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A Derivation of Amortization — Bret D. Whissel This is my derivation of the formula for amortization. The or in English, the sum of all the payments (left side) is goal is to ﬁnd a payment amount, x, which pays oﬀ the equal to the principal borrowed plus all of the interest paid loan principal, P , after a speciﬁed number of payments, with regular payments plus interest paid on the balloon N . We start with some variable deﬁnitions: payment (right side). Note that if there will be no balloon P The principal borrowed payment (B = 0), then the B terms drop out. N The number of payments Now we glue some more pieces together: replace Ij of equa- i The fractional (periodic) interest rate tion (3) using the relationship given by eq. (1) and then Pj The principal part of payment j substitute the recurrence identity of eq. (2): Ij The interest part of payment j B A ﬁnal balloon payment N iB x The regular payment B− + Nx = P + x − (x − iP )(1 + i)j−1 1+i j=1 Assuming that all payments (excluding an optional ﬁnal N iB balloon payment) are the same amount, a payment x con- B− + N x = P + N x − (x − iP ) (1 + i)j−1 sists of its interest part and its principal part: 1+i j=1 N x = Ij + Pj (1) i P −B 1− = (x − iP ) (1 + i)j−1 1+i I1 = iP P1 = x − I1 j=1 I2 = i(P − P1 ) P2 = x − I2 Now we can see our way clear to solve for x: I3 = i(P − P1 − P2 ) P3 = x − I3 , etc. i P −B 1− 1+i This schedule states that the payment x includes interest x= + iP. (4) N on all of the remaining principal, including that which is j=1 (1 + i)j−1 part of the current payment. The ﬁrst payment, therefore, includes an interest payment on the total borrowed, which The series form of eq. (4) can be rewritten without the deﬁnes the minimum payment. (If we are to make any series after a little transformation. First, we separate the progress toward paying oﬀ the loan, we must pay more summation and rewrite its limits: than the amount iP .) i 1 The Pj ’s may be rewritten into a recurrence relation: x= P −B 1− N −1 + iP. 1+i j=0 (1 + i)j P1 = x − iP To simplify the transformation, we can substitute by let- N −1 P2 = x − i(P − P1 ) ting g = 1 + i so that the summation looks like j=0 g j . Next we multiply the series by (1 − g)/(1 − g) so that all = x − i [P − (x − iP )] but the ﬁrst and last terms drop out: = x − iP + ix − i2 P N −1 N −1 N = (x − iP )(1 + i) (1 − g) j=0 gj j=0 gj − j=1 gj 1 − gN = = . 1−g 1−g 1−g P3 = x − i(P − P1 − P2 ) Since the series is originally in the denominator, we invert = x − i P − (x − iP ) − (x − iP + ix − i2 P ) the transformed result, and then undo the substitution: = x + 2ix + i2 x − iP − 2i2 P − i3 P i 1 − (1 + i) = x(1 + i)2 − iP (1 + i)2 x= P −B 1− + iP. (5) 1+i 1 − (1 + i)N 2 = (x − iP )(1 + i) Now we can expand and rearrange to taste: In general, we will ﬁnd that P (1 + i)N B Pj = (x − iP )(1 + i) j−1 . (2) x=i N −1 + . (6) (1 + i) (1 + i) − (1 + i)N +1 If there is to be a balloon payment, then the ﬁnal payment Quod erat demonstrandum (“That which was to be will consist of the ﬁnal principal payment Pf and interest shown”), otherwise known as Q.E.D. — BDW on that principal iPf so that B = Pf + iPf . Rewriting Pf in terms of B gives Pf = B/(1 + i). Equations (5) and (6) solve for the payment amount, but either can be re-arranged to solve for any of the other vari- Next, we deﬁne an equation which uses these ideas: ables, with the exception of i, the periodic interest rate. N To date I have been unable to ﬁnd an analytic solution for B this variable, so the program invokes an iterative method B + Nx = P + Ij + i , (3) j=1 1+i to ﬁnd successive approximations to the solution.

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beat, home, loan, amortization, tables

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posted: | 8/11/2009 |

language: | English |

pages: | 1 |

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