Weighted Unit Cost

[alejo12]

A contractor has the data tabulated relating the size(sf) and cost os seven slab projects.

Project Cost Area

1 $21021 3000

2 $24180 3750

3 $15915 2250

The problem says to apply the weighted unit cost to find the conceptual cost estimates for a slab of a given area(sf):

Do I need a dictionary or where can I find the method for solving such problem ?

Will Appreciate any help.

 

[IlPadrino]

Do I need a dictionary or where can I find the method for solving such problem ?

"Weighted unit cost" implies (to me at least) that they want you to find a weighted average unit cost from the seven projects. Rather than just averaging the seven unit costs [(U1 + U2 + ... + U7) / 7], they want you to weight with the areas: (U1xA1 + U2xA2 + ... + U7xA7) / (A1 + A2 + ... + A7)

That doesn't really make much sense to me from a real-world perspective because if I'm doing a small project I don't really care what the unit cost is of a project ten times the size... but the weighted unit cost would be influenced significantly by the larger project. Though, if I'm doing a larger project, I'd rather avoid too much influence from a very small project, so it does help in that case. To me, much better to bin the size of projects and calculate unit costs for small, medium, and large projects.

Regardless, you sure don't want to overthink a problem on the exam.

Can you post the entire problem?

 

[alejo12]

Sure ,

A contractor has the data tabulated below relating the size, and cost of seven concrete slab projects:

Project, Cost($), Area(sf)

1, 21021, 3000

2, 24180, 3750

3, 15,915, 2250

4, 23115 ,3025

5 ,20476 ,2800

6 ,22360 ,3020

7 ,17250 ,2350

Applying a weighted unit cost method , the conceptual cost estimate for a slab that is 2,750 sf, in dollars, is most nearly:

So, can we just interpolate to get the solution ????????????

 

[IlPadrino]

So, can we just interpolate to get the solution ????????????

If I was to solve applying a weighted unit cost method I get this:

Slab # | Cost ($) | Area (SF) | Unit Cost ($/SF) | Area * Unit Cost

1 $21,021.00 3000 7.01 21021

2 $24,180.00 3750 6.45 24180

3 $15,915.00 2250 7.07 15915

4 $23,115.00 3025 7.64 23115

5 $20,476.00 2800 7.31 20476

6 $22,360.00 3020 7.40 22360

7 $17,250.00 2350 7.34 17250

Sum 20195 144317

Weighted Unit Cost ($/SF) = 144317 ($) / 20195 (SF) = $7.15/SF

Estimate = 2750SF * $7.15/SF = $19652

Assuming, of course, I didn't make any math mistake (I didn't double check!)

 

[IlPadrino]

I'm not sure of the "correct method" for this problem. But, if I was actually estimating a job, and only had the data given, I'd say that a fair conceptual estimate , assuming that these costs are all current, would be about $7.32/foot. This is based on the cost of the 2800 sf slab (close direct comparison), and the trend that costs per foot are basically flat for slabs in this range (linear interpolation). So, without considering inflation, my best guess for the conceptual cost of the 2750 sf slab would be $7.32*2750, or $20130.
I'm not sure that will get you the "correct" answer, but it should get you pretty close.

I don't follow... are you saying you'd pick just one data point (slab 5) which has a unit cost of $7.31/SF and use that unit cost for 2750 SF?

In the real world, this is an easy estimating problem using a standard estimating manual (RS Means for example). But the exam isn't real world and you really should follow the directions as given.

 

[IlPadrino]

You're absolutely right about following the directions for an exam problem. And in that regard your answer is probably correct. As I said before, I wasn't sure of the "correct" method. It actually doesn't make any sense to me to use a weighted average in this situation.

Agreed, it doesn't make sense!

And, yes. In the real world, if I had it, I would use any empirical data given to determine the cost, not an estimating manual or some type of mathematical average, weighted or otherwise. The data given in this instance includes one point that is very close to the point in question, and a linear trend that is nearly flat on either side of the point in question. Using this information it is simple to determine a realistic conceptual cost estimate for a slab of this size.

What's wrong with an estimating manual? The data points aren't even close to linear but they are all within 10% of the average value so your point about "realistic" is well taken.

By the way, what weights are you using? Just the sizes of the slabs?

Yeah... there's nothing else given but the area to weight against.

 

[Samuel]

Agreed, it doesn't make sense!

Yeah... there's nothing else given but the area to weight against.

This logic makes no engineering sense to me. If the question doesn't make sense, how did you decide to use the areas as the weighted parameter?

What's wrong with an estimating manual?

I didn't say there was anything wrong with estimating manuals. But as a rule, empirical data is simply more accurate. If you want to know what a can of beans costs, you can use historical ads from Sunday papers across the country, price indexes, local adjustments, and farming reports if you want. I'll just drive down to my local supermarket and take a look at the canned bean shelf if I can.

And, again, I don't know why the question is set up this way. But if there's a method given on the exam, you should use that method to solve the problem.

Please find the solution to the problem:

UC = (A + 4B + C) / 6 Where UC = Forecast Unit Cost,

A = Mininmum Unit cost of previous projects

B = Average unit cost of previous projects

C = Maximum unit cost of previous projects

Total Unit Cost = (7.01 + 6.45 + 7.07 + 7.64 + 7.31 + 7.40 + 7.34) = $50.22

Average Unit Cost =50.22/7 = $7.17

UC = [7.64 + 4(7.17) + 6.45] / 6 = $7.13