Question

money, anu Juuay pu mom. Alternative Comparison-Different Lives 5.17 Dexcon Technologies, Inc. is evaluating two alter- natives to produce its new plastic filament with tribological (i.e., low friction) properties for creat- ing custom bearings for 3-D printers. The esti- mates associated with each alternative are shown below. Using a MARR of 20% per year, which al- ternative has the lower present worth? Method First cost, $ M&O cost, $/year Salvage value, $ Life, years DDM - 164,000 -55,000 LS -370,000 -21,000 30,000

Solve it in spreadsheet

Answer 1

Since life of DDM and LS are different, it is required to consider two cycles of DDM so that life becomes 4 years

PW_DDM = - 164,000 - 164,000 (P/F, 20%,2) -55,000 (P/A,20%,4) = -164,000 -164,000 *1/(1+0.2)^4 - 55,000* [(1+0.2)^4-1] / (0.2*(1+0.2)^4) = -420,269.29

PW_LS  = - 370,000 +30,000 (P/F,20%,4) -21,000 (P/A, 20%,4) = -370,00 +30,000 * 1/(1+0.2)^4 - 21,000* [(1+0.2)^4-1] / (0.2*(1+0.2)^4) = -409,895.83

Thus PW of LS is lower. So LS shall be selected

	DDM	LS
First Cost	-164,000	-370,000
M&O cost per year	-55,000	-21,000
Salvage	-	30,000
life, years	2	4
MARR	20%
P/F,i%,n=2	0.694444444
P/F,i%,n=4	0.482253086
P/A,i%,n=4	2.588734568
PWDDM	-420,269.29
PWLS	-409,895.83

DDMLS 2 First Cost -164,000 -370,000 3 M&O cost per year -55,000 -21,000 4 Salvage 30,000 life, years 2 6 MARR 20% 7 P/F,1%, n=2 0.694444444 8 P/F,7%, n=4 0.482253086 9 P/A,1%,n=4 2.588734568 10 11 PWDDM -420,269.29 12 PWLS -409,895.83

DDM -164000 -55000 -370000 -21000 30000 2 2 First Cost 3 M&O cost per year 4 Salvage 5 life, years 6 MARR 7 P/F,1%, n=2 8 P/F,i%,n=4 9 P/A,1%, n=4 0.2 =1/(1+B6)^B5 =1/(1+B6)^C5 =((1+B6)^C5-1)/(16*(1+B6) 10 11 PWDDM 12 PWLS =B2+B2*B7+B3*B9 =C2+C3*B9+C4*B8