ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.19 : In the purchase of a unit of engineering office, a loan has been made to a bank with the following details : Term N = 30 years; interest rate R = 8.07 % / year; security : primary residence; present value pv = \$450000; salary = \$75000 / year or \$56000 / year after tax. (a) Let the discounted present value PV = [ 1 - 1 / (1 + r) ^ n ] / r for arrears, where r = interest rate of discount, n = number of payment, ^ = symbol for power. If the loan repayment was made monthly : (i) calculate the value of r where r = R / k and R is in decimal value; (ii) find the value of n where n = kN; (iii) estimate the value of k where k = number of repayment per year; (iv) calculate the value of PV based on the formula of discounted present value. (b) Calculate the monthly repayment of the loan, MR based on the following formula : pv = PV x MR. (c) Find the percentage of salary remains after paying the loan every month.

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.19 : In the purchase of a unit of engineering offi..

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.19 : Value of k has to be found first : 12 months / year, then k = 12. (a) (i) r = R / k = 0.0807 / 12 = 0.006725. (ii) n = kN = 12 x 30 = 360. (iii) k = 12. (iv) PV = [ 1 - 1 / (1 + r) ^ n ] / r = [ 1 - 1 / (1 + 0.006725) ^ 360 ] / 0.006725 = 135.382. (b) MR = pv / PV = \$450000 / 135.382 = \$3323.928. (c) Annual repayment = MR x k = \$3323.928 x 12 = \$39887.136. Percentage of salary remains after paying loan = 100 [ 1 - (annual repayment) / (annual salary after tax) ] = 100 (1 - 39887.136 / 56000) = 28.773 %. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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