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Brooks Clinic is considering investing in new heart-monitoring equipment. It has two options. Option A would have an initial lower cost but would require a significant expenditure for rebuilding after 4 years. Option B would require no rebuilding expenditure, but its maintenance costs would be higher. Since the Option B machine is of initial higher quality, it is expected to have a salvage value at the end of its useful life. The following estimates were made of the cash flows. The companys cost of capital is 5%.

		Option A		Option B
Initial cost		$183,000		$281,000
Annual cash inflows		$70,000		$80,000
Annual cash outflows		$28,000		$25,000
Cost to rebuild (end of year 4)		$48,000		$0
Salvage value		$0		$7,000
Estimated useful life		7 years		7 years

Compute the (1) net present value, (2) profitability index, and (3) internal rate of return for each option. (Hint: To solve for internal rate of return, experiment with alternative discount rates to arrive at a net present value of zero.) (If the net present value is negative, use either a negative sign preceding the number eg -45 or parentheses eg (45). Round answers for present value and IRR to 0 decimal places, e.g. 125 and round profitability index to 2 decimal places, e.g. 12.50. For calculation purposes, use 5 decimal places as displayed in the factor table provided.)

Future Value of 1 Table (FV of 1 Table) FV Factors for a Single Amount of 1.000 (rounded to three decimal places). Note: This table begins with the row n = 0, which is different from most future value of 1 tables.

	i=1%	i=2%	i=3%	i=4%	i=5%	i=6%	i=8%	i=10%	i=12%
n = 0	1.000	1.000	1.000	1.000	1.000	1.000	1.000	1.000	1.000
n = 1	1.010	1.020	1.030	1.040	1.050	1.060	1.080	1.100	1.120
n = 2	1.020	1.040	1.061	1.082	1.103	1.124	1.166	1.210	1.254
n = 3	1.030	1.061	1.093	1.125	1.158	1.191	1.260	1.331	1.405
n = 4	1.041	1.082	1.126	1.170	1.216	1.262	1.360	1.464	1.574
n = 5	1.051	1.104	1.159	1.217	1.276	1.338	1.469	1.611	1.762
n = 6	1.062	1.126	1.194	1.265	1.340	1.419	1.587	1.772	1.974
n = 7	1.072	1.149	1.230	1.316	1.407	1.504	1.714	1.949	2.211
n = 8	1.083	1.172	1.267	1.369	1.477	1.594	1.851	2.144	2.476
n = 9	1.094	1.195	1.305	1.423	1.551	1.689	1.999	2.358	2.773
n = 10	1.105	1.219	1.344	1.480	1.629	1.791	2.159	2.594	3.106
n = 11	1.116	1.243	1.384	1.539	1.710	1.898	2.332	2.853	3.479
n = 12	1.127	1.268	1.426	1.601	1.796	2.012	2.518	3.138	3.896
n = 13	1.138	1.294	1.469	1.665	1.886	2.133	2.720	3.452	4.363
n = 14	1.149	1.319	1.513	1.732	1.980	2.261	2.937	3.797	4.887
n = 15	1.161	1.346	1.558	1.801	2.079	2.397	3.172	4.177	5.474
n = 16	1.173	1.373	1.605	1.873	2.183	2.540	3.426	4.595	6.130
n = 17	1.184	1.400	1.653	1.948	2.292	2.693	3.700	5.054	6.866
n = 18	1.196	1.428	1.702	2.026	2.407	2.854	3.996	5.560	7.690
n = 19	1.208	1.457	1.754	2.107	2.527	3.026	4.316	6.116	8.613
n = 20	1.220	1.486	1.806	2.191	2.653	3.207	4.661	6.727	9.646
n = 21	1.232	1.516	1.860	2.279	2.786	3.400	5.034	7.400	10.804
n = 22	1.245	1.546	1.916	2.370	2.925	3.604	5.437	8.140	12.100
n = 23	1.257	1.577	1.974	2.465	3.072	3.820	5.871	8.954	13.552
n = 24	1.270	1.608	2.033	2.563	3.225	4.049	6.341	9.850	15.179
n = 25	1.282	1.641	2.094	2.666	3.386	4.292	6.848	10.835	17.000
n = 26	1.295	1.673	2.157	2.772	3.556	4.549	7.396	11.918	19.040
n = 27	1.308	1.707	2.221	2.883	3.733	4.822	7.988	13.110	21.325
n = 28	1.321	1.741	2.288	2.999	3.920	5.112	8.627	14.421	23.884
n = 29	1.335	1.776	2.357	3.119	4.116	5.418	9.317	15.863	26.750
n = 30	1.348	1.811	2.427	3.243	4.322	5.743	10.063	17.449	29.960

n= the number of time periods in which the interest is compounded i= the interest rate per period with the interest added and compounded at the end of each period

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