In: Finance
Fresh Foods is considering the purchase of a new packaging system. The system costs $55022. The company plans to borrow three-quarters (3/4) of the purchase price from a bank at 10% per year compounded annually. The loan will be repaid using equal, annualpayments over a 7-year period. The payments will be made at the end of each year for the life of the loan, with the first payment occurring at the end of year 1. The system is expected to last 15 years and have a salvage value of $12506 at the end of its life.
Over the 15-year period, the company expects to pay $1045 per year for maintenance. In addition, the system will require an overhaul at the end of year 9 which will cost $13427. The system will save $3967 per year because of efficiencies. The company uses a MARR of 5% to evaluate investments. What is the equivalent uniform annual worth (EUAW) of this system?
Enter your answer as follows: 1234
Round your answer. Do not use a dollar sign ("$"), any commas (",") or a decimal point (".").
Hints: The loan interest is at a different interest rate than our MARR. How will this impact the problem?
The remainder of the purchase price (the amount we do not take out in the form of a loan) would be considered our initial cost, at year 0
Incorrect Answer: -2558
HINT from when I got it incorrect
Draw a cash flow diagram. This is really important for this problem.
First, calculate the loan payment. Remember that the interest for the loan is not the same as the MARR. Second, find the net present worth of the loan payments, using the MARR.
For the remainder of the problem, separate out those amounts already in annual form and those amounts that will require further calculation to find the annual worth.
Amount of bank loan=0.75*55022 | $41,267 | |||||||||||||
Number of years to pay | 7 | |||||||||||||
Annual interest | 10% | |||||||||||||
Annual Paymenttothe bank | $8,476 | (Using PMT function of excel with Rate=10%, Nper=7, PV=-41267) | ||||||||||||
Present Value (PV) of Cash Flow: | ||||||||||||||
(Cash Flow)/((1+i)^N) | ||||||||||||||
i=Discount Rate=MARR=5%=0.05 | ||||||||||||||
N=Year of Cash Flow | ||||||||||||||
Initial Cash flow for purchase of system | ($13,756) | (55022-41267) | ||||||||||||
Yearwise Cash Flow: | ||||||||||||||
N | A | B | C | D | E | F | G=A+B+C+D+E+F | PV=G/(1.05^N) | ||||||
Year | Net Cash Flow | Year | Initial cash flow | Payment to bank | Maintenance cost | Overhaul cost | Savings | Salvage Value | Net CashFlow | Present Value of Net Cash Flow | ||||
0 | (13,756) | 0 | (13,756) | (13,756) | (13,756) | |||||||||
1 | (5,554) | 1 | (8,476) | (1,045) | 3,967 | (5,554) | (5,290) | |||||||
2 | (5,554) | 2 | (8,476) | (1,045) | 3,967 | (5,554) | (5,038) | |||||||
3 | (5,554) | 3 | (8,476) | (1,045) | 3,967 | (5,554) | (4,798) | |||||||
4 | (5,554) | 4 | (8,476) | (1,045) | 3,967 | (5,554) | (4,570) | |||||||
5 | (5,554) | 5 | (8,476) | (1,045) | 3,967 | (5,554) | (4,352) | |||||||
6 | (5,554) | 6 | (8,476) | (1,045) | 3,967 | (5,554) | (4,145) | |||||||
7 | (5,554) | 7 | (8,476) | (1,045) | 3,967 | (5,554) | (3,947) | |||||||
8 | 2,922 | 8 | (1,045) | 3,967 | 2,922 | 1,978 | ||||||||
9 | (10,505) | 9 | (1,045) | (13,427) | 3,967 | (10,505) | (6,772) | |||||||
10 | 2,922 | 10 | (1,045) | 3,967 | 2,922 | 1,794 | ||||||||
11 | 2,922 | 11 | (1,045) | 3,967 | 2,922 | 1,708 | ||||||||
12 | 2,922 | 12 | (1,045) | 3,967 | 2,922 | 1,627 | ||||||||
13 | 2,922 | 13 | (1,045) | 3,967 | 2,922 | 1,550 | ||||||||
14 | 2,922 | 14 | (1,045) | 3,967 | 2,922 | 1,476 | ||||||||
15 | 15,428 | 15 | (1,045) | 3,967 | 12,506 | 15,428 | 7,421 | |||||||
SUM | $ (35,114) | |||||||||||||
Net Present Value | (35,114) | |||||||||||||
Equivalent Uniform Annual Worth(EUAW) | ($3,383) | (Using PMT function with Rate=5%, Nper=15, PV=35114) | ||||||||||||