Solutions to End−of−Chapter Exercises Check with the MATLAB diff(f) function
Chapter 6 Mathematics of Finance and Economics 6.3 Sinking Funds
6.4 Annuities
Annuity: A series of equal payments or receipts occurring over a specified number of periods.
Ordinary annuity: A series of equal payments or receipts occurring over a specified number of periods with the payments or receipts occurring at the end of each period.
Annuity due: A series of equal payments or receipts occurring over a specified number of periods with the payments or receipts occurring at the beginning of each period.
NOTE: While technically correct, the last two definitions shown above can be a somewhat con- fusing. Whether a cash flow appears to occur at the end or the beginning of a period often depends on our perspective. For example, the end of year is also the beginning of year .
Therefore, the real key to distinguishing between an ordinary annuity and an annuity due is the point at which either a future or present value is to be calculated. Remembering the following characteristics should help us identify the type of annuity that we are dealing with:
1. For an ordinary annuity, future value is calculated as of the last cash flow, while present value is calculated as of one period before the first cash flow.
2. For an annuity due, future value is calculated as of one period after the last cash flow, while present value is calculated as of the first cash flow.
The equations used in the computations of sinking funds apply also to annuities.
Example 6.19
Suppose that on January 1, 2007 our bank account was , and withdrawals of each were made at the end of each year for years. If the account earned annual interest, what would be the balance in the account at the end of 2010 immediately after the last withdrawal is made?
Solution:
The computations are shown in Table 6.11.
R Sn i
1 i+ ( )n–1
--- 150000 0.03 1 0.03+
( )10–1
---
× 150000 0.08723× $13 084.58,
= = = =
2 3
$6000 $500
4 4%
Chapter 6 Mathematics of Finance and Economics
As shown in the table above, the principal at the end of 2010 would be .
In many instances, it is desirable to find the value of these sums of money at the origin date of the annuity. The equation that will compute the amount at the origin date is derived as follows:
We recall from (6.7), Page 6−13, that
(6.19) and from (6.15), Page 6−27, that
(6.20) As stated earlier, these equations apply to both sinking funds and annuities. For convenience, we will denote (6.19) as
(6.21) and (6.20) as
(6.22) By substitution of (6.22) into (6.21) we obtain
(6.23)
Example 6.20
Suppose that an annuity has been established that will enable us to receive at the end of each year for years starting with December 31, 2006. If the interest rate is , what amount can we expect to receive on January 1, 2012 should we decide to close this annuity?
TABLE 6.11 Computations for Example 6.19 Year
Principal at
beginning Interest Earned End−of−year
withdrawal Principal at end of period 2007
2008 2009 2010
$6000.00 6000.00 0.04× = $240.00 $500.00 6000.00 240.00 500.00+ – = $5740.00 5740.00 5740.00 0.04× = $229.60 $500.00 5740.00 229.60 500.00+ – = $5469.60 5469.60 5469.60 0.04× = $218.78 $500.00 5469.60 218.78 500.00+ – = $5188.38 5188.38 5188.38 0.04× = $207.54 $500.00 5188.38 207.54 500.00+ – = $4895.92
$4895.92
n
P–n = P0(1 i+ )–n
Sn R(1 i+ )n–1 ---i
=
PVannuity = FVannuity(1 i+ )–n
FVannuity Payment(1 i+ )n–1 ---i
=
PVannuity Payment(1 i+ )n–1
--- 1 ii ( + )–n Payment1–(1 i+ )–n ---i
= =
$300
5 6%
Annuities
Solution:
Equation (6.23), Page 6−30, is applicable for this example. Thus,
Example 6.21
A business firm must make a disbursement of at the end of each year from 2008 to 2011 inclusive. In order to meet the obligations as they fall due, it must deposit in a fund at the begin- ning of 2008 an amount of money that will be just sufficient to provide the necessary payments. If the interest rate is compounded annually, what sum should be deposited?
Solution:
Let denote the sum to be deposited in the fund on Jan. 1, 2008. Then,
This relationship holds for any valuation date we select; the most convenient date to use is obvi- ously the beginning of 2008, which is both the date of the deposit and the origin date of the annu- ity. Then,
Example 6.22
A father wishes to establish a fund for his newborn son’s college education. The fund is to pay on the 18th, 19th, 20th, and 21st birthdays of the son. The fund will be built up by the deposit of a fixed sum on the son’s 1st to 17th birthdays, inclusive. If the fund earns , what should the yearly deposit into the fund be?
Solution:
Let denote the periodic deposit in the fund. Selecting the end of the 17th year as our valuation date, we have
That is, at the end of the 17th year, the fund must have accumulated . PVannuity 3001–(1 0.06+ )–5
---0.06 300 4.21236× 1263.71
= = =
$800
3%
x
x = Value of deposit = Value of annuity
PVannuity 8001–(1 0.03+ )–3
---0.03 800 3.7171× 2973.68
= = =
$2 000,
2%
R
PVannuity Payment1–(1 i+ )–n
---i 20001–(1 0.02+ )–4
---0.02 2000 3.80773× $7615.46
= = = =
$7615.46
Chapter 6 Mathematics of Finance and Economics
Now, we must find the annual payments that must be deposited in order to accumulate this amount in 17 years. Solving (6.22) for , we obtain
Determining the Interest Rate of an Annuity
In many problems pertaining to annuities that arise in practice, the known quantities are the value of the annuity at either the origin or terminal date and the amount of the periodic payment, while the interest rate by which the given quantities are related is unknown. This is often true, for example, where an asset is purchased on the installment plan. The purchaser knows the purchase price of the asset and the periodic payment he is obligated to make, but he is not directly aware of the interest rate implicit in the loan. Only an approximate solution of this type of problem is pos- sible, and a solution by straight−line interpolation yields results of a sufficient degree of accuracy.
Example 6.23
An asset costing was purchased on the installment plan, the terms of sale requiring that the buyer make a down payment of and five annual payments of each. The first of these periodic payments is to be made year subsequent to the date of purchase. What is the interest rate pertaining to this loan?
Solution:
From (6.23), Page 6−30,
Assuming that is the value at the origin date, this value is and the is . Substitution of these values into the above relation and rearranging yields
Obviously, it is a formidable task to solve this equation for the interest rate . Instead, we con- struct the following spreadsheet where we observe that the interest rate corresponding to the value at the origin is between and .
Payment
Payment FVannuity i
1 i+ ( )n–1
--- 7615.46 0.02· 1 0.02+
( )17–1
--- 7615.46 0.04997× $380.54
= = = =
$5 000,
$2 000, $720
1
PVannuity Payment1–(1 i+ )–n ---i
=
PVannuity 5000 2000– = 3000
Payment $720
1–(1 i+ )–n
---i 3000 ---720
=
i
6% 7%
Annuities
We will use the graph of Figure 6.5 below to apply straight−line interpolation.
Figure 6.5. Graph for Example 6.23
For our example,
Therefore, the interest rate is .
Check with the Microsoft Excel IRR function:
In a blank spreadsheet we enter the values 3000 (5000−2000), −720, −720, −720, −720, and − 720 in A1 through A6 respectively, and we use the formula =IRR(A1:A6) and Excel returns
.
i R n Value at origin date
1% 720 5 3494.47
2% 720 5 3393.69
3% 720 5 3297.39
4% 720 5 3205.31
5% 720 5 3117.22
6% 720 5 3032.90
7% 720 5 2952.14
8% 720 5 2874.75
9% 720 5 2800.55
10% 720 5 2729.37
6.00 3032.90
7.00 2952.14 3000.00
x (%) y ($)
xi 1
slope
---×(yi–y1)+x1 1
m----×(yi–y1)+x1
= =
xy=3000.00 1 slope
---×(3000.00 3032.90– )+0.06 1
2952.14 3032.90– 0.07 0.06– ---
---×(–32.90)+0.06
= =
0.01 80.76
–---×(–32.90)+0.06 = 0.06407
=
6.4%
6.402%
Chapter 6 Mathematics of Finance and Economics