Question 5 12 marks Queens is considering acquiring a new sn

Question 5: (12 marks) Queen\'s is considering acquiring a new snow removal machine with a capital cost of $30,000 in order to realize savings of $5,000 per year. The salvage value of the machine is $3,000 at the end of year 10. The University uses a MARR of 15%. a) What is the present worth of this investment? (3 marks) b) What is its annual worth? (3 marks) c) What is the future worth after 10 years? (3 marks) d) What is the simple payback period (1 mark) e) What is the discounted payback period? Do not exceed the life of the machine. (2 marks)

Solution

(a)

Present worth, PW ($) = - 30,000 + 5,000 x P/A(15%, 10 years) + 3,000 x P/F(15%, 10 years)

= - 30,000 + 5,000 x 5.02 + 3,000 x 0.25

= - 30,000 + 25,100 + 750

= - 4,150

(b)

Annual worth, AW ($) = - 30,000 x A/P(15%, 10 years)** + 5,000 + 3,000 x P/F(15%, 10 years)

= - 30,000 x 0.2 + 5,000 + 3,000 x 0.25

= - 6,000 + 5,000 + 750

= - 250

**A/P(R%, N years) is capital recovery factor.

(c)

Future worth = - 30,000 x F/P(15%, 10 years) + 5,000 x F/A(15%, 10 years) + 3,000

= - 30,000 x 4.05 + 5,000 x 20.3 + 3,000

= - 121,500 + 101,500 + 3,000

= - 17,000

(d)

Payback period (PBP) is the time by when the project\'s cash inflows recover the cash outflow, or, the time when cumulative cash flows equal 0.

So, simple PBP is 6 years.

(e)

Discounted PBP is the time by when cumulative discounted cash flow becomes 0.

So we can see that during project life (10 years), discounted cumulative cash flows remain negative. So discounted PBP falls beyond project life.

Year	Cash Flow ($)	Cumulative Cash Flow ($)
0	-30,000	-30,000
1	5,000	-25,000
2	5,000	-20,000
3	5,000	-15,000
4	5,000	-10,000
5	5,000	-5,000
6	5,000	0
7	5,000	5,000
8	5,000	10,000
9	5,000	15,000
10	8,000	23,000