Transcript Document
Sections 7.4, 7.5
We now consider transactions where each payment may be reinvested at
a rate which may or may not be equal to the original investment rate.
Consider an investment of 1 for n periods at rate i where the interest is
reinvested at rate j. The accumulated value at the end of the n periods is
1 + is –
n|j
Consider an investment of 1 at the end of each period for n periods at
rate i where the interest is reinvested at rate j. The accumulated value
at the end of the n periods is
s– –n
n|j
(Is)
–––
n+i
n–1| j
= n + i ————
j
If the payments of 1 are made at the beginning of each period (instead
of at the end), the accumulated value at the end of the n periods is
s––– – (n + 1)
n+1| j
– j = n + i ———————
n + i (Is) n|
j
Payments of $1000 are invested at the end of each year for 10 years. The
payments earn interest at 7% effective, and the interest is reinvested at
5% effective. Find the
(a) amount in the fund at the end of 10 years,
– 10
s ––
10 | 0.05
1000 10 + 0.07 ——————
0.05
=
12.5779 – 10
1000 10 + 0.07 ——————
0.05
=
$13,609.06
(b) purchase price an investor must pay for a yield rate of 8% effective.
$13,609.06(1.08)–10 = $6303.63
Three loan repayment plans are described for a $3000 loan over a 6-year
period with an effective rate of interest of 7.5%. If the repayments to the
lender can be reinvested at an effective rate of 6%, find the yield rates.
(Total interest paid with each plan was found in a previous example.)
(a) The entire loan plus accumulated interest is paid in one lump sum
at the end of 6 years. This lump sum is 3000(1.075)6 = $4629.90.
Since there is no repayment to reinvest during the 6-year period, the
yield rate is obviously 7.5%.
(b) Interest is paid each at the end of each year as accrued, and the
principal is repaid at the end of 6 years. Each year during the 6-year
period, the payment is 3000(0.075) = $225.
The accumulated value of all payments at the end of the 6-year
period is 3000 + 225 s –
= 3000 + 225(6.9753) = $4569.44.
6 | 0.06
To find the yield rate, we solve 3000(1 + i)6 = 4569.44.
The yield rate is 0.07265 = 7.265%.
(c) The loan is repaid with level payments at the end of each year over
the 6-year period. Each year during the 6-year period, the payment is
R where 3000 = R a ––
. R = $639.13
6 | 0.075
The accumulated value of all payments at the end of the 6-year
period is 639.13 s –
= 639.13(6.9753) = $4458.12.
6 | 0.06
To find the yield rate, we solve 3000(1 + i)6 = 4458.12.
The yield rate is 0.06825 = 6.825%.
Look at formulas (7.8), (7.9), and (7.10) in the textbook, and look at
Example 7.7 on page 264.
In practice, it is common for a fund to be incremented with new principal
deposits, decremented with principal withdrawals, and incremented with
interest earnings many times throughout a period. Since these
occurrences are often at irregular intervals, we devise notation for the
purpose of obtaining an effective rate of interest:
A = the amount in the fund at the beginning of the period
B = the amount in the fund at the end of the period
I = the amount of interest earned during the period
Ct = the net amount of principal (positive or negative) contributed
at time t, where 0 t 1
C=
C = total net amount of principal contributed during the period
t
t
aib
= the amount of interest earned by investing 1 from time b to time
b + a, where a 0, b 0 , and b + a 1
Observe that we must have B = A + C + I.
Understanding that I is received at the end of the period (consistent with
the definition of effective rate of interest), we have
C
t 1 – t it
I = iA +
t
With compound interest, 1 – t it = (1 + i)1–t – 1.
C
t
We then have I = iA +
[(1 + i)1–t – 1] which can be solved for i
t
by iteration, and i is guaranteed to be unique as long as the balance never
becomes negative (from previous results).
Alternatively, an approximate value for i can be obtained by using the
“simple interest” approximation 1 – t it (1 – t)i . This approximate
I
solution is
(This approximation is very
————————
i A + C (1 – t)
good as long as the C s are
t
t
small in relation to A.)
This denominator is often called the exposure associated with i.
t
One useful special case is when we might assume that net principal
contributions occur at time t = 1/2, in which case we have
I
I
2I
I
———————— = ———— = ——————— = ————
i A + C (1 – t)
A + 0.5C
A + 0.5(B – A – I) A + B – I
t
t
If we assume that net principal contributions occur at time t = k (where
of course 0 < k < 1), then
I
I
I
———————— = ————— = ————————— =
i A + C (1 – t)
A + (1 – k)C
A + (1 – k)(B – A – I)
t
t
I
——————————
kA + (1 – k)B – (1 – k)I
Consider again the “simple interest” approximation 1 – t it (1 – t)i on
which these approximate formulas are based. (Observe that t is not
playing the same role that it plays when defining the accumulation
function a(t) = 1 + ti for simple interest.) Let us consider writing an
expression for t i0 when we assume 1 – t it = (1 – t)i .
If 1 is invested for one period, then 1 + i is the value of the investment at
the end of the period. This investment should yield the same amount as
first investing 1 from time 0 to time t, and then investing the amount
from the first investment from time t to time 1. Under the assumption
that 1 – t it = (1 – t)i , we can write
1 + t i0 + (1 + t i0) 1 – t it = 1 + i
1 + t i0 + (1 + t i0)(1 – t)i = 1 + i
Solving for t i0 is what you need to do in Exercise 7-22(a).
Formulas (7.31), (7.32), and (7.33) in the textbook assume payments and
contributions are made continuously (but this situation does not often
occur in practice).
At the beginning of a year, an investment fund was established with an
initial deposit of $3000. At the end of six months, a new deposit of
$1500 was made. Withdrawals of $500 and $800 were made at the end
of four months and eight months respectively. The amount in the fund at
the end of the year is $3876. Using the formula which results from the
approximation 1 – t it (1 – t)i , find the approximate effective rate of
interest earned by the fund during the year.
First, we find I = B – A – C = 3876 – 3000 – (1500 – 500 – 800) = 676
I
i ———————— =
A + Ct (1 – t)
t
676
676
———————————————————————— = ——
=
3000 + (–500)(1 – 1/3) + (1500)(1 – 1/2) + (–800)(1 – 2/3)
3150
0.2146 or 21.46%
Look at Example 7.9 on page 268 in the textbook.