Question

Two insulation thickness alternatives have been proposed for a process steam line subject to severe weather conditions. One alternative must be selected. Estimated savings in heat loss and installation cost are given below:

Q: Useful life of 2 cm Thickness alternative is 10 years and useful life of 5 cm thickness alternative is 5 years. Which thickness would you recommend for a MARR= 0.13 per year and negligible market (salvage) values using a)Present Worth Analysis? b) Annual Worth Analysis? (Use repeatability assumption)

Thickness	Installation cost	Annual Savings	Maintenance once in each 2 years
2 cm	$111000	$81000	$4500
5 cm	$367000	$125000	0

Answer 1

a.

Analysis period = 10 yrs (LCM of 5 & 10)

NPW of 2cm thickness = -111000 + 81000*(P/A,13%,10) -4500*((P/F,13%,2)+(P/F,13%,4)+(P/F,13%,6)+(P/F,13%,8))

= -111000 + 81000*5.426243 -4500*(0.783147+0.613319+0.480319+0.376160)

= 318387.43

NPW of 5cm thickness = -367000 + 125000*(P/A,13%,10) - 367000*(P/F,13%,5)

= -367000 + 125000*5.426243 - 367000*0.542760

= 112087.45

As NPW of 2cm thickness is more, it should be selected

b.

NPW of 2cm thickness = -111000 + 81000*(P/A,13%,10) -4500*((P/F,13%,2)+(P/F,13%,4)+(P/F,13%,6)+(P/F,13%,8))

= -111000 + 81000*5.426243 -4500*(0.783147+0.613319+0.480319+0.376160)

= 318387.43

AW of 2 cm thickness = 318387.43 * (A/P,13%,10)

= 318387.43 * 0.184290 = 58675.62

AW of 5cm thickness = -367000*(A/P,13%,5) + 125000

= -367000*0.284315 + 125000

= 20656.39

As AW of 2cm thickness is more, it should be selected