Computing a Model for Asset Management with Risk - Global Drinking Water Management and Conservation - page 171

d ¼ 0 : 1 ð 8 : 3 Þ

x

p 2

1 þ sin

f ðÞ¼ ð 8 : 4 Þ

2

I ¼ 100 ð 8 : 5 Þ

Maintenance and failure costs are assumed to increase at an increasing rate.
Therefore, functions representing these costs must have positive rst and second

derivatives. The discount rate is 10 percent and the investment cost is 100. The
probability of failure is monotonically increasing,

first at an increasing rate, then at a
decreasing rate. This is due to the fact that the probability of failure asymptotically
approaches 100 percent as the asset ages. Therefore, the first derivative of this

function must be positive and the second derivative must be positive over some
initial range; after that the derivative changes to negative.
The generalized time-dependent service cost was derived in Chap. 7 (Eq. 7.8 ).
Utilizing the parameters outlined above (Eqs. 7.1

7.5 ), Eq. 7.8 becomes:

-

R

x 2 e 0 : 1x dx þ R

t 1 t 1

p ðÞ
2

x

x 3 e 0 : 1x 1 þ sin dx þ 100e 0 : 1t

1 1

C ðÞ¼

ð 8 : 6 Þ

1 e 0 : 1 t þ 1

ð Þ

The objective is to find the time t that minimizes Eq. 8.6 . This will determine the

renewal interval that minimizes the service cost of the asset over an in nite horizon,

discounted to present values. This can be done numerically or graphically.

8.3.1 A Numerical Solution

A numerical solution can be derived with the help of a mathematical software
package. This is standard continuous optimization. The first derivative, with respect

to t, must be set to zero to find the minima and maxima. Then t is isolated to
determine the values of t for which the function is minimized or maximized. The
second derivative must be checked to determine which of these solutions are
minima and which are maxima. If the second derivative is positive, then the cor-
responding value of t is a minimum. If it is negative, then the point is a maximum.
In this case, Eq. 8.6 can be expanded to eliminate the integrals yielding:

C ðÞ¼ 31915 : 78910 e0 : 1 0 : 1t 28955 2895t 145t2 5t3 þ cos ð 0 : 3184713376t ð

1

ð

: 888471338 Þ 171 ð : 0764517 þ 60 : 43230478t 3 : 408162357t2 1 : 429096774t3 Þ

þ sin ð

0 : 4487363871t3 ÞÞ þ 100e 0 : 1t Þ=ð 1 e 0 : 1t 0 : 1 Þ

ð 0 : 3184713376t 1 : 888471338 Þ 136 : 0394311 þ 40 : 37900330t þ 12 : 39192863t2

ð 8 : 7 Þ