 Lecture No.13 Chapter 4 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

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Lecture No.13 Chapter 4 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

Debt Management Credit card debt and commercial loans are among the most significant financial transactions involving interest. Contemporary Engineering Economics, 5th edition, © 2010

Example 4.12 Loan Balance, Principal, and Interest: Tabular Approach  Given: P = \$5,000, i = 12% APR, N = 24 months  Find: A, and loan repayment schedule A = \$5,000(A/P, 1%, 24) = \$235.37 Contemporary Engineering Economics, 5th edition, © 2010

Calculating the Remaining Loan Balance after Making the n th Payment Contemporary Engineering Economics, 5th edition, © 2010 The interest payment in period n is, I n = i x B n-1 = A x (P/A, i, N-n+1) x i

Example 4.13 Loan Balance, Principal, and Interest: Remaining –Balance Method  Given: P = \$5,000, i = 12% APR, N = 24 months  Find: Loan balance, principal, and interest payment for the 6 th payment A = \$5,000(A/P, 1%, 24) = \$235.37 Contemporary Engineering Economics, 5th edition, © 2010

Example 4.13 Continued To compute I 6, we first need to find B 5, B 5 = \$235.37 x (P/A, 1%, 19) = \$ 4,054.44 Then, I 6 = \$ 4,054.44 x 0,01 = \$ 40,54. Note that the principal payment is the remaining part of the total monthly payment amount \$235.37. Thus, P 6 = \$235.37 – 40.54 = \$194.83 The remaining balance after the 6 th payment, B 6, is equal to \$3,859.62 as computed on the previous slide.

Example 4.15 Financing your Vehicle  Given: Three financing options, r = 4.5%, payment period = monthly, and compounding period = monthly  Find: Which option?  Issue: Which interest rate to use in calculating the equivalent cost of financing for each option Option A: Conventional Debt Financing: P debt = \$4,500 + \$736.53(P/A, 4.5%/12, 42) - \$17,817(P/F, 4.5%/12, 42) = \$17,847 Option B: Cash Financing: P cash = \$31,020 - \$17,817(P/F,4.5%/12,42) = \$15,795 Option C: Lease Financing: P lease = \$1,507.76 + \$513.76(P/A, 4.5%/12, 42) + \$395(P/F, 4.5%/12, 42) = \$21,336 Contemporary Engineering Economics, 5th edition, © 2010

Home Mortgage Types of Home Mortgages The Cost of a Mortgage Fixed-rate mortgage Adjustable-rate mortgage Hybrid Mortgage Loan amount Loan term Payment frequency Points (prepaid interest) Fees Types of mortgages Contemporary Engineering Economics, 5th edition, © 2010

Example 4.16 An Interest-Only versus a Fully Amortized Mortgage  Given: P = \$200,000, APR = 6.6% or 0.55% per month, and N = 30 years  Find: (a) monthly payment; (b) interest payments during the first year of ownership of the home.  Option 1: A fully amortized payment option.  Option 2: A five-year interest-only option. (a) Monthly payments. (b) Interest payments: Contemporary Engineering Economics, 5th edition, © 2010