Crap. I can't figure this out. (Math)

A clue may be in looking at items B & C.

B 747 items 41.5 x 38.5 x 23.5
C 830 items 41.4 x 38.5 x 22.o

747= 83 x 9 and 830= 83 x 10

So the common container can have 9 layers of B ( 211.5) or 10 layers of C ( 220) whichever is greater.

83 however is prime and a container that is long a skinney is not all that practical. So look for a number greater than 83 that isn't prime and provides a container that too prime---try 84 (2 x 2 x 3 x 7). From this munber there's 3 x 28, 4 x 21, 6 x 14, and 12 x 7) . 6 x 14 and 7 x 12 seem like they could provide a foot print that is more square and make a more practical container

6 x 14 (249 x 539) or (231 x 581) footprint
7 x 12 (290.5 x 462) or (269.5 x 498)

As someone who has seen packaging before the squarer the betterer. I'd go with a container as near as 290.5cm x 462cm x 220cm as possible.

Now lets see if this container can handle the numbers of items A and D

A 35.5 x 31.5 x 21.5
220/21.5= 10.23..= 10 layers of A
290.5/35.5=8.21 8 rows of A
463/31.5 = 14.69 14 columns of A

8 x 14 x 10= 1120 A need another 40 A so the container

Try the other stacking array

290.5/31.5=9.222 9 rows of A
463/35.5=13.04 13 columns of A

9 x 13 x 10 =1170 and 1170>1160 so the container will work for A

Now do the same thing for item D (47.5 x 45.5 x 27.5)

220/27.5 = 8.00 8 layers of D

290.5/47.5 = 6.11 6 rows of D

462/45.5= 10.62 10 columns of D

8 x 6 x 10 =480 bingo

OK now we have a container size 290.5cm x 462cm x 220cm and we have a constraint

10 layers of A, 9 of B, 10 of C and 8 of B if the container is to be filled completely---the problem isn't solved but it is simplified.

not count items in layers
A= 9x13=117 per layer---10 layers max
B=83 per layer---9 layers max
C=83 per layer---10 layers max
D=6x10=60 per layer---8 layers max

So far that's where I am

Rap

It is a badly defined real world problem, worse because I don't know anything about the physical shape of the container, just that it's a 20 footer.


This came up because, in the past, we've been dealing with large customers. They place an order for 50,000-75,000 pieces of various sized boxes, we (our shipping department) would then tell the customer how many containers (40ft, 40 ft highboy, 20 ft) their order was going to fill.

Now, I have begun working with a few really small customers (who might get bigger) but they want to know if there is a minimum order. The shipping department's response to that question was essentially "Wha?" and they sent me the data as above: size of box and how many of those fill a container. BUT that didn't answer the question the customer asked: (Let me try again)

"If I am going to buy some of every kind, how many of each kind do I have to buy in order to 1) get the most of each kind and 2) fill the container full? "

I am not a good enough mathematician (actually not one at all) to describe the situation any better than that.

I appreciate everyone's help.

Maybe when we get the formula, I'll send it in as a puzzler to Click and Clack the Tappet Brothers.

Joe(We'd have to make it more folkloric)Nation

Went back and looked at the question--the fewest of each type would be 1 layer of A, 1 layer of B, 1 layer of C and the rest of D

1/10+1/9+1/10=9/90+9/90+10/19=28/90=14/45
1-14/45=(45-14)/34=31/45
(31/45)/(1/8)=5.51 = 5 layers of D

too much void

so modify

make 5/8+j/10+k/9 as close to 1 as possible with j and k integers (j>=2), (k>=1).

try j=3, k=1
5/8+3/10+1/9=1.o36 pretty close
do try 5 D, 1 A, 1 B and 1C

5 x 27.5+1 x 21.5+1 x 23.5+2 x 22= 226.5 container height is 220 wont work
so go

5 x 27.5 + 2 x 21.5 + 1 x 23.5 + 2 x 22 =225.5 container height is 220 still won't work

so its 5 x 27.5 + 21.5+23.5+22=204.5

So the least number of packages for a full container as possible is

1 layer of A (117 A)
1 layer of B (83 B)
1 layer of C (83 C)
5 layers of D (5 x 60=300 D)

As I said, not a trivial engineering problem and I'm a Hoosier (not Japanese).

Rap

I'd be happy to solve your problem as a consultant--

Rap

Changed my mind. 'Course, if I was packing the container, I could probably mess up getting the proper amount of boxes jammed in there.

I think Raprap has it. I'll be going over the numbers tonight.

So, what has Raprap won, Karl Castle's voice on his phone?

Joe(hell no, it's just an honor to participate)Nation

The ideal container that I calculated would have internal dimensions 290.5cm wide x 462cm long x 220cm high--that's 9'6.5" wide x 15'2" long and 7'2.5" tall. Now that you've specified what looks like a 20 foot SeaLand the answer may have different results because of specified internal dimension's.

Interestingly it would make the solution less tedious.

Rap

raprap, You may not be Asian, but I'll certify you as an honorary one! You're a goddam genius!

Noo--I'm just easily amuzed and I don't multitask.

Rap

I multi-task and make a whole bunch of mistooks.

I select my answer because I think that method has the advantage of being flexible and readily understood, and is quite practical- you can easily select the proportions of different packages to suit whatever situation.

Works great for liquids. Doesn't necessarily work for packing boxes into containers.

I was thinking the same thing. Also, the question is really the answer. The different size boxes are already fitted into the right size containers; that's the most efficient use of space.

Using internal dimensions from here:
http://en.wikipedia.org/wiki/Intermodal_container
and simply dividing the length of the container into four unequal sections,
I can get:
A: 198
B: 180
C: 180
D: 200

This is based on simply orienting each box of the same type the same way. Improvements may be made by considering packings with differently oriented boxes. Improvements may also be made by slicing the container with two perpendicular planes instead of three parallel planes.

Wow. I am so impressed. Thank you, markr.

We've been busy with other things, but will run your numbers by the shipping folks in the next few days.

That looks right to me.

It is like one of those wooden block puzzles, for every extra D, you have to subtract one or more of the A,B.C s.

If this wasn't work, it would be fun.

Joe(it's still fun)Nation

The problem is easier now that the internal dimensions of the container is specified---in my solution that was another set of constraints to be determined.

Kudos

Rap

Not to mention the fact that the original definition had you packing as inefficiently as possible (fewest boxes) - whatever that means...

Okay have you filled that container yet, with various sized boxes?

I saw a fantastic TV programme once, about shipping containers, entitled "The Box That Changed The World."

I recommend it to you.

My total was 583
117 A's 1 layer
83 B's 1 layer
83 C's 1 layer
300 D's 5 layers

Rap

I read a book written by Simon Winchester called "The Map That Changed the World." Same author?