Newton's Method (as applied to Auto Financing)

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SUMMARY

The discussion focuses on applying Newton's Method to determine the monthly interest rate charged by a car dealer offering a vehicle for $18,000 or $375 monthly for five years. The relevant formula for present value is A = [R(1 - (1 + i))^-60] / i, where A is the present value, R is the monthly payment, and i is the interest rate. The correct approach involves setting up the equation 18000 = 375 * [(1+r)/r - 1/(r(1+r)^{59})] to solve for the monthly interest rate r, utilizing Newton's Method for iterative approximation.

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  • Understanding of present value calculations in finance
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  • Basic algebraic manipulation skills
  • Knowledge of exponential functions and their properties
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  • Learn how to derive and manipulate present value formulas
  • Explore the concept of discounting cash flows in finance
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Finance students, mathematicians, and anyone interested in applying numerical methods to solve real-world financial problems, particularly in auto financing scenarios.

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Homework Statement


A car dealer sells a new car for $18,000. He also offers to sell the same car for monthly payments of $375.00 for five years. What monthly rate is this dealer charging?

Homework Equations


A = [R(1 - (1 + i))^-60] / i

where A = the present value, R = the monthly payment, i = the interest rate, n = the number of monthly payments

Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 + 1 = 0.

The Attempt at a Solution


I thought that, "Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 -1 = 0.", seemed to suggest that they wanted me to solve for i (or x), but in hindsight I don't think that's what I need to do. I don't know where to start now. [edit] Or I suppose rather that I'm just supposed to set it equal to 0 somehow. In which case I guess I don't know what they're doing to (1 + x)^60.[edit]
 
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communitycoll said:

Homework Statement


A car dealer sells a new car for $18,000. He also offers to sell the same car for monthly payments of $375.00 for five years. What monthly rate is this dealer charging?


Homework Equations


A = [R(1 - (1 + i))^-60] / i

where A = the present value, R = the monthly payment, i = the interest rate, n = the number of monthly payments

Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 + 1 = 0.


The Attempt at a Solution


I thought that, "Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 -1 = 0.", seemed to suggest that they wanted me to solve for i (or x), but in hindsight I don't think that's what I need to do. I don't know where to start now. [edit] Or I suppose rather that I'm just supposed to set it equal to 0 somehow. In which case I guess I don't know what they're doing to (1 + x)^60.[edit]

You are using an incorrect formula. If you have 60 monthly payments of R, the payment in month 2 is discounted by 1/(1+r), that in month 3 is discounted by 1/(1+r)^2, ... and the final payment in month 60 is discounted by 1/(1+r)^59. Here, r is the monthly interest rate. So, equating the net present value to the car's price P we have
[tex]P = R \left( 1 + \frac{1}{1+r} + \frac{1}{(1+r)^2} + \cdots + \frac{1}{(1+r)^{59}} \right)\\<br /> = R \left( \frac{1+r}{r} - \frac{1}{r(1+r)^{59}} \right).[/tex]
So, we need to solve
[tex]18000 = 375 \left( \frac{1+r}{r} - \frac{1}{r(1+r)^{59}} \right).[/tex]

RGV
 

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