MATH2805 (Spring 2021) Solutions to Homework #1 1. (a) For the given A(t), we have a(t) = 1 A(t) = (t2 + 2t + 3). A(0) 3 (b) First, a(0) = 1. Second, noting that a0 (t) = (2t + 2)/3 > 0 for any t > 0, a(t) is an increasing function. Third, a(t) is a continuous function. Therefore, a(t) satisfies the three properties of an accumulation function. (c) In = A(n) − A(n − 1) = 2n + 1. 2. For the form a(t) = at2 + b, by letting t = 0 we have a(0) = b. This shows that b = 1. Also by A(3) = 100a(3) = 172, we a(3) = 9a + 1 = 1.72. This leads to a = 0.08 so that the accumulated function is a(t) = 0.08t2 + 1. Thus, the accumulated value at time 10 of $100 invested at time 5 is A(5) = 100a(5) = $300. 3. (a) For A(t) = 100 + 5t, we have A(5) − A(4) 1 = , A(4) 24 A(10) − A(9) 1 = . = A(9) 29 i5 = i10 (b) For A(t) = 100(1.1)t , we have A(5) − A(4) = 0.1, A(4) A(10) − A(9) = = 0.1. A(9) i5 = i10 4. By definition, we have A(7) = A(4)(1 + i5 )(1 + i6 )(1 + i7 ) = 1000(1 + 0.01 ∗ 5)(1 + 0.01 ∗ 6)(1 + 0.01 ∗ 7) = 1190.91. 1 5. (a) Let the interest rate be i. We have 1 500(1 + 2 ∗ i) = 615. 2 This leads to i = 9.2%. (b) Let the number of years be n. We have 500(1 + 7.8% ∗ n) = 630. This leads to n = 3 13 . 6. For simple interest, we have a(t) = 1 + it. The effective rate at time n is in = i 4% a(n) − a(n − 1) = = = 2.5%. a(n − 1) 1 + i(n − 1) 1 + 4%(n − 1) This results in n = 16. 7. For convenience, we define f (i) = (1 + i)t − (1 + it). Then, f (0) = 0 and f 0 (i) = t(1 + i)t−1 − t = t[(1 + i)t−1 − 1]. (a) If 0 < t < 1, then f 0 (i) < 0, i.e. f (i) is decreasing in (0, 1), so f (i) < f (0) = 0, i.e. (1 + i)t < 1 + it. (b) If t > 1, then f 0 (i) > 0, so f (i) is increasing in (0, 1) and f (i) > f (0) = 0, i.e. (1 + i)t > 1 + it. 8. For compound interest, we have 600(1 + i)2 = 600 + 264. This leads to i = 0.2. Consequently, the accumulated value of $2000 invested for three years is 2000((1 + 0.2)3 = $3456. 9. We have 1 = v n + v 2n . This leads to √ n v = Further, (1 + i)2n = (v n )−2 5−1 . 2 √ 2 3 + 5 = (√ )2 = . 2 5−1 10. We have 500(1 + i)30 = 4000. This leads to (1 + i)10 = 2, or equivalently, v 10 = 0.5. Further, the sum of present value is 10, 000(v 20 + v 40 + v 60 ) = 10, 000(0.52 + 0.54 + 0.56 ) = 3281.25. 2 11. From the question, we have Ai = 336 and Ad = 300. Note also that i= d . 1−d Solving the above equations, we have d = 3/28 and A= 300 28 = 300 × = 2800. d 3 12. Define f (d) = (1 − d)t − (1 − dt), then f (0) = 0 and f 0 (d) = −t(1 − d)t−1 + t = t[1 − (1 − d)t−1 ]. (a) If 0 < t < 1, then f 0 (d) < 0, i.e. f (d) is decreasing in (0, 1), so f (d) < f (0) = 0, i.e. (1 − d)t < 1 − dt. (b) If t > 1, then f 0 (d) > 0, so f (d) is increasing in (0, 1) and f (d) > f (0) = 0, i.e. (1 − d)t > 1 − dt. h 13. (a) 1 + i(3) 3 h (b) 1 + i(6) 6 i3 i6 h = 1− d(4) 4 h = 1− d(2) 2 i−4 i−2 − 34 (3) i d(4) = 4 1 − 1 + 3 . − 13 d(2) (6) i =6 1− 2 −1 . ⇒ ⇒ 14. The accumulated value of $100 after two years is 8 6% = 112.65. 100 × 1 + 4 15. By the formula 1 + i(m) /m = (1 − d(m) /m)−1 , we have i(m) d(m) i(m) d(m) − = · m m m m Plugging in the values, 0.1844144 0.1802608 0.1844144 0.1802608 − = · m m m m This leads to m = 8. 0 (t) 16. (a) By definition, we have δt = aa(t) = dtd ln a(t). Then, Z n Z n d δt dt = ln a(t)dt 0 0 dt = ln a(n) − ln a(0) = ln(1 + i)n = − ln v n . 3 (b) As δt = A0 (t) , A(t) n Z 0 we have Z n A0 (t) A(t) A0 (t)dt dt = A(t) 0 = A(n) − A(0) = [A(n) − A(n − 1)] + · · · + [A(1) − A(0)] = I1 + I2 + · · · + In . 17. For the given question, we have (1 + i)3 = (1 − 0.08)−1 (1 − 0.07)−1 (1 − 0.06)−1 . This leads to i = 7.53%. 18. For ease of notation, let wi = si /(s1 + s2 + · · · + sn ) for i = 1, . . . , n. Following the lecture note, the value of t̄ is t̄ = s1 t1 + s2 t2 + . . . + sn tn = w1 t1 + w2 t2 + · · · + wn tn . s1 + s2 + · · · + sn Further, we have v t̄ = v w1 t1 +w2 t2 +···+wn tn = (v t1 )w1 (v t2 )w2 · · · (v tn )wn . On the other side, for the true value of t, vt = s1 v t1 + s2 v t2 + · · · + sn v tn = w1 v t1 + w2 v t2 + · · · + wn v tn . s1 + s2 + · · · + sn Treating v t1 , v t2 , . . . , v tn as the variables, together with the well-known inequality that “the arithmetic mean is always greater than or equals to the geometric mean” (also referred to as the Weighted AM-GM inequality), we have (v t1 )w1 (v t2 )w2 · · · (v tn )wn ≤ w1 v t1 + w2 v t2 + · · · + wn v tn . or equivalently, v t̄ ≤ v t . Finally, noting that 0 < v < 1, we have t̄ ≥ t. 4