Question

A paper mill is considering two types of wastewater reuse equipment: Neutralization Precipitation Initial Investment $ 700,000 $ 500,000 Annual chemical cost $ 220,000 $ 250,000 Annual maintenance cost $ 40,000 $ 110,000 Salvage Value $175,000 $ 125,000 Interest rate 8% 8% Useful life, in years 5 years 5 years a. For the two alternatives, i) Calculate the PW (DO NOT use the AW or FW values); ii) Calculate the AW (DO NOT use the PW or FW values); iii) Calculate the FW (DO NOT use the PW or AW values) b. Which alternative should be purchased?

Answer 1

(a)

(i)

PW, Neutralization = - 700,000 - (220,000 + 40,000) x P/A(8%, 5) + 175,000 x P/F(8%, 5)

= - 700,000 - 260,000 x 3.9927 + 175,000 x 0.6806

= - 700,000 - 1,038,102 + 119,105

= - 1,618,997

PW, Precipitation = - 500,000 - (250,000 + 110,000) x P/A(8%, 5) + 125,000 x P/F(8%, 5)

= - 500,000 - 360,000 x 3.9927 + 125,000 x 0.6806

= - 500,000 - 1,437,372 + 85,075

= - 1,852,297

(ii)

AW, Neutralization = - 700,000 x A/P(8%, 5) - (220,000 + 40,000) + 175,000 x P/F(8%, 5) x A/P(8%, 5)

= - 700,000 x 0.2505 - 260,000 + 175,000 x 0.6806 x 0.2505

= - 175,350 - 260,000 + 29,836

= - 405,514

AW, Precipitation = - 500,000 x A/P(8%, 5) - (250,000 + 110,000) + 125,000 x P/F(8%, 5) x A/P(8%, 5)

= - 500,000 x 0.2502 - 360,000 + 125,000 x 0.6806 x 0.2505

= - 125,100 - 360,000 + 21,311

= - 463,789

(iii)

FW, Neutralization = - 700,000 x F/P(8%, 5) - (220,000 + 40,000) x F/A(8%, 5) + 175,000

= - 700,000 x 1.4693 - 260,000 x 5.8666 + 175,000

= - 1,028,510 - 1,525,316 + 175,000

= - 2,378,826

FW, Precipitation = - 500,000 x F/P(8%, 5) - (250,000 + 110,000) x F/A(8%, 5) + 125,000

= - 500,000 x 1.4693 - 360,000 x 5.8666 + 125,000

= - 734,650 - 2,111,976 + 125,000

= - 2,721,626

(b)

Since Neutralization option has lower negative PW, AW and FW, this should be selected.