Question

2. Suppose a car dealer promotes two options for the purchase of a new $20 000 car. The first option is for the customer to pay up front and receive a $1000 rebate. The second option is “0%-interest financing" where the customer makes 20 monthly payments of $1000 beginning in one month's time. Because of option 1, the effective price of the car is really $19,000, so the dealer really is charging some interest rate i for option 2. We can calculate this value using the formula for the present value of an annuity: Use a while() loop to iteratively calculate the interest rate i which satisfies the fixed- point equation (1 - (1 + iold)-20) inew 19 Use a starting guess of io = 0.006. Stop the calculation when two successive values of the interest rate are less than 0.000001 apart.

Please use RStudio to solve this problem, thank you so much in advance.

Answer 1

Solution:-

Given that

diff = 1 tolerance = 1e-6 i_old = 0.006 i_new = 0. while diff > tolerance: i_new = (1-pow( 1+i_old, -20)) / 19 diff = abs(i_new - i_old) i_old = i_new # print(i_new, i_old, ": ", diff) print("Final interest: ", i_new)

We need to iteratively calculate the new interest i_new from i_old and keep track of their difference as well as updating i_old to the newer calculated value. The loop breaks when the difference is lesser than the given threshold/tolerance.

Code Snippet (R program)

diff = 1
tolerance = 1e-6

i_old = 0.006
i_new = 0.

while (diff >= tolerance) {
i_new = (1 - (1+i_old) ** (-2)) / 19
  
diff = abs(i_new - i_old)
i_old = i_new

print(diff)
}

print("Final Interest")
print(i_new)

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