Question

Q)

Determine the present worth of a geometric gradient series with a cash flow of $38,537 in year 1 and increases of 11% each year begining of year 4 through year 11. The interest rate is 16% per year.

Q) The capital cost (CC), now, when a family want to have $6,004, per year, starting in year 26 and continuing forever, is close to. Assume  rate of return13%

Q)

Alternative X has a first cost of $20000, an operating cost of $9000 per year, and a $5000 salvage value after 6 years. Alternative Y will cost $18,898 with an operating cost of $7,514 per year and a salvage value of $7,270 after 3 years.  Note: i=0.12% refers to  i=12%

At an MARR of 0.12% per year, find the PW of machine Y?

Answer 1

Part 1) We have the following information

Interest rate = 16% per annum or 0.16

It is given that from 4^th year onwards the Year 1 Cash Flow increases by 11% per year

Year	Cash Flow ($)
1	38,537.00
2	-
3	-
4	42,776.07
5	47,481.44
6	52,704.40
7	58,501.88
8	64,937.09
9	72,080.17
10	80,008.98
11	88,809.97

Present Worth = 38,537.00(P/F, 16%, 1) +   42,776.07(P/F, 16%, 4) +   47,481.44(P/F, 16%, 5) + 52,704.40(P/F, 16%, 6) + 58,501.88(P/F, 16%, 7) + 64,937.09(P/F, 16%, 8) + 72,080.17(P/F, 16%, 9) + 80,008.98(P/F, 16%, 10) +   88,809.97(P/F, 16%, 11)

Present Worth = 38,537.00/(1 + 0.16)¹ + 42,776.07/(1 + 0.16)⁴ +   47,481.44/(1 + 0.16)⁵ + 52,704.40/(1 + 0.16)⁶ + 58,501.88/(1 + 0.16)⁷ + 64,937.09/(1 + 0.16)⁸ + 72,080.17/(1 + 0.16)⁹ + 80,008.98/(1 + 0.16)¹⁰ +   88,809.97/(1 + 0.16)¹¹

Present Worth = $196,036.13

Part 2) We have the following information

A family want to have $6,004, per year, starting in year 26 and continuing forever.

Interest rate = 13% or 0.13

Capitalized Cost = 6,004(A/F, 13%, 26)(P/A, 13%, n=infinite)

Capitalized Cost = 6,004 × [0.13/((1 + 0.13)²⁶ – 1)] × (1/0.13)

Capitalized Cost = 6,004 × 0.00565 × 7.69

Capitalized Cost = $260.94

Part 3) We have the following information about Machine Y

Initial Cost ($)	18,898
Annual Operating Cost ($)	7,514
Salvage Value ($)	7,270
Life (n) in years	3
MARR	12% per annum or 0.12

Present Worth (PW) = Initial Cost + Annual Cost(P/A, i, n) – Salvage Value(P/F, i, n)

Present Worth (PW) = 18,898 + 7,514(P/A, 12%, 3) – 7,270(P/F, 12%, 3)

Present Worth (PW) = 18,898 + 7,514[((1+0.12)³ – 1)/0.12(1+0.12)³] – 7,270/(1 + 0.12)³

Present Worth (PW) = 18,898 + 18,048.63 – 5,174.79

Present Worth of Machine Y = $31,771.84