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 nth Payment
The interest payment in period n is, In= i x Bn-1 = A x (P/A, i, N-n+1) x i
Contemporary Engineering Economics, 5th edition, © 2010
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 6th payment
A = $5,000(A/P, 1%, 24) =
$235.37
Contemporary Engineering Economics, 5th edition, © 2010
Example 4.13 Continued
To compute I6, we first need to find B5,
B5 = $235.37 x (P/A, 1%, 19) = $ 4,054.44
Then, I6 = $ 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,
P6 = $235.37 – 40.54 = $194.83
The remaining balance after the 6th payment, B6, is equal to $3,859.62 as
computed on the previous slide.
Contemporary Engineering Economics, 5th
edition, © 2010
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:
Pdebt = $4,500 + $736.53(P/A, 4.5%/12, 42)
- $17,817(P/F, 4.5%/12, 42)
= $17,847
•
Option B: Cash Financing:
Pcash = $31,020 - $17,817(P/F,4.5%/12,42)
= $15,795
 Option C: Lease Financing:
Please = $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
 Loan amount
 Adjustable-rate mortgage
 Loan term
 Hybrid Mortgage
 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
 (b) Interest payments:
 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.
Contemporary Engineering Economics, 5th edition, © 2010
Calculating the Monthly Payments with
Adjustable-Rate Mortgage (ARMs)
 Index – a guide that lenders use to
measure interest rate changes
 Margin – an interest rate that
represents the lender’s cost of doing
business plus the profit
 Adjustment period – the period
between potential interest rate
adjustments (e.g., 3/1 means your
interest rate is fixed for the first 3
years then could be adjusted every
year thereafter.)
 Interest rate cap – a limit on the
amount your interest can change (a
periodic cap and a lifetime cap)
 Payment cap – how much your
monthly payment can increase at
each adjustment
Contemporary Engineering Economics, 5th edition, © 2010
Example 4.18 A 5/1 Hybrid Adjustable Mortgage
Plan
 Monthly payments under 5/1 Hybrid mortgage
 Given: Varying annual
mortgage rates and N = 30
years
 Find: (a) the monthly
payment; (b) the total
interest payment over the
10-year ownership
Contemporary Engineering Economics, 5th edition, © 2010