1 answer

Question 1 (1 point) A 5 year-old tooling kit that was purchased new for $50,000 and...

Question:

Question 1 (1 point) A 5 year-old tooling kit that was purchased new for $50,000 and has a current market (trade-in) value of
Question 1 (1 point) A 5 year-old tooling kit that was purchased new for $50,000 and has a current market (trade-in) value of $8000 and expected O&M costs of $8600, increasing by 1000 per year. The kit is in need of an immediate overhaul (repair) expected to cost $1000. Future market values are expected to decline by 25% annually (going forward). The kit can be used for another 7 years at most. The optimal replacement kit information is below. The new model kit will be needed indefinitely. Assume a unique minimum AECN(15%) for both kits (both the current and replacement kit). The replacement kit has N*=4 and AEC c = 14,000 What is AEC D? a) Between 12,538 - 12,578 Ob) Between 12,458 - 12,498 Oc) Between 12,498 - 12,538 d) None of the answer is correct e) Between 12,578 - 12,618

Answers

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,15%,n) EUAC
A B C D=C*B E F=E+9000 G H=G*B I=F-H J K = I*J
1 0.86957 8600.00 7478.26 7478.26 16478.26 6000.00 5217.39 11260.87 1.15000 12950
2 0.75614 9600.00 7258.98 14737.24 23737.24 4500.00 3402.65 20334.59 0.61512 12508
3 0.65752 10600.00 6969.67 21706.91 30706.91 3375.00 2219.12 28487.79 0.43798 12477
4 0.57175 11600.00 6632.34 28339.25 37339.25 2531.25 1447.25 35892.00 0.35027 12572
5 0.49718 12600.00 6264.43 34603.68 43603.68 1898.44 943.86 42659.82 0.29832 12726
6 0.43233 13600.00 5879.66 40483.33 49483.33 1423.83 615.56 48867.77 0.26424 12913
7 0.37594 14600.00 5488.68 45972.01 54972.01 1067.87 401.45 54570.56 0.24036 13117
Discount factor 1/(1+0.15)^n
(A/P,i,n) i((1 + i)^n)/((1 + i)^n-1)

As Minimum EUAC = 12477, Economic life is 3 yrs

it lies between 12458 - 12498

option B is the correct answer

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,15%,n) EUAC
A B C D=C*B E F=E+9000 G H=G*B I=F-H J K = I*J
1 =1/(1.15)^A127 8600 =C127*B127 =D127 =9000+E127 =8000*0.75 =G127*B127 =F127-H127 =0.15*((1 + 0.15)^A127)/((1 + 0.15)^A127-1) =I127*J127
2 =1/(1.15)^A128 =C127+1000 =C128*B128 =E127+D128 =9000+E128 =G127*0.75 =G128*B128 =F128-H128 =0.15*((1 + 0.15)^A128)/((1 + 0.15)^A128-1) =I128*J128
3 =1/(1.15)^A129 =C128+1000 =C129*B129 =E128+D129 =9000+E129 =G128*0.75 =G129*B129 =F129-H129 =0.15*((1 + 0.15)^A129)/((1 + 0.15)^A129-1) =I129*J129
4 =1/(1.15)^A130 =C129+1000 =C130*B130 =E129+D130 =9000+E130 =G129*0.75 =G130*B130 =F130-H130 =0.15*((1 + 0.15)^A130)/((1 + 0.15)^A130-1) =I130*J130
5 =1/(1.15)^A131 =C130+1000 =C131*B131 =E130+D131 =9000+E131 =G130*0.75 =G131*B131 =F131-H131 =0.15*((1 + 0.15)^A131)/((1 + 0.15)^A131-1) =I131*J131
6 =1/(1.15)^A132 =C131+1000 =C132*B132 =E131+D132 =9000+E132 =G131*0.75 =G132*B132 =F132-H132 =0.15*((1 + 0.15)^A132)/((1 + 0.15)^A132-1) =I132*J132
7 =1/(1.15)^A133 =C132+1000 =C133*B133 =E132+D133 =9000+E133 =G132*0.75 =G133*B133 =F133-H133 =0.15*((1 + 0.15)^A133)/((1 + 0.15)^A133-1) =I133*J133
Discount factor 1/(1+0.15)^n
(A/P,i,n) i((1 + i)^n)/((1 + i)^n-1)
.

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