For the following CFD, at what actual time period is the resulting P when finding P = 2000(P/A,8%,3)? t=1 t=0 t=3 t=5 t=2
I said answer: (e)
Because we need the present amount before
a year before
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Yes, I know.
Then why did you think the answer was e) t = 2 ????
Let me look at something in the text, apparently not understand the relationship between p and A
t= 3
This is a uniform series compound interest right?
Did you calculate the value of P at year 6 ??? I got $6492.80 his is a uniform series compound interest right? I think that is correct/
Why do I need to calculate the P at year 6?
Just for practice.
Oh okay
I can do that
I just want to see if I did it right.
I will even attach an image with my solution.
I am pretty sure I am correct because I even compared using the values we have in the back of the textbook. The present worth factor.
You mind checking my work for the next problem? I have a solution to it.
You did it correctly for a 6 year time period , but wrong for the problem asked. Based on the CFD, the first 3 years P is zero so the ACTUAL time period is 6 - 3 or three so... P when finding P = 2000(P/A,8%,3) is the equation to calculate the P at year 6.
Oh tricky. I see why you asked that.
My mistake even your calculation of the 6 year time period doesn't appear to be correct.... I got $14,671.85 much more than $9245.75 Maybe I not doing it right.
Wait why would it not be 4 years instead? First 3 years P = 0 but from 5 to 6 P = 0 as well
Unless the first period from 0 to 1 doesn't count
The Register Company just purchased a new piece of equipment. At the end of year 4, the company will begin paying maintenance costs of $23473 per year. The company will pay the maintenance costs for 8 years, at which point the equipment will be sold. How much should Register set aside today, in an account paying an annually compounded interest rate of 1%, to pay for the future maintenance?
No, you still have principle and interest from year 5 to 6. The P and interest start a year 3, 4 5 and 6 count for 3 years.
This new question might take a while.
Why does the P start at year 3 instead of 2?
Here lets do another example of a problem like before.
An energy-effcient machine costs $5000 and has a life of 5 years. If the interest rate is 8%, how much must be saved every year to recover the cost of the capital invested in it?
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