Question

Kermit is considering purchasing a new computer system. The purchase price is $128663. Kermit will borrow one-fourth of the purchase price from a bank at 10 percent per year compounded annually. The loan is to be repaid using equal annual payments over a 3-year period. The computer system is expected to last 5 years and has a salvage value of $6897 at that time. Over the 5-year period, Kermit expects to pay a technician $20,000 per year to maintain the system but will save $76807 per year through increased efficiencies. Kermit uses a MARR of 12 percent to evaluate investments. What is the net present worth for this new computer system? 

Question 8

Consider palletizer at a bottling plant that has a first cost of $135893, has operating and maintenance costs of $15800 per year, and an estimated net salvage value of $28768 at the end of 30 years. Assume an interest rate of 8%, what is the future worth of this project?

Answer 1

	Purchase Price	1,28,663
	Loan (1/4 of Purchase Price)	32,166
	Int	10%	Compounded annually
	Loan time period	3
	Salvage Value	6,897.00
	Useful Life	5
	Annual Maintenance Cost	20,000.00
	Savings due to increased efficiencies	76807
	MARR	12%
	Loan Calculation
X	Principal	32,165.75	Formula = (128663-32166)
Y	Interest at 10% for 3 years	10,646.86	Formula = (1*(1+10/100)^3)
	Total Loan to be Paid (X+Y)	42,812.61
	Outflows
	Time period (t)	1	2	3	4	5
A	Initial Investment (Purchase Price - Loan Value)	96,497.3
B	Loan EMI	14,270.9	14,270.9	14,270.9
C	Annual Maintenance Cost	20000	20000	20000	20000	20000
D	Total Outflows (A+B)	1,30,768	34,271	34,271	20,000	20,000
E	Present value Factor @ 12% (1/(1+12/100)^t)	0.89	0.80	0.71	0.64	0.57
F	Present Value of Outflows (D*E)	1,16,757	27,321	24,393	12,710	11,349
	Inflows
G	Annual Savings due to increased efficiencies	76,807	76,807	76,807	76,807	76,807
H	Salvage Value					6,897
I	Total Inflows (G+H)	76,807	76,807	76,807	76,807	83,704
J	Present value Factor @ 12% (1/(1+12/100)^t)	0.89	0.80	0.71	0.64	0.57
K	Present value of Inflows (I*J)	68,578	61,230	54,670	48,812	47,496
L	Total Present value of Outflows	1,92,530
M	Total Present value of Inflows	2,80,786
	NPV (M-L)	88,256

Question 8

	Initial Cost	1,35,893.00
	Annual Maintenance Cost	15,800.00
	Salvage Value at end of 30 Years	28,768.00
	Interest rate % (R )	8
	Future Worth
	Time period (t)	30
	Initial Cost	1,35,893
A	Future value after 30 Years	13,67,445		Formula = 135893*(1+(1+R/100)^t)
B	Future value of Annual Maintenance Cost	1,58,990		Formula = 15800*(1+(1+R/100)^t)
C	Salvage Value at end of 30 Years	28768
	Value of the Project after 30 years (A-B+C)	12,37,223