Transcript Sections 4.1, 4.2, 4.3, 4.4 Suppose the payments for an annuity are level, but the payment period and the interest conversion period.

Sections 4.1, 4.2, 4.3, 4.4
Suppose the payments for an annuity are level, but the payment period
and the interest conversion period differ. One approach to computing the
numerical value of the annuity is the following two-step procedure:
(1) Find the rate of interest which is equivalent to the given rate of
interest and convertible at the same frequency as payments are made.
(2) Using the rate of interest from in step 1, find the value of the annuity.
Find the accumulated value at the end of six years of an investment fund
in which $100 is deposited at the beginning of each quarter for the first
three years and $50 is deposited at the beginning of each quarter for the
second three years, if the fund earns 6% convertible monthly.
The quarterly interest rate equivalent to 6% convertible monthly is
(1.005)3 – 1 = 0.015075 .
The accumulated value at the end of six years is
..
12 + 50 ..
s––
= $2246.98
100 s––
(1.015075)
12 | 0.015075
12 | 0.015075
A loan of $6000 is to be repaid with quarterly installments at the end of
each quarter for four years. If the rate of interest charged on the loan is
8% convertible semiannually, find the amount of each quarterly
payment.
The quarterly interest rate equivalent to 8% convertible semiannually
is (1.04)1/2 – 1 = 0.019804 .
If R denotes the quarterly payments, then the equation of value is
= $6000
R a ––
16 | 0.019804
6000
= $441.21
R = ––––––––––
a ––
16 | 0.019804
Suppose we want to find the annual effective rate of interest i for which
payments of $75 at the end of every quarter accumulate to $3000 at the
end of six years.
(a) Letting j = i(4)/4, where i(4) is equivalent to the desired annual
effective rate of interest, write the equation of value.
75 s––
24 |
j
= 3000
75[(1 + j)24 – 1] – 3000j = 0
(b) Solving the equation in part (a) for j = i(4)/4 is difficult. Use the
TI-84 calculator to solve for j, and then use the Excel file
Interest_Solver to find j. You should find j = i(4)/4 = 0.041803.
Solve for j on the TI-84 calculator by doing the following:
(Note: On the TI-83 calculator, the | 2nd | | FINANCE | keys
should be used in place of the | APPS | key and Finance option.)
Press the | APPS | key, select the Finance option, and select the
TVM_Solver option. Enter the following values for the variables
displayed:
N = 24
I% = 0
PV = 0
PMT = –75
FV = 3000
P/Y = 1
C/Y = 1
Select the BEGIN option for PMT , press the | APPS | key, and
select the Finance option.
Select the tvm_Pmt option, and after pressing the | ENTER | key,
(4)/4 = 0.041803
j
=
i
the desired result should be displayed.
Suppose we want to find the annual effective rate of interest i for which
payments of $75 at the end of every quarter accumulate to $3000 at the
end of six years.
(a) Letting j = i(4)/4, where i(4) is equivalent to the desired annual
effective rate of interest, write the equation of value.
75 s––
24 |
j
= 3000
75[(1 + j)24 – 1] – 3000j = 0
(b) Solving the equation in part (a) for j = i(4)/4 is difficult. Use the
TI-84 calculator to solve for j, and then use the Excel file
Interest_Solver to find j. You should find j = i(4)/4 = 0.041803.
Type the formula =75*((1+j)^24-1)-3000*j in cell A5.
To avoid getting the solution j = 0 for the equation of value, type the
restriction j >= 0.01 instead of j >= 0.
Select options Tools > Solver to solve the equation of value as
indicated below (and if necessary, first use options Tools > Add-Ins)
(c) Using the solution j = i(4)/4 = 0.041803 found in part (b), find the
desired annual effective rate of interest.
4
i(4)
i= 1+— –1
4
= (1 + j)4 – 1 =
0.17799 or 0.178
A second approach with annuities where the payment period and the
interest conversion period differ involves algebraic analysis.
First, consider annuities payable less frequently than interest is
convertible. We let
k = the number of interest conversion periods in one payment period,
n = the term of the annuity measured in interest conversion periods,
i = rate of interest per conversion period.
It follows that
n / k = the number of annuity payments made. (Note that n / k must be
an integer, but k need not be an integer.)
The present value of an annuity which pays 1 at the end of each interval
of k interest conversion periods is
vk + v2k + … + vn = vk[1 + vk + (vk)2 + … + (vk)n/k – 1] =
a –n|
n
n
n) / i
1
–
v
1
–
v
(1
–
v
vk ——k = ————
= ——————
= s–
k
k
1–v
(1 + i) – 1 [(1 + i) – 1] / i
k|
The accumulated value of this annuity immediately after the last
payment is
a –n|
s n|
–
n =
(1
+
i)
s k|
–
s k|
–
The present value of an annuity which pays 1 at the beginning of each
interval of k interest conversion periods is
1 + vk + v2k + … + vn – k = 1 + vk + (vk)2 + … + (vk)n/k – 1 =
a –n|
1 – vn (1 – vn) / i
——k = ————
= a–
k
1–v
(1 – v ) / i
k|
The accumulated value of this annuity k interest conversion periods after
the last payment is
a –n|
s n|
–
n =
(1
+
i)
a –k|
a –k|
…
Conversion Periods 
Payment Periods 
1
2
…
k1
1
k
2
…
k1
1
2k
2
…
k1
2
1
3k
n
1
…
3
k
2
k1
n
k
1
k=n
n
k
Note that in general n / k must be an integer, but k need not be an
integer.
Also, note that for any of
Present Accumulated
the four formulas, putting
Value
Value
double dots (..) above the
symbol in the numerator
Annuity
and above the symbol in
Immediate
the denominator results in
exactly the same formula.
Annuity
Due
…
Conversion Periods 
1
2
…
k1
1
2
k
…
k1
1
2k
2
…
k1
1
2
1
3k
n
Payment Periods 
…
3
k
2
k1
n
k
1
k=n
n
k
Note that in general n / k must be an integer, but k need not be an
integer.
Also, note that for any of
Present Accumulated
the four formulas, putting
Value
Value
double dots (..) above the
symbol in the numerator
s n|
–
a –n|
Annuity
and above the symbol in
s k|
–
s k|
–
Immediate
the denominator results in
exactly the same formula.
Annuity
Due
a –n|
s n|
–
ak–|
ak–|
The present value of a perpetuity-immediate which pays 1 at the end of
each interval of k interest conversion periods is
vk + v2k + v3k + … = vk[1 + vk + (vk)2 + (vk)3 + … ] =
vk
1
1
1/i
1
——k = ————
= ——————
= is–
k
k
1–v
(1 + i) – 1 [(1 + i) – 1] / i
k|
The present value of a perpetuity-due which pays 1 at the beginning of
each interval of k interest conversion periods is
1
1
k
2k
3k
k
k
2
k
3
1 + v + v + v + … = 1 + v + (v ) + (v ) + … = ——k = i a –
1–v
k|
Find the accumulated value at the end of six years of an investment fund
in which $100 is deposited at the beginning of each quarter for the first
three years and $50 is deposited at the beginning of each quarter for the
second three years, if the fund earns 6% convertible monthly.
(Note that this was done earlier using the other approach.)
s 36|
–– 0.005
100 a –
3| 0.005
100
(1.005)36 + 50
s 36|
–– 0.005
a –3| 0.005
39.3361
39.3361
(1.005)36 + 50
2.9702
2.9702
=
=
$2247.01
An investment of $3000 is used to make payments of $500 at the end of
every year for as long as possible with a smaller final payment made at
the same time as the last regular payment. If interest is 8% convertible
semiannually, find the number of payments and the amount of the final
payment.
If the smaller final payment were equal to 0, then the equation of value
a –n| 0.04
would be
500 s –
= 3000
2| 0.04
a –n| 0.04 = 6 s 2|
– 0.04 = 6(2.0400) = 12.24
From the TI-84 calculator, we find that 17 < n < 18 , which implies that
8 regular payments and a smaller 9th payment denoted as R are made.
The equation of value at the end of 8 years is
s 16|
–– 0.04
R + 500 s –
= 3000(1.04)16
R = $269.79
2| 0.04
The amount of the final payment is $500 + $269.79 = $769.79
A series of payments of $5 are made every 3 months forever, with the
first payment made immediately. At what annual effective rate of
interest is the present value of these payments equal to $75?
The equation of value is
75 = 5 + 5v1/4 + 5(v1/4)2 + 5(v1/4)3 + …
15 = 1 + v1/4 + (v1/4)2 + (v1/4)3 + …
1
15 = ——1/4
1–v
1
14
v = —— = —
1 + i 15
15
i= —
14
4
4
– 1 = 0.31781 or 31.781%
Now, consider annuities payable more frequently than interest is
convertible. We let
m = the number of payment periods in one interest conversion period,
n = the term of the annuity measured in interest conversion periods,
i = rate of interest per interest conversion period.
It follows that
mn = the number of annuity payments made.
The present value of an annuity which pays 1/m at the end of each mth
1
of an interest conversion period is (m)
a –n| = — [v1/m + v2/m + … + vn] =
m
n
1/m 1 – vn
v1/m
1
–
v
v
1
— [1 + v1/m + (v1/m)2 + … + (v1/m)mn – 1] = — ——— = — ————— =
m
m 1 – v1/m m (1 + i)1/m – 1
1 – vn
i
——
a –n|
= —
(m)
(m)
i
i
The accumulated value of this annuity immediately after the last
payment is
n–1
(m) 1 – vn
(1
+
i)
i
n
s n|
– = —— (1 + i) = ———— = — s –
i(m)
i(m)
i(m) n|
The present value of an annuity which pays 1/m at the beginning of each
mth of an interest conversion period (similar to the notation and
derivation for an annuity-due) is ..(m)
1 – vn
i
a –n| = —— = — a –
d(m)
d(m) n|
The accumulated value of this annuity 1/mth of an interest conversion
period after the last payment is
i
..(m) (1 + i)n – 1
s n|
– = ————
—(m) s n|
–
=
(m)
d
d
Observe that
..(m)
a –n| = (1 + i)1/m a(m)
–n| =
i(m) i
1+— —
a –n| =
(m)
m i
i
i
—
+ — a –n|
(m)
i
m
i(m) i
1+— —
s n|
–
(m)
m i
i
i
—
+ — s n|
–
(m)
i
m
and that
..(m)
s n|
– = (1 + i)1/m s (m)
– =
n|
=
Appendix 4 (at the end of Chapter 4) in the textbook displays several of
these types of formulas.
The present value of a perpetuity-immediate which pays 1/m at the end
of each mth of an interest conversion period is (m)
1
a|
– = —
i(m)
The present value of a perpetuity-due which pays 1/m at the beginning
of each mth of an interest conversion period is ..(m)
1
a|
– = —
d(m)
Observe that when applying formulas derived for payments of 1/m, we
must multiply by Pm when each actual payment is P.
A loan of $6000 is to be repaid with quarterly installments at the end of
each quarter for four years. If the rate of interest charged on the loan is
8% convertible semiannually, find the amount of each quarterly
payment.
We have m = 2 and n = 8 from which we have
i = 0.04 and i(m) = 2[(1.04)1/2 – 1] = 0.039608 .
If P denotes the quarterly payments, then the equation of value is
We have m = 2 and n = 8 from which we have
i = 0.04
and i(m) = 2[(1.04)1/2 – 1] = 0.039608.
If P denotes the quarterly payments, then the equation of value is
(2)
(P)(2) a –8| 0.04 = $6000
3000
=
P = ––––––––––
i
—
a–
i(2) 8| 0.04
3000
––––––––––––––– = $441.21
(1.0099)(6.732745)
Observe that we may apply the formulas derived for annuities payable
less frequently than interest is convertible to annuities payable more
frequently than interest is convertible by setting the number of interest
conversion periods in one payment period k equal to 1/m .
Observe that we may apply the formulas derived for annuities payable
less frequently than interest is convertible to annuities payable more
frequently than interest is convertible by setting the number of interest
conversion periods in one payment period k equal to 1/m .
A loan of $6000 is to be repaid with quarterly installments at the end of
each quarter for four years. If the rate of interest charged on the loan is
8% convertible semiannually, find the amount of each quarterly
payment.
We have n / k = 16 and k = 1/2 ; therefore, n = 8 . We also have that the
effective interest rate per interest conversion period is 0.04 .
If P denotes the quarterly payments, then the equation of value is
a ––
8 | 0.04
= $6000
P s ––
1/2| 0.04
1.041/2  1
P = $6000 –––––––––
= $441.21
8
1  1.04
A series of payments of $5 are made every 3 months forever, with the
first payment made immediately. At what annual effective rate of
interest is the present value of these payments equal to $75?
..(4)
– = 75
The equation of value is (4)(5) a|
20
——
= 75
(4)
d
4
—— = d (4)
15
14
i= —
15
4
– 1 = 0.31781 or 31.781%