A load of \$1000 is to be repaid by annual payments of \$100 to commence at the end of the 5th year and to continue thereafter for as long as possible. Find the time and amount of the final payment, if the final payment is larger than the regular payments? Assume $i=4.5%$
Is the following solution precisely correct?
The value of the \$1000 loan after 5 years is $1000 * (1+.045)^5 = 1246.18$
The number of payments at 100 a payment = $-ln(\frac{1 - .045 * 1246.18}{100})$ And then we divide this by:
$ln(1 + .045)= \frac{-ln(0.439219)}{ln(1.045)} = \frac{.822757}{.044017} = 18.7$
Since we want the last payment to be larger than the regular payment of \$100, we will finish in 18 payments.
how much do we owe after 18 payments of \$100?
$1246.18 * (1 + .045) ^18 - 100 * \frac{[(1 + .045)^18 - 1]}{.045} = 1246.18 * 2.208 - 100 * \frac{[1.208]}{.045} = 2751.57 - 2684.44 = 67.13$
So we add that to the normal payment of \$100 to get a final payment of \$167.13
