You could either do a linear reduction in hours or use some type of distribution for a more gradual change in hours. Below, I provided an example with both linear and a normal distribution...
# Hours | # Months | Linear Hours | Norm Dist | Norm Hours |
2160 | 18 | 227.3684211 | 0.997355701 | 230.067402 |
| | 214.7368421 | 0.987782386 | 227.859055 |
| | 202.1052632 | 0.959610268 | 221.360384 |
| | 189.4736842 | 0.914430916 | 210.93853 |
| | 176.8421053 | 0.854730735 | 197.167048 |
| | 164.2105263 | 0.783664452 | 180.773664 |
| | 151.5789474 | 0.704779689 | 162.576734 |
| | 138.9473684 | 0.621725982 | 143.418122 |
| | 126.3157895 | 0.537981153 | 124.100084 |
| | 113.6842105 | 0.456622713 | 105.332532 |
| | 101.0526316 | 0.380163455 | 87.6951106 |
| | 88.42105263 | 0.310459981 | 71.6160958 |
| | 75.78947368 | 0.248692847 | 57.3678149 |
| | 63.15789474 | 0.195408462 | 45.0763125 |
| | 50.52631579 | 0.150607239 | 34.7416835 |
| | 37.89473684 | 0.113859882 | 26.2648996 |
| | 25.26315789 | 0.084434128 | 19.4770435 |
| | 12.63157895 | 0.061416875 | 14.167484 |
| | | | |
| Total | 2160 | | 2160 |
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The table starts in cell A1. In col C is the formula, =$A$2*($B$2+2-ROW())/$B$2/SUMPRODUCT(ROW(INDIRECT("1:"&$B$2))/$B$2)
This gives a linear decrease in hours.
In col D is a normal distribution with the formula, =NORMDIST(($B$2+2-ROW())/$B$2,1,0.4,FALSE)
0.4 is the standard deviation I used... this affects the slope of your curve, lower value gives steeper drop-off.
This is just the right hand side of a normal distribution curve.
In col E the normal distribution is applied to the hours with the formula, =$A$2*$D2/SUM($D$2:$D$19)
The bottom row is just a sum of the rows above to verify that the distribution does, indeed, add up to the alloted number of hours.
HTH,
~ Jim