$Trash Algebraic Formula$

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$Trash Algebraic Formula$

Hello, I am trying to figure this out to include in a term paper for a class I'm taking on sustainability. In the trash industry a compactor truck picks up multiple containers of varying cubic yard capacity. Although at the end of the day the net truck weight is calculated at a landfill or other site the individual container weights are not known. I need a calculation that would look something like this but forgive my syntax, etc.

Net truck weight (20 short tons or 40,000 lbs) = (Qty 20, 6 Cubic Yard containers)x + (Qty 20, 8 Cubic Yard containers)y + (Qty 10, 3 Cubic Yard containers)z.

The idea is to allocate and determine a hypothetical weight to each group of containers and therefore to each individual container. So I guess we would solve for x, y, and z but thats where I'm stuck. Its a sustainability paper anyway and not a math paper. Thank you.

In case you're going to say something like I need to find out the general basis weight of the material and then multiply by that to back into the truck weight the whole purpose of the exercise is to develop a hypothetical basis weigh if you will by using the equation. OK hope that makes sense.

I was thinking of putting this into a spreadseet and just selecting weights until the individual weights added up close to the total weight of the truck. Kinda like backing into it. I think it definately would work don't believe its the right way to do it.

Algebra

Vince9272

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OK, so as you know with only one total and three unknowns, the system is underdetermined. As you also mentioned, you need an assumption to allocate the weights and the cleanest/defensible assumption is “same lbs. per cubic yard across all containers”. That gives you a unique and simple solution.

Let W = lbs. per cubic yard

Total cubic yards = $20\times 6 + 20\times 8 + 10\times 3 = 120 + 160 + 30 = 310 yd^{3} .$

Given total net truck weight T = 40,000 lbs., we have:

$W = \frac{T}{310} = \frac{40,000}{310} = 129.03 \frac{lbs}{yd^{3}}$

So, you can now easily calculate per-container weights:

6-yd container: $6 \times W = 774.19 lbs. $

8-yd container: $8 \times W = 1,032.25 lbs. $

3-yd container: $3 \times W = 387.097 lbs. $

You can then calculate the group weights, similarly:

20×6-yd = 15,483.87 lbs.

20×8-yd = 20,645.16 lbs.

10×3-yd = 3,870.97 lbs.

Checks: sums to 40,000 lbs.

For your paper, you can write as below:

Let sizes be $s = [s_{1} , s_{2}, s_{3}] yd^{3} $ and counts $n = [n_{1} , n_{2}, n_{3}]$.

Total volume is:

$V=\sum_{1}^{3} n_{i} s_{i} $

With equal density W, per-container weight for size $s_{1}$ is $s_{1}\times W$, where $W = \frac{T}{V}$

This is the “maximum-entropy” or “least-assumption” allocation and is easy to justify, you can say that without evidence that density differs by container size, treat all volume equally.

Now, I can do this using a different approach and acknowledge uncertainty, by putting some plausible bounds on W and solve a linear program that picks x, y, z within ranges and exactly hits T. But, that seems to be outside the scope of your paper and needs more time to do. I can also do another approach by scaling those proportionaly from a baseline, if you have some rough baseline weights $x_{0} , y_{0}, z_{0}$.

Anyways, there you go.

Kav10

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$Trash Algebraic Formula$

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