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I just started to use Mathematica. Using the following simple command to minimize a function of variables $a,b,c,d$ with parameter $n$, it takes too much time. Do you know if there is a mistake in this command syntax?

Minimize[{((b (b - 1)/2)*(c + d) + 
 b*a*(c + d) + (d (d - 1)/2)*(a + b) + 
 c*d*(a + b))/((a + b)*(a + b - 1)*(c + d)/
   2 + (c + d)*(c + d - 1)*(a + b)/2), a + c == n && b + d == n && n >= 1000 a >= 1 && b >= 0 && c >= 1 && d >= 0}, {a, b, c, d}]
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    $\begingroup$ Just a little curious: is there a comma missed? n >= 1000 a >= 1 or n >= 1000, a >= 1? $\endgroup$ – wuyudi Sep 8 '20 at 15:14
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    $\begingroup$ If I guess right,Minimize[{((b (b - 1)/2)*(c + d) + b*a*(c + d) + (d (d - 1)/2)*(a + b) + c*d*(a + b))/((a + b)*(a + b - 1)*(c + d)/ 2 + (c + d)*(c + d - 1)*(a + b)/2), a + c == n, b + d == n, n >= 1000 , a >= 1, b >= 0, c >= 1, d >= 0}, {a, b, c, d}] returns what you want. $\endgroup$ – wuyudi Sep 8 '20 at 15:16
  • $\begingroup$ Thank you a lot @wuyudi! $\endgroup$ – Penelope Benenati Sep 8 '20 at 15:29
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Assuming[n >= 1000, 
 Minimize[{((b (b - 1)/2)*(c + d) + 
       b*a*(c + d) + (d (d - 1)/2)*(a + b) + 
       c*d*(a + b))/((a + b)*(a + b - 1)*(c + d)/
         2 + (c + d)*(c + d - 1)*(a + b)/2), 
    a + c == n && b + d == n && n >= 1000 && a >= 1 && b >= 0 && 
     c >= 1 && d >= 0}, {a, b, c, d}] // Simplify]

(*{n/(2 (-1 + n)), {a -> 1, b -> n, c -> -1 + n, d -> 0}}*)
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  • $\begingroup$ Thank you yarchik! $\endgroup$ – Penelope Benenati Sep 8 '20 at 17:11

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