Question

Options are under consideration to meet a 12 period need are shown below. Options have different lives so an adjustment for equivalence is required to perform a valid analysis using engineering economic methods. The interest rate for use is 8.0%.

Option	A	B	C	D
Purchase Cost	65,000	41,000	29,000	19,000
First Year Operating Cost	4,800	6,400	7,500	8,100
Operating Cost Increase per Period	500	700	900	1,400
Life/Salvage	6/4,000	4/2,000	3/1,500	2/800

a. What is the PW of Option A?

b. What is the PW of Option B?

c. What is the PW of Option C?

d. What is the PW of Option D?

Answer 1

Formula	Option	A	B	C	D
	Purchase cost (PC)	65,000	41,000	29,000	19,000
	First year operating cost (OC1)	4,800	6,400	7,500	8,100
	Operating cost increase/period (G)	500	700	900	1,400
	Salvage value (SV)	4,000	2,000	1,500	800
	Life (n)	6	4	3	2
	Interest rate (i)	8%	8%	8%	8%
PW is same as purchase cost since it occurs at t = 0	PW of purchase cost (a)	65,000	41,000	29,000	19,000
OC1*(P/A, i, n)2	PW of   uniform operating cost (b)	22,189.82	21,197.61	19,328.23	14,444.44
	(P/G, i, n)	38.49	26.06	21.47	17.69
G*(P/G, i, n)3	PW of operating cost increase/period ('c)	19,242.83	18,244.84	19,318.61	24,760.90
SV/(1+i)^n	PW of salvage value (d)	2,520.68	1,470.06	1,190.75	685.87
(a+b+c-d)	PW of the option	103,911.97	78,972.39	66,456.09	57,519.47

Note:

1). Since operating cost is increasing in arithmetic progression every year, its PW can be found as a separate series apart from the uniform operating cost of 1st year which will occur every year.

2). (P/A, i, n) = [((1+i)^n) -1]/i*(1+i)^n

3). (P/G, i, n) = 1/i{ [[((1+i)^n) -1]/i*(1+i)^n] - [n/(1+i)^n]}