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

Amortization table for a loan of 1000 units at interest rate of 10% per period with payments of 500 units
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$\mathrm{amortization}\left(1000.00\,500.00\,0.10\right)$

$\left[\begin{array}{ccccc}{n}& {\mathrm{Payment}}& {\mathrm{Interest}}& {\mathrm{Principal}}& {\mathrm{Balance}}\\ {0}& {0}& {0}& {}{1000.00}& {1000.00}\\ {1}& {500.00}& {100.0000}& {400.0000}& {600.0000}\\ {2}& {500.00}& {60.000000}& {440.000000}& {160.000000}\\ {3}& {176.0000000}& {16.00000000}& {160.0000000}& {0.}\end{array}\right]$
 (1) 
From this you can see that there will be 3 payments, the last one being of 176 units.
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$\mathrm{amortization}\left(1000.00\,500.00\,0.10\,\mathrm{output}\=\mathrm{cost}\right)$

The cost of the loan is 176 units.
You can make payments to be of 500 units + the interest for that period:
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$\mathrm{amortization}\left(1000.00\,\left(i\,\mathrm{interest}\right)\→500.00\+\mathrm{interest}\,0.10\,\mathrm{output}\=\mathrm{list}\right)$

$\left[\left[{0}{\,}{0}{\,}{0}{\,}{}{1000.00}{\,}{1000.00}\right]{\,}\left[{1}{\,}{600.0000}{\,}{100.0000}{\,}{500.0000}{\,}{500.0000}\right]{\,}\left[{2}{\,}{550.000000}{\,}{50.000000}{\,}{500.000000}{\,}{0.}\right]\right]{\,}{150.000000}$
 (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:
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$\mathrm{compute\_payment}:=\left(i\,\mathrm{interest}\right)\→150.00\+10.00\mathrm{trunc}\left(\frac{i1}{4}\right)\:$

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

$\left[\begin{array}{ccccc}{n}& {\mathrm{Payment}}& {\mathrm{Interest}}& {\mathrm{Principal}}& {\mathrm{Balance}}\\ {0}& {0}& {0}& {}{1000.00}& {1000.00}\\ {1}& {150.00}& {31.37720250}& {118.6227975}& {881.3772025}\\ {2}& {150.00}& {27.65515096}& {122.3448490}& {759.0323535}\\ {3}& {150.00}& {23.81631186}& {126.1836881}& {632.8486654}\\ {4}& {150.00}& {19.85702073}& {130.1429793}& {502.7056861}\\ {5}& {160.00}& {15.77349811}& {144.2265019}& {358.4791842}\\ {6}& {160.00}& {11.24807395}& {148.7519260}& {209.7272582}\\ {7}& {160.00}& {6.580654650}& {153.4193454}& {56.3079128}\\ {8}& {58.0746976}& {1.766784782}& {56.3079128}& {0.}\end{array}\right]$
 (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:
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$\mathrm{amortization}\left(1000.00\,\mathrm{compute\_payment}\,\frac{\mathrm{effectiverate}\left(0.12\,4\right)}{4}\,\mathrm{nperiods}\=4\right)$

$\left[\begin{array}{ccccc}{n}& {\mathrm{Payment}}& {\mathrm{Interest}}& {\mathrm{Principal}}& {\mathrm{Balance}}\\ {0}& {0}& {0}& {}{1000.00}& {1000.00}\\ {1}& {150.00}& {31.37720250}& {118.6227975}& {881.3772025}\\ {2}& {150.00}& {27.65515096}& {122.3448490}& {759.0323535}\\ {3}& {150.00}& {23.81631186}& {126.1836881}& {632.8486654}\\ {4}& {150.00}& {19.85702073}& {130.1429793}& {502.7056861}\end{array}\right]$
 (5) 
It is also possible to display the amortization table as an embedded datatable with the output = embed option:
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$\mathrm{amortization}\left(1000.00\,\mathrm{compute\_payment}\,\frac{\mathrm{effectiverate}\left(0.12\,4\right)}{4}\,\mathrm{nperiods}\=4\,\mathrm{output}\=\mathrm{embed}\right)$

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