0
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I try to solve the following set of equations but i keep getting the error message "This system cannot be solved with the methods available to Reduce". I wonder if the problem is syntax or there is no solution? Thanks

    FullSimplify[
 Reduce[λ*d10 + μ*p*d01 <= (λ + μ + β)*
     d10 && λ*d10 + μ*p*
      k <= (λ + μ + β)*d10 && λbar*
     d10 <= (λbar + β)*d01 && λbar*d10 + 
     x*p*μ (k - d01) <= (λbar + β)*
     d01 && λbar*r + 
     x*p*μ (k - d01) <= (λ + β)*d01 && 
   0 < λ < 1 && 0 < μ < 1 && 0 < p < 1 && 
   0 < β < 1 && λ + λbar + β + μ < 1 &&
    d01 >= 0 && d10 >= 0 && r > 0 && k > 0 && d10 <= r && 
   re <= r && x > 0 && p <= pe && 
   d01 <= (r - re) + (1 - pe)*d01, {d10, 
   d01}]]
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  • 2
    $\begingroup$ FindInstance for $\{\text{d10},\text{d01},e,k,p,r,x,\lambda ,\mu ,\beta ,\lambda ,\mu ,\beta ,\text{$\lambda $bar}\} $ says "FindInstance::nsmet: The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist " $\endgroup$ – user64494 May 10 '17 at 9:40
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    $\begingroup$ It was a good idea to subsitute the Subscript-variables by normal varialbles. Now "FindInstance" yields one result, but when I try, to get more, the kernel crashes due to too less memory. May be it would help to reduce the system by one or two parameters. FindInstance"-results: {{d10 -> 27/32, d01 -> 1, [Lambda] -> 1/32, [Mu] -> 1/2, p -> 1/2, [Beta] -> 1/2, k -> 1/2, [Lambda]bar -> -3, x -> 1, r -> 1, pe -> 2, re -> -1}} $\endgroup$ – Akku14 May 10 '17 at 11:41
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    $\begingroup$ "I wonder if the problem is syntax or there is no solution?" I thought the message was pretty clear. The methods built into Reduce cannot solve this system. That does not mean that there is no solution. It also does not mean that you have a syntax error. $\endgroup$ – Szabolcs May 10 '17 at 12:33
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    $\begingroup$ If you take a much simpler system of inequalities with this many parameters, and try to solve it by hand, going through all the possible special cases, that will give you a hint about why it is not very reasonable to expect a solution at all. $\endgroup$ – Szabolcs May 10 '17 at 12:34

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