# Thread: Optimisation involving if statements and ceiling values?

1. ## Optimisation involving if statements and ceiling values?

$\frac{ln(999/x) } {ln(y/100+1) }+ 1 = N$

How can I minimise N and maximise x given these constraints:

1. $1 =< x =< 315$
2. $1 =< y =< 110$
3. $x+y =< 340$
4. if x >= 150 then y =< 50
5. if x >= 250 then y =< 20
6. if y >= 90 then x =< 60
7. N is a ceiling value, so 3.1 would be the same as 4.

By minimising N and maximising x I mean that lets say x = 200, y = 105 gives 3.24 for N while x = 210, y = 90 gives 3.43 for N. Since N is a ceiling value, both results are the same however the second set of answers is more desirable as x is 210 as opposed to 200. This particular set of values isn't valid since it ignores some of the restraints but it demonstrates what I mean by minimising N and maximising x.

I dont know how to solve this (is it even possible?), it seems like a linear programming type problem or lagrangian multipliers but I'm not sure how to solve it. Also how do you make $1/2$ look like 1 'on' 2?

2. Originally Posted by Corpsecreate
$\frac{ln(999/x) } {ln(y/100+1) }+ 1 = N$

How can I minimise N and maximise x given these constraints:...
Do you mean:

Maximise $x$ such that

$\displaystyle \left\lceil \frac{\ln(999/x) } {\ln(y/100+1) }+ 1 \right\rceil$

is minimised and subject to the constraints ..

CB