Question

Problem 7: Two pumps are being considered for pumping water from a reservoir. Installed cost and salvage value for the two pumps are given below. Pump A Pump B Installed Cost$20,000 $25,000 Salvage Value$2,000 $4,000 Pump A has a service life of 4 years. Determine the service life of pump B at which the two pumps are competitive if the annual effective interest rate is 15%. To be competitive, pump B must have the same capitalized cost as pump A.

Answer 1

Let V_0 rightarrow initial cost of an equivalent (installation cost) V_s rightarrow salvage value after - years n rightarrow service lite V_R rightarrow replacement cost (V_R = V_0 - V_s) Let x rightarrow amount invested to earn interest equal to replacement cost Interest = x (1 + i)^n - x Replacement cost = Interest V_0 = x (1 + i)^n - x x = V_R/(1 + i)^n - 1 = V_0 - V_s/(1 + i)^n - 1 Total capitalized cost (V_cc) V_cc = V_0 + 1 = V_0 + (V_0 + V_s)/(1 + i)^n - 1 V_cc = V_0 + (V_0 + V_s)/(1 + i)^n - 1 For competitiveness capitalized cost of both the pump must be same (V_cc)_A = (V_cc)_B (V_0)_A + (V_0)_A - (V_s)_A/(1 + i)^n_A - 1 = (V_0)_B + (V_0)_B - (V_s)_B/(1 + i)^n_B - 1 \r\n = > 20,000 + 20,000 - 2000/(1 + 0.15)^4 - 1 = 25000 + 25,000 - 4000