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My object function is $\frac{4 d k (1 - 2 r) r + k^2(3 r-1)r + d^2 (1 + r (6 r-5))}{ 2 (1-r)^2 r} + q* \frac{d^2 (2 - 3 r) + 2 d k ( 2 r-1) + k^2 (1 -(3 - r) r)}{(1-r)^2}$.

And I'm trying to find $r\in (\frac{1}{2},1)$ that minimizes the object function under the condition of $d\in (0,1)$, $\frac{2r-1}{r}d < k <d$, and $q>1$.

My code is as follows:

minR = Last@Minimize[{(4 d k (1 - 2 r) r + k^2 r (-1 + 3 r) + d^2 (1 + r (-5 + 6 r)))/(2 (-1 + r)^2 r) + q*(d^2 (2 - 3 r) + 2 d k (-1 + 2 r) + k^2 (1 + (-3 + r) r))/(-1 + r)^2, 1/2 < r < 1, 0 < d < 1, 0 < (2 r - 1)/r d < k < d,q>1}, r]

... and it has been running forever, more than 6 hours. Any comments will be greatly appreciated!

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  • $\begingroup$ Any assumptions on $q$? Also, why didn't you include your constraints on $r$ in the Minimize[] call? $\endgroup$ – J. M. will be back soon Mar 3 at 3:31
  • $\begingroup$ @J.M. is computer-less: Thanks! Just edited with q>1. And I do have the constraint on r in the Minimize[] $\endgroup$ – ppp Mar 3 at 3:34
  • $\begingroup$ Oh, it was there and I missed it, apologies. Then, you should put in the $q$ constraint in Minimize[] as well. Since the constraints are expected to be in the second part of the list input, you should do something like Minimize[{function, 0 < d < 1 && 1/2 < r < 1 && q > 1 && 0 < (2 r - 1)/r d < k < d}, r]. $\endgroup$ – J. M. will be back soon Mar 3 at 3:37
  • $\begingroup$ @J.M. is computer-less: Oh, you are right, I missed to include q>1 in Minimize[]. But it is still running long even when I add q>1 as you suggested. $\endgroup$ – ppp Mar 3 at 3:41
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First, strict inequalities make a problem. To illustrate it, let us consider

Minimize[{x, 1 < x && x < 2}, x]

Minimize::wksol: Warning: there is no minimum in the region in which the objective function is defined and the constraints are satisfied; returning a result on the boundary.
{1,{x->1}}

Second, the problem is too complicated for current CASes. To demostrate it, let us consider a simplified problem with weak inequalities instead of strict ones:

{(4 d k (1 - 2 r) r + k^2 r (-1 + 3 r) +  d^2 (1 + r (-5 + 6 r)))/(2 (-1 + r)^2 r) +  
q*(d^2 (2 - 3 r)+d k (-1 + 2 r) + k^2 (1+(-3 + r) r))/(-1 + r)^2, (2 r - 1)/r d <= k <=  d}/.
{d -> 1/2, q -> 3/2};
Minimize[%, r]

which produces a piecewise expression in k including

$$ \text{Root}\left[2048 \text{$\#$1}^3+\text{$\#$1}^2 \left(-7168 k^2+2048 k-3328\right)+\text{$\#$1} \left(8192 k^4-5120 k^3+9344 k^2-2304 k+1504\right)-3072 k^6+3072 k^5-6784 k^4+2880 k^3-2548 k^2+516 k-117\&,2\right]$$

Addition. The following

ClearAll[d, k, q, r]; {(4 d k (1 - 2 r) r + k^2 r (-1 + 3 r) + 
  d^2 (1 + r (-5 + 6 r)))/(2 (-1 + r)^2 r) + 
 q*(d^2 (2 - 3 r) + d k (-1 + 2 r) + k^2 (1 + (-3 + r) r))/(-1 + r)^2, (2 r - 1)/r d <= k <= d && 
 r > 1/2 && r < 1} /. {d -> 1/2, q -> 3/2};Minimize[%, r]

works. The result is a piecewise expression which is too long to be cited here.

Edit. Missed inequalities r > 1/2 && r < 1 are added.

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  • $\begingroup$ Thanks, user64494. Can you please give me some insight on what the notation # is? And why we have Root[]? $\endgroup$ – ppp Mar 3 at 17:42
  • $\begingroup$ See reference.wolfram.com/language/ref/Root.html for info. $\endgroup$ – user64494 Mar 3 at 17:48
  • $\begingroup$ Thanks, user64494! $\endgroup$ – ppp Mar 3 at 18:14

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