In: Economics
An industrial sewing machine costs $5389 and is expected to have a scrap value of $3570 whenever it is retired. Operating and Maintenance costs are $1585 for the first year and expected to increase by $2270 thereafter. If the MARR is 11%, determine the minimum equivalent uniform annual cost associated with the optimal economic life of the machine. The service life of this machine is 5 years. Note: round your answer to two decimal places, and do not include spaces, currency signs, plus or minus signs, nor commas.
MARR = 11%
Using Excel for Economic life analysis
Year | Discount factor | O&M cost | PV (O&M) | Cumulative (O&M) | Cumulative (O&M) + Initial Cost | Salvage value | PV (Salvage value) | NPV | (A/P,11%,n) | EUAC |
A | B | C | D=C*B | E | F=E+5389 | G | H=G*B | I=F-H | J | K = I*J |
1 | 0.900901 | 1585.00 | 1427.93 | 1427.93 | 6816.93 | 3570.00 | 3216.22 | 3600.71 | 1.110000 | 3996.79 |
2 | 0.811622 | 3855.00 | 3128.80 | 4556.73 | 9945.73 | 3570.00 | 2897.49 | 7048.24 | 0.583934 | 4115.70 |
3 | 0.731191 | 6125.00 | 4478.55 | 9035.28 | 14424.28 | 3570.00 | 2610.35 | 11813.93 | 0.409213 | 4834.41 |
4 | 0.658731 | 8395.00 | 5530.05 | 14565.33 | 19954.33 | 3570.00 | 2351.67 | 17602.66 | 0.322326 | 5673.80 |
5 | 0.593451 | 10665.00 | 6329.16 | 20894.48 | 26283.48 | 3570.00 | 2118.62 | 24164.86 | 0.270570 | 6538.29 |
Discount factor | 1/(1+0.11)^n | |||||||||
(A/P,i,n) | i((1 + i)^n)/((1 + i)^n-1) |
Minimum EUAC = 3996.79 at 1 yr
So Economic service life is 1 yr