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Finance[amortization] - amortization table for a loan
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Calling Sequence
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amortization(amount, payments, rate, nperiods)
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Parameters
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amount
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amount of the loan
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payments
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(number or procedure) size of the payments
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rate
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interest rate
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nperiods
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maximum number of payments (default = infinity). Stops when balance reaches 0
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Description
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The result is a sequence of two elements: an amortization table and the cost of the loan.
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The amortization table consists of lists comprising of 5 elements
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1) The period number
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2) The amount of the payment
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3) The interest for the period
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4) The amount by which the principal is reduced (increased if negative)
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5) The new balance
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The cost of the loan is the sum of the third column.
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The payment can be a procedure. It will be called with two arguments: the period number and the interest to be paid. This can be used, for example to have increasing payments, or to pay down a fixed amount of principal at each period
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Refer to Finance[annuity] for more details on mortgages. Refer to Finance[effectiverate] for selecting the appropriate value for the interest rate to be used.
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The command with(Finance,amortization) allows the use of the abbreviated form of this command.
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Since amortization used to be part of the (now deprecated) finance package, for compatibility with older worksheets, this command can also be called using finance[amortization]. However, it is recommended that you use the superseding package name, Finance instead: Finance[amortization].
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Compatibility
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The Finance[amortization] command was introduced in Maple 15.
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Examples
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Amortization table for a loan of 1000 U at interest rate of 10% per period with payments of 500 U
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![[[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.]], 176.0000000](/support/helpjp/helpview.aspx?si=8934/file01704/math148.png)
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From this we see that there will be 3 payments, the last one being of 176 U. The cost of the loan is 176 U.
Now we make payments to be of 500 U + the interest for that period:
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![[[0, 0, 0, -1000.00, 1000.00], [1, 600.0000, 100.0000, 500.0000, 500.0000], [2, 550.000000, 50.000000, 500.000000, 0.]], 150.000000](/support/helpjp/helpview.aspx?si=8934/file01704/math159.png)
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There are now 2 payments, one of 600 U and one of 550 U. The cost of the loan is 150 U. Now we make quarterly payments of 150 U on a loan of 1000 U at a stated rate of 12%. The payments are increased yearly by 10 U. The amortization table is computed as follows:
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![[[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.]], 138.0746974](/support/helpjp/helpview.aspx?si=8934/file01704/math172.png)
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There were 8 payments altogether with a loan cost of 138 U. Obtaining just the first year, we make use of the fourth argument:
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![[[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]], 102.7056860](/support/helpjp/helpview.aspx?si=8934/file01704/math181.png)
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which indicates the cost of the loan at that point.
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