Suppose I owe (no interest) 300 and 600 dollars with due dates in 12 and 24 months, respectively. The trivial idea is to consider both debts independently and put aside monthly $\frac{300}{12}=25$ dollars for the first debt and $\frac{600}{24}=25$ for the second one. Therefore, I have to put aside \$50 each of the first 12 months and \$25 in the second 12 months which is rather imbalanced.
Intuitively it is clear that the smarter idea is to put aside $\frac{300+600}{24}=37.5$ monthly for the whole 24 months interval: I have $37.5 \cdot 12 = 450$ dollars in 12 months which is enough to pay the first debt ($300$) and I'm left with $450-300=150$ dollar for the second debt. For the second 12 months I get another $37.5 \cdot 12 = 450$ bucks. $150 + 450 = 600$ and I'm good to go.
It is also clear (or not?) that the procedure above makes sense if the "later" debt is larger than the "earlier" one: if I would have to pay \$600 first (in 12 months) there were no chance I put aside less than 50 dollars per month…
Now a more complex example. Suppose I have several debts (just random numbers):
| debt (\$) | due in (months) |
|---|---|
| 600 | 3 |
| 160 | 4 |
| 1200 | 6 |
| 1400 | 7 |
| 900 | 9 |
| 300 | 10 |
| 1800 | 12 |
For each month (from 1 to 12) I want to know the smallest possible amount of money I have to put aside taking all the listed debts into account.
Is it an optimization problem?
If so, to which of the well known ones can it be boiled down?
Best Answer
Let $n$ be the longest term (in months) of any of the loans.
Let $D_1,D_2,...,D_n$ be the amount due in month $n$. (Some of these values may be $0$.)
Let $C_k$ be the total cumulative amount due of all loans through month $k$.
Let $A_k=\frac{C_k}{k}$. This represents the average amount due over the first $k$ months.
In your second example, we have the $D$s being $0,0,600,160,0, 1200,1400,0,900,300,0,1800$. (All amounts dollars.)
So the $C$s (cumulative totals) are $0,0,600,760,760,1960,3360,3360,4260,4560,4560,6360$.
The $A$s are $0,\;0,\;200,\;190,\;152,\;326.67,\;480,\;420,\;473.33,\;456,\;414.55,\;530$
In this example, because the largest average is last, that amount ($530$) represents the amount you should deposit each month.
However, more generally, find the maximum $A_k$ and for the first $k$ months, deposit that amount. At the end of those $k$ months, you will have exactly paid off your loans through those $k$ months (with no leftover money saved for payments). Then recalibrate (in the same way) for the remaining months. Your subsequent deposit amounts will be lower. Depending on how many maximum averages you run into, you may have to recalibrate several more times.