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Initially I modeled my objective function as
$$\arg \min \operatorname{Var}(f(x),g(x)) + \operatorname{Var}(c(x),d(x)) + \cdots$$
where $f$, $g$, $c$, $x$ are linear functions.
To be able to solve the problem with linear solvers I have change the problem as follows ($\operatorname{abs}$ stands for absolute value):
$$\arg \min\; \operatorname{abs}(f(x)-g(x)) + \operatorname{abs}(c(x)-d(x)) + \cdots$$
Is this form correct and has the same meaning as the first model?
Without knowing much about your question it is hard to answer more specifically. So first and foremost - no. These are not equivalent. Minimizing variangce will generally make you converge to some kind of sample mean, whereas minimizing absolute deviation would (generally) make you find some kind of sample median.
A generally favourable property of sample means is fast convergence (at least, when compared to median). A good property of sample medians is robustness - insensitivity to outliers.
