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A small company has a choice between 2 projects. Because the company is highly specialized, it picks its projects carefully. Project A has annual expenses of $591 and annual profit of $22121. The project will last for 6 years. Project B has annual expenses of $4470 and annual profit of $28457. It will require the purchase of specialized equipment of $24820 at the beginning of the project. The equipment will have a $2380 salvage value when the project ends. The contract for the project is for 8 years. Assume an interest rate of 3%, compounded annually. Comparing the two projects, what is the equivalent uniform annual worth (EUAW) of Project A?
@retirEEd
i am working on it.
Wow. Do they just want project A for now?
Is the annual expense for Project A really $591? It seems way too low for that much profit.
I am going to need to watch the lecture vids before proceeding. I will be back in awhile.
ok
I have a better idea now
EUAW = 21,530
What did you get scott?
I got it right. Going to the next question
Trixie is buying a new car at a cost of $19005. She estimates the car will sell for $3127 at the end of year 7. Insurance, fees, maintenance, and gas will be $975 in year 1 and will increase every year by 7%. If she keeps the car for 7 years, what is the equivalent uniform annual worth (EUAW) of owning this car using an interest rate of 2%, compounded annually?
Sorry I was on the phone. Let me restart, since I had to get my book out.
ok
Trixie is buying a new car at a cost of $17977. She estimates the car will sell for $4462 at the end of year 7. Insurance, fees, maintenance, and gas will be $1147 in year 1 and will increase every year by 6%. If she keeps the car for 7 years, what is the equivalent uniform annual worth (EUAW) of owning this car using an interest rate of 5%, compounded annually?
#s changed. I already attempted the problem and got it wrong...
scott what did you get? I have an answer
Sorry I don't know how to do these. I am getting really confused.
I'll try the second problem.
wait... that is the same problem
all these problems are the same or similar
Charleston Industries is trying to decide if expanding to a new product line would be worth the investment. The new line is estimated to have annual expenses of $31749 and annual profits of $91016. Charleston will be required to pay an annual product licensing fee beginning at the end of year 4 through the end of year 10. The fee will be $15144 and it will decrease by $683 each year. Charleston estimates the product line will be viable for 18 years. Charleston uses a MARR of 4% to evaluate investments. What is the equivalent uniform annual worth (EUAW) of this product line?
I am working on this one now and waiting to hear back from my instructor for the other question
On the car question.... I didn't know how to handle the increasing cost of operating the car for an annual basis, so I calculated the entire 7 years of cost and averaged it. With that said the cost to own the car purchase - resale came to $-2335.66 and the average cost of operating the car was $-1375.39 for a total EUAW of $-3711.05 What did you get?
4,709.42
I used the arithmetic gradient to find present worth at time 0, right? So then after finding the present worth at time 0 I used EUAW to find the annual amount.
I dont have (A/P,i,n)
In the cost of operating the car for 7 years, I didn't include any interest. That arithmetic gradient stuff freaks my brain out.
oh lol okay.
What did you get for JUST the cost of the car ?
the first question?
yeah the EUAW for the $17977 car.
look up geometric gradient series and arithmetic gradient series
Trixie is buying a new car at a cost of $17977. She estimates the car will sell for $4462 at the end of year 7. Insurance, fees, maintenance, and gas will be $1147 in year 1 and will increase every year by 6%. If she keeps the car for 7 years, what is the equivalent uniform annual worth (EUAW) of owning this car using an interest rate of 5%, compounded annually?
my bad I got 4,709.42
I'm out. I need to study some more, after dinner here in Texas.
i got to get to class. post whatever you need to say i will check back later
I won't be on until after 630 pacific time
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