HIERARCHAL INDUCTIVE PROCESS MODELING AND ANALYSIS ...

Using integrating factor e (a 15+a 18 )t and Proposition 1 we get,
D u (t) =
( (1 − a10 )a
) (
11
P 0 e αut (a11 + (1 − a 12 )a 13 )(1 − a 10 )
)
+
Z 0 e −δt + C u e −(a 15+a 18 )t
α u + a 15 + a
} {{ 18 −δ + a
}
15 + a
} {{ 18
}
β u1 β u2
Let’s rewrite to simplify the expression a bit,
D u (t) = β u1 P 0 e αut + β u2 Z 0 e −δt + C u e −(a 15+a 18 )t ,
where β u1 = 0.224039 and β u2 = −3.754762.
Assuming D(0) = D 0 and still following Proposition 1 we solve for C u ,
C u = D 0 − β u1 P 0 − β u2 Z 0
Similarly for the lower bound, using Proposition 1, (16) and (20) we get,
Let’s rewrite it as,
D l (t) =
( (1 − a10 )a
)
11
P 0 e −αlt + C l e −(a 15+a 18 )t
−α l + a 15 + a
} {{ 18
}
β l
D l (t) = β l P 0 e −α lt + C l e −(a 15+a 18 )t ,
where β l = 2.9784819 and C l = D 0 − β l P 0 .
Summarizing the bounds,
β l P 0 e −α lt + C l e −(a 15+a 18 )t ≤ D(t) ≤ β u1 P 0 e αut + β u2 Z 0 e −δt + C u e −(a 15+a 18 )t
(21)
This concludes our analysis of Model A; the results will be discussed further on.
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