Find the periodic payment R required to amortize a loan of P dollars over t years with interest charged at the rate of r%/year compounded m times a year. (Round your answer to the nearest cent.)
P = 180,000, r = 5.5, t = 28, m = 4
This question requires application of PV of annuity formula according to which
PV =
PV = $180,000, n = 28 * 4 = 112 periods, r = 5.5%/4 = 1.375%,
180,000 = P * 56.9715
P = $3,159.48
> Here, P is PV not FV, so I rechecked, the above answer for R = 3159.48 ($) is CORRECT. However, it is not P, it is R.
Tulsiram Garg Mon, Dec 27, 2021 2:06 AM
SOLUTION :
Following formula will be applicable for periodic payment $R to accumulate $P in t years at interest rate of r% per year compounded m times a year ;
P = R((1 + r / (100m))^(mt) - 1) / (r / (100m))
For P = $180000 ; r = 5.5% ; t = 28 years ; m = 4 , we have :
=> 180000 = R ( (1 + 5.5 / 400)^(112) - 1 ) / (5.5/ 400)
=> 180000 = R * 262.975115116
=> R = 180000/262.975115116
=> R = 684.48 ($) (ANSWER).
> Please ignore this solution as it is for P as FV. However P should have been PV. Please see correct solution I n answer 3.
Tulsiram Garg Mon, Dec 27, 2021 2:08 AM
SOLUTION :
Following formula will be applicable for periodic payment $R to pay the loan $P in t years at interest rate of r% per year compounded m times a year ;
PV = P = R((1 + r / (100m))^(mt) - 1) / (r / (100m))
For P = $180000 ; r = 5.5% ; t = 28 years ; m = 4 , we have :
=> 180000 = R ( (1 + 5.5 / 400)^(112) - 1 ) / ( (5.5/ 400) *(1 + 5.5 / 400)^(112) )
=> 180000 = R * 56.9714831
=> R = 180000/56.9714831
=> R = 3159.48 ($) (ANSWER).
> In the formula replace the denominator by : ( (r / (100m) (1 + r/(100m))^(tm) )
Tulsiram Garg Mon, Dec 27, 2021 2:15 AM
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> answer is wrong. Even at 0% interest accumulated amount P for R = 3159.48 will be equal to $353861.76. Moreover, author is confused between P an R. R is to be calculated but he finds P against given P.
Tulsiram Garg Mon, Dec 27, 2021 1:56 AM