of the two integers is 76, then the sum of the two integers is _______.
I came up with the formula 4n2 + 3n -76=0. Then I plugged it into the quadratic formula, y= -b(+ or -)(Square root; b2 -4ac) / 2a *
I got the answer 46.41. The sum of the two integers is actually 23, can anyone explain how to get this answer?
*a=first term, b=2nd term, c=third term
It's a system of equations.
a = integer one
b = integer two
a = 4b + 3
a * b = 76
Use the second equation to solve for b in terms of a (b = 76/a). Then, plug that value of b in to the first equation (a = 4[76/a] + 3). Then, solve for a and plug that value back into the second equation to find b.
a = 19
b = 4
a + b = 23
a = integer one
b = integer two
a = 4b + 3
a * b = 76
Use the second equation to solve for b in terms of a (b = 76/a). Then, plug that value of b in to the first equation (a = 4[76/a] + 3). Then, solve for a and plug that value back into the second equation to find b.
a = 19
b = 4
a + b = 23
Jophiel wrote:
I managed to be both retarded and entertaining.
Argh, maths.
My Brain Hurt.
My Brain Hurt.
Overlord Demea wrote:
It's a system of equations.
a = integer one
b = integer two
a = 4b + 3
a * b = 76
Use the second equation to solve for b in terms of a (b = 76/a). Then, plug that value of b in to the first equation (a = 4[76/a] + 3). Then, solve for a and plug that value back into the second equation to find b.
a = 19
b = 4
a + b = 23
a = integer one
b = integer two
a = 4b + 3
a * b = 76
Use the second equation to solve for b in terms of a (b = 76/a). Then, plug that value of b in to the first equation (a = 4[76/a] + 3). Then, solve for a and plug that value back into the second equation to find b.
a = 19
b = 4
a + b = 23
Thank you very much.
I just hope you bastids don't decide to make a bomb or something. 
Sir Weebs wrote:
I just hope you bastids don't decide to make a bomb or something.
Nah, that would require motivation.
The Glorious Atomicflea wrote:
Sir Weebs wrote:
I just hope you bastids don't decide to make a bomb or something.
Nah, that would require motivation.
Meh, what she said.
Jophiel wrote:
I managed to be both retarded and entertaining.
Quote:
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