What is the fortnightly payment for this mortgage?

This problem’s question: {\color{#D61F06}\text{This problem's question:}} What is the fortnightly payment for this mortgage?

This problem’s answer is in integer cents. {\color{#EC7300}\text{This problem's answer is in integer cents.}}

For the benefit of those whom do not recognize the word fortnight , it means two weeks or fourteen days.

The problem is worked in dollars and cents

The loan amount is $ 300 000 300\,000 .

The stated annual percentage rate (APR) of the loan is 3 7 8 % 3\frac78\% . The loan is compounded every fortnight. You will need to compute the effective interest rate per fortnight.

The loan length is 338 fortnights, which is slightly less than 13 years.

The payment has been ceiling-ed to next higher cent so that the final payment will be reduced if necessary.

Amounts for escrow for insurance or taxes are not included in the needed payment amount.

Afterwards, notice what happens when Round or Floor is used.


The answer is 113468.

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1 solution

paymentInterval = Quantity [ 1 , Fortnight ] \text{paymentInterval}=\text{Quantity}[1,\text{Fortnight}]

oneYear = Quantity [ 1 , Year ] \text{oneYear}=\text{Quantity}[1,\text{Year}]

intervalYearFraction = N [ paymentInterval oneYear ] 0.0383561643835616 \text{intervalYearFraction}=N\left[\frac{\text{paymentInterval}}{\text{oneYear}}\right] \Rightarrow 0.0383561643835616

statedAPR = 3 7 8 % \text{statedAPR}=3\frac{7}{8}\%

EffectiveInterest [ rate , fraction ] ( rate × fraction + 1 ) 1 / fraction 1 \text{EffectiveInterest}[\text{rate},\text{fraction}] \Rightarrow (\text{rate}\times\text{fraction}+1)^{1/\text{fraction}}-1

effectiveAPR = EffectiveInterest [ statedAPR , intervalYearFraction ] 0.0394806686994558 \text{effectiveAPR}=\text{EffectiveInterest}[\text{statedAPR},\text{intervalYearFraction}] \Rightarrow 0.0394806686994558

interestPerInterval = effectiveAPR intervalYearFraction 0.00151432701860926 \text{interestPerInterval} = \text{effectiveAPR}\, \text{intervalYearFraction} \Rightarrow 0.00151432701860926

Solve this equation for the payment amount, which gives the payment amount per dollar of loan per payment interval:

payment payment ( interestPerInterval + 1 ) numberOfPayments interestPerInterval = 1 0.00378224586250132 \frac{\text{payment}-\text{payment} (\text{interestPerInterval}+1)^{-\text{numberOfPayments}}}{\text{interestPerInterval}}=1 \Rightarrow 0.00378224586250132

loan = $ 300000 \text{loan}=\$\, 300000

paymentAmount 113468 \text{paymentAmount} \Rightarrow 113468 in cents.

Here is the amortization table summarized by year:

Period Payment Interest Principal Balance 1-26 29501.68 11472.79 18028.89 281971.11 27-52 29501.68 10749.35 18752.33 263218.78 53-78 29501.68 9996.88 19504.80 243713.98 79-104 29501.68 9214.20 20287.48 223426.50 105-130 29501.68 8400.14 21101.54 202324.96 131-156 29501.68 7553.43 21948.25 180376.71 157-182 29501.68 6672.70 22828.98 157547.73 183-208 29501.68 5756.67 23745.01 133802.72 209-234 29501.68 4803.83 24697.85 109104.87 235-260 29501.68 3812.76 25688.92 83415.95 261-286 29501.68 2781.96 26719.72 56696.23 287-312 29501.68 1709.78 27791.90 28904.33 313-338 29498.90 594.57 28904.33 0.00 \begin{array}{||c|r|r|r|r||} \hline \hline \text{Period} & \text{Payment} & \text{Interest} & \text{Principal} & \text{Balance} \\ \hline \hline \text{1-26} & 29501.68 & 11472.79 & 18028.89 & 281971.11 \\ \text{27-52} & 29501.68 & 10749.35 & 18752.33 & 263218.78 \\ \text{53-78} & 29501.68 & 9996.88 & 19504.80 & 243713.98 \\ \text{79-104} & 29501.68 & 9214.20 & 20287.48 & 223426.50 \\ \text{105-130} & 29501.68 & 8400.14 & 21101.54 & 202324.96 \\ \text{131-156} & 29501.68 & 7553.43 & 21948.25 & 180376.71 \\ \text{157-182} & 29501.68 & 6672.70 & 22828.98 & 157547.73 \\ \text{183-208} & 29501.68 & 5756.67 & 23745.01 & 133802.72 \\ \text{209-234} & 29501.68 & 4803.83 & 24697.85 & 109104.87 \\ \text{235-260} & 29501.68 & 3812.76 & 25688.92 & 83415.95 \\ \text{261-286} & 29501.68 & 2781.96 & 26719.72 & 56696.23 \\ \text{287-312} & 29501.68 & 1709.78 & 27791.90 & 28904.33 \\ \text{313-338} & 29498.90 & 594.57 & 28904.33 & 0.00 \\ \hline \hline \end{array}

In the Wolfram Mathematica code below, the "$" characters represent "\" characters that the verbatim processor does not handle correctly.

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balance = 100 QuantityMagnitude[loan];
paymentAmount =$[LeftCeiling]balance paymentPerUnit$[RightCeiling];
amort = Table[
   interest = Round[balance interestPerInterval];
   principal = Round[paymentAmount - interest];
   balance = Round[balance - principal];
   {period, paymentAmount, interest, principal, balance},
   {period, numberOfPayments}
   ];

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