An if-then-else logic whose condition is an inequality

Question

I was hoping to get some help in modelling the following logic. I tried to solve it by "Big-M method" but failed. Thank you in advance!

$a(k,n)$ and $b(k,n)$ are known constants, $\lambda$ and $\mu_k$ are variables used for optimization, and such logic is used at a gp model.

My original logic is as follows: if $a_{k,n}-b_{k,n}\ge\dfrac{\lambda}{\mu_k}$

(To ensure that the following expression has a real number solution)

$$p_{k,n} = \frac{1}{2}\left[\sqrt{\left(\frac1{a_{k,n}}-\frac1{b_{k,n}}\right)^2+\frac{4\mu_k}{\lambda}\left(\frac1{b_{k,n}}-\frac1{a_{k,n}}\right)}-\left(\frac1{a_{k,n}}+\frac1{b_{k,n}}\right)\right]$$

else $p_{k,n} = 0$.

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I first tried to avoid the condition $a_{k,n}-b_{k,n} \ge \frac{\lambda}{\mu_k}$ by changing to the following constraints:

\begin{align}\text{tmp1} &= \left[\left(\frac1{a_{k,n}}-\frac1{b_{k,n}}\right)^2+\frac{4\mu_k}{\lambda}\left(\frac1{b_{k,n}}-\frac1{a_{k,n}}\right)\right]^+\\\text{tmp2}&=\sqrt{\text{tmp1}}-\left(\frac1{a_{k,n}}+\frac1{b_{k,n}}\right)\\p_{k,n}&=\frac{1}{2}[\text{tmp2}]^+\end{align}

where $[x]^+=\max\{0,x\}$, but got the following error

Disciplined convex programming error:
    Cannot perform the operation max({constant},{concave})

due to \begin{align}\text{tmp1} &= \left[\left(\frac1{a_{k,n}}-\frac1{b_{k,n}}\right)^2+\frac{4\mu_k}{\lambda}\left(\frac1{b_{k,n}}-\frac1{a_{k,n}}\right)\right]^+\\&=\max\left\{0,\left(\frac1{a_{k,n}}-\frac1{b_{k,n}}\right)^2+\frac{4\mu_k}{\lambda}\left(\frac1{b_{k,n}}-\frac1{a_{k,n}}\right)\right\}\end{align}

Now I feel that the if-then-else constraint may be converted in a big-M way, but I don't know how to express it; but maybe this intuition is wrong.

Answer 1

As @J.Dionisio suggested, you can:
$a_{k,n}-b_{k,n} - {\lambda \over \mu_{k}} \le Mx_k$ where
$x_k \in\ \{0,1\}$ and $M$ is upper bound of any value possible in the model (constants or vars)
Also since dividing will be problem replace
$y_k \cdot \mu_k = 1$ and
$\lambda \cdot z = 1$

So first constraint turns like
$(a_{k,n}-b_{k,n}) - y_k \lambda \le Mx_k$
$y_k \lambda - (a_{k,n}-b_{k,n}) \le M(1-x_k)$

$2p_{n,k} = x_k \cdot tmp1$
In tmp1, replace $\lambda$ with $z$