E120_SU11_Midterm_Solutions

E120_SU11_Midterm_Solutions - E120 Midterm Solutions 1. (a)...

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Unformatted text preview: E120 Midterm Solutions 1. (a) (1 + r m ) 12 = 1 + EAR = (1 + 12% 4 ) 4 . Solving we get r m = 0 . 990163405%. (b) p 1+ r m + ... + p (1+ r m ) 360 = 100 , 000, we find p = 100 , 000 98 . 0838792 = $1019 . 5355323. (c) If we were to have an advanced calculator, we could have quickly found the real interest rate using the following way: p 1+ˆ r m + ... + p (1+ˆ r m ) 360 = 97 , 000. The above reduces to finding ˆ r m that satisfies 1- ( 1 1+ˆ rm ) 360 ˆ r m = 97 , 000 P = ⇒ 1+ˆ r m = 1 . 02425117 = ⇒ 1+ˆ r quarter = 1 . 03104336 = ⇒ APR = 12 . 4173347%. To show that 14% is an upper bound to the real interest rate is an easier problem, and is doable even without an advanced calculator. To proceed, consider the following function f ( r ): f ( r ) = p 1+ r + ... + p (1+ r ) 360 , where p = $1019 . 5355323. Note that with r = r m = 0 . 9901634% (i.e. APR of 12% with quarterly com- pounding), f ( r m ) = 100 , 000....
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This note was uploaded on 08/15/2011 for the course ENGR 120 taught by Professor Alder during the Summer '11 term at Berkeley.

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E120_SU11_Midterm_Solutions - E120 Midterm Solutions 1. (a)...

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