In this subtopic we look into the case which payment period not equal to interest conversion period. The payment made either increase or decrease A = đ?’—đ?’Œ + đ?’—
And
+đ?’Š
đ?’Œ
� =
đ?’Œ
+â‹Ż +
đ?’Œ
−
đ?’—đ?’Œ + â‹Ż +
+
Now subtracting the first equation from the second �
+đ?’Š
đ?’Œ
−
This can be expressed as
�=
+ đ?’—đ?’Œ + đ?’—
=
đ?’‚ | − đ?’— đ?’‚đ?’Œ | đ?’Œ đ?’Š
đ?’Œ|
đ?’— đ?’Œ đ?’Œ
−đ?’Œ
−
+ đ?’—
đ?’Œ
đ?’—
− đ?’Œ
+â‹Ż+ đ?’—
+
−đ?’Œ
−
đ?’Œ
đ?’— đ?’Œ
−đ?’Œ
đ?’—
Next case which payment are made more frequently than interest is convertible. Two different situation arise depend on whether the rate of payment is constant/varies during each interest conversion period đ?‘°đ?’‚
|
=
đ?’‚
|− đ?’—
đ?’Š
This formula gives the present value of n – period annuity – immediate, payable mthly in which each payment during the first interest conversion period is 1/m, each payment during the second interest conversion period is 2/m, and so forth, until each payment during the nth interest conversion period is n/m.
Example Suppose a deposit of $1000 is made on the first of the months of January, February and March, $1200 at the beginning of each month in the second quarter, $1400 at the beginning of each month in the third quarter and $1600 each month in the forth quarter. If the account has a nominal rate 8% rate of interest compounded quarterly, what is the balance at the end of the year. =
=
= [ .
The present value at t = -1 is = At t = 0 the value is
− .
.
=
−
.
which is the balance at the end of the year
= .
− ]= .
− −
+
.
And its value at t = 12 is
.
$ $
,
. . .
,
−
.
−
−
=$ .
đ?‘Ł=
=$
−
.
,
. ,
−
.
.
=$
,
.
Consider the next situation which the rate of payment changes with each payment period. Increasing annuity is payable at the rate of 1/m per interest conversion period at the end of the first mth of an interest conversion period. 2/m per interest conversion period at the end of the second mth of an interest conversion period, and so forth. First payment will be 2/
, and so forth.
Denoting the present value of such an annuity by đ??ź đ?‘°
đ?’‚
=
|
=
đ?’‚ | − đ?’— đ?’Š
đ?’— +
đ?‘Ž
|
đ?’— + â‹Ż+
đ?’—
In this subtopic we look into an increasing annuity which payment are being made continuously at varying interest rate. The present value of this annuity is denoted by đ??ź đ?‘Ž đ?‘°đ?’‚ | = đ?’—
|
and an expression for it would be
To calculate the present value, some formulae can be used. The first formula is derived from formula đ??ź
đ??˘
Í˘âˆž
đ??ˆ
đ??š
|
= đ??˘
đ?‘Ž
|
đ??š | − đ??Ż đ??˘ Í˘âˆž
=
Another more general formula : đ?‘°đ?’‚
|=
đ?’—
đ?’‚ |− đ?’— đ?œš
Example A one-year deferred continuous varying annuity is payable for 13 years. The rate of payment at time t is − per annum and force of interest at time t is + − . Find the present value. đ?‘°đ?’‚
đ?&#x;’| =
đ?&#x;’
−
−
|=
.
Solve this integration and will obtain : đ??źđ?‘Ž
+ −
Since the differential expression đ?‘Ł moment t.
is the present value of the payment
The expression can be obtained by performing integration by part đ??źđ?‘Ž
|=
=
=
= =
=
đ?‘Łđ?‘Ą ] ln đ?‘Ł
đ?‘Ł
đ?‘Łđ?‘Ą ] ln đ?‘Ł đ?‘Ł − đ?›ż
− −
−
−đ?‘Ł − đ?›ż đ?‘Ž |− đ?‘Ł đ?›ż
đ?‘Łđ?‘Ą ln đ?‘Ł
�� ] ln � � + � � � �
made at exact
Example Find an expression for the present value of a continuously increasing annuity with the term of n years if the force of interest is δ and if the rate of payment at time is per annum. −đ?›ż
=−
= −
= − = −
=
�
�
� � �
−đ?›ż
−đ?›ż −đ?›ż
−đ?›ż
−đ?›ż
] + −
− − đ?›ż
�
�
−đ?›ż
�
�
+
−đ?›ż
−đ?›ż đ?›ż
+
−đ?›ż
] +
− −
�
�
�
�
]
−đ?›ż −đ?›ż
]
−đ?›ż
+
�
SUMMARY OF RESULT Table below summarizes the expressions for present values of level annuities for various payment periods and interest conversion periods. The annuity illustrated pays 60 per annum for 10 years at 12% per annum Payment period Interest conversion
Annual
Quarterly
Monthly
Continuous
period Annual đ?‘Ž
|.
đ?‘Ž
|.
đ?‘Ž
|.
đ?‘Ž
|.
đ?‘Ž
|.
đ?‘Ž
|.
đ?‘Ž
|.
đ?‘Ž
|.
Quarterly
|.
Monthly đ?‘Ž
|.
|.
đ?‘Ž
|.
|.
đ?‘Ž
|.
đ?‘Ž
|.
Continuous − .
− .
−
− .
− .
−
− .
− .
−
− .
− .
Explanation đ?‘Ž
|.
:
 The payment is annually paid in 10 years while the interest is 12% in a year.  N = 10 because the n shows the total number of conversion period in the whole term.  Interest = 12% because the interest is 12% per annum. a |. s |.
:
 The payment is annually paid in a year while the interest conversion period is quarterly.  N = 40 because the n shows the total number of conversion period in the whole term (4×10). The payment made is less frequent than the interest conversion period.  K = 4 since the interest conversion period in one payment (1 year) =4.
đ?‘Ž
|.
:
|.
 The payment is annually paid in a year while the interest conversion period is monthly.  The payment made is less frequent than the interest conversion period.  N = 120 because the n shows the total number of conversion period in the whole term (12Ă—10). − − .
 K = 12 since the interest conversion period in one payment (1 year) = 12 . − payment is annually paid in a year while the interest conversion period is continuous.  The payment made is less frequent than the interest conversion period.  In this case, to find the present value, we use the formula ( be found using the following formula and will get . − . +
 [
đ?‘Ž
    
|.
=
+
đ?‘–
=
�
=
−
−
=
−
đ?‘?
đ?‘?
−đ?‘?
−đ?‘Ł đ?‘–
). The
: The
for this case can
]
:
The payment is quarterly paid in a year while the interest conversion period is annually. The payment made is more frequent than the interest conversion period. (m) = 4 since the payment is made in one interest conversion period is equal to 4. N = 10 because the n shows the total number of conversion period in the whole term . The interest = 12% since the interest conversion period is annually.
đ?‘Ž
|.
đ?‘Ž
|.
:
 The payment made in one year is quarterly paid in a year while the interest conversion period is quarterly.  The amount paid = 15 because the payment is made quarterly in a year which mean 60/4= 15.  N = 40 because the n shows the total number of conversion period in the whole term (4×10).  The interest is equal to 3% since the interest conversion period in one year is quarterly (12%/4) = 3%
|.
:
 The payment made in one year is quarterly paid in a year while the interest conversion period is monthly.  The amount paid = 15 because the payment is made quarterly in a year which mean 60/4= 15.  N = 120 because the n shows the total number of conversion period in the whole term (12×10).  Since the payment is made quarterly in one year while the interest conversion period is monthly K=3. This is because in one payment made for every 3 month in a year the interest conversion period in one payment equal to 3. Hence K = 3. The interest equal to 0.01 since the interest in convertible monthly (0.12/12) = 0.01
− − . . −
:
The payment is made quarterly and the interest given is force of interest. −đ?‘Ł
In this case, to find the present value, we use the formula ( ). đ?‘– The original payment must be divided by 4 to find the amount of payment made for each quarter. So, 60/4 equals to 15. The for this case will be it can be found using formula (1.34) and will get . − . đ?‘Ž
|.
:
The payment is monthly paid in a year while the interest conversion period is annually. The payment made is more frequent than the interest conversion period. (m) = 12 since the payment is made in one interest conversion period is 12. N = 10 because the n shows the total number of conversion period in the whole term. The interest = 12% since the interest conversion period is annually.
đ?‘Ž
|.
:
 The payment made in one year is 12 times in a year while the interest conversion period is quarterly.  The payment made is more frequent than the interest conversion period.  (m) = 3 since the payment made in one interest conversion period is 3.  The amount paid = 15 because the payment is made quarterly in a year which mean 60/4= 15.  N = 40 because the n shows the total number of conversion period in the whole term (4×10).  The interest is equal to 3% because the interest conversion period in one year is quarterly (12%/4) = 3% �
|.
:
 The payment made in one year in monthly and the interest conversion period is monthly.  The amount needs to pay equal to 5 because the payment made in one year equal 12 times (60/12 month) = 5.  N = 120 because the n shows the total number of conversion period in the whole term (12×10).  The interest equal to 0.01 since the interest in convertible monthly (0.12/12) = 0.01
− − . . −
:
 The payment made is monthly and the interest conversion period is continuously.  In this case, to find the present value, we use the formula (
−đ?‘Ł đ?‘–
).
 Since the interest given is convertible continuously, we have to find the denominator by using the formula (1.34) that shows the series of equalities as follows; +
=
+
=
�
=
−
−
 And then we will get ; +
= =
= .
.
−
=
−
đ?‘?
đ?‘?
−đ?‘?
đ?‘Ž
|.
:
đ?‘Ž
|.
:
|.
:
 The payment made is continuous and the interest conversion period is annually.  N = 10 because the n shows the total number of conversion period in the whole term while interest=12% because the interest is 12% per annum.
 The payment made is continuous and the interest conversion period is quarterly.  N = 40 because the n shows the total number of conversion period in the whole term (4 × 10) while interest = 3% because the interest conversion period is quarterly (12%/4) = 3%
đ?‘Ž
 The payment made in one year is continuous and the interest conversion period is monthly.  The amount needs to pay equal to 5 because the payment made in one year equal 12 times (60/12 month) = 5.  N = 120 because the n shows the total number of conversion period in the whole term (12 Ă— 10 ).  The interest equal to 0.01 since the interest in convertible monthly (0.12/12) = 0.01 − − . .
:
 The payment made in one year is continuous and the interest conversion period is continuous. So we just the 4.19 formula (
− − đ?›ż ) đ?›ż
to find the present value.