Best use of steel bar cut for mechanical study
$begingroup$
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.
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
$endgroup$
add a comment |
$begingroup$
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.
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
$endgroup$
3
$begingroup$
This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
$endgroup$
– C. E.
3 hours ago
$begingroup$
Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
$endgroup$
– LCarvalho
2 hours ago
add a comment |
$begingroup$
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.
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
$endgroup$
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.
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
edited 2 hours ago
LCarvalho
asked 3 hours ago
LCarvalhoLCarvalho
5,66942886
5,66942886
3
$begingroup$
This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
$endgroup$
– C. E.
3 hours ago
$begingroup$
Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
$endgroup$
– LCarvalho
2 hours ago
add a comment |
3
$begingroup$
This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
$endgroup$
– C. E.
3 hours ago
$begingroup$
Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
$endgroup$
– LCarvalho
2 hours ago
3
3
$begingroup$
This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
$endgroup$
– C. E.
3 hours ago
$begingroup$
This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
$endgroup$
– C. E.
3 hours ago
$begingroup$
Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
$endgroup$
– LCarvalho
2 hours ago
$begingroup$
Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
$endgroup$
– LCarvalho
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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}}
$endgroup$
add a comment |
$begingroup$
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}
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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}}
$endgroup$
add a comment |
$begingroup$
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}}
$endgroup$
add a comment |
$begingroup$
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}}
$endgroup$
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}}
edited 27 mins ago
answered 2 hours ago
kglrkglr
180k9200413
180k9200413
add a comment |
add a comment |
$begingroup$
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}
$endgroup$
add a comment |
$begingroup$
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}
$endgroup$
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$begingroup$
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}
$endgroup$
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}
answered 35 mins ago
rmwrmw
2247
2247
add a comment |
add a comment |
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3
$begingroup$
This is known as the knapsack problem and there is a built-in function solve it, KnapsackSolve.
$endgroup$
– C. E.
3 hours ago
$begingroup$
Almost this. There are several combinations possible of cut with these lengths that added are "close" of 3000
$endgroup$
– LCarvalho
2 hours ago