2010-03-24 14:18:08 -05:00
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<HTML>
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<HEAD>
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<TITLE>Financial Equations Documentation</TITLE>
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</HEAD>
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<BODY>
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<A NAME="TOP"></A>
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<A HREF="./finutil.html#FinEquation">Return</A>
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<HR>
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<br>
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<dl>
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<dt><A HREF="#BasicEquation">Basic Equation</A></dt>
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<dt><A HREF="#SeriesSum">Series Sum</A></dt>
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</dl>
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<HR>
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<h1>Financial Equation Derivation</h1>
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<p>The financial equation is derived in the following manner:
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<p>Start with the basic equation to find the balance or Present Value, PV[1], after
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one payment period. Note PV[1] is the Present Value after one payment and PV[0]
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is the initial Present Value. PV[0] will be shortened to just PV.
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<p>The interest due at the end of the first payment period is the original present value,
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PV, times the interest rate for the payment period plus the periodic payment times the
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interest rate for beginning of period payments:
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<p>ID[1] = PV * i + X * PMT * i = (PV + X * PMT) * i
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<p>The Present Value after one payment is the original Present Value with the periodic
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payment, PMT, and interest due, ID[1], added:
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<pre>
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PV[1] = PV + (PMT + ID[1])
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PV[1] = PV + (PMT + (PV + X * PMT) * i)
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PV[1] = PV * (1 + i) + PMT * (1 + Xi)
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</pre>
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<p>This equation works for all of the cash flow diagrams shown previously. The Present Value,
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money received or paid, is modified by a payment made at the beginning of a payment
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period and multiplied by the effective interest rate to compute the interest
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due during the payment period. The interest due is then added to the payment
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to obtain the amount to be added to the Present Value to compute the new Present Value.
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<p>For diagram 1): PV < 0, PMT == 0, PV[1] < 0
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<br>For diagram 2): PV == 0, PMT < 0, PV[1] < 0
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<br>For Diagram 3): PV > 0, PMT < 0, PV[1] >= 0 or PV[1] <= 0
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<br>For Diagram 4): PV < 0, PMT > 0, PV[1] <= 0 or PV[1] >= 0
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<p>X may be 0 or 1 for any diagram.
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<p>For the standard loan, PV is the money borrowed, PMT is the periodic payment to repay
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the loan, i is the effective interest rate agreed upon and FV is the residual loan amount
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after the agreed upon number of periodic payment periods. If the loan is fully paid off
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by the periodic payments, FV is zero, 0. If the loan is not completely paid off after the
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agreed upon number of payments, a balloon payment is necessary to completely pay off the loan.
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FV is then the amount of the needed balloon payment. For a loan in which the borrower pays
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only enough to repay the interest due during a payment period, interest only loan, the
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balloon payment is equal to the negative of PV.
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<p>To calculate the Present Value after the second payment period, the above calculation
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is applied iteratively to PV[1] to obtain PV[2]. In fact to calculate the Present Value
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after any payment period, PV[n], the above equation is applied iteratively to PV[n-1]
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as shown below.
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<pre>
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PV[2] = PV[1] + (PMT + (PV[1] + X * PMT) * i)
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= PV[1] * (1 + i) + PMT * (1 + iX)
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= (PV * (1 + i) + PMT * (1 + iX)) * (1 + i) + PMT * (1 + iX)
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= PV * (1 + i)^2 + PMT * (1 + iX) * (1 + i)
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+ PMT * (1 + iX)
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</pre>
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<p>Similarly, PV[3] is computed from PV[2] as:
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<pre>
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PV[3] = PV[2] + (PMT + (PV[2] + X * PMT) * i)
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= PV[2] * (1 + i) + PMT * (1 + iX)
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= (PV * (1 + i)^2 + PMT * (1 + iX) * (1 + i)
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+ PMT * (1+ iX)) * (1 + i)
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+ PMT * (1+ iX)
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= PV * (1 + i)^3 + PMT * (1 + iX) * (1 + i)^2
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+ PMT * (1 + iX) * (1 + i)
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+ PMT * (1 + iX)
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</pre>
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<p>And for the n'th payment, PV[n] is computed from PV[n-1] as:
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<pre>
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PV[n] = PV[n-1] + (PMT + (PV[n-1] + X * PMT) * i)
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PV[n] = PV * (1 + i)^n + PMT * (1 + iX) * (1 + i)^(n-1)
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+ PMT * (1 + iX) * (1 + i)^(n-2) +
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+ PMT * (1 + iX) * (1 + i)
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+ PMT * (1 + iX)
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PV[n] = PV * (1 + i)^n + PMT * (1 + iX) * [(1 + i)^(n-1) + ... + (1 + i) + 1]
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</pre>
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<p>The formula for PV[n] can be proven using mathematical induction.
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<A NAME="BasicEquation">
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<h1>Basic Financial Equation</h1></A>
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<p>As shown above, the basic financial transaction equation is simply:
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<pre>
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PV[n] = PV[n-1] + (PMT + (PV[n-1] + X * PMT) * i)
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= PV[n-1] * (1 + i) + PMT * (1 + iX)
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for: n >= 1
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</pre>
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<p>relating the Present Value after n payments, PV[n] to the previous Present Value, PV[n-1].
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<!--########################################################################-->
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<hr>
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<A NAME="SeriesSum">
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<h1>Series Sum</h1></A>
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<p>The sum of the finite series:
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<p>1 + k + (k^2) + (k^3) + ... + (k^n) = (1-k^(n+1))/(1-k)
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<p>as can be seen by the following. Let S(n) be the series sum. Then
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<p>S(n) - k * S(n) = 1 - k^(n+1)
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<p>and solving for S(n):
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<p>S(n) = (1-k^(n+1))/(1-k) = 1 + k + (k^2) + (k^3) + ... + (k^n)
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<!--########################################################################-->
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<hr>
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<p>Using this in the equation above for PV[n], we have:
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<pre>
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PV[n] = PV * (1 + i)^n + PMT * (1 + iX) * [(1 + i)^(n-1) + ... + (1 + i) + 1]
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= PV * (1 + i)^n + PMT * (1 + iX) * [1 - (1 + i)^n]/[1 - (1 + i)]
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= PV * (1 + i)^n + PMT * (1 + iX) * [1 - (1 + i)^n]/[-i]
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= PV * (1 + i)^n + PMT * (1 + iX) * [(1 + i)^n - 1]/i
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</pre>
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<p>or:
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<pre>
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PV * (1 + i)^n + PMT * [(1 + i)^n - 1]/i - PV[n] = 0
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</pre>
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<p>If after n payments, the remaining balance is repaid as a lump sum, the lump sum
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is known as the Future Value, FV[n]. Since FV[n] is negative if paid and positive
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if received, FV[n] is the negative of PV[n].
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<p>Setting: FV[n] = -PV[n]
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<p>Since n is assumed to be the last payment, FV[n] will be shortened to simply
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FV for the last payment period.
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<pre>
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PV*(1 + i)^n + PMT*(1 + iX)*[(1 + i)^n - 1]/i + FV = 0
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</pre>
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<p>Up to this point, we have said nothing about the value of PMT. PMT can be any value mutually
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agreed upon by the lender and the borrower. From the equation for PV[1]:
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<pre>
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PV[1] = PV + (PMT + (PV + X * PMT) * i),
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</pre>
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<p>Several things can be said about PMT.
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<ol>
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<li>If PMT = -(PV * i), and X = 0 (end of period payments):
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<p>The payment is exactly equal to the interest due and PV[1] = PV. In this case, the borrower
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must make larger future payments to reduce the balance due, or make a single payment, after
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some agreed upon number of payments, with PMT = -PV to completely pay off the loan. This is
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an interest only payment with a balloon payment at the end.
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<li>If |PMT| < |PV * i|, and X = 0 and PV > 0
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<p>The payment is insufficient to cover even the interest charged and the balance due grows
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<li>If |PMT| > |PV * i|, and X = 0 and PV > 0
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<p>The payment is sufficient to cover the interest charged with a residual amount to be
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applied to reduce the balance due. The larger the residual amount, the faster the loan is
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repaid. For most mortgages or other loans made today, the lender and borrower agree upon
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a certain number of repayment periods and the interest to be charged per payment period.
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The interest may be multiplied by 12 and stated as an annual interest rate. Then the
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lender and borrower want to compute a periodic payment, PMT, which will reduce the balance
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due to zero after the agreed upon number of payments have been made. If N is the agreed
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upon number of periodic payments, then we want to use:
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<pre>
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PV * (1 + i)^N + PMT*(1 +iX)*[(1 + i)^N - 1]/i + FV = 0
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</pre>
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<p>with FV = 0 to compute PMT:
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<pre>
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PMT = -[PV * i * (1 + i)^(N - X)]/[(1 + i)^N - 1]
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</pre>
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<p>The value of PMT computed will reduce the balance due to zero after N periodic payments.
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Note that this is strictly true only if PMT is not rounded to the nearest cent as is the
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usual case since it is hard to pay fractional cents. Rounding PMT to the nearest cent has
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an effect on the FV after N payments. If PMT is rounded up, then the final Nth payment
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will be smaller than PMT since the periodic PMTs have paid down the principal faster than
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the exact solution. If PMT is rounded down, then the final Nth payment will be larger than
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the periodic PMTs since the periodic PMTs have paid down the principal slower than the
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exact solution.
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</ol>
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<!--#############################################################################-->
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<p>With a simple alegebraic re-arrangement, The financial Equation becomes:
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<pre>
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2) [PV + PMT*(1 + iX)/i][(1 + i)^n - 1] + PV + FV = 0
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</pre>
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<p>or
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<pre>
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3) (PV + C)*A + PV + FV = 0
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</pre>
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<p>where:
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<pre>
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4) A = (1 + i)^n - 1
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5) B = (1 + iX)/i
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6) C = PMT*B
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</pre>
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<p>The form of equation 3) simplifies the calculation procedure for all five
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variables, which are readily solved as follows:
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<pre>
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7) n = ln[(C - FV)/(C + PV)]/ln((1 + i)
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8) PV = -[FV + A*C]/(A + 1)
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9) PMT = -[FV + PV*(A + 1)]/[A*B]
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10) FV = -[PV + A*(PV + C)]
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</pre>
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<p>Equations 4), 5) and 6) are computed by the functions in the <tt>"fin.exp"</tt> utility:
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<br><tt>_A</tt>
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<br><tt>_B</tt>
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<br><tt>_C</tt>
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<p>respectively. Equations 7), 8), 9) and 10) are computed by functions:
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<br><tt>_N</tt>
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<br><tt>_PV</tt>
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<br><tt>_PMT</tt>
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<br><tt>_FV</tt>
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<p>respectively.
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<p>The solution for interest is broken into two cases:
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<ol>
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<li>PMT == 0
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<p>Equation 3) can be solved exactly for i:
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<pre>
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i = [FV/PV]^(1/n) - 1
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</pre>
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<li>PMT != 0
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<p>Since equation 3) cannot be solved explicitly for i in this case, an
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iterative technique must be employed. Newton's method, using exact
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expressions for the function of i and its derivative, are employed. The
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expressions are:
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<pre>
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12) i[k+1] = i[k] - f(i[k])/f'(i[k])
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where: i[k+1] == (k+1)st iteration of i
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i[k] == kth iteration of i
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and:
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13) f(i) = A*(PV+C) + PV + FV
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14) f'(i) = n*D*(PV+C) - (A*C)/i
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15) D = (1 + i)^(n-1) = (A+1)/(1+i)
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</pre>
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<p>To start the iterative solution for i, an initial guess must be made
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for the value of i. The closer this guess is to the actual value,
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the fewer iterations will have to be made, and the greater the
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probability that the required solution will be obtained. The initial
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guess for i is obtained as follows:
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<ol>
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<li>PV case, PMT*FV >= 0
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<pre>
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| n*PMT + PV + FV |
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16) i[0] = | ----------------|
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| n*PV |
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= abs[(n*PMT + PV + FV)/(n*PV)]
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</pre>
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<li>FV case, PMT*FV < 0
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<ol>
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<li>PV != 0
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<pre>
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| FV - n*PMT |
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17) i[0] = |---------------------------|
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| 3*[PMT*(n-1)^2 + PV - FV] |
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= abs[(FV-n*PMT)/(3*(PMT*(n-1)^2+PV-FV))]
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</pre>
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<li>PV == 0
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<pre>
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| FV + n*PMT |
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18) i[0] = |---------------------------|
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| 3*[PMT*(n-1)^2 + PV - FV] |
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= abs[(FV+n*PMT)/(3*(PMT*(n-1)^2+PV-FV))]
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</pre>
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</ol>
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</ol>
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</ol>
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<HR>
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<A HREF="./finutil.html#FinEquation"><IMG SRC="images/back.png" BORDER=0 HEIGHT=13 WIDTH=7>Return</A>
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</body>
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