Suppose, you take out a \(T\) period, fixed rate (rate = \(i\)) mortgage for a house worth \(B\). We want to find out how much your fixed monthly payment \(D\)?
To find the answer to this problem consider two independent "thought" buckets.
In one bucket, you have your loan $B grow without payment.
In the other bucket, you regularly contribute $D every year (I am using a yearly interest rate - modifications for monthly payments are straightforward).
Left unattended, at the end of \(T\) years, the first bucket balloons to \(S^- = B(1+i)^T\), and the second bucket to \(S^+ = D [(1+i)^T - 1]/i\).
If we set things up just right (choose the correct D), the amount we owe (bucket 1), might be just about equal to the amount we have in bucket 2.
Mathematically, we can impose \(S^+ = S^-\), to obtain this "just right" monthly payment. Solving for \(D\), we get,\[D^* = B \frac{i}{1-(1+i)^{-T}}.\] Suppose \(B\) = 100,000, \(i\) = 0.05/12 (monthy interest rate), and \(T = 360\) (30 years), we get \(D^* = $536.82\).
If you used a higher D, then you would end up with a net excess, and vice versa if you used a smaller D.
To find the answer to this problem consider two independent "thought" buckets.
In one bucket, you have your loan $B grow without payment.
In the other bucket, you regularly contribute $D every year (I am using a yearly interest rate - modifications for monthly payments are straightforward).
Left unattended, at the end of \(T\) years, the first bucket balloons to \(S^- = B(1+i)^T\), and the second bucket to \(S^+ = D [(1+i)^T - 1]/i\).
If we set things up just right (choose the correct D), the amount we owe (bucket 1), might be just about equal to the amount we have in bucket 2.
Mathematically, we can impose \(S^+ = S^-\), to obtain this "just right" monthly payment. Solving for \(D\), we get,\[D^* = B \frac{i}{1-(1+i)^{-T}}.\] Suppose \(B\) = 100,000, \(i\) = 0.05/12 (monthy interest rate), and \(T = 360\) (30 years), we get \(D^* = $536.82\).
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| Bucket 1 |
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| Bucket 2 |
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| Net of Bucket 1 and 2 |
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