# transforming a non-linear equation to a linear equation

 1 I have an equation of the form x/y+z=c, where x,y,z are decision variables (>0) and c is a constant. Is there a possible way of converting this to a linear form? asked 28 May '12, 03:10 abhik 11●1●1●2 accept rate: 0% fbahr ♦ 4.6k●7●16 2 What types are (x) and (y) (continuous, integer, or binary)? (28 May '12, 05:19) Ehsan ♦

 1 $$x/y+z = c \Rightarrow x + zy - cy = 0$$ AIMMS Modeling Guide, section 7.7: Elimination of products of variables answered 28 May '12, 08:14 fbahr ♦ 4.6k●7●16 accept rate: 13% 1 If z occurs nowhere else, replace zy with a new variable u and it is linear. If z does appear elsewhere, substitution will not help, and the equation is quadratic (better than rational, but not linear). (28 May '12, 15:13) Paul Rubin ♦♦ @Paul: Following AIMMS Modeling Guide, the product of two continuous variables, $$x_1 \cdot x_2$$, can be converted into a separable form: Introduce two new (continuous) variables $$y_1$$ and $$y_2$$; defined as: $$y_1 = \frac{1}{2} \cdot (x_1 + x_2);$$ $$y_2 = \frac{1}{2} \cdot (x_1 − x_2)$$ Now, replace $$x_1 \cdot x_2$$ by $$y_1^2 − y_2^2$$ – which can be approximated by using piecewise linearization (AIMMS Modeling Guide, section 7.6) [...at the cost of having an approximate solution, of course]. (28 May '12, 16:05) fbahr ♦ 1 @Florian: Okay. I've seen the difference-of-squares trick before, but (as you note) while it can be approximated with piecewise linear functions, it's not truly linear. TANSTAAFL. (28 May '12, 16:39) Paul Rubin ♦♦
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