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.

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

edited Jan 23 at 11:37

LCarvalho

asked Jan 23 at 11:29

LCarvalhoLCarvalho

5,85642986

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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.

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

edited Jan 23 at 11:37

LCarvalho

asked Jan 23 at 11:29

LCarvalhoLCarvalho

5,85642986

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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.

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

edited Jan 23 at 11:37

LCarvalho

asked Jan 23 at 11:29

LCarvalhoLCarvalho

5,85642986

$endgroup$

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

share|improve this question

edited Jan 23 at 11:37

LCarvalho

asked Jan 23 at 11:29

LCarvalhoLCarvalho

5,85642986

share|improve this question

edited Jan 23 at 11:37

LCarvalho

asked Jan 23 at 11:29

LCarvalhoLCarvalho

5,85642986

share|improve this question

share|improve this question

edited Jan 23 at 11:37

LCarvalho

edited Jan 23 at 11:37

LCarvalho

edited Jan 23 at 11:37

LCarvalho

asked Jan 23 at 11:29

LCarvalhoLCarvalho

5,85642986

asked Jan 23 at 11:29

LCarvalhoLCarvalho

5,85642986

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
  

  
  
  
  

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.
  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

add a comment | 

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

edited Jan 23 at 14:09

answered Jan 23 at 11:42

kglrkglr

188k10204422

$endgroup$

add a comment | 

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

answered Jan 23 at 14:01

rmwrmw

2897

$endgroup$

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

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

edited Jan 23 at 14:09

answered Jan 23 at 11:42

kglrkglr

188k10204422

$endgroup$

add a comment | 

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

edited Jan 23 at 14:09

answered Jan 23 at 11:42

kglrkglr

188k10204422

$endgroup$

add a comment | 

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

edited Jan 23 at 14:09

answered Jan 23 at 11:42

kglrkglr

188k10204422

$endgroup$

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

edited Jan 23 at 14:09

answered Jan 23 at 11:42

kglrkglr

188k10204422

share|improve this answer

share|improve this answer

edited Jan 23 at 14:09

edited Jan 23 at 14:09

edited Jan 23 at 14:09

answered Jan 23 at 11:42

kglrkglr

188k10204422

answered Jan 23 at 11:42

kglrkglr

188k10204422

answered Jan 23 at 11:42

kglrkglr

188k10204422

188k10204422

add a comment | 

add a comment | 

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

answered Jan 23 at 14:01

rmwrmw

2897

$endgroup$

add a comment | 

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

answered Jan 23 at 14:01

rmwrmw

2897

$endgroup$

add a comment | 

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

answered Jan 23 at 14:01

rmwrmw

2897

$endgroup$

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

answered Jan 23 at 14:01

rmwrmw

2897

share|improve this answer

share|improve this answer

answered Jan 23 at 14:01

rmwrmw

2897

answered Jan 23 at 14:01

rmwrmw

2897

answered Jan 23 at 14:01

rmwrmw

2897

2897

add a comment | 

add a comment | 

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