Question

HopHeart Brewery is considering 3 different bottling machines. It is expected that each machine will be replaceable at the same cost when their useful life ends. The details of the machines are as follows: Machine X has a useful life of 6 years. It costs $10,000 to purchase and $2,000 per year to maintain. Machine Y has a useful life of 12 years. It costs $15,000 to purchase, and $1,000 per year to maintain. Machine Z has a useful life of 8 years. It costs $20,000 to purchase, and $200 per year to maintain. a) What is the appropriate planning horizon for analyzing these choices? Correct: Your answer is correct. years b) Using the planning horizon from part a, analyze the present worth of the cost of each alternative if HopHeart has a MARR of 8.5%/year. Machine X _____________ Machine Y _____________ Machine Z _____________

Answer 1

ANSWER:

i am solving this in 2 ways , first in the planning horizon which includes lcm to find the common no of years and secondly the present worth in there actucal years, so that you can understand the difference.

1) Since the planning horizon is years , therefore we will have to find the lcm of the three useful lives that is the lcm of(6,12,8) which is 24.

i = 8.5% , n = 24 years

since the table of 8.5% interest is not available , therefore i will solve it in excel.

cash flows = initial cost in year 0 + maintenance cost in year 0 + ..... + initial cost in year 24 + maintenance cost in year 24

machine x
years	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24
initial cost	10,000						10,000						10,000						10,000
maintenance cost		2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000	2,000
cash flows	10,000	2,000	2,000	2,000	2,000	2,000	12,000	2,000	2,000	2,000	2,000	2,000	12,000	2,000	2,000	2,000	2,000	2,000	12,000	2,000	2,000	2,000	2,000	2,000	2,000
npv	42,397.5067646209
machine y
years	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24
initial cost	15,000												15,000
maintenance cost		1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000
cash flows	15,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	16,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000
npv	30,739.6222562305
machine z
years	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24
initial cost	20,000								20,000								20,000
maintenance cost		200	200	200	200	200	200	200	200	200	200	200	200	200	200	200	200	200	200	200	200	200	200	200	200
cash flows	20,000	200	200	200	200	200	200	200	20,200	200	200	200	200	200	200	200	20,200	200	200	200	200	200	200	200	200
npv	37,856.1418290675

Since the npv of machine b is the lowest therefore we will choose machine y.

2) pw of each machine via there original lives.

machine x
years	0	1	2	3	4	5	6
initial cost	10,000
maintenance cost		2,000	2,000	2,000	2,000	2,000	2,000
cash flows	10,000	2,000	2,000	2,000	2,000	2,000	2,000
npv	19,107.1743390960
machine y
years	0	1	2	3	4	5	6	7	8	9	10	11	12
initial cost	15,000
maintenance cost		1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000
cash flows	15,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000	1,000
npv	22,344.6860696889
machine z
years	0	1	2	3	4	5	6	7	8
initial cost	20,000
maintenance cost		200	200	200	200	200	200	200	200
cash flows	20,000	200	200	200	200	200	200	200	200
npv	21,127.8365936075

Had we not chosen the planning horizon of years , we would have chosen machine x as it has the lowest present worth but since the planning horizon is for years , we have to find the lcm and then choose the option y.

the 2nd part is for your better understanding , please use the values in part 1 for the present worth ( although both are correct but part 1 has taken horizon planning while part 2 is without horizon planning,)

in excel we find the present worth by using the =npv(rate,cash flows of each year) + initial year cash flow