Mixing
Pack boxes with different size inside a large box
$begingroup$
Suppose there are 3 sizes of boxes:
- 10cm x 10cm x 10cm
- 20cm x 20cm x 20cm
- 10cm x 15cm x 4cm
and I have boxes of dimensions 20 x a, 10 x c and 30 x b. How many boxes of 200 x 200 x 100cm do I need to pack them all?
rectangles packing-problem
edited Jan 13 at 15:03
quid♦
37.1k95093
asked Jun 22 '17 at 16:04
Ansary Ans21Ansary Ans21
1012
$endgroup$
add a comment |
$begingroup$
Suppose there are 3 sizes of boxes:
- 10cm x 10cm x 10cm
- 20cm x 20cm x 20cm
- 10cm x 15cm x 4cm
and I have boxes of dimensions 20 x a, 10 x c and 30 x b. How many boxes of 200 x 200 x 100cm do I need to pack them all?
rectangles packing-problem
edited Jan 13 at 15:03
quid♦
37.1k95093
asked Jun 22 '17 at 16:04
Ansary Ans21Ansary Ans21
1012
$endgroup$
1
$begingroup$
"I have boxes of dimensions 20 x a, 10 x c and 30 x b" : I don't get that part. Do you mean you have 20 boxes of type a, 30 of type b and 10 of type c ?
$endgroup$
– Evargalo
Jun 22 '17 at 16:14
1
$begingroup$
1 box will give you plenty enough room!
$endgroup$
– Bram28
Jun 22 '17 at 16:43
$begingroup$
@Evargalo yes correct.
$endgroup$
– Ansary Ans21
Jun 22 '17 at 23:58
add a comment |
$begingroup$
Suppose there are 3 sizes of boxes:
- 10cm x 10cm x 10cm
- 20cm x 20cm x 20cm
- 10cm x 15cm x 4cm
and I have boxes of dimensions 20 x a, 10 x c and 30 x b. How many boxes of 200 x 200 x 100cm do I need to pack them all?
rectangles packing-problem
edited Jan 13 at 15:03
quid♦
37.1k95093
asked Jun 22 '17 at 16:04
Ansary Ans21Ansary Ans21
1012
$endgroup$
Suppose there are 3 sizes of boxes:
- 10cm x 10cm x 10cm
- 20cm x 20cm x 20cm
- 10cm x 15cm x 4cm
and I have boxes of dimensions 20 x a, 10 x c and 30 x b. How many boxes of 200 x 200 x 100cm do I need to pack them all?
rectangles packing-problem
rectangles packing-problem
edited Jan 13 at 15:03
quid♦
37.1k95093
asked Jun 22 '17 at 16:04
Ansary Ans21Ansary Ans21
1012
edited Jan 13 at 15:03
quid♦
37.1k95093
asked Jun 22 '17 at 16:04
Ansary Ans21Ansary Ans21
1012
edited Jan 13 at 15:03
quid♦
37.1k95093
edited Jan 13 at 15:03
quid♦
37.1k95093
edited Jan 13 at 15:03
quid♦
37.1k95093
37.1k95093
asked Jun 22 '17 at 16:04
Ansary Ans21Ansary Ans21
1012
asked Jun 22 '17 at 16:04
Ansary Ans21Ansary Ans21
1012
asked Jun 22 '17 at 16:04
Ansary Ans21Ansary Ans21
1012
1012
1
$begingroup$
"I have boxes of dimensions 20 x a, 10 x c and 30 x b" : I don't get that part. Do you mean you have 20 boxes of type a, 30 of type b and 10 of type c ?
$endgroup$
– Evargalo
Jun 22 '17 at 16:14
1
$begingroup$
1 box will give you plenty enough room!
$endgroup$
– Bram28
Jun 22 '17 at 16:43
$begingroup$
@Evargalo yes correct.
$endgroup$
– Ansary Ans21
Jun 22 '17 at 23:58
add a comment |
1
$begingroup$
"I have boxes of dimensions 20 x a, 10 x c and 30 x b" : I don't get that part. Do you mean you have 20 boxes of type a, 30 of type b and 10 of type c ?
$endgroup$
– Evargalo
Jun 22 '17 at 16:14
1
$begingroup$
1 box will give you plenty enough room!
$endgroup$
– Bram28
Jun 22 '17 at 16:43
$begingroup$
@Evargalo yes correct.
$endgroup$
– Ansary Ans21
Jun 22 '17 at 23:58
1
1
$begingroup$
"I have boxes of dimensions 20 x a, 10 x c and 30 x b" : I don't get that part. Do you mean you have 20 boxes of type a, 30 of type b and 10 of type c ?
$endgroup$
– Evargalo
Jun 22 '17 at 16:14
$begingroup$
"I have boxes of dimensions 20 x a, 10 x c and 30 x b" : I don't get that part. Do you mean you have 20 boxes of type a, 30 of type b and 10 of type c ?
$endgroup$
– Evargalo
Jun 22 '17 at 16:14
1
1
$begingroup$
1 box will give you plenty enough room!
$endgroup$
– Bram28
Jun 22 '17 at 16:43
$begingroup$
1 box will give you plenty enough room!
$endgroup$
– Bram28
Jun 22 '17 at 16:43
$begingroup$
@Evargalo yes correct.
$endgroup$
– Ansary Ans21
Jun 22 '17 at 23:58
$begingroup$
@Evargalo yes correct.
$endgroup$
– Ansary Ans21
Jun 22 '17 at 23:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$1$ box will be far more than enough! To see this, let's pack even more boxes, let's say a nice even:
$100$ of type $a$: $10*10*10$
$50$ of type $b$: $20*20*20$
$100$ of type $c$: $10*15*4$
Now, the $100$ boxes of type $a$ can be packed in a square of $10$ by $10$ of such boxes, giving you a layer of $100*100*10$
The $50$ boxes of type $b$ can be packed by putting them into two $5x5$ squares, thus giving $2$ layers of $100*100*20$, for a total of $100*100*40$
Finally, the $100$ boxed of type $$c are all smaller in every dimension than those of type $b$, so let's just pack 100 more boxes of type b: if we can do that, we can certainly pack $100$ boxes of type $c$.
Well, if we pack $100$ more boxes of type $b$ the same was as the earlier $50$ boxes, then it would take us an additional $100*100*80$ of space.
So, stack all these layers, and you get $100*100*130$, which easily fits into a single $100*100*200$ box. (indeed, a $100*100*100$ box will be just fine, giving that the boxes of type $c$ can be packed with a height of $4$ instead of $20$, so they would only take up $100*100*16$, for a total of $100*100*66$)
Plenty of space!!
answered Jun 22 '17 at 16:52
Bram28Bram28
62.3k44793
$endgroup$
$begingroup$
Yeah it works. Say I am replacing the 200x200x100 box with a 60x60x30 box. So how many boxes of 60x60x30 do I need to pack them all? Thank you for the above answer!
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:30
$begingroup$
Is there any common formula to solve such problems?
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:41
$begingroup$
@AnsaryAns21 Well, I'm no expert in packing problems ... I just try to create nice layers! :). for example, for your 60*60*30 box I would first make a layer of 9 type b boxes in a 3x3 square, so that creates a layer of 60x60x20. Then I'll put together some type c boxes: I can nicely fit their length 4 times into the 60, so I 'll create a 60*8*10 rectangle with 8 pf them. Also, 18 of the type a boxes in a 3x6 patterns takes up 60x30x10, so that leaves me with 60x22x10: plenty to pack the last 2 c boxes and the last two a boxes. So that's the first box. Now I have 21 b boxes left. .. (Continued)
$endgroup$
– Bram28
Jun 23 '17 at 1:12
$begingroup$
Since I already figures out that I can at best get 9 boxes of type b into a 60x60x30 box, it will take me 3 more bix boxes .. So I need 4 big boxes in that case.
$endgroup$
– Bram28
Jun 23 '17 at 1:13
add a comment |
Your Answer
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$begingroup$
$1$ box will be far more than enough! To see this, let's pack even more boxes, let's say a nice even:
$100$ of type $a$: $10*10*10$
$50$ of type $b$: $20*20*20$
$100$ of type $c$: $10*15*4$
Now, the $100$ boxes of type $a$ can be packed in a square of $10$ by $10$ of such boxes, giving you a layer of $100*100*10$
The $50$ boxes of type $b$ can be packed by putting them into two $5x5$ squares, thus giving $2$ layers of $100*100*20$, for a total of $100*100*40$
Finally, the $100$ boxed of type $$c are all smaller in every dimension than those of type $b$, so let's just pack 100 more boxes of type b: if we can do that, we can certainly pack $100$ boxes of type $c$.
Well, if we pack $100$ more boxes of type $b$ the same was as the earlier $50$ boxes, then it would take us an additional $100*100*80$ of space.
So, stack all these layers, and you get $100*100*130$, which easily fits into a single $100*100*200$ box. (indeed, a $100*100*100$ box will be just fine, giving that the boxes of type $c$ can be packed with a height of $4$ instead of $20$, so they would only take up $100*100*16$, for a total of $100*100*66$)
Plenty of space!!
answered Jun 22 '17 at 16:52
Bram28Bram28
62.3k44793
$endgroup$
$begingroup$
Yeah it works. Say I am replacing the 200x200x100 box with a 60x60x30 box. So how many boxes of 60x60x30 do I need to pack them all? Thank you for the above answer!
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:30
$begingroup$
Is there any common formula to solve such problems?
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:41
$begingroup$
@AnsaryAns21 Well, I'm no expert in packing problems ... I just try to create nice layers! :). for example, for your 60*60*30 box I would first make a layer of 9 type b boxes in a 3x3 square, so that creates a layer of 60x60x20. Then I'll put together some type c boxes: I can nicely fit their length 4 times into the 60, so I 'll create a 60*8*10 rectangle with 8 pf them. Also, 18 of the type a boxes in a 3x6 patterns takes up 60x30x10, so that leaves me with 60x22x10: plenty to pack the last 2 c boxes and the last two a boxes. So that's the first box. Now I have 21 b boxes left. .. (Continued)
$endgroup$
– Bram28
Jun 23 '17 at 1:12
$begingroup$
Since I already figures out that I can at best get 9 boxes of type b into a 60x60x30 box, it will take me 3 more bix boxes .. So I need 4 big boxes in that case.
$endgroup$
– Bram28
Jun 23 '17 at 1:13
add a comment |
$begingroup$
$1$ box will be far more than enough! To see this, let's pack even more boxes, let's say a nice even:
$100$ of type $a$: $10*10*10$
$50$ of type $b$: $20*20*20$
$100$ of type $c$: $10*15*4$
Now, the $100$ boxes of type $a$ can be packed in a square of $10$ by $10$ of such boxes, giving you a layer of $100*100*10$
The $50$ boxes of type $b$ can be packed by putting them into two $5x5$ squares, thus giving $2$ layers of $100*100*20$, for a total of $100*100*40$
Finally, the $100$ boxed of type $$c are all smaller in every dimension than those of type $b$, so let's just pack 100 more boxes of type b: if we can do that, we can certainly pack $100$ boxes of type $c$.
Well, if we pack $100$ more boxes of type $b$ the same was as the earlier $50$ boxes, then it would take us an additional $100*100*80$ of space.
So, stack all these layers, and you get $100*100*130$, which easily fits into a single $100*100*200$ box. (indeed, a $100*100*100$ box will be just fine, giving that the boxes of type $c$ can be packed with a height of $4$ instead of $20$, so they would only take up $100*100*16$, for a total of $100*100*66$)
Plenty of space!!
answered Jun 22 '17 at 16:52
Bram28Bram28
62.3k44793
$endgroup$
$begingroup$
Yeah it works. Say I am replacing the 200x200x100 box with a 60x60x30 box. So how many boxes of 60x60x30 do I need to pack them all? Thank you for the above answer!
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:30
$begingroup$
Is there any common formula to solve such problems?
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:41
$begingroup$
@AnsaryAns21 Well, I'm no expert in packing problems ... I just try to create nice layers! :). for example, for your 60*60*30 box I would first make a layer of 9 type b boxes in a 3x3 square, so that creates a layer of 60x60x20. Then I'll put together some type c boxes: I can nicely fit their length 4 times into the 60, so I 'll create a 60*8*10 rectangle with 8 pf them. Also, 18 of the type a boxes in a 3x6 patterns takes up 60x30x10, so that leaves me with 60x22x10: plenty to pack the last 2 c boxes and the last two a boxes. So that's the first box. Now I have 21 b boxes left. .. (Continued)
$endgroup$
– Bram28
Jun 23 '17 at 1:12
$begingroup$
Since I already figures out that I can at best get 9 boxes of type b into a 60x60x30 box, it will take me 3 more bix boxes .. So I need 4 big boxes in that case.
$endgroup$
– Bram28
Jun 23 '17 at 1:13
add a comment |
$begingroup$
$1$ box will be far more than enough! To see this, let's pack even more boxes, let's say a nice even:
$100$ of type $a$: $10*10*10$
$50$ of type $b$: $20*20*20$
$100$ of type $c$: $10*15*4$
Now, the $100$ boxes of type $a$ can be packed in a square of $10$ by $10$ of such boxes, giving you a layer of $100*100*10$
The $50$ boxes of type $b$ can be packed by putting them into two $5x5$ squares, thus giving $2$ layers of $100*100*20$, for a total of $100*100*40$
Finally, the $100$ boxed of type $$c are all smaller in every dimension than those of type $b$, so let's just pack 100 more boxes of type b: if we can do that, we can certainly pack $100$ boxes of type $c$.
Well, if we pack $100$ more boxes of type $b$ the same was as the earlier $50$ boxes, then it would take us an additional $100*100*80$ of space.
So, stack all these layers, and you get $100*100*130$, which easily fits into a single $100*100*200$ box. (indeed, a $100*100*100$ box will be just fine, giving that the boxes of type $c$ can be packed with a height of $4$ instead of $20$, so they would only take up $100*100*16$, for a total of $100*100*66$)
Plenty of space!!
answered Jun 22 '17 at 16:52
Bram28Bram28
62.3k44793
$endgroup$
$1$ box will be far more than enough! To see this, let's pack even more boxes, let's say a nice even:
$100$ of type $a$: $10*10*10$
$50$ of type $b$: $20*20*20$
$100$ of type $c$: $10*15*4$
Now, the $100$ boxes of type $a$ can be packed in a square of $10$ by $10$ of such boxes, giving you a layer of $100*100*10$
The $50$ boxes of type $b$ can be packed by putting them into two $5x5$ squares, thus giving $2$ layers of $100*100*20$, for a total of $100*100*40$
Finally, the $100$ boxed of type $$c are all smaller in every dimension than those of type $b$, so let's just pack 100 more boxes of type b: if we can do that, we can certainly pack $100$ boxes of type $c$.
Well, if we pack $100$ more boxes of type $b$ the same was as the earlier $50$ boxes, then it would take us an additional $100*100*80$ of space.
So, stack all these layers, and you get $100*100*130$, which easily fits into a single $100*100*200$ box. (indeed, a $100*100*100$ box will be just fine, giving that the boxes of type $c$ can be packed with a height of $4$ instead of $20$, so they would only take up $100*100*16$, for a total of $100*100*66$)
Plenty of space!!
answered Jun 22 '17 at 16:52
Bram28Bram28
62.3k44793
answered Jun 22 '17 at 16:52
Bram28Bram28
62.3k44793
answered Jun 22 '17 at 16:52
Bram28Bram28
62.3k44793
answered Jun 22 '17 at 16:52
Bram28Bram28
62.3k44793
62.3k44793
$begingroup$
Yeah it works. Say I am replacing the 200x200x100 box with a 60x60x30 box. So how many boxes of 60x60x30 do I need to pack them all? Thank you for the above answer!
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:30
$begingroup$
Is there any common formula to solve such problems?
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:41
$begingroup$
@AnsaryAns21 Well, I'm no expert in packing problems ... I just try to create nice layers! :). for example, for your 60*60*30 box I would first make a layer of 9 type b boxes in a 3x3 square, so that creates a layer of 60x60x20. Then I'll put together some type c boxes: I can nicely fit their length 4 times into the 60, so I 'll create a 60*8*10 rectangle with 8 pf them. Also, 18 of the type a boxes in a 3x6 patterns takes up 60x30x10, so that leaves me with 60x22x10: plenty to pack the last 2 c boxes and the last two a boxes. So that's the first box. Now I have 21 b boxes left. .. (Continued)
$endgroup$
– Bram28
Jun 23 '17 at 1:12
$begingroup$
Since I already figures out that I can at best get 9 boxes of type b into a 60x60x30 box, it will take me 3 more bix boxes .. So I need 4 big boxes in that case.
$endgroup$
– Bram28
Jun 23 '17 at 1:13
add a comment |
$begingroup$
Yeah it works. Say I am replacing the 200x200x100 box with a 60x60x30 box. So how many boxes of 60x60x30 do I need to pack them all? Thank you for the above answer!
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:30
$begingroup$
Is there any common formula to solve such problems?
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:41
$begingroup$
@AnsaryAns21 Well, I'm no expert in packing problems ... I just try to create nice layers! :). for example, for your 60*60*30 box I would first make a layer of 9 type b boxes in a 3x3 square, so that creates a layer of 60x60x20. Then I'll put together some type c boxes: I can nicely fit their length 4 times into the 60, so I 'll create a 60*8*10 rectangle with 8 pf them. Also, 18 of the type a boxes in a 3x6 patterns takes up 60x30x10, so that leaves me with 60x22x10: plenty to pack the last 2 c boxes and the last two a boxes. So that's the first box. Now I have 21 b boxes left. .. (Continued)
$endgroup$
– Bram28
Jun 23 '17 at 1:12
$begingroup$
Since I already figures out that I can at best get 9 boxes of type b into a 60x60x30 box, it will take me 3 more bix boxes .. So I need 4 big boxes in that case.
$endgroup$
– Bram28
Jun 23 '17 at 1:13
$begingroup$
Yeah it works. Say I am replacing the 200x200x100 box with a 60x60x30 box. So how many boxes of 60x60x30 do I need to pack them all? Thank you for the above answer!
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:30
$begingroup$
Yeah it works. Say I am replacing the 200x200x100 box with a 60x60x30 box. So how many boxes of 60x60x30 do I need to pack them all? Thank you for the above answer!
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:30
$begingroup$
Is there any common formula to solve such problems?
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:41
$begingroup$
Is there any common formula to solve such problems?
$endgroup$
– Ansary Ans21
Jun 23 '17 at 0:41
$begingroup$
@AnsaryAns21 Well, I'm no expert in packing problems ... I just try to create nice layers! :). for example, for your 60*60*30 box I would first make a layer of 9 type b boxes in a 3x3 square, so that creates a layer of 60x60x20. Then I'll put together some type c boxes: I can nicely fit their length 4 times into the 60, so I 'll create a 60*8*10 rectangle with 8 pf them. Also, 18 of the type a boxes in a 3x6 patterns takes up 60x30x10, so that leaves me with 60x22x10: plenty to pack the last 2 c boxes and the last two a boxes. So that's the first box. Now I have 21 b boxes left. .. (Continued)
$endgroup$
– Bram28
Jun 23 '17 at 1:12
$begingroup$
@AnsaryAns21 Well, I'm no expert in packing problems ... I just try to create nice layers! :). for example, for your 60*60*30 box I would first make a layer of 9 type b boxes in a 3x3 square, so that creates a layer of 60x60x20. Then I'll put together some type c boxes: I can nicely fit their length 4 times into the 60, so I 'll create a 60*8*10 rectangle with 8 pf them. Also, 18 of the type a boxes in a 3x6 patterns takes up 60x30x10, so that leaves me with 60x22x10: plenty to pack the last 2 c boxes and the last two a boxes. So that's the first box. Now I have 21 b boxes left. .. (Continued)
$endgroup$
– Bram28
Jun 23 '17 at 1:12
$begingroup$
Since I already figures out that I can at best get 9 boxes of type b into a 60x60x30 box, it will take me 3 more bix boxes .. So I need 4 big boxes in that case.
$endgroup$
– Bram28
Jun 23 '17 at 1:13
$begingroup$
Since I already figures out that I can at best get 9 boxes of type b into a 60x60x30 box, it will take me 3 more bix boxes .. So I need 4 big boxes in that case.
$endgroup$
– Bram28
Jun 23 '17 at 1:13
add a comment |
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$begingroup$
"I have boxes of dimensions 20 x a, 10 x c and 30 x b" : I don't get that part. Do you mean you have 20 boxes of type a, 30 of type b and 10 of type c ?
$endgroup$
– Evargalo
Jun 22 '17 at 16:14
1
$begingroup$
1 box will give you plenty enough room!
$endgroup$
– Bram28
Jun 22 '17 at 16:43
$begingroup$
@Evargalo yes correct.
$endgroup$
– Ansary Ans21
Jun 22 '17 at 23:58