Question

Q1

Eb's Eggs just bought a new egg sorting machine for $124201. The machine will save $33343 in year 1, $30661 in year 2, $15526 in year 3, and $7650 per year from year 4 until the machine is salvaged at the end of year 11. At the end of year 11 it will have a salvage value of $2001. Eb uses a MARR of 8% to make decisions.

What is the payback period (PBP) for this machine?

Enter your answer as 12.34

Q2

Professor Xavier is creating a budget for his recently awarded 9-year research grant. His research requires machinery that has an initial cost of $44104. He will need to pay for maintenance on the machinery beginning at the end of year 4. He estimates the first year of maintenance will cost $3370, with the maintenance costs increasing by $511 per year each subsequent year.  How much should he put in the budget today for maintenance costs? Assume he will pay maintenance until the end of the research grant. He is using a MARR of 4%.

Enter your answer as follows: 1234

Round your answer. Do not use a dollar sign ("$"), any commas (","), or a decimal point (".").

Answer 1

We have the following information

Period	Cashflow ($)	Present Value of Net Cash Flow at MARR of 8%	Discounted Cumulative Cash Flow
0	-124201	-124201	-124201
1	33343	30873.15	-93327.9
2	30661	26286.87	-67041.0
3	15526	12325.04	-54715.9
4	7650	5622.978	-49093.0
5	7650	5206.461	-43886.5
6	7650	4820.798	-39065.7
7	7650	4463.702	-34602.0
8	7650	4133.057	-30469.0
9	7650	3826.905	-26642.0
10	7650	3543.43	-23098.6
11	2001	858.1946	-22240.4

Following present value formula has been used for calculating Present Value of Net Cash Flow at MARR of 8%

P = F/(1 + i)ⁿ

P = Present Value

F = Future Cash Flow

i = Interest Rate

n = Number of interest periods

So, we can say that with the current savings and salvage value the initial investment cost from the discounted net cash flows cannot be recovered.

Part b) We have the following information

Initial Cost ($)	0	44104
Maintenance Cost ($)	Year 1	0
Maintenance Cost ($)	Year 2	0
Maintenance Cost ($)	Year 3	0
Maintenance Cost ($)	Year 4	3370
Maintenance Cost ($)	Year 5	3881
Maintenance Cost ($)	Year 6	4392
Maintenance Cost ($)	Year 7	4903
Maintenance Cost ($)	Year 8	5414
Maintenance Cost ($)	Year 9	5925

Minimum Acceptable Rate of Return (MARR) = 4% per annum

We will use the following formula

P = F/(1 + i)ⁿ

P = F(P/F, i, n)

(P/F, i, n) = Single payment present worth factor

F = Future amount at the end of year n

P = Principal amount

n = Number of interest periods

i = Interest rate = 8% or 0.08

P = 44104 + 3370(P/F, 8%, 4) + 3881(P/F, 8%, 5) + 4392(P/F, 8%, 6) + 4903(P/F, 8%, 7) + 5414(P/F, 8%, 8) + 5925(P/F, 8%, 9)

P = 44104 + 3370/(1 + 0.08)⁴ + 3881/(1 + 0.08)⁵ + 4392/(1 + 0.08)⁶ + 4903/(1 + 0.08)⁷ + 5414/(1 + 0.08)⁸ + 5925/(1 + 0.08)⁹

P = 44104 + 2477.05 + 2641.34 + 2767.71 + 2860.85 + 2925.02 + 2963.98

P = 60739.95