Thank you for shedding some light and for posting the Guidelines Summary - I hadn't read that post yet and it was informative.
I'm a little confused by your formula but I'm going to apply it to my example above. Man-Day is just a productivity unit like miles per hour except Man-Day in my field / this example is square feet of material installed per eight hours. Each layer has its own difficulty-rating so one layer might be installed at 800 square-feet in 8 hours and another really difficult layer might be installed at 400 square feet in 8 hours. My productivity formula above was an attempt to capture this by saying:
Layer #1 - 400 SF / 8 Hours (50 SF / Hour)
Layer #2 - 500 SF / 8 Hours (62.5 SF / Hour)
Layer #3 - 700 SF / 8 Hours (87.5 SF / Hour)
Layer #4 - 700 SF / 8 Hours (87.5 SF / Hour)
Layer #5 - 800 SF / 8 Hours (100 SF / Hour)
I attempted to reverse the formula instead of solving for square-feet per 8 hours I tried to look at the 8 hour time-block as a unit of measurement that could be divided evenly among the layers so: [MATH]8 \div 5 = 1.6[/MATH] for simplicity sake I'm going to say the layers get installed in equal parts of the 8 hour period. If I take the square-feet per hour of installation productivity and multiply each individually by 1.6 it looks like this:
Layer #1 - [MATH](50 \times 1.6) = 80[/MATH] square-feet of layer #1 in 8 hours
Layer #2 - [MATH](62.5 \times 1.6) = 100[/MATH] square-feet of layer #2 in 8 hours
Layer #3 - [MATH](87.5 \times 1.6) = 140[/MATH] square-feet of layer #3 in 8 hours
Layer #4 - [MATH](87.5 \times 1.6) = 140[/MATH] square-feet of layer #4 in 8 hours
Layer #5 - [MATH](100 \times 1.6) = 160[/MATH] square-feet of layer #5 in 8 hours
Consequently [MATH]80+100+140+140+160 = 620[/MATH] square-feet of all five layers in equal parts installed per 8 hours.
Does that make mathematical sense? I think I ended up just confusing myself even more. It doesn't make sense in my mind when I think of the practical application. A man standing on a roof is going to physically install 620 square-feet of all 5 layers of roof in 8 hours. I guess in my mind that seems much higher than I think it should be. In my mind it seems more plausible that a person could install closer to 150 square-feet of the 5-layer assembly in 8 hours than 600 square-feet. Does that make sense?
Your formula is a little confusing for me:
man days required [MATH]m=(n \times s)/600[/MATH] because it's solving for man days required. But if I spin your formula and solve for square feet - using the square-feet per 8 hours from my process above - and make man days = 1 then it looks like this:
[MATH]1 = \dfrac{5 \times s}{620} = 620 =5 \times s = \dfrac{620}{5} = 124[/MATH]
I'm taking the 620 from my formula above which hopefully shows how many square-feet I can install of all 5-layers in one 8-hour time-block. And using your formula I think we just solved for m=1 - so in my assembly above (1) man can install (124) SF of all 5-layers of roofing material in (1) 8-hour time block. Does that make sense?
Great.
I misunderstood your first post. My incorrect
assumption that there was no difference in productivity for different layers did not reflect all of your first post, but was based on the words "all (3) of them are installed at the same productivity (600) SF / MD." I did ask if that was correct. My formula works
ONLY if that assumption is correct.
There are probably different ways to achieve what you want, but I am going to use what I think is simplest on the assumption that the area to be covered is the
same for each layer.
[MATH]\text {Let:}[/MATH]
[MATH]m = \text {man days required;}[/MATH]
[MATH]s = \text {area to be covered in square feet, and}[/MATH]
[MATH]L = \text {number of levels required.}[/MATH]
[MATH]\text {If } L = 1 \text {, then } m = \dfrac{s}{400}.[/MATH]
[MATH]\text {If } L = 2 \text {, then } m = \dfrac{s}{400} + \dfrac{s}{500} = \dfrac{9s}{2000}.[/MATH]
[MATH]\text {If } L = 3 \text {, then } m = \dfrac{9s}{2000} + \dfrac{s}{700} = \dfrac{83s}{14000}.[/MATH]
[MATH]\text {If } L = 4 \text {, then } m = \dfrac{83s}{14000} + \dfrac{s}{700} = \dfrac{103s}{14000}.[/MATH]
[MATH]\text {If } L = 5 \text {, then } m = \dfrac{103s}{14000} + \dfrac{s}{800} = \dfrac{120.5s}{14000}.[/MATH]
Let's see why that may work.
Suppose we want one layer for 2800 square feet. This will obviously take 7 man days because only 400 square feet (or 1/7th of the total area) can be done in one many day.
Our formula says 2800/400 = 7, which is correct.
Suppose we want two layers for 2800 square feet. The first layer will take 7 man days as we saw above. But for the second layer, we can do 500 square feet a day, which means that we can do the second layer in 5.6 man days. So the total is 12.6 man days.
Our formula says 9 * 2800/2000 = 12.6, which is correct.
Suppose we want three layers for 2800 square feet. The first two layers will take 12.6 man days as we saw above. But for the third layer, we can do 700 square feet a day, which means that we can do the third layer in 4 man days. So the total is 16.6 man days.
Our formula says 83 * 2800/14000 = 16.6, which is correct.
Suppose we want four layers for 2800 square feet. The first three layers will take 16.6 man days as we saw above. But for the fourth layer, we can do 700 square feet a day, which means that we can do the fourth layer in 4 man days. So the total is 20.6 man days.
Our formula says 103 * 2800/14000 = 20.6, which is correct.
Suppose we want five layers for 2800 square feet. The first four layers layers will take 20.6 man days as we saw above. But for the fifth layer, we can do 800 square feet a day, which means that we can do the fifth layer in 3.5 man days. So the total is 24.1 man days.
Our formula says 120.5 * 2800/14000 = 24.1, which is correct.
Does this work?
If productivity changes in the future, you can see how I came up with the factors. It is a problem in adding fractions and so a problem in getting common denominators. They teach you the mechanics in fourth grade, but they do not teach you how to figure out when you need it.
