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You value a car to be $30,000. If you plan to make continuous payments over 5 years and at an interest rate of r = 10%.

1) How much should you pay per year so that the present value of your total payments in 30; 000?

The formula that i think i should use is

$$PV = PMT\frac{1-(1+i)^{-n}}{i}$$

So solving for $PMT$ I got 7913.48. Did i do that correctly?

2)What if instead you decided to let your payments increase with time and pay at a rate of $6000 + t*1000$ per year, where t is measured in years. How long would it take you to pay off the car ? (Note the equation you get might be difficult to solve, so you can use a graphing calculator to estimate.)

I'm not too sure how to approach this. Do i use the same equation and solve for t? I assume the $t$ is the same as the $n$ I used in my $pv$ formula, is this correct?

Thanks

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  • $\begingroup$ I apologize for the poor formatting of the equation $\endgroup$ – user71317 Apr 16 '13 at 3:22
  • $\begingroup$ is this the wrong place to ask this question? $\endgroup$ – user71317 Apr 16 '13 at 4:00
  • $\begingroup$ would it still be a discrete time because the payments are over a 5 year time period? $\endgroup$ – user71317 Apr 16 '13 at 4:19
  • $\begingroup$ There are different ways to pay an annuity, which is simply a regular payment schedule. It could be daily, monthly, yearly, or continuous which involves $e$. Since in your question you have said continous payments, we must assume you mean not monthly, not yearly, not daily, not weekly, etc. These numbers will change the value of your present value. $\endgroup$ – Eleven-Eleven Apr 16 '13 at 4:30
  • $\begingroup$ Never mind...I saw you need yearly...I will change it. What is the interest rate? 1%? $\endgroup$ – Eleven-Eleven Apr 16 '13 at 4:35
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For a yearly payment, we need $n=5, i=.1$, and a present value, which is 30,000. $$30,000=PMT\frac{1-\frac{1}{1.1^5}}{.1}.$$ Dividing by the right hand side give us a yearly payment of 7,913.92

For an increasing annuity (you might not have the machinery for solving this with simple calculus), you would use the formula for an increasing annuity which is $$P\frac{1-\nu^n}{i}+Q\frac{\frac{1-\nu^n}{i}-n\nu^n}{i}$$ .where P is your 6,000 dollars and Q is your 1,000 dollars increasing in t, and $\nu$ is defined as $\frac{1}{1+i}$.

If you don't have that machinery, you can simply look at your payment stream. At time $1$, we have $6,000+1*1,000 = 7,000$, at time $2$, we have $6,000+2*1,000=8,000$. $$year (1): 7,000$$ $$year (2): 7,000+1,000$$ $$year (3): 7,000+1,000+1,000$$ $$year (4): 7,000+1,000+1,000+1,000$$ $$year (5): 7,000+1,000+1,000+1,000+1,000$$ Writing it this way you can see you have 5 different equal sets of payments. You have a 5 year annuity with 7,000, a 4,3,2, and 1 year annuity with 1,000. You just have to discout the 4 year annuity once, the 3 year annuity twice, etc. Hence, $$PV=7,000\frac{1-\frac{1}{1.1^5}}{.1}+1,000\frac{1-\frac{1}{1.1^4}}{.1}\left(\frac{1}{1.1}\right)+1,000\frac{1-\frac{1}{1.1^3}}{.1}\left(\frac{1}{1.1^2}\right)+1,000\frac{1-\frac{1}{1.1^2}}{.1}\left(\frac{1}{1.1^3}\right)+1,000\frac{1-\frac{1}{1.1}}{.1}\left(\frac{1}{1.1^4}\right)$$

I'll leave the calculation for you.

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I do not know what $r$ reprents, so will assume that it is the effective annual rate.

If that is the case, then a dollar grows in a year to $e^k$ dollars, where $e^k=1.10$. Thus $k=\ln(1.10)$. But we will keep on using $k$.

Let $c$ be the amount paid per year. So we are paying at a yearly rate of $c$. The present value of these payments is $$\int_0^5 c e^{-kt}\,dt.$$

Integrate. The result should be $30000$. Thus we obtain the equation $$c\frac{1}{k}\left(1-e^{-5k}\right)=30000.$$ In the above equation, we know everything except $c$, so can solve for $c$.

For the second problem, the setup is roughly similar, except that we do not know the number $y$ of years, but we do know the rate of payment. We end up with the equation $$\int_0^y (6000+1000t)e^{-kt}\,dt=30000.$$ We can do the integration using integration by parts. However, we end up with an equation in $y$ that we cannot solve analytically. However, the equation can be solved by using a numerical method. It can also be solved to adequate accuracy with a graphing calculator.

Remark: Conceivably the $r$ could refer to the force of interest. In that case we would have $k=0.10$. I chose to interpret it as the effective annual rate, since that seemed more plausible.

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