Chapter 15 The Cost of Home Ownership Copyright © 2011 by the McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill/Irwin

List the types of mortgages available 2. Utilize an amortization chart to compute monthly mortgage payments 3. Calculate the total cost of interest over the life of a mortgage The Cost of Home Ownership #15 Learning Unit Objectives Types of Mortgages and the Monthly Mortgage Payment LU15.1

Calculate and identify the interest and principal portion of each monthly payment 2. Prepare an amortization schedule The Cost of Home Ownership #15 Learning Unit Objectives Amortization Schedule -- Breaking Down the Monthly Payment LU15.2

15-4 Table Amortization Chart (PARTIAL) (Mortgage principal and interest per $1,000)

15-5 Computing the Monthly Payment for Principal and Interest Gary bought a home for $200,000. He made a 20% down payment. The 9% mortgage is for 30 years (30 x 12 = 360 payments). What are Gary’s monthly payment and total cost of interest?

15-6 Step 2. Look up the rate (9%) and the term (30 years) in the amortization chart. At the intersection is the table factor. ($8.05) Step 3. Multiply Step 1 by the factor in Step 2 $160 x $8.05 = $1, Step 1. Divide the amount of the mortgage by $1,000 $160,000 = $160 $1,000 Computing Monthly Payment by Using an Amortization Chart

15-7 Computing the Monthly Payment for Principal and Interest $160,000 = 160 x $8.05 (table rate) = $1, $1,000 Total payments Mortgage Total interest $463,680 - $160,000 = $303,680 ($1, x 360) Monthly Payment

15-8 Table Effect of Interest Rates on Monthly Payments 9% 11% Difference Monthly payment$1,288.00$1, $ (160 x $8.05) (160 x $9.53) Total cost of interest$303,680$388,828 $85,248 ($1, x 360) - $160,000 ($ x 360) ($1, x 360) - $160,000

15-9 The Effect of Loan Types on Monthly Payments Suppose Gary chose a 15-year mortgage vs. a 30-year mortgage. What would be the effect? 15 Year30 YearDifference Monthly Payment$1,624.00$1,288.00$ Total Interest$100,912$303,680($202,768) ($1, x 180) -$140,000 ($1, x 360) - $160,000

15-10 Hidden Cost in Purchasing a Home Closing Costs - Cost associated with the passing of property from the seller to buyer. Include: lawyer’s fees, title search, points, etc. A point is a one-time charge that is a percent of the mortgage. Escrow Amount - A special interest bearing account in which the buyer is required to deposit 1/12 of the insurance cost and 1/12 of the real estate taxes each month Repairs and Maintenance - The cost of keeping the property up. Includes: paint, wallpaper, landscaping, etc.

15-11 Step 2. Calculate the amount used to reduce the principal: Principal reduction = Monthly payment - Interest (Step 1.) $1, $ = $88.00 Step 3. Calculate the new principal: Current principal - Reduction of principal (Step 2) = New Principal $160,000 - $88.00 = $159, Step 1. Calculate the interest for a month (use current principal): Interest = Principal x Rate x Time $160,000 x.09 x 1/12 = $1, Calculating Interest, Principal, and New Balance of Monthly Payment

15-12 Step 2. Principal reduction = Monthly payment - Interest (Step 1.) $1, $1, = $88.66 Step 3. Current Principal - Reduction of principal (Step 2) = New Principal $159, $88.66 = $159, Step 1. Interest = Principal x Rate x Time $159, x.09 x 1/12 = $1, Calculating Interest, Principal, and New Balance of Monthly Payment 2nd Month

15-13 Table Partial Amortization Schedule Payment Principal Principal Balance of number (current) Interest reduction principal 1 $160,000 $ $88.00 $159, ($160,000 x.09 x 1/12) ($1, – 1,200) ($160,000 - $88.00) 2 $159, $1, $88.66 $159, ($159,912 x.09 x 1/12) ($1,288 – 1,199.34) ($159,912 - $88.66) 3 $159, $1, $89.32 $159, $159, $1, $89.99 $159, $159, $1, $90.67 $159,553.36

15-14 Problem 15-9: $215,000 x 0.2 = $43,000 $215,000 - $43,000 = $172,000 $172,000/$1,000 = 172 x $5.37 = $923.64

15-15 Problem 15-10:$50,000,000 X.20 (down payment) - 10,000,000 $ 10,000,000 (down)$40,000,000 mortgage payment $40,000,000 $1,000 = $40,000 x $6.66 = $266,400 monthly payment

15-16 Problem 15-11: $140,000 - $28,000 = $112,000/$1,000 = 112 x $5.68 = $ x 360 = $229, $112,000 = $117, total interest

15-17 Problem 15-13: Payment Portion to - Balance of loan Number Interest Principal outstanding Monthly payment is: 1 $1, $47.00 $119, $120,000 ($1,422 -$1,375) ($120,000 - $47.00) $1,000 =120 x $11.85 = $1,422 2 $1, $47.54 $119, $120,000 x.1375 x 1/12 ($1,422 -$1,374.46) ($119, $47.54) = $1, $1,373.92$48.08 $119, $119, x.1375 x 1/12 = $1, $119, x.1375 x 1/12 = $1,373.92

15-18 Problem 15-14: a. 40 x $10.17 = $ x 300 = $122,040 - $40,000 = $82,040 b. 40 x $10.91 = $ x 300 = $130,920 - $40,000 = $90,920 c. 40 x $11.66 = $ x 300 = $139,920 - $40,000 = $99,920 d. 40 x $12.81 = $ x 300 = $153,720 - $40,000 = $113,720 e. $113,720 - $82,040 = $31,680 difference f. 40 x $12.65 = $506 x 360 = $182,160 - $40,000 = $142, x $9.91= $ x 360 = $142,704 - $40,000 = - 102,704 difference $ 39,456