ACTS 4308 Instructor: Natalia A. Humphreys SOLUTION TO HOMEWORK 4 Section 7: Annuities whose payments follow a geometric progression. Section 8: Annuities whose payments follow an arithmetic progression. Problem 1 Suppose you buy a perpetuity-due with varying annual payments. The first 5 payments are constant and equal to 12. Starting the sixth payment, the payments start to increase so that each year’s payment is K% larger than the previous years payment. At an annual effective interest rate of 7%, the perpetuity has a present value of 305. Calculate K, given K < 7. A) 3.1 B) 3.3 C) 3.5 D) 3.7 E) 3.9 Solution. 12 0 12 1 12 2 12 3 12 4 12 · (1 + K) 12 · (1 + K)2 5 6 ... ... This perpetuity consists of two cash flows: (1) Annuity-due of 12 for 4 years (2) Annuity whose payments follow a geometric progression starting with payment of 12 at the beginning of the 5th year Hence, 12 9.7956 P V = 12¨ a4 i + · 1.07−3 = 43.4918 + = 305 0.07 − K 0.07 − K 9.7956 = 261.5082 ⇔ 0.07 − K = 0.037458 ⇔ K = 0.032542 ≈ 3.3% ⇒ B 0.07 − K Problem 2 The price of a stock is $100 per share. Annual dividends are paid at the end of each year forever; the first dividend is $K and the expected growth rate for the dividends is 2% per year. The annual effective interest rate is 5%. Calculate K. Solution. 100 = K K = ⇒K=3 0.05 − 0.02 0.03 Problem 3 Morris makes a series of payments at the end of each year for 19 years. The first payment is 100. Each subsequent payment through the tenth year increases by 5% from the previous payment. After the tenth payment, each payment decreases by 5% from the previous payment. Calculate the present value of these payments using an annual effective rate of 7%. Solution. 0 100 1 100 · 1.05 2 100 · 1.052 3 We split our cash flow into two cash flows: ... ... 100 · 1.059 10 100 · 1.058 11 ... ... 100 19 ACTS 4308. AU 2014. SOLUTION TO HOMEWORK 4. (1) Annuity whose payments follow a geometric progression starting with payment of 100 at time 1, with multiplier q = 1.05 for 10 years; (2) Annuity whose payments follow a geometric progression starting with payment of 100 · 1.058 1 at time 11, with multiplier q = 1.05 for 9 years, discounted for 10 years. The present value of the first cash flow is: 10 1 − 1.05 1.07 P V1 = 100 · = 859.7628 0.07 − 0.05 Let us now calculate the present value of the second cash flow. Let K = 100 · 1.058 , q = 1/1.05, v = 1/1.07. Then, P V2 = v 10 Kv + Kqv 2 + Kq 2 v 3 + · · · + Kq n−1 v n = v 10 · Kv(1 + qv + (qv)2 + · · · + (qv)n−1 ) = 1 − (qv)n 10 ,⇒ Kv · =v 1 − qv 9 1 1 − 1.05·1.07 1 −10 8 P V2 = (1.07) · 100 · 1.05 · = (1.07)−10 · 815.7041 = 414.6626 · 1 1.07 1 − 1.05·1.07 Hence, P V = P V1 + P V2 = 859.7628 + 414.6626 = 1274.4254 ≈ 1274 Alternatively, 0 100 1 100 · 1.059 · 0.95 11 100 · 1.05 2 100 · 1.059 · 0.952 12 100 · 1.052 3 100 · 1.059 10 ... ... ... ... 100 · 1.059 · 0.959 19 Again, We split our cash flow into two cash flows: (1) Annuity whose payments follow a geometric progression starting with payment of 100 at time 1, with multiplier q = 1.05 for 10 years; (2) Annuity whose payments follow a geometric progression starting with payment of 100 · 1.059 · 0.95 at time 11, with multiplier q = 0.95 for 9 years, discounted for 10 years. The present value of the first cash flow is: 1.05 10 1 − 1.07 P V1 = 100 · = 859.7628 0.07 − 0.05 Let us now calculate the present value of the second cash flow. Let K = 100 · 1.059 · 0.95, q = 1/1.05, v = 1/1.07. Then, P V2 = v 10 Kv + Kqv 2 + Kq 2 v 3 + · · · + Kq n−1 v n = v 10 · Kv(1 + qv + (qv)2 + · · · + (qv)n−1 ) = 1 − (qv)n 10 ,⇒ =v Kv · 1 − qv 9 1 − 0.95 1 −10 9 1.07 P V2 = (1.07) · 100 · 1.05 · 0.95 · · = 410.29 0.95 1.07 1 − 1.07 Hence, P V = P V1 + P V2 = 859.7628 + 410.29 = 1270.05 Problem 4 Common stock S pays a dividend of 35 at the end of the first year, with each subsequent annual dividend being 4% greater than the preceding one. Mary purchases the stock at a theoretical price Copyright ©Natalia A. Humphreys, 2014 Page 2 of 5 ACTS 4308. AU 2014. SOLUTION TO HOMEWORK 4. to earn an expected annual effective yield of 8%. Immediately after receiving the 12th dividend, Mary sells the stock for a price of P . Her annual effective yield over the 12-year period was 6.75%. Calculate P . Solution. The initial price of the stock is S= 35 · 1.042 3 35 · 1.04 2 35 1 0 35 = 875 0.08 − 0.04 35 · 1.0411 12 ... ... Therefore, −12 P · 1.0675 1.04 12 1 − 1.0675 = 0.4567P + 342.2178 = 875 ⇒ P = 1166.7138 ≈ 1166.71 + 35 · 0.0675 − 0.04 Problem 5 Vernon buys a 30-year decreasing annuity-immediate with annual payments of 30, 29, 28, . . . , 1. On the same date, Elizabeth buys a perpetuity-immediate with annual payments. For the first 31 years, payments are 1, 2, 3, . . . , 31. Thereafter, payments remain constant at 31. At an annual effective interest rate of i, both annuities have a present value of X. Calculate X. A) 272 B) 281 C) 288 D) 292 E) 298 Solution. Vernon: 30 1 0 29 2 28 3 P VV = (Da) 30 ... ... = 2 29 30 − a30 i 1 30 =X Elizabeth: 1 1 0 2 2 ... ... 28 28 29 29 30 30 31 30 31 30 31 ... a ¨ − 30v 30 31v 30 a ¨ a ¨ + v 30 31 v 30 = 30 + = 30 + = 30 =X 30 i i i i i i Equating these two values, we obtain: 30 − a30 = a ¨30 + v 30 . n Recall that: a ¨n = an + 1 − v . Hence, P VE = (Ia) + (1 + i)−30 · a ¨30 = a30 + 1 − v 30 ⇒ 30 − a30 = a30 + 1 − v 30 + v 30 ⇒ 2a30 = 29 a30 = 14.5 ⇒ i = 5.5211% ⇒ X = 30 − 14.5 = 280.7431 ≈ 281 ⇒ B 0.05521 Problem 6 You are given two series of payments. Series A is a perpetuity with payments of 1 at the end of each of the first 2 years, 2 at the end of each of the next 2 years, 3 at the end of each of the next 2 years, and so on. Series B is a perpetuity with payments of K at the end of each of the first 3 years, 2K at the end of each of the next 3 years, 3K at the end of each of the next 3 years, and so on. The present values of the two series of payments are equal. Calculate K. A) 3i 2 B) 3d 2 C) a 3 a 2 D) a 3 a ¨ 2 E) s¨ 3 s¨ 2 Solution. Series A: Copyright ©Natalia A. Humphreys, 2014 Page 3 of 5 ACTS 4308. AU 2014. SOLUTION TO HOMEWORK 4. 1 1 0 1 2 2 3 2 4 3 5 3 6 ... ... i(0.5) 0.5 1+j (1 + i)2 (1 + i)2 1 1 = = = + 2 = j j j2 ((1 + i)2 − 1)2 (2 + i)2 i2 P VA = (Ia)∞ j + (1 + i) (Ia)∞ j , where j = (Ia)∞ j (1 + j)0.5 = 1 + i ⇒ j = (1 + i)2 − 1 = (1 + i − 1)(1 + i + 1) = i(2 + i) ⇒ (1 + i)2 (1 + i)2 P VA = (2 + i) (Ia)∞ j = (2 + i) · = (2 + i)2 i2 (2 + i)i2 Series B can be thought as a payment of Ks3 i , 2Ks3 i , 3Ks3 i , . . . at the end of years 3, 6, 9, etc. K 1 0 K 2 K 3 2K 4 2K 5 2K 6 ... ... Therefore, P VB = Ks3 i 1 1 + 2 j j 1 , where j = i( 3 ) 1 3 1 , (1 + j) 3 = 1 + i, 1 + j = (1 + i)3 , j = (1 + i)3 − 1 (1 + i)3 − 1 1 1 1+j (1 + i)3 , + 2 = = i j j j2 ((1 + i)3 − 1)2 (1 + i)3 (1 + i)3 (1 + i)3 − 1 · = K P VB = K · i ((1 + i)3 − 1)2 ((1 + i)3 − 1)i s3 i = Since the present values of the two series of payments are equal (1 + i)2 (1 + i)3 1 1+i (1 + i)3 − 1 = K ⇔ = K ⇒ K = (2 + i)i2 ((1 + i)3 − 1)i (2 + i)i (1 + i)3 − 1 (2 + i)(1 + i)i 3 3 s3 i (1 + i) − 1 (1 + i) − 1 i (1 + i)2 − 1 = (2 + i)i ⇒ K = = · = ((1 + i)2 − 1)(1 + i) i ((1 + i)2 − 1)(1 + i) s¨2 i P VA = P VB ⇔ K= a (1 + i)3 − 1 ((1 + i)3 − 1)v 3 1 − v3 1 − v3 = = = = 3i ⇒C 2 2 3 2 2 2 ((1 + i) − 1)(1 + i) ((1 + i) − 1)(1 + i)v ((1 + i) − 1)v 1−v a2 i Problem 7 At an annual effective interest rate of i, the present value of a perpetuity-immediate starting with a payment of 500 in the first year and increasing by 20 each year thereafter is 1475.56. Calculate i. A) 32.5% B) 35.0% C) 37.5% D) 40.0% E) 42.5% Solution. This perpetuity can be represented as the sum of two perpetuities: a level perpetuityimmediate of 500, plus an increasing perpetuity with the first payment of 20 at the end of the second year increasing by 20 in each subsequent year: 500 1 0 500 + 20 500 + 30 2 3 ... ... = 0 500 1 500 2 500 3 Copyright ©Natalia A. Humphreys, 2014 ... ... + 0 1 20 2 40 3 60 3 ... ... Page 4 of 5 ACTS 4308. AU 2014. SOLUTION TO HOMEWORK 4. Therefore, 500 500 1 20 1 −1 PV = + (1 + i) · 20 · (Ia)∞ i = + + 2 = i i 1+i i i 500 20 500 20 1 + i = + · 2 = + 2 = 1475.56 ⇔ 73.778i2 − 25i − 1 = 0 i 1+i i i i 25 + 30.3333 D = 625 + 4 · 73.778 = 920.112 = 30.33332 , i = = 0.375 = 37.5% ⇒ C 2 · 73.778 Problem 8 Payments of 100 are made at the end of each month for a year. These payments earn interest at a nominal rate of j% convertible monthly. The interest is immediately reinvested at a nominal rate of 12% convertible monthly. At the end of the year, the accumulated value of the 12 payments and the reinvested interest is 1295.55. Calculate j. Solution. Interest payments are: I = 100 · 100 · 12 + I · (Is) 11 j 12 0 100 1 100 2 100 3 0 1 I 2 2I 3 100 12 14I 14 = 1295.55 ⇔ 1200 + I · 1% 68.2503I = 95.55 ⇒ I = 1.4 ⇒ j = 11I 12 s11 1% − 12 0.01 = 1295.55, s11 1% = 12.6825 ⇒ 12 · 1.4 = 0.168 100 Problem 9 Payments of X are made at the beginning of each year for 15 years. These payments earn interest at the end of each year at an annual effective rate of 8%. The interest is immediately reinvested at an annual effective rate of 5%. At the end of 15 years, the accumulated value of the 15 payments and the reinvested interest is 4000. Calculate X. A) 147 B) 152 C) 157 D) 162 E) 167 Solution. X 0 X 1 X 2 X 14 15 0 I 1 2I 2 14I 14 15I 15 I = 0.08X, 15X + I · (Is) 15 = 4000 ⇔ 15X + 0.08X · s16 5% − 16 5% 0.05 = 4000, s16 5% = 23.6575 ⇒ 27.252X = 4000 ⇒ X = 146.78 ≈ 147 ⇒ A Problem 10 Mike receives a cash flow of 100 today, 200 in two years, and 100 in four years. The present value of this cash flow is 378 at an annual effective rate of interest i. Calculate i. A) 2.91% B) 3.91% C) 4.91% D) 5.91% E) 6.91% Solution. Copyright ©Natalia A. Humphreys, 2014 Page 5 of 5 ACTS 4308. AU 2014. SOLUTION TO HOMEWORK 4. 100 0 1 200 2 3 100 4 100 + 200v 2 + 100v 4 = 378, x = v 2 ⇒ x2 + 2x + 1 = 3.78 ⇔ (1 + x)2 = 3.78, x = 0.94422 v = 0.9717 ⇒ i = 2.91% ⇒ A Copyright ©Natalia A. Humphreys, 2014 Page 6 of 5