Computing a Model for Asset Management with Risk - Global Drinking Water Management and Conservation - page 172

first-order conditions for a min-
imum (the resulting equation is too large to display here). Since the maintenance
cost clearly exceeds the investment cost after t=10, the range over which the

Equation 8.7 is differentiated to determine the

rst

derivative can be set to zero must be between 0 and 10. Setting the first derivative

equal to zero and solving for t yields t=4.5099 (after rounding). Checking the
second order condition yields 39.2497 the function is convex at t=4.5099, and

therefore this value of t is the minimum. If the asset is renewed at t=4.5099, then
the service cost will be minimized.

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8.3.2 A Graphical Solution

An alternative approach to determining the optimal renewal period for assets is by
plotting the cost function over an appropriate interval. In the example above, this
interval would be t =0 10, since maintenance exceeds the renewal cost after t=10.

The function is shown in Fig. 8.1 :
It is obvious that the minimum occurs somewhere between t=4 and t=5.
Through an iterative process the range of the graph is decreased until a value for
t has been determined at an appropriate level of precision. Figure 8.2 demonstrates
this iterative process and shows that the result is the same as with the numerical
solution in Sect. 8.3.1 .
This process demonstrates that the minimum cost of service occurs approxi-
mately at t=4.51, i.e. the same as in Sect. 8.3.1 . The asset in this example should
be renewed at approximately every 4.5 years (assuming time is measured in years).

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Fig. 8.1 The service cost of
this asset is minimized if it is
renewed approximately every
4 5 time periods

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