Question

1. You were asked by the Manager of Engineering to propose a solution for a current production line problem. After analyzing the issues you came up with two solutions, both of which will solve the problem for the next five years. Solution A would initially cost $90,000, have annual O&M costs of $22,000 and will generate annual savings of $48,000, while solution B will need an initial $62,000, $17,000 for annual O&M costs, and will generate annual savings of $36,000. Both will have salvage values, $15,000 for solution A, and $10,000 for B. Your company’s marginal income tax rate is 40%, and its MARR is 10%. The proposed equipment for both options is subject to a five-year MACRS property class. Answer the following questions:

1. Which option would you recommend? Show the work that backs your choice.
2. What value of MARR would make each solution break even?
3. What would be the amount of additional revenue that Solution A should generate to make the Manager of Engineering indifferent to the choice between the two options.

Answer 1

Solution A:

Formula	Year (n)	0	1	2	3	4	5
	Initial cost (IC)	(90,000)
	Annual savings (S)		48,000	48,000	48,000	48,000	48,000
	Annual O&M costs		(22,000)	(22,000)	(22,000)	(22,000)	(22,000)
	Depreciation rate ('r)		20%	32%	19.20%	11.52%	11.52%
	Depreciation (D)		(18,000)	(28,800)	(17,280)	(10,368)	(10,368)
(S-O&M-D)	EBIT		8,000	(2,800)	8,720	15,632	15,632
(40%*EBIT)	Tax @ 40%		(3,200)	1,120	(3,488)	(6,253)	(6,253)
(EBIT-Tax)	Net income (NI)		4,800	(1,680)	5,232	9,379	9,379
	Add: depreciation (D)		18,000	28,800	17,280	10,368	10,368
(NI+D)	Operating Cash Flow (OCF)		22,800	27,120	22,512	19,747	19,747
	Salvage value (sv)						15,000
(IC - accumulated depreciation)	Book value (bv)						5,184
(sv - 40%*(sv-bv))	After-tax salvage value (SV)						11,074
(IC + OCF + SV)	Free Cash Flow (FCF)	(90,000)	22,800	27,120	22,512	19,747	30,821
1/(1+MARR)^n	Discount factor @ 10%	1.000	0.909	0.826	0.751	0.683	0.621
(FCF*Discount factor)	PV of FCF	(90,000.00)	20,727.27	22,413.22	16,913.60	13,487.60	19,137.29
Sum of all PVs	NPV	2,678.99

Solution B:

Formula	Year (n)	0	1	2	3	4	5
	Initial cost (IC)	(62,000)
	Annual savings (S)		36,000	36,000	36,000	36,000	36,000
	Annual O&M costs		(17,000)	(17,000)	(17,000)	(17,000)	(17,000)
	Depreciation rate ('r)		20%	32%	19.20%	11.52%	11.52%
	Depreciation (D)		(12,400)	(19,840)	(11,904)	(7,142)	(7,142)
(S-O&M-D)	EBIT		6,600	(840)	7,096	11,858	11,858
(40%*EBIT)	Tax @ 40%		(2,640)	336	(2,838)	(4,743)	(4,743)
(EBIT-Tax)	Net income (NI)		3,960	(504)	4,258	7,115	7,115
	Add: depreciation (D)		12,400	19,840	11,904	7,142	7,142
(NI+D)	Operating Cash Flow (OCF)		16,360	19,336	16,162	14,257	14,257
	Salvage value (sv)						10,000
(IC - accumulated depreciation)	Book value (bv)						3,571
(sv - 40%*(sv-bv))	After-tax salvage value (SV)						7,428
(IC + OCF + SV)	Free Cash Flow (FCF)	(62,000)	16,360	19,336	16,162	14,257	21,685
1/(1+MARR)^n	Discount factor @ 10%	1.000	0.909	0.826	0.751	0.683	0.621
(FCF*Discount factor)	PV of FCF	(62,000.00)	14,872.73	15,980.17	12,142.45	9,737.70	13,464.95
Sum of all PVs	NPV	4,197.99

a). Solution B should be chosen as it has a higher NPV.

b). Using Solver, solution A NPV = 0 with MARR = 11.1%

solution B NPV = 0 with MARR = 12.6%

c). Using Solver again, if annual savings of A becomes 48,667.85 then NPV for both solutions will be same and the Manager of Engineering will be indifferent to the choice. Additional savings needed for A = 48,667.85 - 48,000 = 667.85