>

$\mathrm{with}\left(\mathrm{Finance}\right)\:$

Amortization table for a loan of 1000 units at interest rate of 10% per period with payments of 500 units
>

$\mathrm{amortization}\left(1000.00\,500.00\,0.10\right)$

$\left[\begin{array}{ccccc}n& \mathrm{Payment}& \mathrm{Interest}& \mathrm{Principal}& \mathrm{Balance}\\ 0& 0& 0& 1000.00& 1000.00\\ 1& 500.00& 100.0000& 400.0000& 600.0000\\ 2& 500.00& 60.000000& 440.000000& 160.000000\\ 3& 176.0000000& 16.00000000& 160.0000000& 0.\end{array}\right]$
 (1) 
From this you can see that there will be 3 payments, the last one being of 176 units.
>

$\mathrm{amortization}\left(1000.00\,500.00\,0.10\,\mathrm{output}\=\mathrm{cost}\right)$

The cost of the loan is 176 units.
You can make payments to be of 500 units + the interest for that period:
>

$\mathrm{amortization}\left(1000.00\,\left(i\,\mathrm{interest}\right)\→500.00\+\mathrm{interest}\,0.10\,\mathrm{output}\=\mathrm{list}\right)$

$\left[\left[{0}{\,}{0}{\,}{0}{\,}{\mathrm{1000.00}}{\,}{1000.00}\right]{\,}\left[{1}{\,}{600.0000}{\,}{100.0000}{\,}{500.0000}{\,}{500.0000}\right]{\,}\left[{2}{\,}{550.000000}{\,}{50.000000}{\,}{500.000000}{\,}{0.}\right]\right]{,}{150.000000}$
 (3) 
There are now 2 payments, one of 600 units and one of 550 units. The cost of the loan is 150 units. Now, if you make quarterly payments of 150 units on a loan of 1000 units at a stated rate of 12%, the payments are increased yearly by 10 units. The amortization table is computed as follows:
>

$\mathrm{compute\_payment}\u2254\left(i\,\mathrm{interest}\right)\→150.00\+10.00\mathrm{trunc}\left(\frac{i1}{4}\right)\:$

>

$\mathrm{amortization}\left(1000.00\,\mathrm{compute\_payment}\,\frac{\mathrm{effectiverate}\left(0.12\,4\right)}{4}\right)$

$\left[\begin{array}{ccccc}n& \mathrm{Payment}& \mathrm{Interest}& \mathrm{Principal}& \mathrm{Balance}\\ 0& 0& 0& 1000.00& 1000.00\\ 1& 150.00& 31.37720250& 118.6227975& 881.3772025\\ 2& 150.00& 27.65515096& 122.3448490& 759.0323535\\ 3& 150.00& 23.81631186& 126.1836881& 632.8486654\\ 4& 150.00& 19.85702073& 130.1429793& 502.7056861\\ 5& 160.00& 15.77349811& 144.2265019& 358.4791842\\ 6& 160.00& 11.24807395& 148.7519260& 209.7272582\\ 7& 160.00& 6.580654650& 153.4193454& 56.3079128\\ 8& 58.0746976& 1.766784782& 56.3079128& 0.\end{array}\right]$
 (4) 
There were 8 payments altogether with a loan cost of 138 units. For obtaining just the first year, you can make use of the nperiods argument:
>

$\mathrm{amortization}\left(1000.00\,\mathrm{compute\_payment}\,\frac{\mathrm{effectiverate}\left(0.12\,4\right)}{4}\,\mathrm{nperiods}\=4\right)$

$\left[\begin{array}{ccccc}n& \mathrm{Payment}& \mathrm{Interest}& \mathrm{Principal}& \mathrm{Balance}\\ 0& 0& 0& 1000.00& 1000.00\\ 1& 150.00& 31.37720250& 118.6227975& 881.3772025\\ 2& 150.00& 27.65515096& 122.3448490& 759.0323535\\ 3& 150.00& 23.81631186& 126.1836881& 632.8486654\\ 4& 150.00& 19.85702073& 130.1429793& 502.7056861\end{array}\right]$
 (5) 
It is also possible to display the amortization table as an embedded datatable with the output = embed option:
>

$\mathrm{amortization}\left(1000.00\,\mathrm{compute\_payment}\,\frac{\mathrm{effectiverate}\left(0.12\,4\right)}{4}\,\mathrm{nperiods}\=4\,\mathrm{output}\=\mathrm{embed}\right)$

${''amortizationtable0''}$
 (6) 