In: Finance
OPTION 1: The initial purchase price of the machine is $30,000. The salvage value at the end of the useful life will be $4,000. Maintenance costs are $2,000 for the first year and are estimated to increase by $200 per year. OPTION 2: The machine is leased for an initial payment of $2,000 plus annual payments of $3,500. There is no salvage value. Annual maintenance cost is $10,000. Put down the present value of each of the items specified for each machinery option in the table below. Assume an interest rate of 6% and a useful lifetime of 8 years. Show your calculations in the spaces below the table.
a.) Please show the steps.
Present Value (PV) of Item |
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Option 1 |
Option 2 |
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PV of Purchase/Lease |
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PV of Salvage Value |
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PV of Maintenance |
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Total PV |
b.) What is the annual equivalent cost of capital (capital recovery), over the 8-year lifetime, for the machine with the lowest present value calculated in the above table?
c.) What is the amount of money that you should invest now if you want to receive payments of $1,000 at the end of each year for ten years with the receipt of the first payment starting three years from now? Assume interest rate is 5% compounded annually.
d.) Provide explanation on your calculation in part c(i). Include in your explanation the specific type of cash flow given and the reason of why you have to derive the present or future equivalent value at various time points.
Present Value(PV) of Cash Flow: | ||||||||||||
(Cash Flow)/((1+i)^N) | ||||||||||||
i=interest rate=6%=0.06 , N=Year of cash flow | ||||||||||||
N | Year | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |||
A | Maintenance Cost Option1 | $2,000 | $2,200 | $2,400 | $2,600 | $2,800 | $3,000 | $3,200 | $3,400 | SUM | ||
B=A/(1.06^N) | Present Value | $1,887 | $1,958 | $2,015 | $2,059 | $2,092 | $2,115 | $2,128 | $2,133 | $16,388 | ||
(OPTION 1) | Present Value of Maintenance Cost | $16,388 | ||||||||||
PV of Salvage Value=4000/(1.06^8) | $2,510 | |||||||||||
0 | ||||||||||||
CASH FLOW OF OPTION-2 (LEASING) | ||||||||||||
N | Year | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ||
C | Annual Leasing Cost | $2,000 | $3,500 | $3,500 | $3,500 | $3,500 | $3,500 | $3,500 | $3,500 | $3,500 | SUM | |
D=C/(1.06^N) | Present Value | $2,000 | $3,302 | $3,115 | $2,939 | $2,772 | $2,615 | $2,467 | $2,328 | $2,196 | $23,734 | |
Present Value of Lease | $23,734 | |||||||||||
N | Year | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |||
E | Maintenance Cost Option2(Lease) | $10,000 | $10,000 | $10,000 | $10,000 | $10,000 | $10,000 | $10,000 | $10,000 | SUM | ||
F=E/(1.06^N) | Present Value | $9,434 | $8,900 | $8,396 | $7,921 | $7,473 | $7,050 | $6,651 | $6,274 | $62,098 | ||
(OPTION 2) | Present Value of Maintenance Cost | $62,098 | ||||||||||
Present Value (PV) of Item | ||||||||||||
Option 1(Purchase) | Option 2(Lease) | |||||||||||
PV of Purchase/Lease | $30,000 | $23,734 | ||||||||||
PV of Salvage Value | $2,510 | $0 | ||||||||||
PV of Maintenance | $16,388 | $62,098 | ||||||||||
Total PV | $48,898 | $85,832 | ||||||||||
b | Annual Equivalent Cost | $7,874.26 | (Using PMT function of excel with Rate=6%, Nper=8,Pv=-48898) | |||||||||
c | ||||||||||||
Rate | Interest Rate | 5% | ||||||||||
Nper | Number of years | 10 | ||||||||||
Pmt | Annual Payment | $1,000 | ||||||||||
PV | Amount of money to invest | $7,721.73 | (Using PV function of excel with Rate=5%, Nper=10,Pmt=-1000) | |||||||||