Question

44 You are evaluating whether to rce eectric motors at a treatment plant. Using an inflation rate of 3.5% and a salva wastewater 15% of the initial costs, determine which of these pumps is the most nomical from a life cycle perspective: a wastewaler Option A Option B $000 2250 Annual energy cost 11,000 $6700 S500 Input Option C Option D $21,500 0 $5500 S Cost Sssoo $1000 $11,000 Annual maintenance $500 $500 Annual inspection/ 2500 2500 $2500 $2500 certification cost Life 12

Answer 1

We need to calculate the life cycle cost of each option using NPV method.

Option A

Initial Cost is $5000

Total Annual Cost = Energy Cost + Maintenance cost + Inspection cost = 11000 + 500 + 2500 = $14000

Salvage Value = 15% of initial cost = 15% * 5000 = $750

Present Value of Annual Costs will be PV = P * (1-(1+r)^-n)/r where P = $14000, r= 3.5%, n = 8

Hence PV = 14000 *(1- (1+3.5%)^-8 ) / 3.5%

= 14000 *(1-1.035)^-8)/0.035

= 14000 * (1- 0.7594)/0.035

= 14000 * 0.2406/ 0.035

= $96235.38

Hence Present Value of Annual Costs = $96235.38

Present Value of Salvage Value = Savage Value / (1+r)^n

= 750/ (1+3.5%)^8

=750/ (1.035)^8

= 750/1.3168

Present Value of Salvage Value=$569.56

Hence total lifecycle cost of option A is Initial cost + Annual cost - Salvage value= 5000 + 96235.38- 569.56 = $100665.82

Option B

Initial Cost is $2250

Total Annual Cost = Energy Cost + Maintenance cost + Inspection cost = 6700 + 500 + 2500 = $9700

Salvage Value = 15% of initial cost = 15% * 2250 = $337.5

Present Value of Annual Costs will be PV = P * (1-(1+r)^-n)/r where P = $9700, r= 3.5%, n = 6

Hence PV = 9700 *(1- (1+3.5%)^-6 ) / 3.5%

= 9700 *(1-1.035)^-6)/0.035

= 9700 * (1- 0.8135)/0.035

= 9700 * 0.1865/ 0.035

= $51686.964

Hence Present Value of Annual Costs = $51686.964

Present Value of Salvage Value = Savage Value / (1+r)^n

= 337.5/ (1+3.5%)^6

=337.5/ (1.035)^6

= 337.5/1.2293

Present Value of Salvage Value=$274.55

Hence total lifecycle cost of option B is Initial cost + Annual cost - Salvage value= 2250 + 51686.964- 274.55 = $53662.414

Option C

Initial Cost is $21500

Total Annual Cost = Energy Cost + Maintenance cost + Inspection cost = 5500 + 1000 + 2500 = $9000

Salvage Value = 15% of initial cost = 15% * 21500 = $3225

Present Value of Annual Costs will be PV = P * (1-(1+r)^-n)/r where P = $9000, r= 3.5%, n = 12

Hence PV = 9000 *(1- (1+3.5%)^-12 ) / 3.5%

= 9000 *(1-1.035)^-12)/0.035

= 9000 * (1- 0.6618)/0.035

= 9000 * 0.3382/ 0.035

= $86965.71

Hence Present Value of Annual Costs = $86965.71

Present Value of Salvage Value = Savage Value / (1+r)^n

= 3225/ (1+3.5%)^12

=3225/ (1.035)^12

= 3225/1.5111

Present Value of Salvage Value=$2134.25

Hence total lifecycle cost of option C is Initial cost + Annual cost - Salvage value= 21500 + 86965.71- 2134.25 = $106331.46

OptionD

Initial Cost is 0

Total Annual Cost = Energy Cost + Maintenance cost + Inspection cost = 11000 + 500 + 2500 = $14000

Salvage Value = 15% of initial cost = 15% * 0= 0

Present Value of Annual Costs will be PV = P * (1-(1+r)^-n)/r where P = $14000, r= 3.5%, n = 5

Hence PV = 14000 *(1- (1+3.5%)^-5 ) / 3.5%

= 14000 *(1-1.035)^-5)/0.035

= 14000 * (1- 0.842)/0.035

= 14000 * 0.158/ 0.035

= $63210.73

Hence Present Value of Annual Costs = $63210.73

Hence total lifecycle cost of option D is Initial cost + Annual cost - Salvage value= 0+63210.73-0 = $63210.73

Total Life cycle cost of all options are as follows:

A- $100665.82

B- $53662.414

C-$106331.46

D- $63210.73

Since option B has lowest total life cycle cost it is most economical