Question

National Laboratory developed a special alloy material used in solar cells. The new material is expected to have a 7 years life. To produce this alloy $950,000 is invested now. The expected revenue from this investment is $450,000 starting year 3, and amounts increasing by $50,000 per year through year 7. National Laboratory MARR is 6% per year. Determine the Discounted Payback period 0--950,000+ [450,000 (P/A, 6%, np - 1) + 50,000(P/G, 6%, np - 1)] (P/F, 1%, 2) np = 4.43 years O 0-.950,000 + [450,000 (PA, 6%, np - 1) +50,000(PG, 6%, np - 1)] (P/F,i%, 2) np = 5.43 years 0 = -950,000+ [450,000 (P/A, 6%, np - 2) +50,000(PG, 6%, np - 2) (PF, 1%, 2) np = 4.43 years 0 = -950,000 + [450,000 (PA, 6%, np - 2) - 50,000(PG, 6º, np - 2)] (PF, 1%, 2) np = 5.43 years

Answer 1

The correct equation and payback period is the one below.

0 = -950,000 + [450,000 (P/A, 6%, np - 2) +50,000(PG, 6%, np - 2)) (PF, 1%, 2) np = 4.43 years

The equation should have years (np-2) because it is starting from the 3rd year. 7-2 = 3

A 2 1 year 0 1 B CF PVIF @6% PV Cumulative PV -950000 1.0000 -950000.00 -950000.00 0.9434) 0.00 -950000.00 0.8900 0.00 -950000.00 450000 0.8396 377828.68 -572171.32 500000 0.7921 396046.83 -176124.49 550000 0.7473 410992.00 234867.50 600000 0.7050 422976.32 657843.83 650000 0.6651 432287.12 1090130.95 3 4 5 6 Discounted Payback period

1 year 20 -950000 31 PVIF @6% =1/1.06^A2 =1/1.06^A3 =1/1.06^A4 =1/1.06^A5 =1/1.06^A6 450000 =B5+50000 =B6+50000 =B7+50000 =B8+50000 PV =B2*C2 =B3*C3 =B4*C4 =B5*C5 =B6*C6 =B7*C7 =B8*C8 =B9*C9 Cumulative PV =D2 =E2+D3 =E3+D4 =E4+D5 =E5+D6 =E6+D7 =E7+D8 -E8+D9 =1/1.06^A8 =1/1.06^A9 Discounted Payback period =A6+(-E6/07)

Please don't forget to rate the answer if its helpful, thank you.