Nominal rates of interest and discount


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1 1 Nominal rates of interest and discount i (m) : The nominal rate of interest payable m times per period, where m is a positive integer > 1. By a nominal rate of interest i (m), we mean a rate payable mthly, i.e. the rate of interest is i (m) /m for each mth of a period and not i (m). Thus, 1 + i = [1 + i(m) m ] m
2 2 and i = ] m [1 + i(m) 1 m i (m) = m[(1 + i) 1 m 1] d (m) : The nominal rate of discount payable m times per period. Thus, 1 d = [1 d(m) m ] m
3 3 and d = 1 [1 d(m) m ] m d (m) = m[1 (1 d) 1 m] = m[1 v 1 m] Generally, ] m [1 + i(m) = m [1 d(p) p ] p
4 4 If m = p, ] [1 + i(m) m = [1 d(m) m ] 1 = i(m) m d(m) m = i(m) m d(m) m.
5 5 Force of interest and discount δ t : the force of interest at time t, is defined as δ t = a (t) a(t) = A (t) A(t) = d dt ln A(t) = d dt ln a(t) Thus, t 0 δ r dr = t 0 d dr ln A(r)dr = ln A(r) t 0 = ln A(t) A(0)
6 6 Hence, e t 0 δ rdr = A(t) A(0) = a(t) a(0) = a(t) Note that n 0 A(t)δ t = n 0 A (t)dt = A(n) A(0). If δ t = δ, 0 t n, then e n 0 δ tdt = e nδ = a(n) = (1 + i) n
7 7 so that e δ = 1 + i = i = e δ 1 = δ = ln(1 + i) δ t: the force of discount at time t, is defined by δ t = d dt a 1 (t) a 1 (t). δ t = d dt a 1 (t) a 1 (t)
8 8 = a 2 (t) d dt a(t) a 1 (t) = a 2 (t)a(t)δ t a 1 (t) = δ t
9 9 Some properties of force of interest For simple interest δ t = = = d dt a(t) a(t) d dt (1 + it) 1 + it i, for 0 t. 1 + it
10 10 For simple discount δ t = δ t = = = d dt a 1 (t) a 1 (t) d dt d 1 dt (1 dt) 1 dt for 0 t < 1 d. i > δ i = e δ 1 = δ + δ2 2! + δ3 3! + δ4 4! +
11 11 lim m i(m) = δ ] m [1 + i(m) = e δ m i (m) [ ] = m em δ 1 [ δ = m m + 1 [ ] 2 δ + 1 [ ] 3 δ + ] 2! m 3! m = δ + δ2 2!m + δ3 3!m 2 +
12 lim m d(m) = δ 12
13 13 Varying interest Let i k denote the rate interest applicable for period k. We consider first the present value of an nperiod annuityimmediate. a n = (1 + i 1 ) 1 + (1 + i 1 ) 1 (1 + i 2 ) (1 + i 1 ) 1 (1 + i 2 ) 1 (1 + i n ) 1 n t = Π (1 + i s ) 1 s=1 t=1 The second pattern would be to compute present values using rate i k for the payment made at time k over all k
14 14 periods. In this case the present value becomes a n = (1 + i 1 ) 1 + (1 + i 2 ) (1 + i n ) n n = (1 + i t ) t t=1 Similarily, s n = (1 + i n ) + (1 + i n )(1 + i n 1 ) + n + (1 + i n )(1 + i n 1 ) (1 + i 1 ) = t=1 t Π (1 + i n s+1 ). s=1
15 15 Alternately, if the payment made at time k earns at rate i k over the rest of the accumulation period, we have s n = (1 + i n ) + (1 + i n 1 ) (1 + i 1 ) n n = (1 + i n s+1 ) t. t=1
16 16 Linear interpolation of (1 + i) n+k, v n+k (1 + i) n+k. = (1 k)(1 + i) n + k(1 + i) n+1 = (1 + i) n [(1 k) + k(1 + i)] = (1 + i) n (1 + ki) Similarily, v n+k = (1 d) n+k. = (1 k)(1 d) n + k(1 d) n+1 = (1 d) n [(1 k) + k(1 d)] = v n (1 kd)
17 17 Method of equated time Let amounts s 1, s 2,, s n be paid at times t 1, t 2,, t n respectively. The problem is to find t, such that s 1 + s s n paid at time t is equivalent to the payments. of s 1, s 2,, s n made separately. (s 1 + s s n )v t = s 1 v t 1 + s 2 v t s n v t n As a first approximation, t is calculated as a weighted average of the various times of payment, where the
18 18 weights are the various amounts paid, i.e. t = s 1t 1 + s 2 t s n t n s 1 + s s n = n s k t k k=1 n k=1 s k This approximation is denoted by t and is often called using the method of equated time. Consider s 1 quantities each equal to v t 1, s 2 quantities each equal to v t 2, and so forth until there are s n
19 19 quantities each equal to v t n. The arithmetic mean of these quantities is s 1 v t 1 + s 2 v t s n v t n s 1 + s s n The geometric mean of these quantities is v s 1 t 1 +s 2 t 2 + +s ntn s 1 +s 2 + +sn = v t And, s 1 v t 1 + s 2 v t s n v t n s 1 + s s n > v t
20 20 = s 1 v t 1 + s 2 v t s n v t n > (s 1 + s s n )v t This means that the value of t is always greater than the true value of t.
21 21 Basic annuities annuityimmediate a n = v + v v n 1 + v n = v 1 vn 1 v = v 1 vn iv = 1 v i s n = 1 + (1 + i) + (1 + i) (1 + i) n 1
22 22 = (1 + i)n 1 (1 + i) 1 = (1 + i)n 1 i 1 = ia n + v n s n = a n (1 + i) n 1 a n = 1 s n + i 1 s n + i = i (1 + i) n 1 + i
23 23 = i + i(1 + i)n i (1 + i) n 1 i = 1 v n = 1 a n annuitydue ä n = 1 + v + v v n 1 = 1 vn 1 v
24 24 = 1 vn iv = 1 vn d s n = (1 + i) + (1 + i) (1 + i) n = (1 + i) (1 + i)n 1 (1 + i) 1 = (1 + i)n 1 iv
25 25 = (1 + i)n 1 d s n = ä n (1 + i) n 1 = 1 + d ä n s n ä n = a n (1 + i) s n = s n (1 + i) ä n = 1 + a n 1 s n = s n+1 1
26 26 Perpetuities a = v + v 2 + v 3 + v = 1 v = v iv = 1 i Alternatively, we have 1 v n a = lim a n n = lim n i = 1 i.
27 27 For a perpetuitydue, we have ä = 1 d.
28 28 Nonstandard terms a n+k, 0 < k < 1 a n+k = 1 vn+k i = 1 vn + v n v n+k i [ (1 + i) = a n + v n+k k ] 1 i = a n + Xv n+k
29 29 Yiele rate Consider a situation in which an investor makes deposits or contributions into an investment of C 0, C 1,, C n at times 0, 1, 2,, n. Thus, we can denote the returns as R 0, R 1,, R n at times 0, 1,, n. Then we have R t = C t for t = 0, 1,, n. Assume that the rate of interest per period is i. Then the net present value at rate i of investment returns by the discounted cash flow technique is denoted be P (i) and is
30 30 given by P (i) = n t=0 v t R t. An important special case of this formula is the one in which P (i) = 0, P (i) = n t=0 v t R t = 0. The rate of interest i which satisfies P (i) = 0 is called the yield rate on the investment. Stated in words:
31 The yield rate is that rate of interest at which the present value of returns from the investment is equal to the present value of contributions into the investment. It is often called the internal rate of return. 31
32 32 Dollarweighted rate of interest Consider finding the effective rate of interest earned by a fund over one measurement period. We make the following definitions: A = the amount in the fund at the beginning of the period B = the amount in the fund at the end of the period I = the amount of interest earned during the period
33 33 C t = the net amount of principal contributed at time t (positive or negative), where 0 t 1 C = the total net amount of principal contributed during the period C = t C t a i b = the amount of interest earned by 1 invested at time b over the following period of length a, where a 0, b 0, and a + b 1
34 34 Note that B = A + C + I I = ia + t C t 1 t i t. Assuming compound interest throughout the period, we have 1 ti t = (1 + i) 1 t 1. = (1 t)i.
35 35 Hence, i. = A + t I C t (1 t). Assume that the net principal contributions occur at time t = 1, we have 2 i. = I A +.5C = I A +.5(B A I) = 2I A + B I If it is known that net principal contributions occur at
36 36 time k on the average, then i. = I ka + (1 k)b (1 k)i.
37 37 Timeweighted rate of interest Let the amount of the net contribution to the fund(positive or negative) at time t k be denoted by C k for k = 1, 2,, m 1. Let the fund values immediately before each contribution to the fund be denoted by B k for k = 1, 2,, m 1. Also the fund value at the beginning of the year is denoted by B 0 = B 0, while the fund value at the end of year is denoted by B m = B 1. The yield rates over the m subintervals by the
38 38 timeweighted method are given by 1 + j k = B k B k 1 + C k 1 The overall yield rate for the entire year is then given by 1 + i = (1 + j 1 )(1 + j 2 ) (1 + j m ) i = (1 + j 1 )(1 + j 2 ) (1 + j m ) 1
39 39 Finding the outstanding loan balance L = B 0 : The original loan balance. B t : The outstanding loan balance at time t. The methods of finding the outstanding loan balance Consider a loan of L = a n at interest rate i per period being repaid with payments of 1 a the end of each period for n period.
40 40 1. Prospective method 2. Retrospective method B p t = a n t B r t = a n (1 + i) t s t
41 41 Amortization schedules L = B 0 = a n = ia n = 1 v n I 1 P 1 = 1 (1 v n ) = v n B 1 I 2 P 2 B 2 = a n v n = a n 1 = ia n 1 = 1 v n 1 = 1 (1 v n 1 ) = v n 1 = a n 1 v n 1 = a n 2
42 42 I t P t B t. = ib t 1 = ia n t+1 = 1 v n t+1 = 1 (1 v n t+1 ) = v n t+1 = a n t+1 v n t+1 = a n t
43 Payment Interest Principal Outstanding Period amount paid repaid loan balance 0 a n v n v n a n v n 1 v n 1 a n t 1 1 v n t+1 v n t+1 a n t..... n v 2 v 2 a 1 n 1 1 v v a 1 v = 0 Total n n a n a n 43
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