1802_0910_LN2[1]

1802_0910_LN2[1] - 2010 THE UNIVERSITY OF HONG KONG...

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Unformatted text preview: 2010 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1802 Financial Mathematics LN2: Chapters 2-4 Chapter 2: Examples of Interest Problems Problem 1. In return for payments of 2,000 at the end of four years and 5,000 at the end of ten years, an investor agrees to pay 3,000 immediately and to make an additional payment at the end of three years. Find the amount of the additional payment if i (4) = . 06. Soln: (Unknown amount of payment) 3 , 000 + Xv 3 = 2 , 000 v 4 + 5 , 000 v 10 with v = (1 + i ) − 1 and 1 + i = (1 + i (4) 4 ) 4 which yields X = 1 , 593. Note that if other comparison dates are chosen, the same answer is obtained. But the same cannot be said of simple interest. For instance, we can use the following equations of values to compute X : or 3 , 000(1 + i ) 3 + X = 2 , 000 v + 5 , 000 v 7 , or 3 , 000(1 + i (4) 4 ) 12 + X = 2 , 000(1 + i (4) 4 ) − 4 + 5 , 000(1 + i (4) 4 ) − 28 . Problem 2. Find how long 1,000 should be left to accumulate at 6% effective in order that it will amount to twice the accumulated value of another 1,000 deposited at the same time at 4% effective. Soln: (Unknown time) 1 , 000(1 . 06) t = 2 × 1 , 000(1 . 04) t ⇒ (1 . 06 / 1 . 04) t = 2 ⇒ t ln(1 . 06 / 1 . 04) = ln 2 ⇒ t = ln 2 / ln(1 . 06 / 1 . 04) = 36 . 39 Problem 3. Find the nominal rate of interest convertible semiannually at which the accumulated value of 1,000 at the end of 15 years is 3,000. Soln: (Unknown rate of interest) 1 , 000(1 + i (2) 2 ) 30 = 3 , 000. Let j = i (2) / 2. Then, we have 1 , 000(1 + j ) 30 = 3 , 000. Define f ( j ) = 1 , 000(1 + j ) 30- 3 , 000 . Find j such that f ( j ) = 0. From the interest tables, we obtain f (0 . 035) = 1 , 000(1 . 035) 30- 3 , 000 = 2 , 806 . 79- 3 , 000 =- 193 . 21, f (0 . 04) = 1 , 000(1 . 04) 30- 3 , 000 = 3 , 243 . 4- 3 , 000 = 243 . 4. 1 Linear interpolation in the interest tables leads to j = 0 . 035 + 0 . 005 ( 0 + 193 . 21 243 . 4 + 193 . 21 ) = 0 . 03721 , which gives i (2) = 2 j = 2(0 . 03721) = 0 . 07442. Note that there are other methods to find the unknown interest rate. For example, successive approximation and Newton-Raphson method . Method of Equated Time The timing of future payments is important in financial analysis. Some indices are needed to measure the timing of future payments. Let amounts s 1 ,s 2 , ··· ,s n be paid at times t 1 ,t 2 , ··· ,t n , respectively. The problem is to find time t such that s 1 + s 2 + ··· + s n paid at time t is equivalent to the payments of s 1 ,s 2 , ··· ,s n made separately. ¯ t = s 1 t 1 + s 2 t 2 + ··· + s n t n s 1 + s 2 + ··· + s n = n ∑ k =1 s k t k n ∑ k =1 s k , which is a weighted average of the various times of payment with weight being the amount paid. Note that the exact method gives ( s 1 + s 2 + ··· + s n ) v t = s 1 v t 1 + s 2 v t 2 + ··· + s n v t n ....
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1802_0910_LN2[1] - 2010 THE UNIVERSITY OF HONG KONG...

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