Question

Apply the present worth method for machine A, B, and C. The useful life for machine A, B, and C is 6 years. MARR=6%. Show formula used.

	machine A	machine B	machine C
first cost	$170000	160000	300000
salvage value	10000	none	10000
operating cost	15000/yr	10000	3000
economic life	3 yrs	6 yrs	9 yrs
maintenance cost	4500/yr	1000-first year and increases by 1500/yr	3300 for third year, sixth year, seventh year, ninth year

Answer 1

Analysis period will be for 18 yrs (LCM of 3,6 & 9)

NPW of machine A = -170000 - (170000-10000)*((P/F,6%,3)+(P/F,6%,6)+(P/F,6%,9)+(P/F,6%,12)+(P/F,6%,15)) - (15000+4500)*(P/A,6%,18) + 10000*(P/F,6%,18)

= -170000 - (170000-10000)*(0.839619+0.704961+0.591898+0.496969+0.417265) - (15000+4500)*10.827603 + 10000*0.350344

= -865748.74

NPW of machine B = -160000 - 160000 *0.704961 - 160000*(P/F,6%,12) - 1000*(P/A,6%,6) - 1500*(P/G,6%,6) - 1000*(P/A,6%,6)*(P/F,6%,6) - 1500*(P/G,6%,6)*(P/F,6%,6) - 1000*(P/A,6%,6)*(P/F,6%,12) - 1500*(P/G,6%,6)*(P/F,6%,12)

= -160000 - 160000 *0.704961 - 160000*0.496969 - 1000*4.917324 - 1500*11.459351 - 1000*4.917324*0.704961 - 1500*11.459351*0.704961 - 1000*4.917324*0.496969 - 1500*11.459351*0.496969

= -400985.44

NPW of machine C = -300000 - (300000-10000) *(P/F,6%,9) + 10000*(P/F,6%,18) - 3000*(P/A,6%,18) - 3300 * ((P/F,6%,3) - (P/F,6%,6) - (P/F,6%,7) - (P/F,6%,9)) - 3300 *(P/F,6%,9)* ((P/F,6%,3) - (P/F,6%,6) - (P/F,6%,7) - (P/F,6%,9))

= -300000 - (300000-10000) *0.591898 + 10000*0.350344 - 3000*10.827603 - 3300 * (0.839619 - 0.704961 - 0.665057 - 0.591898) - 3300 *0.591898* (0.839619 - 0.704961 - 0.665057 - 0.591898)

= -494734.07

As present cost of machine B is lowest it should be selected