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 [$1760+ $11007*(In R)], which will be paid immediately. Maintenance on the actuator will occur in year 6 and in year 12. In each year the maintenance cost as a function of reliability R is [$1350/ln(1.45R)]. Assume the interest rate is 10%.
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

Development and installation cost = $1760 + $11007*(ln R)

Maintenance occurs in year 6 and year 12

Maintenance cost = $1350/ln(1.45R)

Interest rate = 10%

R(T) = 0.91*exp(0.027T)

Taking log on both sides

ln(R) = ln(0.91* exp(0.027T)) = ln(0.91) + 0.027T

ln(1.45R) = ln(1.45) + ln(R) = ln(1.45) + ln(0.91) + 0.027T = ln(1.45*0.91) + 0.027T

                 = ln(1.3195) + 0.027T

The discounted lifecycle costs of the actuator:

LC = Development & Installation cost + PV of maintenance Cost

     = $1760 + $11007*(ln R) + [$1350/ln(1.45R)]*(1+10%)^(-6) + [$1350/ln(1.45R)]*(1+10%)^(-12)

     = $1760 + $11007*(ln R) + [$1350/ln(1.45R)]*[1.1^(-6) + 1.1^(-12)]

Substituting for ln(R) and ln(1.45R)

LC = $1760 + $11007*[ ln(0.91) + 0.027T] + [$1350/(ln(1.3195) + 0.027T)]*[1.1^(-6) + 1.1^(-12)]

To find the optimal torque design we will have to take the first derivative w.r.t T and set it equal to zero

dLC/dT = 11007*0.027 – 0.027*1350/[ln(1.3195) + 0.027T]² * [1.1^(-6) + 1.1^(-12)]

              = 297.189 – 32.1892/[0.277253 + 0.027T]²

Equating to zero,

297.189 = 32.1892/[0.277253 + 0.027T]²

[0.277253 + 0.027T]² = 0.1083

0.277253 + 0.027T = 0.3291

T = (0.3291 – 0.277253)/0.027 = 1.92

So, T = 1.92 Newton-meters