IOE+201+Homework+3+Solutions

IOE+201+Homework+3+Solutions - IOE 201 Economic Decision...

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1 IOE 201 - Economic Decision Making Homework #3 - Solutions 1. Equal payments series: Mortgage amount P = \$260,000 - \$50,000 = \$210,000 Interest rate per month i = 0.85/12 = 0.0070833. Number of monthly payments N = 15 ×12 = 180 months. Monthly payment A is: - + + = 1 ) 1 ( ) 1 ( N N i i i P A - + + = 1 )] 12 085 . 0 ( 1 [( )] 12 085 . 0 ( 1 )[ 12 085 . 0 ( 000 , 210 \$ 180 180 = \$ 2,067.95 2. (a) Linear gradient series of payments: Amount needed now (present worth, P ) is + - - + = N N i i iN i G P ) 1 ( 1 ) 1 ( 2 = + - - + 10 2 10 ) 07 . 0 1 ( ) 07 . 0 ( 1 10 ) 07 . 0 ( ) 07 . 0 1 ( 200 \$ = \$5,543.11 (b) Let X = amount required. Company sets aside \$ X now and \$ X at end of first year. From (a), present worth is \$5543.11 = \$ + + 07 . 0 1 X X = (1.9346) X Hence, X = \$(5543.11)/(1.9346) = \$2,865.28 3. (a) Geometric gradient series of withdrawals. First withdrawal A 1 = W : For i g , P = \$2000 = \$ - + + - - ) ( ) 1 ( ) 1 ( 1 1 g i i g A N N = \$ - + + - - ) 05 . 0 08 . 0 ( ) 08 . 0 1 ( ) 05 . 0 1 ( 1 7 7 W = (5.9657) W . Hence, W = \$(2000)/(5.9657) = \$335.25 (b) For i = g , P = \$2000 = \$ + i N A 1 1 = \$ 05 . 1 7 W . Hence, W = \$ 7 05 . 1 2000 = \$300 4. Price of a \$100 item increases to \$100(1 + 0.05)(1 + 0.08) =

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