Question

An aircraft control actuator is part of the flight control system that enables an airplane to fly. It is important that the actuators are reliable. Reliability can be defined as the probability a system is functioning, so it is a number between 0 and 1. One way to increase reliability of the actuator is to increase the torque. The reliability R as a function of the maximum torque T (in Newton-meters) is given by R(T)=0.91*exp(0.027T) where torque T is between 0.1 and 3.5. The development and installation cost as a function of reliability equals [$1980+ $14030*(In R)], which will be paid immediately. Maintenance on the actuator will occur in year 3 and in year 7. In each year the maintenance cost as a function of reliability R is [$1160/ln(1.45R)]. Assume the interest rate is 12%. What is the optimal torque design (a number between 0.1 and 3.5 rounded to the nearest tenth) of the actuator that minimizes the discounted lifecycle costs of the actuator? (You do not need to calculate annual equivalent cost, but you do need to calculate the present value of the costs.)

Answer 1

NOTE: One needs to generate an algebraic expression for reliability followed by inputing that expression into the intsallation & development cost and maintenance costs occuring in year 3 and year 7. Summing the present value of the two types of costs will give the total cost function. The same needs to be optimized using an EXCEL Function or differentiation.

Reliability: R(T) = 0.91 x exp(0.027T) (where T is between 0.1 and 0.35)

Development and Installation Cost Function = 1980 + 14030 x (ln R) = 1980 + 14030 x (ln(0.91 x e^0.027T))

= 1980 + 14030 x [ln (0.91) + 0.027T] = 656.82 + 378.81T

Maintenance Cost = (1160 / ln 1.45R) (in year 3 and year 7)

Discount Rate =12%

PV of Maintenance Cost of year 3 = (1160/ln1.45R) x (1/(1.12)^(3)) = [1160 x 0.712] / [ln 1.45 + ln(0.91x0.027T]

= [825.66 / (0.277 + 0.027T]

PV of Maintenance Cost of year 7 = (1160/ln1.45R) x (1/(1.12)^(7)) = [524.72 / (ln 1.45 + (ln(0.91 x 0.027T)))

= [524.72 / (0.277 + 0.027T)]

Total Cost Function =

PV of Maintenance Cost of year 3 + PV of Maintenance Cost of year 7 + Development and Installation Cost Function

=[825.66 / (0.277 + 0.027T] + [524.72 / (0.277 + 0.027T)] + 656.82 + 378.81T

= [1350.38 / (0.277 + 0.027T)] + (656.82 + 378.81T)

= [(1532.32 + 122.66T + 10.23T^(2))/(0.277+0.027T)]

Solving the above equation using the EXCEL function SOLVER we get Total Minimal Cost = $ 5503.593 at T=0.35