Menger Sponge Cost Analysis: A Great Sponge or The Greatest Sponge?

On Monday, October 20th, we looked at the Menger Sponge project currently going on. Sure it seemed interesting and figuring out the different properties of the fractal was fun, but I had a different question. How much does this thing cost, both in resources and man-hours required to build this thing?

I ran a computer simulation to figure it out. I determined that we need to include the costs of
treatment for Carpal Tunnel Syndrome. We also need to budget for a Creeper Defense System.

When I started to think about it, I realized that the cost of the fractal grows exponentially. As the cube gets larger, the number of cards and the amount of time required to assemble it gets MUCH larger. This reminded me of my favorite episode of Mythbusters where they built a balloon out of lead and actually got it to fly.

This made me wonder: how large would we have to build this cube so that it were worth it's weight in gold?

I pity the fool who tries to calculate this!

OK, so first things first. I need to define a few functions I am going to use. Let E(n) be the expenses to build a Menger Sponge and W(n) be the weight of the Menger Sponge. W(n) looks easier, so I will start there.

To calculate the total weight of a lot of small objects, you can multiply the weight of one of the objects by the total number of objects. Now by looking at various shipping companies, they ship business cards in boxes of 100, and on average 5 of those boxes is about 2 lbs. So (ignoring the weight of the 5 boxes), the weight of 1 business card is 1/250 lbs. We also know that there are 6 business cards per cube, so the total number of business cards is 6 times the number of cubes.

I can't help but think we missed out on a better cube opportunity here...

One way to think about going from the n sponge to the n+1 sponge is to recognize that you are making a giant sponge out of sponges one size smaller. By looking at the geometry of the structure (8 corner boxes and 12 side boxes), each new larger box requires 20 of the smaller boxes. So 20ⁿ describes the number of boxes required to build Sponge n. This means that there are 6*20ⁿ cards required to build Sponge n. Now we know that:

W(n) = (Weight of 1 Card) * (Number of Cards to Build Sponge n)

W(n) = 1lb/250cards * 6*20ⁿ cards

W(n) = 6/250*20ⁿ (lbs)

Now let's look at the expenses.

The expenses incorporates both the cost of materials (let's call this M(n)) and production (P(n)). Well the cost of materials should be easy since we already know the number of cards required to build Sponge n. After looking around the internet, it seems that 100 plain white cards approximately the size of a business card costs $10. This means that each card costs $0.10.

M(n) = (Cost of 1 Card) * (Number of Cards to Build Sponge n)

M(n) = $0.10/card * 6*20ⁿ cards

M(n) = 6/10*20ⁿ dollars

Minimum wage varies in different areas, but for the sake of my own sanity, I will say it is $10/hour.

For the number of hours, I am going to make some estimates off of anecdotal information. In class on Monday, somebody mentioned that 3 people were able to make 60 cubes (enough cubes for 3 Sponges n=1) in 15 minutes. This means one person can make 20 cubes in 15 minutes, or 80 cubes/hour.

But there is more to building this Sponge than just making small cubes!

We need to attach them to each other. In my own practice, I was able to attach around 3 cubes per minute. But I feel pretty confident that with more practice (and whoever is making this giant Sponge is going to have more practice than...well...I'm not sure what analogy I could use to convey just how much practice this will be. Let's just say it's a lot.) I could have comfortably got 6 attachments per minute. That comes out to 360 attachments/hour.

The number of attachments is trickier. I am going to assume any 2 cubes that are touching are connected. So another way to think about how many connections there are in the sponge is to think about how many cards are facing other cards. Since these cards can only either face other cards or face the outside, the number of cards facing other cards plus the number of cards facing the outside of the sponge should equal the total number of cards. So the total number of cards facing each other equals the total number of cards minus the surface area of the sponge. Since I know the total number of cards, I just need to figure out the surface area!

The sponges of size n can be broken into 2 types of regions: corner pieces size n-1 and edge pieces size n-1. Corner pieces have the same surface as the surface area of a sponge size n-1 minus the surface area of 3 faces of the sponge (since the corner piece is connected on 3 of the 6 sides). In class we talked about how the area of a surface that follows this pattern is 8ⁿ. So this means that the Surface are of the corner pieces of the sponge of size n is SA(Corner)_n = SA(Sponge)_n-1 -3*8^n-1. Similarly, the surface area of the edge piece is equal to the surface area of the sponge n-1 minus the surface area of 2 faces (since the edge pieces are all connected on 2 sides). So SA(Edge)_n = SA(Sponge)_n-1 -2*8^n-1. Since every sponge size n is made of 8 corner pieces and 12 edge pieces, the equation for surface area is

Surface Area = 8*SA(Corner)_n + 12*SA(Edge)_n

SA_n = 8(SA_n-1 -3*8^n-1) + 12(SA_n-1 -2*8^n-1)

SA_n = 20SA_n-1 - 48*8^n-1

SA_n = 20SA_n-1 - 6*8ⁿ

Party on, Weierstrass. Part on, Gauss.

Now that we have a recurrence relation, we can find an equation for the surface area of a Menger Sponge of size n, or SA(n). Since this recurrence relation is a non-homogeneous relation, we need to find both the homogeneous solution and the particular solution (If you have not done any work with discrete mathematics, check out this slide show. What I am doing starts on slide 24. WARNING: Math Ahead. I'm realizing this is a little late for the warning...)

a_n = h_n + p_n

Let p_n be in the form of c*8ⁿ + d

p_n = 20p_n-1 - 6*8ⁿ

c*8ⁿ + d = 20(c*8^n-1 + d) - 6*8ⁿ

0 = 20c*8^n-1 - c*8ⁿ - 6*8ⁿ + 20d - d

0 = 8^n-1(20c - 8c - 6*6) + 19d

0 = 8^n-1(12c-48) + 19d

c = 4 and d = 0.

Therefore p_n = 4*8ⁿ

h_n = 20h_n-1

0 = x - 20

The characteristic equation is (x-20).

Therefore h_n = α20ⁿ

a_n = α20ⁿ + 4*8ⁿ

Since SA(0) = 6, we know that 6 = α20⁰ + 4*8⁰

6 = α + 4

α = 2

Therefore SA(n) = 2*20ⁿ + 4*8ⁿ

If you've never dealt with recurrence relations, this was probably your reaction...

Now that we know the number of cards and the surface area, we have a formula for the number of cards facing other cards. And since there is 1 attachment for every 2 cards facing each other, the number of attachment is 1/2 of the amount of cards minus the surface area.

# of Attachments = (Number of Cards - Surface Area)/2

Attachments = (6*20ⁿ - 2*20ⁿ - 4*8ⁿ)/2

Attachments = 2*20ⁿ - 2*8ⁿ

Finally, for a Menger Sponge of size n, we know (1) how many cubes we have to make, (2) how fast we can build the cubes, (3) how many attachments we need to make, and (4) how fast we can attach cubes. Therefore the equation for the number of hours required to build a Menger Sponge of size n is

Hours = (# of Cubes)/(Cubes/Hours) + (# of Attachments)/(Attachments/Hour)

Hours = (20ⁿ)/80 + (2*20ⁿ - 2*8ⁿ)/360

And since the hourly wage is $10.00, we know the wages are

Wages = (20ⁿ)/8 + (2*20ⁿ - 2*8ⁿ)/36

Wages(n) = (13*20ⁿ - 4*8ⁿ)/72

And since all the expenses come from materials and production, we know that

E(n) = (Cost of Cards) + (Wages to Build)

E(n) = (6*20ⁿ)/10 + (13*20ⁿ - 4*8ⁿ)/72

E(n) = (281*20ⁿ - 20*8ⁿ)/360 (in dollars)

Finally! We know the expenses, we know the weight, and knowing that gold costs about $20,000 per pound, we can compare our results. Below is a graph of E(n) (in green) and 20,000*W(n) (in, of course, gold). (For the following graphs, the x-axis represents the xth Menger Sponge, the y-axis represents dollars, and all of the graphs can be seen here).

Well, this cube definitely won't be worth it's weight in gold for small sizes, but maybe if we kept going for a while...

Nope! It looks like the expenses will never catch up to the amount of gold. In fact, looking at 20,000*W(n) - E(n) shows the graph (in red) below.

This gap is just going to keep going up and up. This cube will never cost it's weight in gold. But it could be worth it's weight in something else...

Let's look at how we graphed the equations. We graphed E(n), and we graphed the price of gold/lb multiplied by W(n). We want to know the material that costs x dollars/pounds such that
E(n) = x*W(n). So let's look at E(n)/W(n) (graphed below in blue).

This graph seems to level off around 30-35 dollars/lb.

I took a look at the price of various materials. Precious metals and minerals were all much higher than this. Common metals (like lead, aluminum, tin, copper, etc.) were all lower. Seafood and meat were all lower. But there was one material that cost around $35/lb.

Uranium.

So if you know anyone who wants to make a doomsday device powered by uranium, let them know that the money they spend on uranium could be spent on a pretty awesome arts and crafts project...

Good news everyone! I invented a machine that makes you read this caption in your head in my voice!