Question

Company X must purchase a new milling machine. It is considering four mills, each of
which has a life of 10 years with no scrap value. The values below are in their respective order.

Mill Type 1, 2, 3, 4
First cost $110 000, $160 000, $210 000, $265 000
Annual savings 26 000, 33 000, 49 000, 52 000

Given a MARR of 12 percent, which alternative should be taken to obtain the largest present worth?

Answer 1

PW, Mill 1 = - 110,000 + 26,000 x P/A(12%, 10) = - 110,000 + 26,000 x 5.6502 = - 110,000 + 146,905 = 36,905

PW, Mill 2= - 160,000 + 33,000 x P/A(12%, 10) = - 160,000 + 33,000 x 5.6502 = - 160,000 + 186,457 = 26,457

PW, Mill 3 = - 210,000 + 49,000 x P/A(12%, 10) = - 210,000 + 49,000 x 5.6502 = - 210,000 + 276,860 = 66,860

PW, Mill 4 = - 265,000 + 52,000 x P/A(12%, 10) = - 265,000 + 52,000 x 5.6502 = - 265,000 + 293,810 = 28,810

Since Mill 3 has highest PW, Mill 3 should be selected.