5.2 Extensibility
5.2.3 Hypothetical
Let us atart with a formular we had before.
Let P be invested at r% compound interest for n years. The accrued amount A is given by
A=P(1 +r)n (3.1.1)
which is rewriten as
P = A
(1 +r)n (3.1.2)
P is called the present value (PV) of A.
The idea is if N100 is invested for 2years at 10% compound interest. The amount A is a= 100
1 + 10 100
2
=N121
Then N100 is called the present value of N121 after 2years at the investment rate of 10%.
The above demonstrates the concept of the present value of a future sum. The investment rate, used in this context, is called the discount rate.
3.2 Use of Discounting Tables
We have intentionally avoided the use of a particular mathematical equipment for the quick evaluation or simplification of calculations. We introduce the use of discounting table here to ease your calculation. You have to purchase a discounting table now. In (3.1.2), the factor
1
(1 +r)n (3.2.1)
is the present value factor or discount factor. There is a range of values of n and r. You have to learn how to use the table. The main column has the n values while the row has the r values.
Example : To evaluate
1 (1 + 10014)4
look at the column for n=4 and then go along the row to r=14, the intersection gives the value 0.5921.
P = 1
(1 + 14)4 = 0.5921
Then the PV of N100 in 4 years at 14% is P = 100
(1.14)4 = 100× 1 (1.14)4
= 100×0.5921 =N59.21 Example :
Use the discounting table to determine the PV of N1, 500 in 6 years time at a discount rate of 19% .
Solution:
On the table, for n=6 and r=19%, the value is 0.3521. So P V = 1,500
(1 + 10019)6 = 1,500×(0.3521)
=N528.15 3.3 Annuity
We shall now consider certain sseries of equal payments or receipts at equal intervals.
This is not new to us we had considered the sum of such a sequence in unit 4.
3.3.1 Definition:
An annuity is a sequence of fixed equal payments (or receipts) made over uniform time intervals.
Common examples of annuities are
* weekly wages or monthly salaries.
* insurance premiums.
* house purchase morgage payment (see morgage Bank policy).
* hire purchase payments.
Remark:
Annuity are use in all areas of business and commerce as: * loans are normally repaid with an annuity.
* investment funds are setup to meet fixed further commitment . (e.g asset replacement) by the payment of annuity.
3.3.2 Types of Annuity (a) Annuity may be paid:
(i) at the end of payment intervals (an ordinary annuity) or (ii) at the begining of payment intervals (a due annuity).
(b) The term of an annuity may:
(i) begin and end on fixed dates (a certain annuity)
(ii)depend on some event that cannot be fixed (a contingent annuity).
(c) a perpertual annuity is one that carries on indefinitely.
3.3.3 Some Aspects and Examples of annuities
(a) The most common form of an annuity is certain and ordinary. That is, the annu-ity will be paid at the end of the payment interval ( or ”in arrears”) and will begin and end on fixed dates. For example, some domestic hire purchase contracts will involve the payment of an intial deposit and then equal payments, payable at the end of each month, up to a fixed date.
Persoal loans from banks or finance houses are paid off in a similar manner, but certainly without the intial deposite.
(b) Annuities that are being invested however are often due, that is, paid ”in advance of the interval. For example a saving scheme, paid as an annuity (with a bank , building society or insurance company) will not be deemed to have started untill the first payment has been made.
(c) Standaerd persion or superannuiation schemes can be thought of as two stages annu-ities. the first stage involves a due, certain annuitu ( i.e regular payment into the fund up to retirement age), the second stage being the receipt of a contingent annuity ( i.e regular receipts untill death ).
Example :
Determine the value of an annuity of three annual payments of N12,000 invested in a fund that pays 12%.
Solution :
The value of the first payment is
12,000 (1.12)3 The value of the second payment is
12,000 (1.12)2 and The value of the third payment is
12,000 (1.12) The total value of the invested monthly is
A= 12,000 (1.12)3+ 12,000 (1.12)2+ 12,000 (1.12) 12,000(1.12)h1 + (1.12) + (1.12)2i
This is a G.P with sum
12,000(1.12)
"
(1.12)3−1 0.12
#
=N45,351.60 The accrued amount is N45, 351.60
Example :
Determine accrued amount of 12 monthly payments of N100 into a building society ac-count which pays a fixed norminal rate of 10.75% compounded monthly.
Solution : (You have worked a similar problem before).
The accreued amounts of the monthly payments is A= 100
1 + 0.1075 12
12
+ 100
1 + 0.1075 12
11
+...+ 100
1 + 0.1075 12
A= 100 (1.00896)12+ 100 (1.00896)11+...+ 100 (1.00896)
= 100(1.00896)h1 + (1.00896) + (1.00896)2...+ (1.00896)11i This is a G.P with sum
= 100(1.00896)
"
(1.00896)12−1 0.0896
#
= 100(12.7223) =N1272.23
So the total accrued amounti N1272.23 3.4 Sum of Present Values
We consider the sum of PVs. Suppose A is the annual payment with present value P at r% discounting rate.
Denote 1 + 100r by R .
After n years, the sum of the present values is P = A
R + A R2 + A
R3 +...+ A
Rn (3.4.1)
= A R
1 + 1 R + 1
R2 +...+ 1 Rn−1
This is a G.P with common ratio R1. Thus P = A
R
1−
1 R
n
1− 1 R = A
R
h1− R1n
i
R−1 R
= A
R−1 =
1− 1 Rn
(3.4.2)
= A
(r+ 1)−1 =
"
1− 1
(r+ 1)n
#
A r =
"
1− 1
(r+ 1)n
#
(3.4.3) (3.4.3) is called the Net present Value (NPV) of annuity A over n years and subject to a discounting rate of r%.
Definition :
An annuity which is payment forever is called perpetuity.
In this case, you will consider the number of years to be very large i.e you consider n to tend to infinity. Denoted infinity by the symbol ∞ and write n→ ∞.
From (3.4), the PV of an annuity A is given as P V = A
r =
"
1− 1
(r+ 1)n
#
(3.9.1) When n → ∞, (r+1)1 n →0 So (3.9.1) becomes
P V = A r which is then the total present Value of A to infinity.
Remark :
For conciseness, you have r = 100r in (3.4.3) Example :
Determine the NPV of an annuity N125 payable at the end of each of fiveyears and subject to a discount rate of 8%.
Solution : Here r=8%, so let
R =
1 + r 100
= 1.08 Since A=125, using (3.4.1)
N P V = A R−1
1− 1 Rn
= 125 0.08
"
1− 1 (1.08)5
#
= 125(3.9926) =N499.08 Example :
The tenants of a rented house have their rent fixed at N1650 per year in advance with immediate effect. They plan to stay in the property for 15 years. Find the total value of the payment (in today’s terms). If the average discount rate is estimated at 10%.
Hint: Notice that this is a due anniuty.
Solution :
The NPV of the rents is
1650 + 1650
(1.10) + 1650
(1.10)2 +... 1650 (1.10)14
= 1650
"
1 + 1
(1.10) + 1
(1.10)2 +... 1 (1.10)14
#
This is a G.P with common ratio 1.101 . So N P V = 1650
h1− (1.10)1 15
i
1− 1.101 1650
"
1− 1
(1.10)150.10
#
×1.10 1650(1.10)
0.10
"
1− 1
(1.10)15
#
= 1650(8.3671) =N13805.72 So the total present value of the payment is N13, 805.72.
Let us consider a debt paying problem where the value of the initial payment is differenct from the value of the instalmental payment.
Example :
A department store advertises goods at N700 deposiote and three further equal annual payment of N500. If the discount rate is 7.5%, calculate the present value of the value of the goods.
Solution :
There are four payment to consider, but the first (N700) is payable now and thus its present value is the same as its face value . The other three payments of N500 have to be discounted over 1, 2 and 3 years respectively, using discount rate 7.5%. Using the formular in (3.4.1)
P V = 700 + 500
1 + 1007.5 + 500
1 + 1007.52
+ 500
1 + 1007.53
= 700 + 500
(1.075) + 500
(1.075)2 + 500 (1.075)3
= 700 + 500(0.9302) + 500(0.8653) + 500(0.8050)
= 700 + 500 [(0.9302) + (0.8653) + (0.8050)]
700 + 500(2.6005) =N2,000.25
Thus, the present value isN2, 00.25. i.e N2, 000.25 could be considered as the present cash price of the goods.
3.5 Present Value of an Interest-Bearing Debt
Everybody is in business in order to make some profit. Giving loan to people is a form of business. So when an amount of money is borrowed. It will attracted interest at an appropriate borrowing rate. If no intermadiate payment are made, at the end of a period, the debt plus interest must be repaid as a lump sum. The PV of such amount is often
required.
Let P be the amount borrowed at r% compounded for n years. The debt will amount to P(1 +r)n. If the investment (discounted) rate is R%, the debt will have to be discounted at this rate back over the n years.
Thus the present value of the debt is
P V =P (1 +r)n
(1 +R)n (3.5.1)
Note that, since the borrowing rate will always be greater than the investment (discount) rate, r > R, hence
(1 +r)n
(1 +R)n >1⇒P V > P Example :
Find the PV of a debt of N2, 500 taken out over 4 years (with no intermediate repayment) where the borrowing rate is 12% and the worth of money (discount rate ) is 9.5%˙Solution :
The value of the debt in 4 years is 2,500
1 + 12 100
4
= 3933.80 The present value of N3933.80 is
P V = 3933.80
1 + 1009.5 = 3933.80 (1.095)4 N2736.25
In otherwords, the original debt of N2,500 will cost N2736.25 (in today’s money terms ) to repay.
Thus, it can be considered that the real cost of the debt is N(2736.25-2, 500)=N236.25 (in today’s money terms ).
3.6 Amortasation Annuity
In the last example, you considered the PV of a debt in which there are no interme-diate repayment. Most debt repayments do not follow this procedure.
If an amount of money is borrowed over a period of time, one way of repaying the debt is by paying an amortisation anniuty. This consists of a regular annuity (ordinary and certain ) in which each payment accounts for both repayment of capital and interest.
The debt is said to be amortised if this method is used.
Many of the loan issued by banks and building societies for house purchase are of this type where it is known as a repayment mortgage.
the annuity payment, A, which will amortise the debt in exactly n years.
In this csae, you allow P to be the Present Value of the series of repayment A over n years. Thus
P = A
(1 +r)+ A
(1 +r)2 + A
(1 +r)3 +...+ A (1 +r)n
= A r
"
1− 1
(1 +r)n
#
Thus, the annual repayment Ais given by
A= P r
h1−(1+r)1 n
i (3.6.1)
Example:
ABC company negotiates a loan of N200, 000 over 15 years at 10.5% p.a. Calculate the annual payment necessary to amortise the debt.
Solution: Here p=200, 000 and R=10.5%. Let A be the amortisation annuity, then A= 200,000×0.105
h1−(1.105)1 15
i =N27,050
Thus the annual payment necessary to amorsatise the debt is N27, 050.
Definition :
The dfference between amount of installment and interes is called Amortisation.
3.7 Sinking Fund Definition :
A sinking fund is the money accumulated at compound interest by setting aside a fixed amount at regular intervals. That is sinking fund is an annuity interest in order to meet a known commitment at some future date.
Remark :
Sinking funds are commnly used for the following purposes:
(a) repayment of debts.
(b) to provide funds to purchase a new asset when the existing one is fully depreciated.
Example : (Debt repayment using a sinking fund)
Suppose a debt P incurred over n years subject to a given borrowing rate R%. A sinking fund must be set up to mature to the outstanding amount of the debt.
The value of the debt is P(1 +R)n. Set up an annuity A at investment rate r%. then P(1 +R)n =A(1 +r)n−1+A(1 +r)n−2A(1 +r)n−3+...+A(1 +r)2+A(1 +r) +A
=Ah1 + (1 +r) +...+A(1 +r)n−1i
=A[(1 +r)n−1]
(1 +r)−1 = A
r [(1 +r)n−1]
⇒A= P(1 +R)nr [(1 +r)n−1]
Example :
The sum of N25, 000 is borrowed over 3 years at 12%compounded. Determine the regular fixed amount paid into a sinking fund to pay the sum, at investment rate of 9.5%.
Solution :
The value of te debt in 3 years is 25,000
1 + 12 100
3
= 35123.20
If payment into the sinking fund are in arrears with an annuity A, then 35123 =A(1.093)2+A(1.093) +A
=Ah(1.095)2+ (1.095) + 1i
=A[3.2940]
⇒ 35123.20
3.2940 = 10662.78
That is, the annuity payment into the sinking fund is N10, 662.78 (which will produce, at 9.5%, N35, 123.20 at the end of 3 years).
Example :
ABC company borrows N46, 000 which is compounded at 15% to finance a new production line. The debt will be discharged at the annual payment into a sinking fund which pays 11.25%. Calculate the annual payment into the fund , assuming that the first payment into the fund is made at the end of the first year.
Solution:
The payments into the the fund in this case form an ordinary annuity, at the end of 5% years, the debt amount to
46,000
1 + 15 100
5
= 46,000(1.15)5 = 92522.43 The payment into the fund will mature to (with annuity A)
A(1.1125)4+A(1.1125)3+A(1.1125)2+A(1.1125) +A A[1.53179 + 1.37689 + 1.23766 + 1.1125 + 1 + 1] = A(6.25884)
⇒ 92522.43
6.25884 =N14782.68
3.8 Sinking Fund Method of Depreciation
You will recall that we had treated two forms of depreciation methods : straight line and reducing balance method. Now
(a) one reason for depreciating an asset is to take proper account of its re-placement. Thus one can consider the periodic book payments in respect of depreciation, the depreciation charge, as forming a pool that, at the end of the assets useful life, will fund a replacement or alternative.
(b) The sinking fund method of depreciation consider the depreciation charge payment as being available for investment into a fund (a depreciation fund, paying a market rate of interest) which will mature to some predetermine value. The book value of the respective assets at the end of any year can be determined by subtracting the curent amount in fund from the origional book value of the asset.
Example :
A mechane value at N12, 500, wth a 6 years life is estimated to have a scrap value of N450. If the depreciation fund earns 8%: You are required to use the sinking fund method based on an ordinary annuity to fund the annual deposit into the fund (Depreciation charge).
Example :
The difference between the original value=12, 500=450=12, 050
which must be the value of the depreciation fund after 6 years, interest paid at 8%.
Let the annuity=A. Then
12,050 =Ah1 + 1.08 + (1.08)2+ (1.08)3+ (1.08)4 + (1.08)5i
=A[(1.08)6−1]
(1.08)−1 =A(7.36)
⇒A= 12050
7.36 = 164.60 Thus the annuity depreciation large is N1642.60 3.9 Student Self Assessment Exercise
(1) Sanya, the son of chief Ajetunmobi is just 5 years old. In order to provide a sum of N50, 000 when sanya is 21 years, Chief Ajetunmobi has decided to invest a fixed sum of money every year begining immmediately and the last payment will be when sanya is 21 years . Find the fixed sum of money assum-ing investment rate is 8%.
(2) Find the interest on N1,000 for 10 years at 4%, interest being paid quar-terly.
(3) A man borrrows N1, 000 and repays the loan by yearly instalment of N100, the first instalment being paid one year afyter the loa. After how many years, will he be out of debt, interest being reckoned through at 4%.
(4) a man borrows N20, 00 and agrees to pay the borrowed amount in 10 equal annual instalment at the rate of 6% . Find the amount of each instalments, the first being paid one years after the money was borrowed.
Hint : You nmust work through the above problems before looking at the prefered
Solution :
(1) Let the fixed sum of money = Ny.
The money will be paid for (21-4)=17 year. The sum of the value of the instalment is
y(1.08)16+y(1.08)15+y(1.08)14+...+y This sum must equal N50, 000. So
50,000 =yh(1.08)16+ (1.08)15+ (1.08)14+...+ 1i This is now a G.P. So
50,000 = y[(1.08)17−1]
(0.08
⇒y= 50,000×(0.08)
(1.08)17−1 = 50,000×(0.08) 3.70−1
⇒y= 50,000×(0.08)
2.70 =N1481 So, the fixed sum of money is N1, 481.
(2) The amount, A is given by
A=P(1 +r)n where r=4%4 = 1%
A = 1,000(1.01)40 since there are 40 quarters i 10 years, A=N1486 Interest =A-P=1486-1, 000=N480.00
(3) The sum of the present values of the instalment must equal to N1, 000, after n years, so,
1,000 = 100
1.04+ 100
(1.04)2 +... 100 (1.04)n
= 100 1.04
"
1 + 100
(1.04) +... 100 (1.04)n−1
#
which is now a G.P . Hence
1,000 = 100 1.04
h1− 1.04)1 n
i
1−(1.04)1 1,000 = 100
1.04
"
1− 1 1.04)n
#
⇒ 1000×0.04
100 = 1− 1
(1.04)n
⇒ 1
(0.04)n = 0.6
⇒(0.04)n= 10
6 = 1.667
⇒nlog 0.04 = log 1.667
⇒n = log 1.667 log 0.04
⇒n= 13.05 years Since n is integral
⇒n= 14 years (4) Let the amount of each instalment=y.
The sum of the PV of each instalment must equal to 20, 000. So 20,000 = y
(1.06) + y
(1.06)2 +...+ y (1.06)10
= y
(1.06)
"
1 + 1
(1.06)2 +...+ 1 (1.06)9
#
This is again a G.P, so
20,000 = y 1.06×
h1−(1.06)1 10
i
1− (0.06)1.062
= y
1.06
"
1−
(1.06)10
#
×1− 1 (1.06)2 y
1.06
"
1− 1
(1.06)10
#
⇒yh1−(1,06)−10i= 20,000×0.06 = 1200
y = 1200
1−(1.06)−10 = 1200 1−0.5584
1200
0.4416 = 2718.
4.0 CONCLUSION :
You can now
* define and give the formular for present Value (Pv) of a future amount of money.
* use the discounting table.
* calculate NPV.
* define annuities as a series of fixed amount of money.
* work problems on Amortisation.
* work problems on sinking fund.
More importantly, you have only used the formula for G.P.
SUMMARY Recall that (1) Amount,
A=P(1 +r)n (2)
P = A
(1 +r)n is the P V (3)
1 (1 +r)n
is the discounting factor which is got from the discounting table for various values of n and r .
(4)
N P V = A r
"
1− 1
(1 +r)n
#
(5)
P V = P(1 +r)n (1 +R)n is the present value of an interest-bearing debt P.
5.0 TUTOR MARKED ASSIGNMENT :
(1)A man puts N10 at the end of every year in the saving Bank at 212% compound interest. How much will his saving amount to , in 15 years.