An individual is borrowing $165,000 for a 25 year loan at 4.0% per year compounded monthly. Compute the monthly payment. Immediately after his 108th monthly payment he decides to refinance at a lower rate of 3% per year compounded monthly for 100 monthly payments. What would be his new monthly payment and how much interest will he save? Draw the cash flow diagrams.

Please use one of the following formulas to solve this problem. I solved the monthly payment, but I am confused on how to solve the new monthly payment.

19166000 2. P=$165000 | 70- 15 gr5 = 30 m2 1 = 4% 7 yr = 0.33% lwo A = ? 30 60 90 120 50 150 2110 240 20 30 AF 169000 (AVP, 0.83,300) com (0.0083 (1+0.0033)) $870.93 As 165,000 (1+0.0033)30-1 Formulas Compound Amount: To find F, given P (F/P, i, n) F= P(1 + i)" Present Worth: To find P, given F (P/F, i, n) P=F(1+i)" Series Compound Amount: To find F, given A Sinking Fund: To find A, given F "-1 (FIA, i,n) F = 44+1)*= (A/F, i, n) A-Fi |(1+i)" 1] Capital Recovery: To find A, given P Series Present Worth: To find P, given A (3P,1,1) A-Pf449 (P/A, i, n) PEA (1 + i)" - 1] | i(l+i)" ] Arithmetic Gradient Uniform Series: To find A, given G (AIG,i,m) A-09 or 1-0-2-5--1] Arithmetic Gradient Present Worth: To find P, given G (PIC,i,m) pa je Geometric Gradient: To find P, given A1, g (P/G, g, i, n) P= A (1+i)- [1- (1+g)(1+i)" when i=g when ing Continuous Compounding at Nominal Rater Single Payment: Uniform Series: unstorm seriet: P-Fk". super"] -74 ANe-11 A-ple(e" - 11 viii Formulas Compound Interest i = Interest rate per interest period. n = Number of interest periods. P = A present sum of money. F = A future sum of money. A = An end-of-period cash receipt or disbursement in a uniform series continuing for n periods. G = Uniform period-by-period increase or decrease in cash receipts or disbursements. 8 = Uniform rate of cash flow increase or decrease from period to period; the geometric gradient. r = Nominal interest rate per interest period. m = Number of compounding subperiods per period. Effective Interest Rates For non-continuous compounding: lefror i = 1+ where r = nominal interest rate per year m= number of compounding periods in a year OR iyor i,= (1 + i)" - 1 where i = effective interest rate per period m= number of compounding periods in a year For continuous compounding: igori,= 621 where r= nominal interest rate per year Values of Interest Factors When n Equals Infinity Single Payment: Uniform Payment Series: (FP, i, 0) = 00 (P/F, 1, 0) = 0 (A/F, 1, 0) = 0 (A/P, 1, 0) =i (FIA, I, 0) = 0 (P/A, 1, 0) = 1 Arithmetic Gradient Series: (A/G, I, 0) = 1/i (P/G, i, 0) = 1/7