Determine the optimal point satisfying the given condition












0












$begingroup$



We are given the following linear problem:
$maxleft{5x_1-2x_2mid3x_1+x_2leq7,4x_1-2x_2leq3,x_1geq0,x_2geq0right}$.
If there exists an optimal point where at least one of its coordinates
is not zero, then $x_1$ and $x_2$ shall differ by at least $1$ whereby
the second coordinate mustn't be lower than the first coordinate.




I don't know how this should all be possible.



First I tried to solve the problem without respecting the conditions, just by using the Simplex algorithm:



enter image description here



Thus optimal point is $x=
begin{pmatrix}
x_1\
x_2
end{pmatrix}=
begin{pmatrix}
frac{17}{10}\
frac{19}{10}
end{pmatrix}$
, clearly not satisfying the conditions above...



How is it possible to determine an optimal point satisfying these conditions? I don't know but I believe I somehow need to apply these conditions while doing the Simplex algorithm? Maybe in the middle table, increase $frac{19}{4}$ by $1$ and continue with that? Or a complete different attempt? :o










share|cite|improve this question











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    0












    $begingroup$



    We are given the following linear problem:
    $maxleft{5x_1-2x_2mid3x_1+x_2leq7,4x_1-2x_2leq3,x_1geq0,x_2geq0right}$.
    If there exists an optimal point where at least one of its coordinates
    is not zero, then $x_1$ and $x_2$ shall differ by at least $1$ whereby
    the second coordinate mustn't be lower than the first coordinate.




    I don't know how this should all be possible.



    First I tried to solve the problem without respecting the conditions, just by using the Simplex algorithm:



    enter image description here



    Thus optimal point is $x=
    begin{pmatrix}
    x_1\
    x_2
    end{pmatrix}=
    begin{pmatrix}
    frac{17}{10}\
    frac{19}{10}
    end{pmatrix}$
    , clearly not satisfying the conditions above...



    How is it possible to determine an optimal point satisfying these conditions? I don't know but I believe I somehow need to apply these conditions while doing the Simplex algorithm? Maybe in the middle table, increase $frac{19}{4}$ by $1$ and continue with that? Or a complete different attempt? :o










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      We are given the following linear problem:
      $maxleft{5x_1-2x_2mid3x_1+x_2leq7,4x_1-2x_2leq3,x_1geq0,x_2geq0right}$.
      If there exists an optimal point where at least one of its coordinates
      is not zero, then $x_1$ and $x_2$ shall differ by at least $1$ whereby
      the second coordinate mustn't be lower than the first coordinate.




      I don't know how this should all be possible.



      First I tried to solve the problem without respecting the conditions, just by using the Simplex algorithm:



      enter image description here



      Thus optimal point is $x=
      begin{pmatrix}
      x_1\
      x_2
      end{pmatrix}=
      begin{pmatrix}
      frac{17}{10}\
      frac{19}{10}
      end{pmatrix}$
      , clearly not satisfying the conditions above...



      How is it possible to determine an optimal point satisfying these conditions? I don't know but I believe I somehow need to apply these conditions while doing the Simplex algorithm? Maybe in the middle table, increase $frac{19}{4}$ by $1$ and continue with that? Or a complete different attempt? :o










      share|cite|improve this question











      $endgroup$





      We are given the following linear problem:
      $maxleft{5x_1-2x_2mid3x_1+x_2leq7,4x_1-2x_2leq3,x_1geq0,x_2geq0right}$.
      If there exists an optimal point where at least one of its coordinates
      is not zero, then $x_1$ and $x_2$ shall differ by at least $1$ whereby
      the second coordinate mustn't be lower than the first coordinate.




      I don't know how this should all be possible.



      First I tried to solve the problem without respecting the conditions, just by using the Simplex algorithm:



      enter image description here



      Thus optimal point is $x=
      begin{pmatrix}
      x_1\
      x_2
      end{pmatrix}=
      begin{pmatrix}
      frac{17}{10}\
      frac{19}{10}
      end{pmatrix}$
      , clearly not satisfying the conditions above...



      How is it possible to determine an optimal point satisfying these conditions? I don't know but I believe I somehow need to apply these conditions while doing the Simplex algorithm? Maybe in the middle table, increase $frac{19}{4}$ by $1$ and continue with that? Or a complete different attempt? :o







      discrete-mathematics optimization linear-programming






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      share|cite|improve this question




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      edited Dec 3 '18 at 18:50









      independentvariable

      13210




      13210










      asked Dec 3 '18 at 17:20









      cnmesrcnmesr

      2,16341842




      2,16341842






















          1 Answer
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          $begingroup$

          The optimal solution is at $x = [1.7,1.9]^T$, so your answer is correct. Also, the second coordinate (I believe this is $x_2$) is greater than the first one.



          Maybe "they shall differ at least 1" should be 0.1. But the answer is correct.



          There is also no alternative optima since the slope of your objective function is not the same with one of the edges of your set.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "Differ by $1$" is clearly $1$ and not $0.1$. I'm not sure if that is really correct your answer because it seems too easy and then I also wonder why there is a condition at all in the given example.
            $endgroup$
            – cnmesr
            Dec 3 '18 at 18:57










          • $begingroup$
            To convince you, I am sharing the solver output (by Yalmip): >>x = sdpvar(2,1) >>optimize([3*x(1)+x(2) <= 7, 4*x(1)-2*x(2)<= 3, x>= 0], -5*x(1)+2*x(2) ) Solved in 1 iterations and 0.04 seconds Optimal objective -4.700000000e+00 >> value(x) ans = 1.7000 1.9000
            $endgroup$
            – independentvariable
            Dec 3 '18 at 18:59












          • $begingroup$
            Also, in the feasible region, you have 4 vertices. They are (0,7), (1.7,1.9), (0,0),(0.75,0). The optimal point has to be for sure at a vertex. Compare the values.
            $endgroup$
            – independentvariable
            Dec 3 '18 at 19:03











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The optimal solution is at $x = [1.7,1.9]^T$, so your answer is correct. Also, the second coordinate (I believe this is $x_2$) is greater than the first one.



          Maybe "they shall differ at least 1" should be 0.1. But the answer is correct.



          There is also no alternative optima since the slope of your objective function is not the same with one of the edges of your set.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "Differ by $1$" is clearly $1$ and not $0.1$. I'm not sure if that is really correct your answer because it seems too easy and then I also wonder why there is a condition at all in the given example.
            $endgroup$
            – cnmesr
            Dec 3 '18 at 18:57










          • $begingroup$
            To convince you, I am sharing the solver output (by Yalmip): >>x = sdpvar(2,1) >>optimize([3*x(1)+x(2) <= 7, 4*x(1)-2*x(2)<= 3, x>= 0], -5*x(1)+2*x(2) ) Solved in 1 iterations and 0.04 seconds Optimal objective -4.700000000e+00 >> value(x) ans = 1.7000 1.9000
            $endgroup$
            – independentvariable
            Dec 3 '18 at 18:59












          • $begingroup$
            Also, in the feasible region, you have 4 vertices. They are (0,7), (1.7,1.9), (0,0),(0.75,0). The optimal point has to be for sure at a vertex. Compare the values.
            $endgroup$
            – independentvariable
            Dec 3 '18 at 19:03
















          1












          $begingroup$

          The optimal solution is at $x = [1.7,1.9]^T$, so your answer is correct. Also, the second coordinate (I believe this is $x_2$) is greater than the first one.



          Maybe "they shall differ at least 1" should be 0.1. But the answer is correct.



          There is also no alternative optima since the slope of your objective function is not the same with one of the edges of your set.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "Differ by $1$" is clearly $1$ and not $0.1$. I'm not sure if that is really correct your answer because it seems too easy and then I also wonder why there is a condition at all in the given example.
            $endgroup$
            – cnmesr
            Dec 3 '18 at 18:57










          • $begingroup$
            To convince you, I am sharing the solver output (by Yalmip): >>x = sdpvar(2,1) >>optimize([3*x(1)+x(2) <= 7, 4*x(1)-2*x(2)<= 3, x>= 0], -5*x(1)+2*x(2) ) Solved in 1 iterations and 0.04 seconds Optimal objective -4.700000000e+00 >> value(x) ans = 1.7000 1.9000
            $endgroup$
            – independentvariable
            Dec 3 '18 at 18:59












          • $begingroup$
            Also, in the feasible region, you have 4 vertices. They are (0,7), (1.7,1.9), (0,0),(0.75,0). The optimal point has to be for sure at a vertex. Compare the values.
            $endgroup$
            – independentvariable
            Dec 3 '18 at 19:03














          1












          1








          1





          $begingroup$

          The optimal solution is at $x = [1.7,1.9]^T$, so your answer is correct. Also, the second coordinate (I believe this is $x_2$) is greater than the first one.



          Maybe "they shall differ at least 1" should be 0.1. But the answer is correct.



          There is also no alternative optima since the slope of your objective function is not the same with one of the edges of your set.






          share|cite|improve this answer









          $endgroup$



          The optimal solution is at $x = [1.7,1.9]^T$, so your answer is correct. Also, the second coordinate (I believe this is $x_2$) is greater than the first one.



          Maybe "they shall differ at least 1" should be 0.1. But the answer is correct.



          There is also no alternative optima since the slope of your objective function is not the same with one of the edges of your set.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 18:45









          independentvariableindependentvariable

          13210




          13210












          • $begingroup$
            "Differ by $1$" is clearly $1$ and not $0.1$. I'm not sure if that is really correct your answer because it seems too easy and then I also wonder why there is a condition at all in the given example.
            $endgroup$
            – cnmesr
            Dec 3 '18 at 18:57










          • $begingroup$
            To convince you, I am sharing the solver output (by Yalmip): >>x = sdpvar(2,1) >>optimize([3*x(1)+x(2) <= 7, 4*x(1)-2*x(2)<= 3, x>= 0], -5*x(1)+2*x(2) ) Solved in 1 iterations and 0.04 seconds Optimal objective -4.700000000e+00 >> value(x) ans = 1.7000 1.9000
            $endgroup$
            – independentvariable
            Dec 3 '18 at 18:59












          • $begingroup$
            Also, in the feasible region, you have 4 vertices. They are (0,7), (1.7,1.9), (0,0),(0.75,0). The optimal point has to be for sure at a vertex. Compare the values.
            $endgroup$
            – independentvariable
            Dec 3 '18 at 19:03


















          • $begingroup$
            "Differ by $1$" is clearly $1$ and not $0.1$. I'm not sure if that is really correct your answer because it seems too easy and then I also wonder why there is a condition at all in the given example.
            $endgroup$
            – cnmesr
            Dec 3 '18 at 18:57










          • $begingroup$
            To convince you, I am sharing the solver output (by Yalmip): >>x = sdpvar(2,1) >>optimize([3*x(1)+x(2) <= 7, 4*x(1)-2*x(2)<= 3, x>= 0], -5*x(1)+2*x(2) ) Solved in 1 iterations and 0.04 seconds Optimal objective -4.700000000e+00 >> value(x) ans = 1.7000 1.9000
            $endgroup$
            – independentvariable
            Dec 3 '18 at 18:59












          • $begingroup$
            Also, in the feasible region, you have 4 vertices. They are (0,7), (1.7,1.9), (0,0),(0.75,0). The optimal point has to be for sure at a vertex. Compare the values.
            $endgroup$
            – independentvariable
            Dec 3 '18 at 19:03
















          $begingroup$
          "Differ by $1$" is clearly $1$ and not $0.1$. I'm not sure if that is really correct your answer because it seems too easy and then I also wonder why there is a condition at all in the given example.
          $endgroup$
          – cnmesr
          Dec 3 '18 at 18:57




          $begingroup$
          "Differ by $1$" is clearly $1$ and not $0.1$. I'm not sure if that is really correct your answer because it seems too easy and then I also wonder why there is a condition at all in the given example.
          $endgroup$
          – cnmesr
          Dec 3 '18 at 18:57












          $begingroup$
          To convince you, I am sharing the solver output (by Yalmip): >>x = sdpvar(2,1) >>optimize([3*x(1)+x(2) <= 7, 4*x(1)-2*x(2)<= 3, x>= 0], -5*x(1)+2*x(2) ) Solved in 1 iterations and 0.04 seconds Optimal objective -4.700000000e+00 >> value(x) ans = 1.7000 1.9000
          $endgroup$
          – independentvariable
          Dec 3 '18 at 18:59






          $begingroup$
          To convince you, I am sharing the solver output (by Yalmip): >>x = sdpvar(2,1) >>optimize([3*x(1)+x(2) <= 7, 4*x(1)-2*x(2)<= 3, x>= 0], -5*x(1)+2*x(2) ) Solved in 1 iterations and 0.04 seconds Optimal objective -4.700000000e+00 >> value(x) ans = 1.7000 1.9000
          $endgroup$
          – independentvariable
          Dec 3 '18 at 18:59














          $begingroup$
          Also, in the feasible region, you have 4 vertices. They are (0,7), (1.7,1.9), (0,0),(0.75,0). The optimal point has to be for sure at a vertex. Compare the values.
          $endgroup$
          – independentvariable
          Dec 3 '18 at 19:03




          $begingroup$
          Also, in the feasible region, you have 4 vertices. They are (0,7), (1.7,1.9), (0,0),(0.75,0). The optimal point has to be for sure at a vertex. Compare the values.
          $endgroup$
          – independentvariable
          Dec 3 '18 at 19:03


















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