Sea Ice

As mentioned earlier it takes n/2 · (n − 1) evaluations per projection bundle to
evaluate all the pairs of future projections, where n is the number of key factors.
Now how many projection bundles are there for a number of n key factors?
Say each of the n key factors has m future projections. In this case there are
m^n possible projection bundles.
Example 5
Assume we have the key factors A, B and C with their respective future projections
A1 and A2, B1 and B2, C1 and C2, i.e. three key factors with two future
projections each. In this case we would expect that there are 23 = 8 projection
bundles, namely
• A1B1C1
• A1B1C2
• A1B2C1
• A1B2C2
• A2B1C1
• A2B1C2
• A2B2C1
• A2B2C2
Now, all these projection bundles need to be checked for their overall consistency
value, by performing n/2 · (n − 1) calculations per projection bundle. To
obtain the smallest number of calculations required to find all projection bundles
and their respective consistency values we multiply m^n · (n/2) · (n − 1), this is a
very fast growing number.

Example 6
For a project with 10 key factors and 2 future projections each the required overall
number of calculations amounts to 2^10 · 5 · 9 = 46, 080.
This is already a number which requires computational support.
For a project with 25 key factors and 3 future projections each this number
amounts to about 254 trillion calculations. This illustrates how fast the number
of calculations grows.
Even with an above average desktop computer a project with about 30 key
factors is not solvable (in a decent amount of time) with the classical algorithms
used for the consistency analysis.