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 $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

Answer 1

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)