Math 618 Notes Chapter 1 • Interest rate j for a period: X = value at the beginning, Y = value at the end. Then Y −X X = j or Y = (1 + j)X. Example The interest rates are i1 for the first year, i2 the second year, and i3 the third year. If deposit $100 in the beginning, what is the accumulated amount A at the end of the third year? Answer: A = 100(1 + i1)(1 + i2)(1 + i3) If i1 = i2 = i3 = i, then A = 100(1 + i)3. • Effective annual rate Example If the monthly rate is .5%, the original amount P becomes P(1 + .005)12 = P(1.0616778) at the end of the year. This is equivalent to an annual rate of 6.16778%, the effective annual rate. • Simple (annual) rate i: α(t) = 1 + ti, A(t) = A(0)(1 + ti) where t is usually m/12 (in months) or d/365 (in days). • Example (#1.1.4) At the end of the 10th year, Joe’s amount is AJ = 10(1 + 10·0.11) + 30(1 + 5·0.11) = 67.5 and Tina’s amount is AT = 10(1.0915)10−n + 30(1.0915)10−2n AT = AJ =⇒ 10(1.0915)10−n + 30(1.0915)10−2n = 67.5 =⇒ (1.0915)10x + 3(1.0915)10x2 = 6.75 (where x = (1.0915)−n) =⇒ ··· =⇒ x = .8157901066 (the other solution is x = −1.149123440). So, (1.0915)−n = .8157901066 =⇒ n = 2.325430527. • Accumulated amount A(t) and rate: – A(t) = A(0)(1 + i)t, A(t) = A(0)(1 + t·i), in general A(t) = α(t)A(0) – The rate for the time period from t1 to t2 is A(t2)−A(t1)A(t1) . For example if t is in year, then in+1 = A(n+1)−A(n)A(n) is the (n + 1)st year’s rate. Example An account is paying simple annual rate of 6%. If you deposit certain amount in this account for 5 years, what is the the effective rate for year 5? A(t) = A(0)(1 + .06t), i5 = A(5)−A(4)A(4) = 0.30−0.241.24 = 0.048387... ≈ 4.84% • Present value: A(t) = A(0)(1 + i)t =⇒ A(0) = A(t)(1 + i)−t = A(t)vt where v = 11+i. Examples (#1.2.5 c)) (#1.2.9) • Nominal rate i(m): j = i(m)/m Example $2500 deposited for 10 years at 4% nominal rate, find the accumulated amount and the effective annual rate if interest is a) compounded quarterly. j = .04/4, t = 40, A = 2500(1 + .04/4)40 = 3722.16. 2500(1 + i)10 = 3722.16 =⇒ i = .04060... b) compounded monthly. j = .04/12, t = 120, A = 2500(1 + .04/12)120 = 3727.08. 2500(1 + i)10 = 3727.08 =⇒ i = .04074... c) compounded daily. j = .04/365, t = 365×10 = 3650, A = 2500(1 + .04/365)3650 = 3729.48. 2500(1 + i)10 = 3729.48 =⇒ i = .040808... • Annual rate of discount d: X = Y (1−d), or d = Y −XY Y = X(1 + i) =⇒ 1−d = 11 + i or 1 + i = 11−d. Solving, i = d1−d or d = i1 + i. • Simple rate of discount: X = Y (1−td) • Nominal rate of discount d(m): X = Y (1−d(m)/m)m. Example (1.4.4) • Force of interest: diamondmath i(m) = mA(t + 1 m)−A(t) A(t) diamondmath i(∞) = limm→∞mA(t + 1 m)−A(t) A(t) = Aprime(t) A(t) = δt diamondmath A prime(t) A(t) = δt =⇒ lnA(t1)−lnA(t0) = integraldisplay t1 t0 δt dt =⇒ A(t1) = A(t0)e R t1 t0 δt dt Example (1.5.3)