Distribution Date: Monday, February 15th, 2010 Corrected on 01 March, 2001; subject to further corrections.
Solutions were to be submitted by Monday, February 08th, 2010 (subject to correction)
1. Short, direct calculations: Show all your work in each case.
(a) Determine whether it is better to pay 10,200 cash, or 900 per month for one year at a nominal annual rate of 10%, compounded monthly
i. if the monthly payments are at the beginning of the month, beginning imme-diately; and
ii. if the monthly payments are at the end of the month, first payment one month from now.
(b) A home has just been bought, and the purchaser has signed a mortgage agreeing to pay 1,600 at the end of each half-month for 15 years, where i(24) = 6.5%. Determine
i. the amount of the mortgage loan; and
ii. the cash price of the home, assuming that the down payment was 60,000.
Solution:
(a) i. The present value of the monthly payments is 900¨a12 10
12% = 900 · 1 + 0.1
ii. The present value of the monthly payments is 900a12 10
12%= 900·1 −
Thus both of the payment schemes cost more than the immediate cash payment.
(b) i. The face value of the mortgage is the present value of the mortgage payments, 1600 · a15×2×12 .065
ii. The cash price of the home will be the sum of the down payment and the value of the mortgage, i.e., 60, 000 + 367, 641.6941 = 427, 641.69.
2. On his granddaughter’s 10th birthday her grandfather deposits a total of 100,000 into a fund that accumulates at a rate of 6%, compounded quarterly.
(a) The fund is to pay the child equal amounts on each of her 20th and her 21st birth-days. Determine the amounts of the equal payments.
(b) Suppose that the fund is to pay twice as much on the 21st birthday as on the 20th birthday. Determine the amount of each of the payments.
(c) Suppose that the payment on the 20th birthday is to be 90,000, and that the balance is to be paid on the 21st birthday. Determine the amount of the 21st-birthday payment.
(d) When the granddaughter reaches age 1514 a baby brother is born. The grandfather revises his plans. He has no more capital available, but he wishes that the fund should now pay each of the children a payment on the children’s respective 20th and 21st birthdays; all four payments are to be equal. Determine the amount of each of those four equal payments.
Solution:
(a) Denote the amount of the equal payments by X. An equation of value as of the time the fund is established is
100000 = X
(1.015)−4×(20−10)+ (1.015)−4×(21−10)
⇒ X = 100000
1.015−40+ 1.015−44 = 93400.94421 .
(b) Denote the amounts of the payment at age 20 by X, so that the payment at age 21 will be 2X. An equation of value as of the time the fund is established is
100000 = X
1 · (1.015)−4×(20−10)+ 2 · (1.015)−4×(21−10)
⇒ X = 100000
1.015−40+ 2 · 1.015−44 = 62891.35502 .
Thus the payments at ages 20 and 21 will respectively be 62,891.36 and 125,782.71.
(c) Denote the amount of the payment on the 21st birthday by X. An equation of value as of the time the fund is established is
100000 = 90000 · (1.015)−4×(20−10)+ X · (1.015)−4×(21−10)
⇒ X = 100000 − 90000(1.015−40)
1.015−44 = 100000(1.015)44− 90000(1.0154)
⇒ X = 97010.58231 .
(The second equation shows an equation of value on the child’s 21st birthday.)
(d) The grandson is born at the end of 4 ×
1514 − 10
= 21 interest periods. The payments to the granddaughter occur at the ends of the 40th and 44th periods; and the payments to the grandson are at the ends of the 101st and 105th periods. The relative times are shown in the following table (which could have been represented by lines in a time diagram):
Quarter-year 0 21 40 44 101 105
Granddaughter’s age 10 1514 20 21 3514 3614 Grandson’s age 0 434 534 20 21
Let the amounts of the four equal payments be denoted by X. An equation of value as of the date of establishment of the fund is
100000 = X
3. (a) 70,000 is placed in an account which pays annual interest of 4% compounded monthly. If a withdrawal is made between interest payment dates, it earns no inter-est from the last payment date. Determine the payment date when the account will contain at least 95,000.
(b) A loan today of 15000 is to be repaid by one payment of 10000 n years from now, and a payment of 20000 2n years from now; n need not be an integer. If future payments are discounted at an effective annual rate of 5%, determine n.
(c) A deposits 9000 in a “special interest” account, which, on interest-compounding dates, pays 1% interest effective per quarter when the balance is less than 10000, and 112% effective per quarter when the balance is at least 10000. Determine the amount in the account after 10 years.
Solution:
(a) Let n denote the number of months required for the balance to be not less than 95,000. An inequality of value as of the date of deposit is 70000 1 + 1
300
!n
≥ 95000, which is equivalent to 1 + 1
300
!n
≥ 95000
70000. We can solve this inequality by taking logarithms of both sides; since the logarithm function is increasing, it preserves the inequality, giving n ≥ ln 95 − ln 70
ln 1.00333333333 = 91.76711031. Thus
the first payment date when the account exceeds 95,000 is d91.76711031e = 92 months7after the original deposit, i.e., 723 years after the deposit.
(b) An equation of value as of today is 15000 = 10000(1 − d)n+ 20000(1 − d)2n, which is a quadratic equation in (1 − d)n; solving this, we obtain (1 − d)n= −1 ± √
13 4 . Of the two solutions, the one with the − is extraneous, as it yields a negative value for the exponential. Hence (1 − d)n = −1 + √
13
4 = 0.6513878188. Taking (natural) logarithms yields
n = ln 0.6513878188
ln 0.95 = 8.356844523 years.
(c) Let n be the number of quarter-years that need to pass for the balance to exceed 10000. Then n is the smallest integer solution to the inequality 9000(1.01)n ≥ 10000 ⇒ (1.01)n ≥ 10000
9000 ⇒ n ≥ ln 10 − ln 9
ln 1.01 = 10.58864445, from which it follows that n = 11 quarter-years.8 The balance does not exceed 10,000 until after the d10.58864445eth, i.e., 11th payment. It is only after this 11th payment that the account qualifies for the higher interest rate, which is applicable first at the 12th payment. After 40 quarter-years the account will have grown to 9000 · (1.01)11 · (1.015)40−11= 15, 462.96762.
4. The force of interest δtis known to be of the form δt = k√
t, where t is time, in year.
(a) Determine the value of k such that money will triple in 9 years.
(b) With this force of interest how long will it take for money invested at time t = 0 to double?
(c) With this force of interest how long will it take for money invested at time t = 1 to double?
(d) What would be the equivalent effective constant annual rate of interest which would cause money to double in the same time as with the current variable force of inter-est?
8The previously posted draft solution was corrected at this point.
(b) Let n denote the length of time for money invested at time t = 0 to double. Then
(c) Let n denote the length of time for money invested at time t = 1 to double. Then 2 = a(n + 1) hence n = 5.877212176 years.
(d) With n as before, we solve (1 + i)n = 2, to obtain i = 26.6205872761 − 1 = 0.110372700 = 11.03727%.
5. Joan is borrowing 10,000 today from the bank.
(a) If she repays the loan over a 5-year period, by regular monthly payments on the first of the month, beginning immediately, what will be the amount of the monthly payment at a nominal discount rate of 6%, compounded monthly?
(b) Joan finds the monthly payments inconvenient, as she receives her salary cheque every half-month. Determine the semi-monthly payment she would have to pay, in advance, which would be equivalent to the preceding monthly payment.
(c) What effective annual interest rate will Joan be paying?
Solution:
(a) Denote the monthly payment in advance by M. An equation of value today is M ¨a5×12 d=6
12% = 10000 which implies that M = 10000 Note that the symbol I have used, with a subscript that reads d = 126, is non-standard.9
9Mathematicians are accustomed to making their own notation as they need it; but actuaries establish precise notational conventions and play by the agreed rules.
(b) Denote the equivalent semi-monthly payment by N. The effective semi-monthly discount rate which is equivalent to an effective monthly discount rate of 12% is 1 − √
0.995 = 0.0025031328, or 0.25031328%. We can use this rate to solve the problem as previously, M = 10000
¨a5×24 d=0.25031328%
= 10000d
1 − v120 = 10000d 1 − (1 − d)120 = 10000(0.0025031328)
1 − (0.995)60 = 96.37106451. Alternatively, we can solve the equation N(1 + (1 − d)) = M ⇒ N = 192.5009023
2 − 0.0025031328 = 96.37106595. We could also have used, instead of 1 + (1 − d), ¨a2 d=0.25031328%. (The slight difference in the values is due to rounding errors.)
(c) The effective monthly discount rate is d = 0.005%, so the effective monthly interest rate is d
1 − d = 0.005
0.995 = 0.005025125628; the effective annual interest rate is 1 + 0.005
0.995
!
− 1 = 0.061996372 = 6.1996372% .
6. If i(3) = 1.4%, find the present value of payments of 100 every 4 months, starting 4 months from now, and continuing through 5 years from the present, together with pay-ments of 200 every 8 months thereafter, and continuing for 6 years.
Solution: The effective interest rate per one-third year is 1.43 %. The present value of the payments of 100 is
For the payments of 200 the effective interest rate per payment interval of 8 months is
1 + 0.0143 2
− 1 = 0.009355112 = 0.9355112%; as of the date of the last pay-ment of 100, the value of the paypay-ments of 200 is 200a6
23
0.9355112% = 200
0.009355112 ·
1 − (1.009355112)−9
= 1718.612932 . The value of these payments 5 years earlier is thus
(1.014)−5×3· 1718.612932 = 1602.685007 ,
so the total value now of all the payments is 1445.450441+1602.685007 = 3, 048.135448.