Monday, April 12, 2021 |
Scott, I read with interest your article, "Amortized Loans." At the end of the article you indicate the explanation of how to derive the formulae as being covered in another article. Can you please tell me where I can find this? Thank you very much, Luke, Here is the derivation of the loan formula from scratch: F --> Future Value #1) F0=P #2) F1=P+Pi-A=P(1+i)-A #3)
F2=F1(1+i)-A=P((1+i)^2)-A(1+i)-A #4)
F3=F2(1+i)-A=P((1+i)^3)-A((1+i)^2)-A(1+i)-A #5)
F4=F3(1+i)-A=P((1+i)^4)-A((1+i)^3)-A((1+i)^2)-A(1+i)-A #6)
Fn=P((1+i)^n)-A[((1+i)^(n-1))+((1+i)^(n-2))+((1+i)^(n-3))+....+1] The first part of #6, P((1+i)^n), is easy to work with because it is a closed form, however, we need to find the closed form for the remaining part of that formula, therefore, for the general case of [((1+i)^(n-1))+((1+i)^(n-2))+((1+i)^(n-3))+....+1]: #7)
y=1+a^1+a^2+a^3+a^4+a^5+a^6+....a^k #8)
y-1=a^1+a^2+a^3+a^4+a^5+a^6+....a^k #9)
(y-1)/a=1+a^1+a^2+a^3+a^4+a^5+....a^(k-1) #10) (y-1)/a+a^k=1+a^1+a^3+a^3+a^4+a^4+a^5+....a^k #11)
1+a^1+a^2+a^3+a^4+a^5+....a^k=y Therefore, #12) (y-1)/a+a^k=y #13) y-1+a^(k+1)=ya #14) y-ya=1-a^(k+1) #15) y(1-a)=1-a^(k+1) #16) y=(1-a^(k+1))/(1-a) Going back to the general case for
the loan formula in #6: #18)
y=[((1+i)^(n-1))+((1+i)^(n-2))+((1+i)^(n-3))+....+1] #19) y=(1-a^(k+1))/(1-a) #20) y=(1-((1+i)^n))/(-i) #21) y=(((1+i)^n)-1)/I Back to #6: #23) Fn=P((1+i)^n)-A[y] #24) Fn=P((1+i)^n)-A[(1-((1+i)^n))/i] Note: Fn=0, the last future value of the loan, when the loan is repaid, therefore, #25) 0=P((1+i)^n)-A[(((1+i)^n)-1)/i]
(at period n) #26) A[(((1+i)^n)-1)/i]=P((1+i)^n) #27) A=[P((1+i)^n)][i/(((1+i)^n)-1)] #27) A=Pi/(1-((1+i)^(-n))) We have it! Equation #27 is the formula used to calculate the periodic payment (A) for a loan of principal (P) at the periodic rate (i) for n periods. Here's a quick example: Solution: Formula is: A=10000(0.02)/(1-((1+0.02)^(-60))) A=200/(1-0.30478)=200/0.69522 A=287.68 Therefore, the monthly payment is $287.68. I hope you enjoyed this derivation! Regards, |
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