Solutions+Problem+Set+4-1

# Solutions+Problem+Set+4-1 - E 120 Problem Set 4 Solutions...

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E 120 : Problem Set 4 Solutions - Fall 2010 Problem 1 (a) We have that L = \$2000 0 . 06 12 × h 1 - ( 1 + 0 . 06 12 ) - 180 i + \$2000 0 . 07 12 × h 1 - ( 1 + 0 . 07 12 ) - 180 i × ( 1 + 0 . 06 12 ) - 180 = \$327 , 676 . 73 (b) Closing costs are given by Closing costs = Loan Amount - (3 points) - (Present Value of Monthly Payments) = L - (0 . 03 × L ) - \$2000 0 . 07 12 × h 1 - ( 1 + 0 . 07 12 ) - 360 i = 0 . 97 × \$327 , 676 . 73 - \$300 , 615 . 14 = \$17 , 231 . 29 Problem 2 True. In this question, the subscript denotes the unit of time. Hence, given a positive interest rate of r % per unit time, it is suﬃcient for us to show that the present value of the ﬁrst cash ﬂow at time 1 is greater than the present value of the second cash ﬂow at time 1, i.e. a 1 + a 2 1 + r > b 1 + b 2 1 + r ⇐⇒ a 1 (1+ r )+ a 2 > b 1 (1+ r )+ b 2 ⇐⇒ ( a 1 + a 2 )+ ra 1 > ( b 1 + b 2 )+ rb 1 which is clearly true, since a 1 > b 1 , a 1 + a 2 > b 1 + b 2 and r > 0. Problem 3

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Solutions+Problem+Set+4-1 - E 120 Problem Set 4 Solutions...

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