bar code for C# So the number of payments on a 30-year loan is in VS .NET Creation PDF-417 2d barcode in VS .NET So the number of payments on a 30-year loan is

So the number of payments on a 30-year loan is using none toconnect none in web,windows applicationc# barcode detection 30*12 ans = 360 Overview of GS1 General Specification and an annual percentage rate of 8% comes out to a monthly rate of 8*percent*peryear ans = 0.0067 Now consider what ha ppens with each monthly payment. Some of the payment is applied to interest on the outstanding principal amount, P, and some of the payment is applied to reduce the principal owed. The total amount, R, of the monthly payment remains constant over the life of the loan.

So if J denotes the monthly interest rate, we have R = J P + (amount applied to principal), and the new principal after the payment is applied is P + J P R = P (1 + J) R = P m R, where m = 1 + J. So a table of the amount of the principal still outstanding after n payments is tabulated as follows for a loan of initial amount A, for n from 0 to 6:. syms m J P R A P = A; for n = 0:6, disp([n, P]), P = simplify(-R + P*m); end 9: Applications [ 0 , A] [ 1, -R+A*m] [ 2, -R-m*R+A*m 2] [ 3, -R-m*R-m 2*R+A*m 3] [ 4, -R-m*R-m 2*R-m 3*R+A*m 4] [ 5, -R-m*R-m 2*R-m 3*Rm 4*R+A*m 5] [ 6, -R-m*R-m 2*R-m 3*Rm 4*R-m 5*R+A*m 6]. We can write this in none none a simpler way by noticing that P = A mn+ (terms divisible by R). For example, with n = 7 we have. factor(p - A*m 7) an s = -R*(1+m+m 2+m 3+m 4+m 5+m 6). But the quantity inside the parentheses is the sum of a geometric series mk = mn 1 . m 1 So we see that the p rincipal after n payments can be written as P = A mn R (mn 1)/(m 1). Now we can solve for the monthly payment amount R under the assumption that the loan is paid off in N installments, that is, P is reduced to 0 after N payments:. syms N; solve(A*m N none for none - R*(m N - 1)/(m - 1), R) ans = A*m N*(m-1)/(m N-1) R = subs(ans, m, J + 1) R= A*(J+1) N*J/((J+1) N-1). For example, with an initial loan amount A = $150,000 and a loan lifetime of 30 years (360 payments), we get the following table of payment amounts as a function of annual interest rate:. Mortgage Payments fo none for none rmat bank; disp( Interest Rate Payment ) for rate = 1:10, disp([rate, double(subs(R, [A, N, J], [150000, 360,...

rate*percent*peryear]))]) end Interest Rate 1.00 2.00 3.

00 4.00 5.00 6.

00 7.00 8.00 9.

00 10.00 Payment 482.46 554.

43 632.41 716.12 805.

23 899.33 997.95 1100.

65 1206.93 1316.36.

Note the use of form at bank to write the oating point numbers with two digits after the decimal point. There s another way to understand these calculations that s a little slicker and that uses MATLAB s linear algebra capability. Namely, we can write the fundamental equation Pnew = Pold m R in matrix from as vnew = Bvold , where v is the column vector ( P ) and B is the square matrix 1 m R .

0 1 We can check this using matrix multiplication:. syms R P; B = [m -R; none none 0 1]; v = [P; 1]; B*v ans = [ m*P-R] [ 1]. 9: Applications which agrees with th none none e formula we had above. (Note the use of syms to reset R and P to unde ned symbolic quantities.) Thus the column vector [P; 1] resulting after n payments can be computed by left-multiplying the starting vector [ A; 1] by the matrix B n.

Assuming m > 1, that is, a positive rate of interest, the calculation. [eigenvectors, diago nalform] = eig(B) eigenvectors = [ 1, 1] [ 0, (m-1)/R] diagonalform = [ m, 0] [ 0, 1]. shows us that the ma none for none trix B has eigenvalues m and 1, and corresponding eigenvectors [1; 0] and [1; (m 1)/R] = [1; J/R]. Now we can write the vector [A; 1] as a linear combination of the eigenvectors: [ A; 1] = x[1; 0] + y[1; J/R]. We can solve for the coef cients:.

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