Best use of steel bar cut for mechanical study












6












$begingroup$


I have available a steel bar with length of 3000 mm.



I intend to get the best use of cutting out of this bar.



I need to cut this steel bar in the following lengths: 230mm, 260mm, 320mm, 350mm and 650mm.



enter image description here



How to get cut options "close" to the length of 3000 mm.



Example:



230 + 230 + 260 + 260 + 320 + 320 + 350 + 350 + 650;


-> 2970



I have the code below. I know I will have to use the "Nearest" function, but I am not able to apply it:



lista={230, 260, 320, 350, 650};

resultados=Join@@DeleteCases[IntegerPartitions[#,{1,[Infinity]},lista]&/@Range[3000],{}];

And@@Thread[Total/@resultados<=3000];

contagem=Map[Lookup[Counts[#],lista,0]&,resultados];

contagem//Short


In the my example, i get groups that are smaller that 3000, but wish the values that are "nearest" of the value of 3000






list-manipulation







share|improve this question











$endgroup$








  • 3




    $begingroup$
    This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
    $endgroup$
    – C. E.
    Jan 23 at 11:34












  • $begingroup$
    Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
    $endgroup$
    – LCarvalho
    Jan 23 at 11:43










  • $begingroup$
    @C.E. KnapsackSolve[{230, 260, 320, 350, 650}, 3000] gives me {13, 0, 0, 0, 0}, which only has total cost 2990. I realize large knapsack problems might be difficult, but surely the algorithm should find one of the maximal answers for this small example... any idea if I'm doing something wrong?
    $endgroup$
    – MassDefect
    Jan 23 at 16:46










  • $begingroup$
    @MassDefect I don't think you're doing anything wrong. It would appear it isn't better than that.
    $endgroup$
    – C. E.
    Jan 23 at 16:54










  • $begingroup$
    @C.E. Ah, okay. I'm surprised by that. I thought for sure it would be good enough and that I was doing something silly. Thanks!
    $endgroup$
    – MassDefect
    Jan 23 at 16:56
















6












$begingroup$


I have available a steel bar with length of 3000 mm.



I intend to get the best use of cutting out of this bar.



I need to cut this steel bar in the following lengths: 230mm, 260mm, 320mm, 350mm and 650mm.



enter image description here



How to get cut options "close" to the length of 3000 mm.



Example:



230 + 230 + 260 + 260 + 320 + 320 + 350 + 350 + 650;


-> 2970



I have the code below. I know I will have to use the "Nearest" function, but I am not able to apply it:



lista={230, 260, 320, 350, 650};

resultados=Join@@DeleteCases[IntegerPartitions[#,{1,[Infinity]},lista]&/@Range[3000],{}];

And@@Thread[Total/@resultados<=3000];

contagem=Map[Lookup[Counts[#],lista,0]&,resultados];

contagem//Short


In the my example, i get groups that are smaller that 3000, but wish the values that are "nearest" of the value of 3000






list-manipulation







share|improve this question











$endgroup$








  • 3




    $begingroup$
    This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
    $endgroup$
    – C. E.
    Jan 23 at 11:34












  • $begingroup$
    Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
    $endgroup$
    – LCarvalho
    Jan 23 at 11:43










  • $begingroup$
    @C.E. KnapsackSolve[{230, 260, 320, 350, 650}, 3000] gives me {13, 0, 0, 0, 0}, which only has total cost 2990. I realize large knapsack problems might be difficult, but surely the algorithm should find one of the maximal answers for this small example... any idea if I'm doing something wrong?
    $endgroup$
    – MassDefect
    Jan 23 at 16:46










  • $begingroup$
    @MassDefect I don't think you're doing anything wrong. It would appear it isn't better than that.
    $endgroup$
    – C. E.
    Jan 23 at 16:54










  • $begingroup$
    @C.E. Ah, okay. I'm surprised by that. I thought for sure it would be good enough and that I was doing something silly. Thanks!
    $endgroup$
    – MassDefect
    Jan 23 at 16:56














6












6








6


1



$begingroup$


I have available a steel bar with length of 3000 mm.



I intend to get the best use of cutting out of this bar.



I need to cut this steel bar in the following lengths: 230mm, 260mm, 320mm, 350mm and 650mm.



enter image description here



How to get cut options "close" to the length of 3000 mm.



Example:



230 + 230 + 260 + 260 + 320 + 320 + 350 + 350 + 650;


-> 2970



I have the code below. I know I will have to use the "Nearest" function, but I am not able to apply it:



lista={230, 260, 320, 350, 650};

resultados=Join@@DeleteCases[IntegerPartitions[#,{1,[Infinity]},lista]&/@Range[3000],{}];

And@@Thread[Total/@resultados<=3000];

contagem=Map[Lookup[Counts[#],lista,0]&,resultados];

contagem//Short


In the my example, i get groups that are smaller that 3000, but wish the values that are "nearest" of the value of 3000






list-manipulation







share|improve this question











$endgroup$




I have available a steel bar with length of 3000 mm.



I intend to get the best use of cutting out of this bar.



I need to cut this steel bar in the following lengths: 230mm, 260mm, 320mm, 350mm and 650mm.



enter image description here



How to get cut options "close" to the length of 3000 mm.



Example:



230 + 230 + 260 + 260 + 320 + 320 + 350 + 350 + 650;


-> 2970



I have the code below. I know I will have to use the "Nearest" function, but I am not able to apply it:



lista={230, 260, 320, 350, 650};

resultados=Join@@DeleteCases[IntegerPartitions[#,{1,[Infinity]},lista]&/@Range[3000],{}];

And@@Thread[Total/@resultados<=3000];

contagem=Map[Lookup[Counts[#],lista,0]&,resultados];

contagem//Short


In the my example, i get groups that are smaller that 3000, but wish the values that are "nearest" of the value of 3000






list-manipulation




list-manipulation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 23 at 11:37







LCarvalho

















asked Jan 23 at 11:29









LCarvalhoLCarvalho

5,85642986




5,85642986








  • 3




    $begingroup$
    This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
    $endgroup$
    – C. E.
    Jan 23 at 11:34












  • $begingroup$
    Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
    $endgroup$
    – LCarvalho
    Jan 23 at 11:43










  • $begingroup$
    @C.E. KnapsackSolve[{230, 260, 320, 350, 650}, 3000] gives me {13, 0, 0, 0, 0}, which only has total cost 2990. I realize large knapsack problems might be difficult, but surely the algorithm should find one of the maximal answers for this small example... any idea if I'm doing something wrong?
    $endgroup$
    – MassDefect
    Jan 23 at 16:46










  • $begingroup$
    @MassDefect I don't think you're doing anything wrong. It would appear it isn't better than that.
    $endgroup$
    – C. E.
    Jan 23 at 16:54










  • $begingroup$
    @C.E. Ah, okay. I'm surprised by that. I thought for sure it would be good enough and that I was doing something silly. Thanks!
    $endgroup$
    – MassDefect
    Jan 23 at 16:56














  • 3




    $begingroup$
    This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
    $endgroup$
    – C. E.
    Jan 23 at 11:34












  • $begingroup$
    Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
    $endgroup$
    – LCarvalho
    Jan 23 at 11:43










  • $begingroup$
    @C.E. KnapsackSolve[{230, 260, 320, 350, 650}, 3000] gives me {13, 0, 0, 0, 0}, which only has total cost 2990. I realize large knapsack problems might be difficult, but surely the algorithm should find one of the maximal answers for this small example... any idea if I'm doing something wrong?
    $endgroup$
    – MassDefect
    Jan 23 at 16:46










  • $begingroup$
    @MassDefect I don't think you're doing anything wrong. It would appear it isn't better than that.
    $endgroup$
    – C. E.
    Jan 23 at 16:54










  • $begingroup$
    @C.E. Ah, okay. I'm surprised by that. I thought for sure it would be good enough and that I was doing something silly. Thanks!
    $endgroup$
    – MassDefect
    Jan 23 at 16:56








3




3




$begingroup$
This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
$endgroup$
– C. E.
Jan 23 at 11:34






$begingroup$
This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
$endgroup$
– C. E.
Jan 23 at 11:34














$begingroup$
Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
$endgroup$
– LCarvalho
Jan 23 at 11:43




$begingroup$
Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
$endgroup$
– LCarvalho
Jan 23 at 11:43












$begingroup$
@C.E. KnapsackSolve[{230, 260, 320, 350, 650}, 3000] gives me {13, 0, 0, 0, 0}, which only has total cost 2990. I realize large knapsack problems might be difficult, but surely the algorithm should find one of the maximal answers for this small example... any idea if I'm doing something wrong?
$endgroup$
– MassDefect
Jan 23 at 16:46




$begingroup$
@C.E. KnapsackSolve[{230, 260, 320, 350, 650}, 3000] gives me {13, 0, 0, 0, 0}, which only has total cost 2990. I realize large knapsack problems might be difficult, but surely the algorithm should find one of the maximal answers for this small example... any idea if I'm doing something wrong?
$endgroup$
– MassDefect
Jan 23 at 16:46












$begingroup$
@MassDefect I don't think you're doing anything wrong. It would appear it isn't better than that.
$endgroup$
– C. E.
Jan 23 at 16:54




$begingroup$
@MassDefect I don't think you're doing anything wrong. It would appear it isn't better than that.
$endgroup$
– C. E.
Jan 23 at 16:54












$begingroup$
@C.E. Ah, okay. I'm surprised by that. I thought for sure it would be good enough and that I was doing something silly. Thanks!
$endgroup$
– MassDefect
Jan 23 at 16:56




$begingroup$
@C.E. Ah, okay. I'm surprised by that. I thought for sure it would be good enough and that I was doing something silly. Thanks!
$endgroup$
– MassDefect
Jan 23 at 16:56










2 Answers
2






active

oldest

votes


















7












$begingroup$

x = {x1, x2, x3, x4, x5};

Maximize[{x.lista, x.lista <= 3000, ## & @@ Thread[x >= 0]}, x, Integers]



{3000, {x1 -> 0, x2 -> 0, x3 -> 5, x4 -> 4, x5 -> 0}}




If you have to use at least one of each piece:



Maximize[{x.lista, x.lista <= 3000, ## & @@ Thread[x >= 1]}, x, Integers]



{3000, {x1 -> 2, x2 -> 1, x3 -> 4, x4 -> 1, x5 -> 1}}




All solutions that use all of 3000mm:



FrobeniusSolve[lista, 3000]



{{0, 0, 0, 3, 3}, {0, 0, 5, 4, 0}, {0, 1, 2, 6, 0}, {0, 4, 3, 1,
1}, {0, 5, 0, 3, 1}, {1, 0, 1, 7, 0}, {1, 2, 5, 0, 1}, {1, 3, 2, 2,
1}, {2, 1, 4, 1, 1}, {2, 2, 1, 3, 1}, {3, 0, 3, 2, 1}, {3, 1, 0, 4,
1}, {4, 3, 0, 0, 2}, {4, 8, 0, 0, 0}, {6, 0, 1, 0, 2}, {6, 5, 1, 0,
0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {9, 1, 1, 1, 0}, {10, 0, 0, 2, 0}}




Or use IntegerPartitions as follows:



Map[Lookup[Counts[#], lista, 0] &, IntegerPartitions[3000, All, lista]]



{{0, 0, 0, 3, 3}, {6, 0, 1, 0, 2}, {4, 3, 0, 0, 2}, {3, 1, 0, 4,
1}, {2, 2, 1, 3, 1}, {0, 5, 0, 3, 1}, {3, 0, 3, 2, 1}, {1, 3, 2, 2,
1}, {2, 1, 4, 1, 1}, {0, 4, 3, 1, 1}, {1, 2, 5, 0, 1}, {1, 0, 1, 7,
0}, {0, 1, 2, 6, 0}, {0, 0, 5, 4, 0}, {10, 0, 0, 2, 0}, {9, 1, 1, 1,
0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {6, 5, 1, 0, 0}, {4, 8, 0, 0, 0}}




You can also use Solve and Reduce:



x /. Solve[{x.lista == 3000, ## & @@ Thread[x >= 0]}, x, Integers]

Reduce[{x.lista == 3000, ## & @@ Thread[x >= 0]}, x, Integers][[All, All, -1]] /. 

 Or | And -> List



{{0, 0, 0, 3, 3}, {0, 0, 5, 4, 0}, {0, 1, 2, 6, 0}, {0, 4, 3, 1,
1}, {0, 5, 0, 3, 1}, {1, 0, 1, 7, 0}, {1, 2, 5, 0, 1}, {1, 3, 2, 2,
1}, {2, 1, 4, 1, 1}, {2, 2, 1, 3, 1}, {3, 0, 3, 2, 1}, {3, 1, 0, 4,
1}, {4, 3, 0, 0, 2}, {4, 8, 0, 0, 0}, {6, 0, 1, 0, 2}, {6, 5, 1, 0,
0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {9, 1, 1, 1, 0}, {10, 0, 0, 2, 0}}







share|improve this answer











$endgroup$





















    1












    $begingroup$

    lst = {230, 260, 320, 350, 650};
    
vars = {c1, c2, c3, c4, c5};
    
eq = Inner[Times, lst, vars, Plus]
    

    
(sol = FindInstance[eq == 3000 && vars > 0, vars, Integers, 10]) // Column
    
{c1 -> 1, c2 -> 3, c3 -> 2, c4 -> 2, c5 -> 1}
    
{c1 -> 2, c2 -> 1, c3 -> 4, c4 -> 1, c5 -> 1}
    
{c1 -> 2, c2 -> 2, c3 -> 1, c4 -> 3, c5 -> 1}
    

    
eq /. sol
    
{3000, 3000, 3000}





    share|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      x = {x1, x2, x3, x4, x5};
      
Maximize[{x.lista, x.lista <= 3000, ## & @@ Thread[x >= 0]}, x, Integers]



      {3000, {x1 -> 0, x2 -> 0, x3 -> 5, x4 -> 4, x5 -> 0}}




      If you have to use at least one of each piece:



      Maximize[{x.lista, x.lista <= 3000, ## & @@ Thread[x >= 1]}, x, Integers]



      {3000, {x1 -> 2, x2 -> 1, x3 -> 4, x4 -> 1, x5 -> 1}}




      All solutions that use all of 3000mm:



      FrobeniusSolve[lista, 3000]



      {{0, 0, 0, 3, 3}, {0, 0, 5, 4, 0}, {0, 1, 2, 6, 0}, {0, 4, 3, 1,
      1}, {0, 5, 0, 3, 1}, {1, 0, 1, 7, 0}, {1, 2, 5, 0, 1}, {1, 3, 2, 2,
      1}, {2, 1, 4, 1, 1}, {2, 2, 1, 3, 1}, {3, 0, 3, 2, 1}, {3, 1, 0, 4,
      1}, {4, 3, 0, 0, 2}, {4, 8, 0, 0, 0}, {6, 0, 1, 0, 2}, {6, 5, 1, 0,
      0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {9, 1, 1, 1, 0}, {10, 0, 0, 2, 0}}




      Or use IntegerPartitions as follows:



      Map[Lookup[Counts[#], lista, 0] &, IntegerPartitions[3000, All, lista]]



      {{0, 0, 0, 3, 3}, {6, 0, 1, 0, 2}, {4, 3, 0, 0, 2}, {3, 1, 0, 4,
      1}, {2, 2, 1, 3, 1}, {0, 5, 0, 3, 1}, {3, 0, 3, 2, 1}, {1, 3, 2, 2,
      1}, {2, 1, 4, 1, 1}, {0, 4, 3, 1, 1}, {1, 2, 5, 0, 1}, {1, 0, 1, 7,
      0}, {0, 1, 2, 6, 0}, {0, 0, 5, 4, 0}, {10, 0, 0, 2, 0}, {9, 1, 1, 1,
      0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {6, 5, 1, 0, 0}, {4, 8, 0, 0, 0}}




      You can also use Solve and Reduce:



      x /. Solve[{x.lista == 3000, ## & @@ Thread[x >= 0]}, x, Integers]
      
Reduce[{x.lista == 3000, ## & @@ Thread[x >= 0]}, x, Integers][[All, All, -1]] /. 
      
 Or | And -> List



      {{0, 0, 0, 3, 3}, {0, 0, 5, 4, 0}, {0, 1, 2, 6, 0}, {0, 4, 3, 1,
      1}, {0, 5, 0, 3, 1}, {1, 0, 1, 7, 0}, {1, 2, 5, 0, 1}, {1, 3, 2, 2,
      1}, {2, 1, 4, 1, 1}, {2, 2, 1, 3, 1}, {3, 0, 3, 2, 1}, {3, 1, 0, 4,
      1}, {4, 3, 0, 0, 2}, {4, 8, 0, 0, 0}, {6, 0, 1, 0, 2}, {6, 5, 1, 0,
      0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {9, 1, 1, 1, 0}, {10, 0, 0, 2, 0}}







      share|improve this answer











      $endgroup$


















        7












        $begingroup$

        x = {x1, x2, x3, x4, x5};
        
Maximize[{x.lista, x.lista <= 3000, ## & @@ Thread[x >= 0]}, x, Integers]



        {3000, {x1 -> 0, x2 -> 0, x3 -> 5, x4 -> 4, x5 -> 0}}




        If you have to use at least one of each piece:



        Maximize[{x.lista, x.lista <= 3000, ## & @@ Thread[x >= 1]}, x, Integers]



        {3000, {x1 -> 2, x2 -> 1, x3 -> 4, x4 -> 1, x5 -> 1}}




        All solutions that use all of 3000mm:



        FrobeniusSolve[lista, 3000]



        {{0, 0, 0, 3, 3}, {0, 0, 5, 4, 0}, {0, 1, 2, 6, 0}, {0, 4, 3, 1,
        1}, {0, 5, 0, 3, 1}, {1, 0, 1, 7, 0}, {1, 2, 5, 0, 1}, {1, 3, 2, 2,
        1}, {2, 1, 4, 1, 1}, {2, 2, 1, 3, 1}, {3, 0, 3, 2, 1}, {3, 1, 0, 4,
        1}, {4, 3, 0, 0, 2}, {4, 8, 0, 0, 0}, {6, 0, 1, 0, 2}, {6, 5, 1, 0,
        0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {9, 1, 1, 1, 0}, {10, 0, 0, 2, 0}}




        Or use IntegerPartitions as follows:



        Map[Lookup[Counts[#], lista, 0] &, IntegerPartitions[3000, All, lista]]



        {{0, 0, 0, 3, 3}, {6, 0, 1, 0, 2}, {4, 3, 0, 0, 2}, {3, 1, 0, 4,
        1}, {2, 2, 1, 3, 1}, {0, 5, 0, 3, 1}, {3, 0, 3, 2, 1}, {1, 3, 2, 2,
        1}, {2, 1, 4, 1, 1}, {0, 4, 3, 1, 1}, {1, 2, 5, 0, 1}, {1, 0, 1, 7,
        0}, {0, 1, 2, 6, 0}, {0, 0, 5, 4, 0}, {10, 0, 0, 2, 0}, {9, 1, 1, 1,
        0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {6, 5, 1, 0, 0}, {4, 8, 0, 0, 0}}




        You can also use Solve and Reduce:



        x /. Solve[{x.lista == 3000, ## & @@ Thread[x >= 0]}, x, Integers]
        
Reduce[{x.lista == 3000, ## & @@ Thread[x >= 0]}, x, Integers][[All, All, -1]] /. 
        
 Or | And -> List



        {{0, 0, 0, 3, 3}, {0, 0, 5, 4, 0}, {0, 1, 2, 6, 0}, {0, 4, 3, 1,
        1}, {0, 5, 0, 3, 1}, {1, 0, 1, 7, 0}, {1, 2, 5, 0, 1}, {1, 3, 2, 2,
        1}, {2, 1, 4, 1, 1}, {2, 2, 1, 3, 1}, {3, 0, 3, 2, 1}, {3, 1, 0, 4,
        1}, {4, 3, 0, 0, 2}, {4, 8, 0, 0, 0}, {6, 0, 1, 0, 2}, {6, 5, 1, 0,
        0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {9, 1, 1, 1, 0}, {10, 0, 0, 2, 0}}







        share|improve this answer











        $endgroup$
















          7












          7








          7





          $begingroup$

          x = {x1, x2, x3, x4, x5};
          
Maximize[{x.lista, x.lista <= 3000, ## & @@ Thread[x >= 0]}, x, Integers]



          {3000, {x1 -> 0, x2 -> 0, x3 -> 5, x4 -> 4, x5 -> 0}}




          If you have to use at least one of each piece:



          Maximize[{x.lista, x.lista <= 3000, ## & @@ Thread[x >= 1]}, x, Integers]



          {3000, {x1 -> 2, x2 -> 1, x3 -> 4, x4 -> 1, x5 -> 1}}




          All solutions that use all of 3000mm:



          FrobeniusSolve[lista, 3000]



          {{0, 0, 0, 3, 3}, {0, 0, 5, 4, 0}, {0, 1, 2, 6, 0}, {0, 4, 3, 1,
          1}, {0, 5, 0, 3, 1}, {1, 0, 1, 7, 0}, {1, 2, 5, 0, 1}, {1, 3, 2, 2,
          1}, {2, 1, 4, 1, 1}, {2, 2, 1, 3, 1}, {3, 0, 3, 2, 1}, {3, 1, 0, 4,
          1}, {4, 3, 0, 0, 2}, {4, 8, 0, 0, 0}, {6, 0, 1, 0, 2}, {6, 5, 1, 0,
          0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {9, 1, 1, 1, 0}, {10, 0, 0, 2, 0}}




          Or use IntegerPartitions as follows:



          Map[Lookup[Counts[#], lista, 0] &, IntegerPartitions[3000, All, lista]]



          {{0, 0, 0, 3, 3}, {6, 0, 1, 0, 2}, {4, 3, 0, 0, 2}, {3, 1, 0, 4,
          1}, {2, 2, 1, 3, 1}, {0, 5, 0, 3, 1}, {3, 0, 3, 2, 1}, {1, 3, 2, 2,
          1}, {2, 1, 4, 1, 1}, {0, 4, 3, 1, 1}, {1, 2, 5, 0, 1}, {1, 0, 1, 7,
          0}, {0, 1, 2, 6, 0}, {0, 0, 5, 4, 0}, {10, 0, 0, 2, 0}, {9, 1, 1, 1,
          0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {6, 5, 1, 0, 0}, {4, 8, 0, 0, 0}}




          You can also use Solve and Reduce:



          x /. Solve[{x.lista == 3000, ## & @@ Thread[x >= 0]}, x, Integers]
          
Reduce[{x.lista == 3000, ## & @@ Thread[x >= 0]}, x, Integers][[All, All, -1]] /. 
          
 Or | And -> List



          {{0, 0, 0, 3, 3}, {0, 0, 5, 4, 0}, {0, 1, 2, 6, 0}, {0, 4, 3, 1,
          1}, {0, 5, 0, 3, 1}, {1, 0, 1, 7, 0}, {1, 2, 5, 0, 1}, {1, 3, 2, 2,
          1}, {2, 1, 4, 1, 1}, {2, 2, 1, 3, 1}, {3, 0, 3, 2, 1}, {3, 1, 0, 4,
          1}, {4, 3, 0, 0, 2}, {4, 8, 0, 0, 0}, {6, 0, 1, 0, 2}, {6, 5, 1, 0,
          0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {9, 1, 1, 1, 0}, {10, 0, 0, 2, 0}}







          share|improve this answer











          $endgroup$



          x = {x1, x2, x3, x4, x5};
          
Maximize[{x.lista, x.lista <= 3000, ## & @@ Thread[x >= 0]}, x, Integers]



          {3000, {x1 -> 0, x2 -> 0, x3 -> 5, x4 -> 4, x5 -> 0}}




          If you have to use at least one of each piece:



          Maximize[{x.lista, x.lista <= 3000, ## & @@ Thread[x >= 1]}, x, Integers]



          {3000, {x1 -> 2, x2 -> 1, x3 -> 4, x4 -> 1, x5 -> 1}}




          All solutions that use all of 3000mm:



          FrobeniusSolve[lista, 3000]



          {{0, 0, 0, 3, 3}, {0, 0, 5, 4, 0}, {0, 1, 2, 6, 0}, {0, 4, 3, 1,
          1}, {0, 5, 0, 3, 1}, {1, 0, 1, 7, 0}, {1, 2, 5, 0, 1}, {1, 3, 2, 2,
          1}, {2, 1, 4, 1, 1}, {2, 2, 1, 3, 1}, {3, 0, 3, 2, 1}, {3, 1, 0, 4,
          1}, {4, 3, 0, 0, 2}, {4, 8, 0, 0, 0}, {6, 0, 1, 0, 2}, {6, 5, 1, 0,
          0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {9, 1, 1, 1, 0}, {10, 0, 0, 2, 0}}




          Or use IntegerPartitions as follows:



          Map[Lookup[Counts[#], lista, 0] &, IntegerPartitions[3000, All, lista]]



          {{0, 0, 0, 3, 3}, {6, 0, 1, 0, 2}, {4, 3, 0, 0, 2}, {3, 1, 0, 4,
          1}, {2, 2, 1, 3, 1}, {0, 5, 0, 3, 1}, {3, 0, 3, 2, 1}, {1, 3, 2, 2,
          1}, {2, 1, 4, 1, 1}, {0, 4, 3, 1, 1}, {1, 2, 5, 0, 1}, {1, 0, 1, 7,
          0}, {0, 1, 2, 6, 0}, {0, 0, 5, 4, 0}, {10, 0, 0, 2, 0}, {9, 1, 1, 1,
          0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {6, 5, 1, 0, 0}, {4, 8, 0, 0, 0}}




          You can also use Solve and Reduce:



          x /. Solve[{x.lista == 3000, ## & @@ Thread[x >= 0]}, x, Integers]
          
Reduce[{x.lista == 3000, ## & @@ Thread[x >= 0]}, x, Integers][[All, All, -1]] /. 
          
 Or | And -> List



          {{0, 0, 0, 3, 3}, {0, 0, 5, 4, 0}, {0, 1, 2, 6, 0}, {0, 4, 3, 1,
          1}, {0, 5, 0, 3, 1}, {1, 0, 1, 7, 0}, {1, 2, 5, 0, 1}, {1, 3, 2, 2,
          1}, {2, 1, 4, 1, 1}, {2, 2, 1, 3, 1}, {3, 0, 3, 2, 1}, {3, 1, 0, 4,
          1}, {4, 3, 0, 0, 2}, {4, 8, 0, 0, 0}, {6, 0, 1, 0, 2}, {6, 5, 1, 0,
          0}, {7, 4, 0, 1, 0}, {8, 2, 2, 0, 0}, {9, 1, 1, 1, 0}, {10, 0, 0, 2, 0}}








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Jan 23 at 14:09

























          answered Jan 23 at 11:42









          kglrkglr

          188k10204422




          188k10204422























              1












              $begingroup$

              lst = {230, 260, 320, 350, 650};
              
vars = {c1, c2, c3, c4, c5};
              
eq = Inner[Times, lst, vars, Plus]
              

              
(sol = FindInstance[eq == 3000 && vars > 0, vars, Integers, 10]) // Column
              
{c1 -> 1, c2 -> 3, c3 -> 2, c4 -> 2, c5 -> 1}
              
{c1 -> 2, c2 -> 1, c3 -> 4, c4 -> 1, c5 -> 1}
              
{c1 -> 2, c2 -> 2, c3 -> 1, c4 -> 3, c5 -> 1}
              

              
eq /. sol
              
{3000, 3000, 3000}





              share|improve this answer









              $endgroup$


















                1












                $begingroup$

                lst = {230, 260, 320, 350, 650};
                
vars = {c1, c2, c3, c4, c5};
                
eq = Inner[Times, lst, vars, Plus]
                

                
(sol = FindInstance[eq == 3000 && vars > 0, vars, Integers, 10]) // Column
                
{c1 -> 1, c2 -> 3, c3 -> 2, c4 -> 2, c5 -> 1}
                
{c1 -> 2, c2 -> 1, c3 -> 4, c4 -> 1, c5 -> 1}
                
{c1 -> 2, c2 -> 2, c3 -> 1, c4 -> 3, c5 -> 1}
                

                
eq /. sol
                
{3000, 3000, 3000}





                share|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  lst = {230, 260, 320, 350, 650};
                  
vars = {c1, c2, c3, c4, c5};
                  
eq = Inner[Times, lst, vars, Plus]
                  

                  
(sol = FindInstance[eq == 3000 && vars > 0, vars, Integers, 10]) // Column
                  
{c1 -> 1, c2 -> 3, c3 -> 2, c4 -> 2, c5 -> 1}
                  
{c1 -> 2, c2 -> 1, c3 -> 4, c4 -> 1, c5 -> 1}
                  
{c1 -> 2, c2 -> 2, c3 -> 1, c4 -> 3, c5 -> 1}
                  

                  
eq /. sol
                  
{3000, 3000, 3000}





                  share|improve this answer









                  $endgroup$



                  lst = {230, 260, 320, 350, 650};
                  
vars = {c1, c2, c3, c4, c5};
                  
eq = Inner[Times, lst, vars, Plus]
                  

                  
(sol = FindInstance[eq == 3000 && vars > 0, vars, Integers, 10]) // Column
                  
{c1 -> 1, c2 -> 3, c3 -> 2, c4 -> 2, c5 -> 1}
                  
{c1 -> 2, c2 -> 1, c3 -> 4, c4 -> 1, c5 -> 1}
                  
{c1 -> 2, c2 -> 2, c3 -> 1, c4 -> 3, c5 -> 1}
                  

                  
eq /. sol
                  
{3000, 3000, 3000}






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Jan 23 at 14:01









                  rmwrmw

                  2897




                  2897






























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