Question

In: Economics

13-24 A 2,000-pound, counterbalanced, propane forklift can be purchased for $30,000. Due to the intended service...

13-24

A 2,000-pound, counterbalanced, propane forklift can be purchased for $30,000. Due to the intended service use, the forklift’s market value drops 20% of its prior year’s value in Years 1 and 2 and then declines 15% until Year 10 when it will have a scrap/market value of $1,000. Maintenance of the forklift is $400 per year during Years 1 and 2 while the warranty is in place. In Year 3 it jumps to $750 and increases by $200 per year thereafter. What is the optimal life of the forklift using i = 10%?

Solutions

Expert Solution

We need to calculate economic life of the forlift

I = 10%

Initial cost = 30000

Salvage value = 20% drop in first year from initial cost, 20% in next year. 15% drop per year afterwards

Maintenance cost = 400 in year 1 and 2, 750 in yr 3 and increase by 200 per year after wards

using excel

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,10%,n) EUAC
A B C D=C*B E F=E+30000 G H=G*B I=F-H J K = I*J
1 0.90909 400.00 363.64 363.64 30363.64 24000.00 21818.18 8545.45 1.10000 9400.00
2 0.82645 400.00 330.58 694.21 30694.21 19200.00 15867.77 14826.45 0.57619 8542.86
3 0.75131 750.00 563.49 1257.70 31257.70 16320.00 12261.46 18996.24 0.40211 7638.67
4 0.68301 950.00 648.86 1906.56 31906.56 13872.00 9474.76 22431.80 0.31547 7076.58
5 0.62092 1150.00 714.06 2620.62 32620.62 11791.20 7321.41 25299.22 0.26380 6673.87
6 0.56447 1350.00 762.04 3382.66 33382.66 10022.52 5657.45 27725.21 0.22961 6365.91
7 0.51316 1550.00 795.40 4178.06 34178.06 8519.14 4371.67 29806.39 0.20541 6122.40
8 0.46651 1750.00 816.39 4994.45 34994.45 7241.27 3378.11 31616.34 0.18744 5926.29
9 0.42410 1950.00 826.99 5821.44 35821.44 6155.08 2610.35 33211.08 0.17364 5766.79
10 0.38554 2150.00 828.92 6650.35 36650.35 1000.00 385.54 36264.81 0.16275 5901.93
Discount factor 1/(1+0.10)^n
(A/P,i,n) i((1 + i)^n)/((1 + i)^n-1)

As EUAC is lowest in yr 9, so optimum life is 9 years

Showing formula in excel

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,10%,n) EUAC
A B C D=C*B E F=E+30000 G H=G*B I=F-H J K = I*J
1 =1/(1.1)^A66 400 =C66*B66 =D66 =30000+E66 =30000*0.8 =G66*B66 =F66-H66 =0.1*((1 + 0.1)^A66)/((1 + 0.1)^A66-1) =I66*J66
2 =1/(1.1)^A67 400 =C67*B67 =E66+D67 =30000+E67 =G66*0.8 =G67*B67 =F67-H67 =0.1*((1 + 0.1)^A67)/((1 + 0.1)^A67-1) =I67*J67
3 =1/(1.1)^A68 750 =C68*B68 =E67+D68 =30000+E68 =G67*0.85 =G68*B68 =F68-H68 =0.1*((1 + 0.1)^A68)/((1 + 0.1)^A68-1) =I68*J68
4 =1/(1.1)^A69 =C68+200 =C69*B69 =E68+D69 =30000+E69 =G68*0.85 =G69*B69 =F69-H69 =0.1*((1 + 0.1)^A69)/((1 + 0.1)^A69-1) =I69*J69
5 =1/(1.1)^A70 =C69+200 =C70*B70 =E69+D70 =30000+E70 =G69*0.85 =G70*B70 =F70-H70 =0.1*((1 + 0.1)^A70)/((1 + 0.1)^A70-1) =I70*J70
6 =1/(1.1)^A71 =C70+200 =C71*B71 =E70+D71 =30000+E71 =G70*0.85 =G71*B71 =F71-H71 =0.1*((1 + 0.1)^A71)/((1 + 0.1)^A71-1) =I71*J71
7 =1/(1.1)^A72 =C71+200 =C72*B72 =E71+D72 =30000+E72 =G71*0.85 =G72*B72 =F72-H72 =0.1*((1 + 0.1)^A72)/((1 + 0.1)^A72-1) =I72*J72
8 =1/(1.1)^A73 =C72+200 =C73*B73 =E72+D73 =30000+E73 =G72*0.85 =G73*B73 =F73-H73 =0.1*((1 + 0.1)^A73)/((1 + 0.1)^A73-1) =I73*J73
9 =1/(1.1)^A74 =C73+200 =C74*B74 =E73+D74 =30000+E74 =G73*0.85 =G74*B74 =F74-H74 =0.1*((1 + 0.1)^A74)/((1 + 0.1)^A74-1) =I74*J74
10 =1/(1.1)^A75 =C74+200 =C75*B75 =E74+D75 =30000+E75 1000 =G75*B75 =F75-H75 =0.1*((1 + 0.1)^A75)/((1 + 0.1)^A75-1) =I75*J75
Discount factor 1/(1+0.10)^n
(A/P,i,n) i((1 + i)^n)/((1 + i)^n-1)

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