Simple Interest 1
Simple Interest 1
Simple Interest 1
Solution:
I=prt
I=50000(.1425) (7/12)
I=4156.25
Maturity value =p + i
=50000+4156.25
=₱54156.25
Io = Pr ( 360D )
Exact Interest – 365 days in one year
Ie = Pr ( 365D )
Where D is the number of days of a given term.
Illustrative Example:
1. Find the ordinary and exact interest on ₱15,000 if it is invested at 12% for 60 days.
a. Ordinary Interest
P= ₱15,000 r=12% D=60 days
Solution:
Io = Pr ( 360D )
60
= (₱15,000) (.12) ( 360 )
= (₱15,000) (.12) ( 16 )
=₱300
b. Exact Interest
P=₱15,000 r=12% = .12D=60 days
Solution:
Ie = Pr ( 365D )
60
= (₱15,000) (.12) ( 365 )
=₱295.89
[ ( )]
F= P 1+ r 360
D
[ ( )]
= ₱ 22, 000 1+.10 360
122
= ₱22, 000(1.033888889)
= ₱22,745.56
2. Actual time
Solution:
1. May (31-4) 27 2. September 6 249
June 30 May 4 124
July 31 125 days
August 31
September 6
125 days
To find now the total amount paid by Julie on September 6, 2002, Using actual time,
we have
F= P 1+ r
[ ( 360D )]
[ ( )]
= ₱ 22, 000 1+.10 360
125
= ₱22, 000(1.034722222)
= ₱22,763.89
Exercise 1.2
Exercises on Approximate and actual time
Exercise 1.2.1
Application Problems on Ordinary and Exact Interest,
Approximate and actual time
1. Find the Ordinary and exact interest if ₱6,500 is borrowed at 24% for 120 days.
Solution:
Io= 360 Ie = 365
120 120
= ₱6,500 (.24) ( 360 ¿ = ₱6,500 (.24) ( 365 ¿
= ₱520 = ₱512.88
2. Allan Castro borrowed ₱3,785 from a friend and promised to repay him in 90 days plus
12% interest. How much will he pay when the loan matures using exact interest?
Solution:
I = Prt MV = I + P
70
= ₱3,785 (.12) ( 365 ¿ = ₱11.99 +₱3785
= ₱111.99 = ₱3896.99
3. On what day will ₱8,000 earn ₱180 interest when invested on April 25, 2002 at 9%
simple interest? Use ordinary interest and actual time.
Solution:
I
t= Pr
180
= 8,000 ( .09 )
= .25
4. On September San Juan went to ABC bank to borrow ₱230,000 at 9% interest. Jody
repaid the loan on January 27, 2003. Assuming the loan is on exact time, ordinary interest,
how much did Jody repay on the maturity date?
Solution:
I=prt MV = I + P
137
= ₱230,000 (.09) ( 360 ¿ = ₱230,000 +₱7,877.5
=₱7,877.5 = ₱237,877.5
5. In problem number 4, suppose Kaye Reyes met Jody San Juan at ABC bank and
suggested she consider the loan on exact time, exact I, recalculate the loan for Jody Under
this consumption. What is your recommendation to Jody?
Solution:
I=prt MV = I + P
137
= ₱230,000 (.09) ( 365 ¿ = ₱230,000 + ₱7,769.5
6. Borris buying a van from the interest at his time deposit in a bank. His April monthly
interest at 12% was ₱125,000. What was Borris’ principal balance at the beginning of
April? Use 365 days.
Solution:
I
P= rt
125,000
30
= 12 ( )
365
= 12,626,262.6
1
7. Using approximate time, find the exact interest on ₱75,000 at 8 4 % from March 8,
I
I=prt t= Pr
I 700
= 10,000 t = Pr = 10,000 ( .12 )
= ₱700 = 213
9. What is the exact simple interest rate using exact time if ₱6,000 grew to ₱6,168 from
February 24, 2001 to May, 2001?
Solution:
I
r= Pt
168
= 83
6,000 ( )
365
= 12.31%
10. On the first day of October 2002, Albert Flores had ₱21,455 in an account that earns
1
10 2 % interest. How much interest was added to his account at the end of December of
the same year if he made no conditions or withdrawals during this time? Use the banker’s
rule.
Solution:
I=prt MV = I + P
89
= ₱21, 455(.1050) ( 360 ¿ =₱556.94 + ₱21, 455
=₱556.94 = ₱22,011.94
Exercise 1.3
Exact and Approximate Time between Two Dates
Date of investment Date of Maturity Exact No. of Days
Approximate No. Days
1. July 15, 2001 March 10, 2002 238 235
2. February 2, 2003 August12, 2003 191 190
3. April 20, 2003 November 6, 2003 200 196
4. June 5, 2003 October 12, 2003 129 127
5. March 28, 2003 September 25, 2003 181 177
6. May 8, 2003 December 10, 2004 582 572
7. July 6, 2003 January 4, 2004 182 178
8. January 17, 2003 July 8, 2003 172 171
9. August 24, 2003 February 28, 2004 188 184
10. Sept. 30, 2003 April 15, 2004 198 195
Exercise 1.3.1
Problem Solving on Ordinary and Exact Interest; Exact and Approximate
Time
3
1. Find the ordinary interest and the amount on ₱6.200 at 7 4 %for 65 days.
Solution:
I =Prt
65
= ₱6,200 (.0775) ( 360 ¿
= ₱86.76
2. What principal invested on July 6, 2003 will amount to ₱12,000 on December 12, 2003 if
the rate is 9.5% and exact simple interest for the exact time is used?
Solution:
I
P= rt
12,000
= .0095 159
365 ( )
= ₱289,970.21
3. If ₱15,000 is borrowed on April 20, 2003 at 6% on what day will payment of ₱15,600
repay the debt and interest?
Solution:
I
t= Pr
600
= 15,000 ( .06 )
= .6667 or 240 days
4. What is the ordinary simple interest rate if ₱3,000 earns ₱90 from February 12, 2003 to
august 8, 2003?
Solution:
I
r= Pt
90
177
= 3,000
360( )
= 0.0610
= 6.10%
5. Using approximate time, find the exact interest on ₱5,800 at 11% from May 15, 2002 to
July 20, 2003? Solution:
I =Prt
425
= ₱5,800(.11)( 365 ¿
= ₱742.88
6. Find the ordinary simple interest on ₱4,200 at 8.25% using approximate time from
March 3, 2003 to June 5, 2004.
Solution:
I =Prt
452
= ₱4,200 (.00825)( 360 ¿
= ₱435.05
7. Find the exact interest and the amount on ₱2,500 at 9% for 40 days.
Solution:
I =Prt
452
= ₱4,200 (.00825) ( 360 ¿
= ₱435.05
8. Using the banker’s rule, find the simple interest and the maturity value if ₱9,000 is
invested at 12% from June 12, 2002 to May 6, 2003.
Solution:
I =Prt MV = ₱984 + ₱9,000
328
= ₱9,000 (.12) ( 360 ¿ =₱9,984
= ₱984
3
9. Using exact time, find the ordinary interest on ₱6,400 at 9 4 from January 25, 2002
= ₱677.73
10. What is the exact simple interest rate using exact time if ₱1,500 earns ₱200 from July
10, 2002 to August 20, 2003.
Solution:
I
r= Pt
200
406
= 1,500 ( )
365
= .1199 or 11.99%
1.4 Simple Discount
If (I) is calculated on the principal (P) at the start of the interest period, discount (D) is
calculated on the amount (F) at the end of the period.
Discount (D) is a deduction from the maturity amount (F) of an obligation allowed to
paying it currently.
Formulas:
D = Fdt, Where D= discount To find P, use
F= amount P= F – D or P = F (1− dt)
d= discount rate
t= time or term of discount
Derived Formulas:
D D D
d= , t= and F= ,
Ft Fd dt
Example:
1. Find the present value of ₱2, 000 which is due at the end of 90 days at 5% simple
discount.
1
Given: F = ₱2,000 t = 4 d = .05
Solution:
D = Fdt P=F−D
1
= ₱2,000 (.05) ( ¿ = ₱2,000 − ₱25
4
= ₱25 = ₱1.975
Alternative solution:
D = F (1−dt)
1
Given: P = ₱4,200 t = 2 4 years d = .065
Solution:
P
F= 1−dt
₱ 4,200
= 1−( .65 ) 2 ( 14 )
= ₱4, 919.47
Exercise 1.4
1. Find the present value of ₱5,000 due at the end of 6 months if the discount
3
rate is 5 4 %.
P= FDT
6
=5, 000( 12 )(.0575)
=₱ 143.75
P = ₱5,000 – 143.75
1
2. Discount ₱6,500 for 120 days at 4 2 % simple discount.
P = Fdt P=
1
4. Accumulate ₱3,200 for 2 years and 6 months at 5 2 % simple discount.
1
5. Accumulate ₱3,000 for 1 year and 9 months at 6 4 simple discount.
6. If ₱2,500 is due on September 20, 2003, find its value on June 22, 2003 if the
discount rate is 6 %.
7. Find the discount rate if ₱7,000 is the present value of ₱7,500 which is due at the
end of 6 months.
8. If ₱2,421 is the present value of ₱2,500 which is due in 6 months, find the discount
rate.
9. Find the amount due on February 3, 2004 if the value on October 6, 2003 is ₱5,250
1
at 6 4 % simple discount.
10. How long will it take ₱5,850 to amount to ₱6,500 at simple discount rate of 5%?
11. When will ₱3,000 be the present value of ₱3,150 if the discount rate is 6%?
F=P+I
Proceeds = F [1−d ( 360 ) ]
days
Md = Dn + tn or
Td = Md – Dd
Proceeds = F – D or f (1−d td)
Where:
r, rate of interest Dd, date of discount
d, rate of simple discount tn, term of the note
tn, term of the note in years Dn, date of the note or initial date
td, term of discount in years Md, maturity date or final date
I, amount of simple interest Pr, amount of proceeds, amount
D, amount of simple discount payable at the discount date
P, Principal or face value
F, final amount or maturity of the
note
Illustrative Example:
Mr. Mananquil has a five month note of ₱15,900 dated February 6, 2002 bearing
3
interest at 8 5 %. If she sells the note on April 10, 2002at a bank discount rate at 8%,
F= [
P 1+ r ( 12tn )]
= [
₱ 15,900 1+.086 ( 125 )]
= ₱16,469.75
Md = Dn + tn
= February 6, 2002 + 5 months
= July 6, 2002
Td = Md – Dd
= July 6, 2002 – April 10, 2002
=187 – 100
= 87 days
Proceeds = [
F 1−d
360 )]
( days
[ ( )]
= ₱ 16, 469.75 1−.08 360
87
= ₱16, 151.33
Exercise 1.5
Problem Solving Involving Promissory Note
1. A 150−day note for ₱12,500 dated June 3, 2002 with interest at 11% is discounted on
1
September 3, 2002 at 8 2 % simple discount. Find the
a. maturity date
Solution:
June 3, 2002 154
150
304 days
b. maturity value
Solution:
F=P+I
= 180/360
c. term of the discount
Solution:
150
I = ₱12,500 × .11 × 360
=₱172.92 + ₱12,500
d. discount
Solution:
td
D =F × d × 360
58
=₱13072.92 × .0850 × 360
=₱ 179.03
e. proceeds
Solution:
P=F–d
= ₱13,072 - ₱179.03
= ₱12,893.89
2. A note for ₱9,000 dated August 12, 2002 is due on October 20, 2002 with simple
interest at 9%. If the note is discounted on August 20, 2002 at 13%, how much should the
proceeds be?
Chapter 2: Compound Interest
2.1 Finding the Compound Amount
̶ Compound Interest is the sum by which the original principal has been increased
by the end of the contract. The original principal plus the compound interest, is called the
Compound Amount.
Annually m=1
Semi-annually m=2
Quarterly m=4
Monthly m=12
Example 1
Find the compound amount of ₱20,000 compounded semi-annually for 2 years at
12%.
12
m=2 i= 2 =6%
t=2 years n=m x t = 2 x 2 = 4
Solution:
F=P (1+ mj ) mt
and I = F –P
Or
F = P (1 + i)n
Example:
1
Accumulate ₱6,500 for 4 years and 5 months at 5 2 % compounded semi-
r .055
t = 4 yrs and 5 mos I = m = 2 = .0275
m=2
1
r=5 2 % = 5.5% = .055
F = P (1 + i)1/n
=₱6,500 (1 = .0275)8.8333…
=₱6,500 (1.0275)8.8333…
=₱6,500 (1.27077572)
F = ₱8,260.04
Example :
Find the amount at the end of 13 years of ₱5,500 which is invested at the rate of
12% compounded quarterly in the first 4 years, 10% compounded semi-annually in the
next 7 years, and 8% effective in the remaining years.
Given: j=12% j =10% j=8%
m=4 m=2 m=1
i=.03 i=.05 i=.08
t=4yrs t=7yrs t=1yr
n=16 n=14 n=1
The Present Value is defined as the principal which you would have invest now at a
given interest rate, so that will amount to some predetermined future sum of money. The
Compound Discount is the difference between the future value and the present value. The
formula for the present value or discounted value of an amount is given by
Example:
1
If money can be invested at 3 2 % compounded quarterly, find the present value
.035
Let F =₱6,500 i= 4 = .00875
Example 1
Find the present value of ₱7,600 due at the end of 4 years and 4 months, if money
is worth 6% compounded quarterly.
.06
Let F = ₱7,600 I= 4 = .015
1
P = ₱7,600 (1 + .015)- 17
3
1
=₱7,600 (1.015)- 17
3
=₱7,600 ( 0.772580047)
P = ₱5,871.61
Illustration:
At j1 = 8.4116% and m = 4, if P = 1,000 and t = 2 yrs, then the accumulate value is
Diagram:
= 1,000 (1.04205)4
= ₱1,181.15
Diagram:
J1 = m 1
W= (1+ mj ) m
–1
Hence:
Or Jc =
Jc =
By logarithmic Method
To find the periodic rate of interest, j, begin from the formula
F = P ( 1+ i)n
log F = log P + n log ( 1+ 1)
n log ( 1 + i ) = log F – log P
log F−log P
log ( 1 + i ) = n
j
and finally, from i = m , the unknown periodic rate is j = (m)(i)
by Exponential Method
Again, to find the periodic rate of interest j, begin from the formula
F = P (1 + i )n
F
P = ( 1 + i )n
1+i= ( FP )
The formula for j is given by
J=m ( FP ) 1n - 1]
¿
Example:
At what rate compounded quarterly will ₱7,000 become ₱18,500 at the end of 11 years?
Solution:
Let P = 7,000 F = 18,500 n = 4 (11) = 44 m=4 and j=?
Substituting the given values to the formula,
By Logarithmic Method
= antilog ( 4.26717173−3.84509804
44 )−1
= antilog ( 0.00959258) – 1
i = 0.02233348
Thus the unknown nominal rate is
J = 4 ( 0.02233348 )
= 0.08933392
= 8.93%
2.9 Finding the Time or Term of Investment
It is very important for an investor to know how long it will take his money to
accumulate into his desired amount. The term of investment or the length of time, t, for a
given amount P to accumulate into an amount, F, at a given interest rate (j, m), may be
obtained through a Logarithmic Method.
To derive the formula for time, t, begin from the formula
F = P(1 + i)n
Hence, n=
log ( FP ) or n=
log F−log P
log(1+i)
log ( 1+i)
Example:
On January 25,2003 Mrs Sally Lorenzo had ₱400,000 on a trust fund which earns
14% converted quarterly. She plans to put up a cake and pastry business as soon as the
fund contains an amount of ₱500,000. On what date will that amount be available?
Solution:
Let P = 400,000 F = 500,000 j = 0.14 m=4 and
t=?
Then from the formula
F=P ( mj ) 4t
1.25 = (1.035)4t
log 1.25 = 4t log 1.035
log 1.25
4t = log 1.035
0.09691001
= 0.01494035
t = 1.6216 years or
t = 1 year, 7 months and 14 days
ILLUSTRATION
Then
x = P = 15,000(1.0475)-14
= 15,000 (0.52220804)
x = ₱7,813.21
b. Let F= 15,000 i = 0.0475 n = 2(3 years) = 6 and x=?
ILLUSTRATION
Then
x = P = 15,000(1.0475)-6
= 15,000 (0.75696502)
x = ₱11,354.48
c. Let P = 15,000 I = 0.0475 n = 2(2) = 4 and x=?
ILLUSTRATION
Then
x = F = 15,000 (1.0475)4
= 15,000 (1.20397128)
x = ₱18,059.57
Solution:
0.09
Let P= 7,000i = 4 = 0.0225 n = 4(2) = 8
Then
F = 7,000(1.0225)8
= 7,000 (1.19483114)
= ₱8,363.82. . . . This is the value of obligation (b) at its maturity
date. If comparison date is at t = 2 then discount obligation (a) for n = 2(1) at
0.08
i= 2 = 0.04. Therefore,
P = 8,500 (1.04)-2
= 8,500 (0.092455621)
= ₱7,858.73
Example:
A debtor owes ₱20,000 at the end of 2 years and ₱50,000 at the end of 8 years. If
interest is at 7% effective rate, what single payment at the end of 6 years would
liquidate both debts?
Solution:
Let x be the single payment. Draw a time diagram and locate the two old
obligations and the new single payment.
Diagram:
The focal date may be chosen arbitrarily but for convenience, choose end of
6 years as focal date. Thus ₱20,000 must be accumulated for 4 years at 7%
effective rate and ₱50,000 must be discounted for 2 years at 7% effective rate. The
single payment x which lies at the focal date is neither accumulated nor discounted.
The equation of values may now be expressed as
New payment = Sum of Old Payments (at focal date –end of 6 years)
x = 20,000 (1.07)4 + 50,000 (1.07)-2
=20,000 (1.310790601) + 50,000 (0.873438728)
=26,215.92 + 43,671.94
x = ₱69,887.86
The following symbols will be used in dealing with ordinary annuity formulas:
R - periodic payment of the annuity
n - total number of payments
i - interest per conversion period
S - amount of an annuity
A - present value of an annuity
0 1 2 3
₱10,000 =P10,000
₱10,000(1 +.10)1 = 11,000
₱10,000(1 +.10)2 = 12,100
S = P33,000
To find the present value of an annuity, add theTerm
discounted payments.
0 1 2 3
₱10,000(1 + .10)-1 = ₱9,090.91
₱10,000(1 +.10)-2 = 8,264.46
₱10,000(1 +.10)-3 = 7,513.15
A = ₱24,868.52
Hence, from the relation
A= S(1 + i)-n and S = A(1 + i)n
A= 33,100(1 + .10)-3 and S = 24,868.52(1 + .10)3
A= ₱24,868.52 S = ₱33,100
R R R R R
0 1 2 3 n-1 n (periods)
R
R(1 + i)1
.
.
.
R(1 + I)n-3
S = sum of the accumulated values of RR(1
at + I)n-2
the end of the
term
R(1
n-2+ I)n-1
1 S = R + R(1 + i) + . . . R(1 + i) n-3 + R(1 + i) + R(1 + i) n-1
N = mt = number of payments
[ ]
n
( 1+i ) −1
i (1 +
A= R
if we let n¬ i = [
1−(1+i)−n
1−(1+i )
i
−n
]
then the formula i of A can be written as
A=
Ran¬ i
where A is the present value of the annuity an¬ I, is read as ”a angle n at i”
an¬ i =
−n
1−(1+i )
i
Illustrative Example 1
Find the amount and present value of an annuity of ₱1,500 payable for
2 years if money is worth 105 compounded semi-annually.
Given R = P1,500 j .10
j = 10% - .10 i = m = 2 =.05
i = 2 years
m=2 n = mt = 2(2) = 4
periods
To find S
S = Rsn¬ i
S = 1,500svn¬ .05
S = 1,500(4.310125)
S = P6,465.19
[ ]
4
( 1+.05) −1
or S=R i
[ ]
4
( 1+.05) −1
s = 1,500 .05
s = 1,500(4.310125)
s = ₱6,465.45
to find A
A = Ran¬ i
or by using the relation
A = S(1 + i)-n
A = 1,5004¬ .05
A = 6,465.19(1 + .05)-4
A = 1,500(3.54595)
A = P5,318.93
A = P5,318.93
Illustrative Example 2
Mrs. Cruz purchased a house and lot. If she paid ₱400,000 as down
payment and promised to pay P5,000 every 3 months for the next 10 years
at 15% compounded quarterly, find the cash value of the house.
Solution:
The down payment is not part of the annuity. The cash value of the
house is ₱400,000 plus the present value of the ordinary annuity of 40
monthly payments of P5,000 each. The cash value is
Solution:
Given: S = ₱300,000
t = 15 years
n = 60 periods
j 8
i = m = 4 = .02
Find R:
S
R= s n¬ i
300,000
R= s 50 ¬.02
300,000
R= 114,05154
R = ₱2,630.38
Illustrative Example 2
Mr. Tuy bought a refrigerator that cost ₱19,500. He paid ₱6,000 as down
payment and the balance will ne paid in 36 equal monthly payments. Find the
monthly payment if money is 15% compounded monthly.
Solution:
A = Cash Value – Down
A = ₱19,500 – ₱6,000 = ₱13,500
n = 36
.15
i= 12 = .0125
Find R:
A 13,500
R= an ¬i = a.36¬.0125
13,500
R= 1−( 1+.0125 )−36
.0125
13,500
R= 28.84726737
Exercise 3.1
1 Analyn bought a dining set. She paid ₱2,500 as down payment and promised
to pay ₱800 at the end of each month for one year. What is the cash
equivalent of the set if the interest rate is 10% compounded monthly?
2 Find the amount present value of an annuity of ₱5,240 payable at the end of
1
every 6 months for 5 years and m6 months at 5 2 converted semi-
annually.
3 What is the present value of ₱1,230 due at the end of every three months for
3
3 years and 6 months, if money is worth 9 4 % compounded quarterly?
5 If Cherry will deposit P1,500 every end of each quarterly in 9 years at the
1
rate of 6 2 compounded quarterly, how much will be the amount of her
8 A man deposits ₱12,500 at the end of every six months in a bank which
credits interest at 9% compounded semi-annually. How much will he have
after 19 years?
11 A man purchased a DVD player worth ₱8,500. He paid P2,000 cash and
agreed to make 15 monthly payments. If interest is 18% (m=12), find the
monthly payment.
12 Marga borrowed P250,00 and agreed to repay her obligation by making equal
quarterly at the end of every months for 4 years. What is the periodic
payment, if money is worth 12.5% compounded quarterly?
1
13 How much should be invested each year in a fund paying 10 2 effective to
1
15 How much should one deposit monthly in a fund that earns 12 4
nR
b and (n - 1)i + 6(n + 1)i + 12 ( 1− S )
2 2
=0
Example 1
A refrigerator can be purchased for ₱9,000 cash down and P630 a month for
18 months. Find the interest rate charged if it is compounded monthly.
Solution:
A = 9,00 – 900 – 8,100
R = 630
n = 18
m = 12
Formula:
Since A is known
(1− nRA ) = 0
(n2 - 1)i2 + 6(n + 1)i + 12
i = -114 ±
√ (114 )2−4 (323)(−4.8)
2( 323)
114 ±138.555404
i= 646
i = -.3909526 or .03801146
Example 2:
Payments of ₱750 each are made every 6 months in 3 years. At what rate
compounded semi-annually will these payments amount to ₱5,100?
Solution:
S = ₱5,100
R = ₱750
n = mt = 2(3) = 6
Formula: Since S is known
Exercise 3.2
1 in purchasing a washing machine worth P21,000, a buyer pays ₱3,500 cash
and agrees to make 12 monthly payments. If the monthly payment is ₱1,580,
find the interest rate m charge.
2 Deposit of ₱8,380 are made every end of six months. At what rate
compounded semi-annually will these deposits amount to P112,400 in 14
years?
4 A man invests ₱9,375 every six months. If he has ₱200,000 in five years, at
what nominal rate compounded semi-annually did hi investment earn?
[
Example 1log 1− Ac
R ]
−log (1+1)
Victoria borrows ₱15,000 with interest at 15% compounded quarterly. She
will discharge the debt by paying P950 quarterly.
a How many payments of ₱950 are required?
b How much would the final payment be if it is made the day after the last
P950 payment?
c How much would the final payment be if it is made 3 months after the last P950
payment?
Solution:
Given: A = ₱15,000 R = 950 i = .0375
a Substitute the given values into the formula
n=
[
log 1−
Ac
R ]
log ( 1+ i)
n=
[
log 1−
15,000(.0375)
950 ]
log ( 1+.0375)
log.407894737
n= log ( 1.0375)
.389451898
n= .015988105
n = 24.35885291
This means that there will be 25 quarterly payments of ₱950 and a
final irregular payment.
0950950950 950950
1 2 3 23 24
. . .
Focal date
x(1.0375)-24 + 950 a24 ¬ .0375 = 15,000
x(.413319095) + 14,862.58 = 15,000
x = P332.47
Therefore the final 24th payment is
= 24th payment + final irregular payment
= 950 + 332.47
= P1,282.47
1 2 3 23 24
. . .
950 950 950 950 950 950 y
0 1 2 .3 . . 22 23 24 25
Exercise 3.3
1 Lulu borrows ₱240.000 with interest at 12% compounded monthly. She will
pay the debt by paying ₱5,300 monthly
a Find the number of regular payments to be made.
b How much will be the final payment if it is made one month after the last
₱5,300 payment?
4 A car is worth ₱730,000. The buyer paid ₱350,000 down payment and
3
promised to pay ₱7,660 every month, if he is charged 7 4 compounded
monthly,
a How many full payments must be made?
b How much will be the final payment?
0 1 2 3 ... n-1 n
R R R R ... R
The formula in finding the present value of an annuity due is:
A = (1st payment) + (present value of remaining payments)
A =A +
Ran−1¬ i
Example 1
Marlon purchased a car. He paid ₱150,000 down payment and ₱10,000
payable at the beginning of each month for 5 years. If money is worth 12%
compounded monthly, what is the equivalent cash price of the car?
Given:
R = ₱10,000 j = 12%
12
Down payment = ₱150,000 i = 12 – 1%
m = 12 A = ? Cash Equivalent - ?
t = 5 years
n = 60
A =R+ Ran−1¬ i
A = ₱10,500 + ₱ 10,50060−1 ¬1
[ ]
−59
1−(1.01)
A = ₱10,500 + ₱10,500 .01
A = ₱10,500 + ₱466,248.18
A = ₱476,748,18 (Total Installment Payment)
Cash Equivalent = Down Payment + Total Installment Payments
Cash Equivalent = ₱150,000 + ₱476,748.18
Cash Equivalent = ₱626,748.18
0 1 2 3 ... n-1 n
R R R R ... R
The formula in finding the amount of an annuity due is:
S = {Value of an annuity of (n +1) payments on the last payment date} – R
S = Rsn+1 ¬ i−R
Example 1
If ₱500 is deposited in a bank at the beginning of each 3 months for 10
years and money is worth 8% compounded, how much is in the fund at the end
of 10 years?
Given:
R = P500 j = 8%
8
M=4 i= 4 = 2%
t = 10 years S=?
n = 40
S= Rsn+1 ¬ i−R
S = ₱500 [ (1.02)41−1
.02 ] -₱500
S = ₱31,305.01 – ₱500
S = 30,805.01
Example 2
Emil invests ₱5,000 at the beginning of each six months. He makes his
first deposit on January 19, 2003. How much will be in his account on January
19, 2015, if money is worth 9% compounded semi-annually?
R = ₱5,000
m=2
t = January 19, 2003 – January 19, 2015 = 12 yrs.
n = 24
j = 9%
9 1
i= 2 =4 2 %
S=?
S= Rsn+1 ¬ i−R
₱ 5,000 s 1 −₱ 5,000
S= 24+1 ¬ 4
2
₱ 5,000 s 1 −₱ 5,000
S= 25 ¬4
2
[ ]
13
(1.045) −1
S= ₱ 5,000 -₱5,000
.045
S = ₱22,826.05 – ₱5,000
S = ₱217,826.05
Where:
A = present value of an annuity due
S = amount of an annuity due
R = periodic payment of the annuity due
i = interest per conversion period
n = total number of payments
Example 1
What equal deposits should be placed in a fund at the beginning of each
year for 15 years in order to have ₱1,500,000 in the fund at the end of 15
years, if money accumulates 12%?
Given:
S = ₱1,500,000 j = 12%
m=1 i = 12%
t = 15 R=?
n = 15
A
R= S n+1 ¬i−1
P 1,500,000
R= S 15+1 ¬12 −1
P 1,500,000
R= S16 ¬12 −1
P 1,500,000
R= 41.75328042
R = ₱35,925.32
Exercise 3.4
For each problem draw a diagram illustrating the data and the solution
1 If money is worth 16% compounded quarterly, find the present value and the
amount of annuity due of ₱1,500 payable quarterly for 10 years.
2 An investment of ₱5,000 is made at the beginning of each month for 8 years
and 7 months. If interest is 12% compounded monthly, how much will the
investment be worth at the end of the term?
4 Mr. Jason Raymundo agrees to pay ₱2,000 at the beginning of each 6 months
1
for 5 years. If money is worth 12 2 % compounded semi-annually, find a)
5 A house and lot is bought for ₱1,000,000 down payment and ₱5,000 payable
at the beginning of each month for 5 years. What is the equivalent cash price
of the house and lot, if interest rate is 14% compounded monthly?
6 On January 19, 2003, Mr. Berlindo Santos opened a saving account for his
wife with an initial deposit of ₱10,000 in a bank paying 6% compounded
quarterly. If Mr. Santos continues to make quarterly deposits of the same
amount until July 19, 2008, how much will be in the account at the end of the
term? (Hint: Count the time by months.)
quarterly?
8 What equal deposits should be placed in a fund at the beginning of each year
for 20 years in order to have ₱2,500,000 in the fund at the end of 20 years, if
1
the money accumulates 10 2 %?
quarterly payment.
Example 1
A sequence of quarterly payments of ₱3,500 each, with the first one due
at the end of 3 years and the last of the end of 10 years. Find the present
value of the deferred annuity, if money is worth 16% compounded quarterly.
Given:
R= ₱3,500 j = 16% m=4
Solution:
16
i= 4 = 4%
d = 3 x 4 = 12; 12 – 1 = 11
n = 7 x 4 = 28; 26 + 1 = 29
40
11 29
RRRRRRRRRRRR
Ad RRRRR
‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ Sd
0 1
0
2 yrs
2
Ad
1
3
2
4
3 7
5
... 6 7 8
d d1 d2
9
...10
d+n-2 Sdd+n-1 d+n
0 1
10deferment
2 3(d) ...
period n periods
d Payments ...
d1 d2are made d+n-2 d+n-1 d+n
No payments
deferment period n periods
(d) Payments are made
No payments
Ad = Ran+d ¬i−Rad ¬ i
Ad = ₱ 3,500 a29 +11¬4 −₱ 3,500 a11¬ 4
Ad = ₱ 3,500 a 40¬ 4 −₱ 3,500 a11 ¬4
Ad = ₱3,500 [ 1−(1.04)−40
.04 ] - ₱3,500 [ 1−(1.04)−11
.04 ]
Ad = ₱69,274.71-₱30,661.67
Ad = ₱38,613.04
Example 2
Find the cash equivalent of an item that sells for ₱20,000 down payment
and 20 semi-annual payments of ₱5,500 each, the first is due at the end of 3
years, if money is worth 10% compounded semi-annually.
Given:
R = ₱5,500 Down payment = ₱20,000 j = 10% m=2
Solution:
10
i= 2 = 5%
d = (3 x 2) – 1 = 6 – 1 = 5
n = 20
Ad
25
5 20
R R R R R R R R R R R R R R R R R R R R R R
‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘
0 1 2 3 4 5 6 7 8 9 10 11 12 13
2 yrs
Ad = Ran+d ¬i−Rad ¬ i
Ad = ₱ 5,500 a20 +5¬ 5 −₱ 5,500 a5 ¬5
Ad = ₱ 5,500 a25 ¬5 −₱ 5,500 a 5¬ 5
[ ] [ ]
−25 −5
Ad 1−(1.05) 1−(1.05)
= ₱5,500 .05 - ₱5,500 .05
Ad = ₱77,516.70 – ₱23,812.12
Ad = ₱73,704.58
Given:
Ad = ₱100,00
J = 12%
m =2
n = 10
d = 10
12
i = 2 = 6%
Solution:
Ad
R= ad +n ¬i −ad ¬ i
P 100,000
R= a20+6 ¬ i−a 10¬ 6
P100,00
[ ][ ]
−20
R= 1−(1.06) 1−10
−
.06 .06
P 100,000
R= 11,46992122−7.360087051
P 100,000
R= 4.109834169
R = ₱24,331.88
Exercise 3.5
For each problem, draw a line diagram illustrating the data and the solution
Find the present values of the deferred annuity which is described, at the
specific interest rate.
1 A farm costs ₱750,000 cash. Paolo will pay ₱150,000 cash and a sequence of
15 annual payments, the first due at the end of 5 years. If money is worth
12%, find the annual payment.
2 A house and lot is offered for sale ₱1,500,000 cash, or ₱500,000 cash and a
sequence of 20 semi-annual payments of 120,000 each, the first is due at
the end of 3 years, if money is worth 14% compounded semi-annually, find
the cash value of the house and lot.
3 Find the present value of a series of quantity payments of ₱2,500 each, the
first payment is due at the end of 4 years and 6 months, and the last at the
end of 10 years and 3 months, if money is worth 16% compounded quarterly.
4 Find the cash equivalent of a computer set that sells for ₱10,000 cash and a
sequence of 10 semi-annual payments of ₱3,500 each, the first is due at the
1
end of 3 years and 6 months, if money is worth 10 2 % compounded semi-
annually.
5 Jason borrows ₱50,000 and agrees to pay his obligation by making 15 equal
annual payments, the first is due at the end of 3 years . Find the annual
payment, if money is worth 12%.
7 Find the present value of the pension of a man, now 60 years old, and who
will receive a pension of ₱10,000 per month for 15 years, with the first
payment to occur one month after his 65th birthday, if money is worth 18%
compounded monthly.
8 An investment in a stock market will yield no operating profit until the end of
5 years, when investor will receive P100,000 after that, he will receive
₱100,000 at the end of each quarter for 10 more years. Find the present
1
value of this income if money is worth12 2 % compounded quarterly,
9 On November 20, 2002 a house and lot was bought for ₱500,000 down and
12 quarterly payments of ₱50,000 each, the first is due on November 20,
2003. Find the cash value of the house and lot, if money is worth 15%
compounded quarterly.
10 Aling Nelie won ₱25,000,000 in Mega Lotto. She invest winnings at 16%
compounded semi-annually with the conditions that she receive 25 semi-
annual withdrawals starting at the end of 2 years. Find the size of the
withdrawals.
annually, find the sum of the values of these payment at the end of a)2 years,
1
b)4 2 years, c) 6 years, d)the actual present value of the payments.
Given;
R = ₱2, 500
j = 12%
m=2
12
i= 2 = 6%
n=5
Time Diagram:
R R R R R
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 yr. 2 yrs. 3 yrs. 4 yrs. 5 yrs. 6 yrs. 7 yrs.
Solution;
1
a) At the end of 2 years b) At the end of 4 years
2
A₂ = ₱ 2, 500 a5 ¬ 6 S = ₱2,500 a5 ¬ 6
c) To find the value of the annuity at the end of 6 yrs, accumulate a) for 4 years, or
1
the result of b) for 1 2 years.
a
¿
Actual Present Value = ₱2,500 ¿ )(1.06)-4
¿
=₱8,341.47
or
Actual Present Value = ₱2,500( a5 ¬ 6 )(1.06)3
=₱8,341.47
Exercise
For each problem in the exercise, draw a time diagram showing the payment dates,
and the beginning and the end of the term of any ordinary annuity involved.
1. A sequence of 16 quarterly payments of ₱5,000 each will start with a payment at the
1
end of 5 2 years. If money is worth 18% compounded quarterly, find the sum of
1
the values of these payments at the end of a) 4 years b) 8 2 years c) 10 years d)
3. If the fund accumulation at 15% compounded semi-annually, how much will be in the
fund at the end of 30 years if ₱5,000 is deposited in it at the end of each 6 months
for the first 25 years?
5. The first payment of 10,000 by Paul is due in 2 years. Payment continues every 3
months until the 15th year. If money is worth 20% compounded quarterly, find the
annuity value at the beginning of the first year.
CHAPTER 4
Amortization and Sinking Fund
Amortization
Amortization is a means of repaying a debt by a series of equal time interval.
the periodic payments form an annuity in which the present value is the principal
of an interest-bearing debt. Hence we use the following annuity formulas;
A=R [ 1−(1+i) ⁻ ⁿ
i ]
R= [ Ai
1−(1+i) ⁻ⁿ ]
Where: A = principal
R = periodic payment
i = interest per period
n = total number of payment periods
Examples:
1. An obligation of ₱21,000 with interest of 8% compounded semi-annually
must be paid at end of every 6 months for 4 years. a) Find the size of periodic
payment. b) Find the remaining liability just after making the 5th payment. c)
Prepare the amortization table.
Solution:
A = 21, 000 m=2 i = 0.04
j = 8% t = 4 years n=8
Ai
a) R = 1−(1+i )⁻ⁿ
21,000 (0.04)
= 1−(1.04)⁻⁸
= ₱3,119.08
b)
R R R R R R R R
yrs
0 1 2 3 4
The remaining liability after the 5th payments is the present value of the
3,119.09
remaining periodic payments.
3,119.09
3,119.09
−n
1−(1+i)
A=R ⌈ i
⌉
−3
1−(1.04)
A = 3,119.08 ⌈ 0.04
⌉
= ₱8,655.73
c) Amortization schedule
Payment for
Period Balance Payment Interest Paid Principal
1. ₱21, 000.00 ₱3,119.08 ₱840.00 ₱2,279.08
2. 18, 720.92 3,119.08 748.84 2,370.24
3. 16, 350.68 3,119.08 654.03 2, 465.05
4. 13, 885.63 3, 119.08 555.43 2,563.65
5. 11, 321.98 3, 119.08 452.88 2, 666.20
6. 8, 655.78 3, 119.08 346.23 2, 772.85
7. 5, 882.93 3, 119.08 235.31 2, 883.76
8. 2, 999.16 3, 119.08 119.97 2, 999.11
Total ₱24, 952.64 ₱3, 953.10 ₱21, 000˟
˟The actual value is ₱20, 999.94. This is due to rounding error.
Exercise 4.1
1. A debt of ₱70, 000.00 with 8% converted quarterly must be paid at the end
of every 3 months for 3 years.
a) Find the size of each payment.
b) Find the remaining liability after the 4th payment.
c) Construct the amortization schedule.
3. A loan of ₱100, 000 with interest at 10% converted quarterly will be settled
by paying ₱12,000 at the end of every 3 months.
a) How many payments amounting ₱12, 000 will be made?
b) What final smaller payment is needed?
c) Construct the amortization schedule.
5. An item worth ₱33, 700 is purchased for a down payment of ₱5, 000. Interest
is computed at 6% converted monthly and the balance to be settled with
payment of ₱1, 500 at the end of each month.
a) Determine the number of ₱1, 500-payment needed.
b) Find the size of final payment.
c) Construct the amortization schedule.
8. The cash value of motorcycle is ₱32, 500. Andro brought the item by
installment. He pays 7,000 down payment and promised to settle the balance
with equal payments due at the end of every month for 1 year. Interest is
converted monthly at 12%
a) How large should each installment payment be?
b) After 1 year, how much of his balanced is reduced?
c) Construct the amortization schedule.
9. Prepare the amortization table for the extinction of a debt of ₱12,000 with
interest at 8% compounded bi-monthly in equal bi-monthly installment for 1
year.
R [ ( 1+i ) ⁿ −1 ]
S= i
S (i)
R= 1−(1+i)⁻ⁿ
Illustrative Examples
1. A fund is created by making equal monthly deposits of ₱3,000 at 9%
converted monthly.
a. Determine the sum after half year.
b. What is the amount in the fund after the 4th deposit?
c. Construct the sinking fund schedule for a 6-month period.
Solution:
[ (1.0075 )6 −1 ]
a) S=3,000
.0075
S=3,000(6.113631847)
S=₱18, 340.89
b) n = 4
[ (1.0075) ⁴−1 ]
S=3,000 .0075
S=₱12,135.68
2. Three years from now, Mr. T needs ₱30,000 to liquidate a certain debt, at 6%
converted semi-annually.
a. How much must he deposit at the end of every 6 months to provide for the
payment of the debt?
b. Prepare a sinking fund table showing the growth of the fund for 3 years.
a Given:
S = ₱30,000
j = 6% = .06
m=4
30,000(.05)
R= [ (1.015)6−1 ]
450
R= .093443263
R = ₱4,815.75
2 Mr. Imperial will deposit ₱10,000 at the QUAK Bank at the end of each
quarter for 2 years. If the banks gives out 9% compounded quarterly, find
the amount to his credit just after the last deposit.
5 Prepare a sinking fund table that shows the growth of fund in 5 months if
₱1,300 is to be paid at the rate of 12% converted monthly.
Illustrative Example
1 The principal of a loan ₱40,000 will be at the end of 15 years by the
accumulation of a sinking fund by quarterly deposits, and interest will be
payable on the debt quarterly at the rate of 9%.
a Find the quarterly expense of the loan to debtor if his sinking fund is
invested at 8% compounded quarterly.
b Find the book value of the debt just after the 4th deposit.
c Find the quarterly expense of the debt if he discharge all the liability as to
principal and interest by paying his creditor equal sums at the end of each
3 months for 2 years.
Given:
S = ₱45,000
n=6
Solution:
S (i)
R= [ (1+i)n −1 ]
40,000(.02)
R= [ (1.02)n6 −1 ]
R = ₱6,341.03
Chapter 5
Depreciation, Capitalization and Perpetuities
Objectives: At the end of the lesson students should be able to:
a. calculate the rate of depreciation, periodic payments using different methods
b. participate actively in class discussion
5.1 Types of depreciation
1. Normal Depreciation
a. Physical Depreciation− is due to the lessening of the ability of a property to
produce a result
b. Functional Depreciation−is due to the lessening in the demand for the function,
which the property was designed to render.
2. Depreciation Due to Changes in Price Levels
3. Depletion−refers to the decrease in the value of a property due to the gradual
extraction of its contents.
Formula: W=P−L ; B = P− E where: W=wearing value B = book value
P=price E = accrued value
L=salvage value
5.2 Three Most Common Types of Depreciation
1. Straight-line Method
Denote:
P= original cost
L= scrap value/resale/salvage value
N= number of year/s (estimated life)
D= annual depreciation
d= 1/n
E=accrued value (tD)
B= book value
Formulas:
P–L
D= n or D= d (W)
Where dn is the rate depreciation to be applied to the wearing value, to obtain the
depreciation charge Dn=dnW since the sum of the numerations;
K
d1+d2+…dn ¿ K =1 , then, D1+D2+… Dn=(d1+d2+…dn)=W
Example: An equipment costing ₱4,500 will have a salvage value of ₱3,000 when retired at
the end of 5 years. Solve the total amount of depreciation every year.
Given: P=₱4,500; L=₱3,000; n=5
n ( n+ 1 )
Solution: Since n=5 then, k ¿ 2
n ( n+ 1 ) 5 (6 )
K ¿ 2 = 2
5 ( 5+1 ) 30
= 2 = 2
=15
Alternative solution: Since n=5 then, k=1+2+3+4+5=15
Solve for the wearing value W.
W =P-L
=₱4,500 - ₱3,000 5 =₱500
=₱1,500 = 15 n−1
n d2 = 15
d1 ¿ , where k=15 =.333333
k
D =d1 (W)
andn=5 s =.333333 (₱1,500
5−1 =.2 =₱200
= 15 D = d3 (W) n−4
=.2 (₱1,500) d5 = 15
=.2666666 =₱300
D=d2 (W) n−3 ( 5−4 )
d4 = 15 = 15
=.2666666(₱1,500)
5−3 =.0333333
=₱400 = 15 D = d5 (W)
n−2
d3= 15 =.1333333 =.0333333(₱1,500)
D = d4 (W) =₱100
5−2
= 15 =.1333333(₱1,500)
A. Depreciation schedule
B. Find the depreciation charge at the end of 4 years and five years without using the
schedule.
Solution:
E=Rsk¬ i
=₱3,257.55s4¬ .05
Using Geometric Progression
2
d= n (Federal revenue act of 1954)
Example:
1. By the method using a fixed percentage of declining book value with rate equal or
twice the natural rate, find the depreciation charges and book value during the life
of ₱4,000 with life of 3 years.
Solution:
1 1
The natural rate of depreciation is n which is = 3
2 2
Hence d = n , d = 3 = .66666 or 67 %
D = .67 (₱4, 000) – the depreciation charge at the end of first year.
D = ₱2,680
The leaving book value:
₱4,000 − ₱2,680 = ₱1,320
Exercise 5.1
2. A sewing machine that costs ₱20,000 has an estimated life of 10 years and has a
salvage value at ₱500. Compute for annual depreciation charge and rate of depreciation.
Given: solution:
₱ 19,500
P=₱20,000 W=P−L d= 10
L=₱500 =₱20,000−₱500 =₱1,950
n=10 years =₱19,500
3. A bicycle depreciates from original costs of ₱8,000 to a salvage value of ₱400 in 6
years. What is the annual depreciation? Construct a schedule of depreciation.
Given: solution:
₱ 7,600
P=₱8,000 a. W=P−L b. d ¿ 6
L=₱400 =₱8,000−₱400 =₱1,266.67
n=6 years =₱7,600
Depreciation schedule
Yearly depreciation (d Accrued depreciation (D n) Book Value (B)
Year(n) n)
0 ₱0 ₱0 ₱8,000
1 1,266.67 1,266.67 6,733.33
2 1,266.67 2,533.34 5,466.66
3 1,266.67 3,800.01 4,199.99
4 1,266.67 5,066.68 2,933.32
5 1,266.67 6,333.35 1,666.65
6 1,266.67 7,600.02 399.98
Exercise 5.1
1. An article costing ₱10,000 depreciates to ₱600 in 12 years. What is the annual
depreciation?
2. A sewing machine that costs ₱20,000 has an estimated life of 10 years and has a
salvage value of ₱500. Compute for the annual depreciation charge and the rate of
depreciation.
3. A bicycle depreciates from the original cost of ₱8,000 to a salvage value of ₱400 in 6
years. What is the annual depreciation? Construct a schedule of depreciation.
5.4 Perpetuity
− considered a form of annuity, considering a Perpetuity of R payable at the end of
each period and let I be the first interest rate per conversation period. Let A be the present
value of the perpetuity. Apparently if A is invested at the rate I, the interest alone on this
fund must provide R at the end of each interest period, so that the capital of the fund may
remain and permit the payments to continue unchanged forever.
Formula No 1:
R j
A= i , 1= m (Present Value of Simple Ordinary Perpetuity)
Example 1
If money is worth 6% compounded quarterly, find the present value
a. Of a perpetuity of ₱100 payable quarterly
b. Of an annuity of ₱100 payable quarterly for 20 years
Solution:
Given:
J = 6% m=4 R = ₱100 t = 20 years
R ₱ 100
a. A = i = .015
=₱6,666.67
b. To solve the annuity use Ran¬ i
Where an¬ i = 1−(1+ I )-n
i
K = ₱100 a80¬ .015
1−(1+ I )
K = ₱100 1−
.015
s k¬ I
A = __W__ (present value of general perpetuity)
s k¬ I
Example:
1. Find the present value of perpetuity of ₱200 payable semi-annually if money is worth
4% compounded quarterly.
Solution:
A = _R_. __1__
i s k¬i
A = _₱200_. __1__
1.1 s 2¬.01
A = (₱20,000) (.49751437)
A = ₱9,950.58
2. To maintain a car in good condition, ₱10,000 will be needed at the end of each year and
annually thereafter. What is the present value of all future maintenance at a rate of 10%
compounded semi -annually?
Solution:
A = _R_. __1__
i s k¬i
12months
K= 6 moonths =2
.10
i= 2 = .05
A = _₱10,000_. __1__
.05 s 2¬.05
A = ₱200,000 (.487804878)
A = ₱97,560.98 (round to the nearest centavo)
Exercise 5.2
1. If money is worth 6 % compounded quarterly, find the present value
a. of a perpetuity of ₱50 payable quarterly
b. of an annuity of ₱50 payable semi –annually for 30 years.
2. How much payment at the end of each year forever can be provided by endowment of
₱1,000 invested at 7%?
3. If money is worth 6%, obtain the present value of a perpetuity of ₱10,000 payable
annually, with the first payment due at the end of 5 years.
5.6 Capitalization
− It is the process of finding the present value of all future earnings.
Capitalized value of an enterprise − is the present value of all earnings similarly to speak
of capitalizing any sequence of equal payments due in the future means to find present
value.
Capitalized cost – It is the sum of the original cost and the present value of all
replacement costs which are assumed to continue forever.
Formulas:
2. Replacement cost R per period that could equitably replace the amount W at
the end of K periods:
R = _1__
s k¬ i
3. Capitalized value (assumed to continue forever) = pr. Value of the perpetual
income.
A = _R_, when future earnings from ordinary perpetuity
i
A = _R_. __1__, when future earnings from general perpetuity
i s k¬i
4. Capitalized cost = Original cost + present value of infinitely many future
replacements
R
C= P+ , when replacement cost per period from an ordinary perpetuity
i
5. Capital (C) as present value of the perpetuity of W payable at the end of K
periods
C= P+¿ _R_. __1__, when replacement costs per period from general
perpetuity
i s k¬ i
Example:
1. It is estimated that the maintenance of a certain section of a LRT railroad will
required ₱2,000 per kilometer at the end of every 4 years. If money is worth
10%, find the capitalized cost of the maintenance per kilometer.
Solution:
Replacement cost per period:
R =__W_ =_₱2,000_
s k¬ I s 4¬ 10%
₱ 2,000
= 4.641
= ₱430.9416074
Capitalized cost per kilometer:
R ₱ 430.9416074
C= I = 0.1
C = ₱ 4,309.42
C = __W__ =__₱2,000__
is k¬ i 0.1(4.641)
C = ₱4,309.42
Formula:
Annual investment cost:
M = Pi +__W_ , W = P − L
S n¬ i
P = Original cost
W = Wearing value
L = Scrap value
Capitalized costs Based Annual Investment Cost of the Asset
Example:
2. Find annual investment cost (M) and also the capitalized cost (K) of the machine
by first finding the value of K and then using M = Ki. Money worth 4% machine
will cost ₱15,000 = life 8 years
Final salvage value ₱3,000
Solution:
Method 1
W=P–L P = ₱15,000 L =₱3,000
W = ₱15,000 − ₱3,000
W = ₱12,000
K = P + =__W__
is k¬ i
K = ₱15,000 + ___₱12,000_____
(0.04)(s8¬ 0.04)
K = ₱15,000 + ₱32,558.34961
K = ₱47,558.35
M = Ki, Annual investment cost
M = (₱47,558.35) (0.04)
M = ₱1,902.33
Method 2
M = Pi + __W , W = P − L
sn¬i
M = (₱15,000)(0.04) + _₱12,000_
s8¬ 0.04
M = ₱6,000 + ₱1,302.33
M = ₱1,902.33 Annual investment cost
At M = Ki
M ₱ 1,902.33
K= I = 0.04
K = ₱47,558
Exercise 5.3
1. It is estimated that a certain mine will yield a net profit of ₱200,000 at the end of each
year practically forever. Approximately, what is the capitalized value of this mine if money
is worth 7%?
2. How much is the capitalized value of an asset of ₱2,500 payable at the end of each
month, indefinitely, if money is worth 6% (m = 12)
3. At 5 % find the capitalized cost of an asset whose cost is ₱50,000, life is 10 years, and
final salvage value is ₱5,000.
The following symbols will be used in dealing with ordinary annuity formulas:
R - periodic payment of the annuity
n - total number of payments
i - interest per conversion period
S - amount of an annuity
A - present value of an annuity
0 1 2 3
₱10,000 =P10,000
₱10,000(1 +.10)1 = 11,000
₱10,000(1 +.10)2 = 12,100
S = P33,000
To find the present value of an annuity, add theTerm
discounted payments.
0 1 2 3
₱10,000(1 + .10)-1 = ₱9,090.91
₱10,000(1 +.10)-2 = 8,264.46
₱10,000(1 +.10)-3 = 7,513.15
A =
Hence, from the relation ₱24,868.52
A = S(1 + i)-n and S = A(1 + i)n
A = 33,100(1 + .10)-3 and S = 24,868.52(1 + .10)3
A = ₱24,868.52 S = ₱33,100
R R R R R
0 1 2 3 n-1 n (periods)
R
R(1 + i)1
.
.
.
R(1 + I)n-3
R(1 + I)n-2
R(1 + I)n-1
S = sum of the accumulated values of R at the end of the term
(1) S = R + R(1 + i) + . . . R(1 + i) n-3 + R(1 + i)n-2 + R(1 + i)n-1
multiplying (1) by (1+i)
(2) (1 + i)S = R(1 + i) + R(1 + i)2 . . . R(1 + i)n-3 + R(1 + i)n-1 + R(1 + i)n
N = mt = number of payments
[ ]
n
( 1+i ) −1
A= R (1 +
i
[ ]
−n
1−(1+i)
if we let n¬ i = i
−n
1−(1+i )
then the formula i of A can be written as
A=
Ran¬ i
where A is the present value of the annuity an¬ I, is read as ”a angle n at i”
an¬ i =
1−(1+i)−n
i
Illustrative Example 1
Find the amount and present value of an annuity of ₱1,500 payable for
2 years if money is worth 105 compounded semi-annually.
Given R = P1,500 j .10
i = m = 2 =.05
n = mt = 2(2) = 4
periods
j = 10% - .10
i = 2 years
m=2
To find S
S = Rsn¬ i
S = 1,500svn¬ .05
S = 1,500(4.310125)
S = P6,465.19
or S=R [ ( 1+.05)4 −1
i ]
[ ]
4
( 1+.05) −1
s = 1,500 .05
s = 1,500(4.310125)
s = ₱6,465.45
to find A
A = Ran¬ i
or by using the relation
A = S(1 + i)-n
A = 1,5004¬ .05
A = 6,465.19(1 + .05)-4
A = 1,500(3.54595)
A = P5,318.93
A = P5,318.93
Illustrative Example 2
Mrs. Cruz purchased a house and lot. If she paid ₱400,000 as down
payment and promised to pay P5,000 every 3 months for the next 10 years
at 15% compounded quarterly, find the cash value of the house.
Solution:
The down payment is not part of the annuity. The cash value of the
house is ₱400,000 plus the present value of the ordinary annuity of 40
monthly payments of P5,000 each. The cash value is
Solution:
Given: S = ₱300,000
t = 15 years
n = 60 periods
j 8
i = m = 4 = .02
Find R:
S
R= s n¬ i
300,000
R= s 50 ¬.02
300,000
R= 114,05154
R = ₱2,630.38
Illustrative Example 2
Mr. Tuy bought a refrigerator that cost ₱19,500. He paid ₱6,000 as down
payment and the balance will ne paid in 36 equal monthly payments. Find the
monthly payment if money is 15% compounded monthly.
Solution:
A = Cash Value – Down
A = ₱19,500 – ₱6,000 = ₱13,500
n = 36
.15
i= 12 = .0125
Find R:
A 13,500
R= an ¬i = a.36¬.0125
13,500
R= 1−( 1+.0125 )−36
.0125
13,500
R= 28.84726737
Exercise 3.1
1. Analyn bought a dining set. She paid ₱2,500 as down payment and promised
to pay ₱800 at the end of each month for one year. What is the cash
equivalent of the set if the interest rate is 10% compounded monthly?
2. Find the amount present value of an annuity of ₱5,240 payable at the end of
1
every 6 months for 5 years and m6 months at 5 2 converted semi-
annually.
3. What is the present value of ₱1,230 due at the end of every three months for
3
3 years and 6 months, if money is worth 9 4 % compounded quarterly?
5. If Cherry will deposit P1,500 every end of each quarterly in 9 years at the
1
rate of 6 2 compounded quarterly, how much will be the amount of her
8. A man deposits ₱12,500 at the end of every six months in a bank which
credits interest at 9% compounded semi-annually. How much will he have
after 19 years?
11. A man purchased a DVD player worth ₱8,500. He paid P2,000 cash and
agreed to make 15 monthly payments. If interest is 18% (m=12), find the
monthly payment.
12. Marga borrowed P250,00 and agreed to repay her obligation by making
equal quarterly at the end of every months for 4 years. What is the periodic
payment, if money is worth 12.5% compounded quarterly?
1
13. How much should be invested each year in a fund paying 10 2
Example 1
A refrigerator can be purchased for ₱9,000 cash down and P630 a month for
18 months. Find the interest rate charged if it is compounded monthly.
Solution:
A = 9,00 – 900 – 8,100
R = 630
n = 18
m = 12
Formula:
Since A is known
(1− nRA ) = 0
(n2 - 1)i2 + 6(n + 1)i + 12
i = -114 ±
√
(114 )2−4 (323)(−4.8)
2( 323)
114 ±138.555404
i= 646
i = -.3909526 or .03801146
Here we get two values of i, we disregard the negative value of i. Therefore,
the periodic interest rate is
i = 3.80% and the nominal interest rate charged is
i = ixm
i = (.03801146)12%
i = 45.61%
Example 2:
Payments of ₱750 each are made every 6 months in 3 years. At what rate
compounded semi-annually will these payments amount to ₱5,100?
Solution:
S = ₱5,100
R = ₱750
n = mt = 2(3) = 6
Formula: Since S is known
Exercise 3.2
1. in purchasing a washing machine worth P21,000, a buyer pays ₱3,500 cash
and agrees to make 12 monthly payments. If the monthly payment is ₱1,580,
find the interest rate m charge.
2. Deposit of ₱8,380 are made every end of six months. At what rate
compounded semi-annually will these deposits amount to P112,400 in 14
years?
3. A television set costs P10,440. It is purchased by a down payment of ₱4,440
and monthly payments of P 382.84 for one year and 6 months. Find the rate
of interest converted monthly. Use either interpolation or the given
derivation of i.
4. A man invests ₱9,375 every six months. If he has ₱200,000 in five years, at
what nominal rate compounded semi-annually did hi investment earn?
[
Example 1log 1−
Victoria borrows
Ac
]
R ₱15,000 with interest at 15% compounded quarterly. She
will discharge (1+1)
−log the debt by paying P950 quarterly.
a) How many payments of ₱950 are required?
b) How much would the final payment be if it is made the day after the last
P950 payment?
c) How much would the final payment be if it is made 3 months after the last P950
payment?
Solution:
Given: A = ₱15,000 R = 950 i = .0375
a) Substitute the given values into the formula
n=
[
log 1− c
A
R ]
log ( 1+ i)
n=
[
log 1−
15,000(.0375)
950 ]
log ( 1+.0375)
log.407894737
n= log ( 1.0375)
.389451898
n= .015988105
n = 24.35885291
This means that there will be 25 quarterly payments of ₱950 and a
final irregular payment.
0950950950 950950
1 2 3 23 24
. . .
Focal date
x(1.0375)-24 + 950 a24 ¬ .0375 = 15,000
x(.413319095) + 14,862.58 = 15,000
x = P332.47
Therefore the final 24th payment is
= 24th payment + final irregular payment
= 950 + 332.47
= P1,282.47
Exercise 3.3
1. Lulu borrows ₱240.000 with interest at 12% compounded monthly. She will
pay the debt by paying ₱5,300 monthly
a. Find the number of regular payments to be made.
b. How much will be the final payment if it is made one month after the last
₱5,300 payment?
0950950950 950950
1 2 3 23 24
. . .
b. How much would the final payment he if it is made on the day of the last
₱18,500 payment?
4. A car is worth ₱730,000. The buyer paid ₱350,000 down payment and
3
promised to pay ₱7,660 every month, if he is charged 7 4 compounded
monthly,
a. How many full payments must be made?
b. How much will be the final payment?
0 1 2 3 ... n-1 n
R R R R ... R
The formula in finding the present value of an annuity due is:
A = (1st payment) + (present value of remaining payments)
A =A +
Ran−1¬ i
Example 1
Marlon purchased a car. He paid ₱150,000 down payment and ₱10,000
payable at the beginning of each month for 5 years. If money is worth 12%
compounded monthly, what is the equivalent cash price of the car?
Given:
R = ₱10,000 j = 12%
12
Down payment = ₱150,000 i = 12 – 1%
m = 12 A = ? Cash Equivalent - ?
t = 5 years
n = 60
A =R+ Ran−1¬ i
A = ₱10,500 + ₱ 10,50060−1 ¬1
[ ]
−59
1−(1.01)
A = ₱10,500 + ₱10,500 .01
A = ₱10,500 + ₱466,248.18
A = ₱476,748,18 (Total Installment Payment)
Cash Equivalent = Down Payment + Total Installment Payments
Cash Equivalent = ₱150,000 + ₱476,748.18
Cash Equivalent = ₱626,748.18
0 1 2 3 ... n-1 n
R R R R ... R
The formula in finding the amount of an annuity due is:
S = {Value of an annuity of (n +1) payments on the last payment date} – R
S = Rsn+1 ¬ i−R
Example 1
If ₱500 is deposited in a bank at the beginning of each 3 months for 10
years and money is worth 8% compounded, how much is in the fund at the end
of 10 years?
Given:
R = P500 j = 8%
8
M=4 i= 4 = 2%
t = 10 years S=?
n = 40
S= Rsn+1 ¬ i−R
S= ₱ 500 s40+ 1¬ 2 −₱ 500
S = ₱500 [ (1.02)41−1
.02 ] -₱500
S = ₱31,305.01 – ₱500
S = 30,805.01
Example 2
Emil invests ₱5,000 at the beginning of each six months. He makes his
first deposit on January 19, 2003. How much will be in his account on January
19, 2015, if money is worth 9% compounded semi-annually?
R = ₱5,000
m=2
t = January 19, 2003 – January 19, 2015 = 12 yrs.
n = 24
j = 9%
9 1
i= 2 =4 2 %
S=?
S= Rsn+1 ¬ i−R
₱ 5,000 s 1 −₱ 5,000
S= 24+1 ¬ 4
2
₱ 5,000 s 1 −₱ 5,000
S= 25 ¬4
2
[ ]
13
(1.045) −1
S= ₱ 5,000 -₱5,000
.045
S = ₱22,826.05 – ₱5,000
S = ₱217,826.05
Where:
A = present value of an annuity due
S = amount of an annuity due
R = periodic payment of the annuity due
i = interest per conversion period
n = total number of payments
Example 1
What equal deposits should be placed in a fund at the beginning of each
year for 15 years in order to have ₱1,500,000 in the fund at the end of 15
years, if money accumulates 12%?
Given:
S = ₱1,500,000 j = 12%
m=1 i = 12%
t = 15 R=?
n = 15
A
R= S n+1 ¬i−1
P 1,500,000
R= S 15+1 ¬12 −1
P 1,500,000
R= S16 ¬12 −1
P 1,500,000
R= 41.75328042
R = ₱35,925.32
Exercise 3.4
For each problem draw a diagram illustrating the data and the solution
1. If money is worth 16% compounded quarterly, find the present value and the
amount of annuity due of ₱1,500 payable quarterly for 10 years.
4. Mr. Jason Raymundo agrees to pay ₱2,000 at the beginning of each 6 months
1
for 5 years. If money is worth 12 2 % compounded semi-annually, find a)
6. On January 19, 2003, Mr. Berlindo Santos opened a saving account for his
wife with an initial deposit of ₱10,000 in a bank paying 6% compounded
quarterly. If Mr. Santos continues to make quarterly deposits of the same
amount until July 19, 2008, how much will be in the account at the end of the
term? (Hint: Count the time by months.)
quarterly?
8. What equal deposits should be placed in a fund at the beginning of each year
for 20 years in order to have ₱2,500,000 in the fund at the end of 20 years, if
1
the money accumulates 10 2 %?
quarterly payment.
Example 1
A sequence of quarterly payments of ₱3,500 each, with the first one due
at the end of 3 years and the last of the end of 10 years. Find the present
value of the deferred annuity, if money is worth 16% compounded quarterly.
Given:
R= ₱3,500 j = 16% m=4
Solution:
16
i= 4 = 4%
d = 3 x 4 = 12; 12 – 1 = 11
n = 7 x 4 = 28; 26 + 1 = 29
40
11 29
RRRRRRRRRRRR RRRRR
‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘
0 1 2 3 4 5 6 7 8 9 10
2 yrs 7
10
Ad = Ran+d ¬i−Rad ¬ i
Ad = ₱Ad3,500 a29 +11¬4 −₱ 3,500 a11¬ 4 Sd
A d 0= ₱ 3,500
1 ...
2 a 40¬34 −₱ 3,500 a11 ¬4d d1 d2 ... d+n-2 d+n-1 d+n
deferment −40 period n periods
[ ] [ ]
−11
Ad 1−(1.04)
(d) 1−(1.04)
Payments are made
= ₱3,500 No payments
.04 - ₱3,500 .04
Ad = ₱69,274.71-₱30,661.67
Ad = ₱38,613.04
Example 2
Find the cash equivalent of an item that sells for ₱20,000 down payment
and 20 semi-annual payments of ₱5,500 each, the first is due at the end of 3
years, if money is worth 10% compounded semi-annually.
Given:
R = ₱5,500 Down payment = ₱20,000 j = 10% m=2
Solution:
10
i= 2 = 5%
d = (3 x 2) – 1 = 6 – 1 = 5
n = 20
Ad
25
5 20
R R R R R R R R R R R R R R R R R R R R R R
‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘ ‘
0 1 2 3 4 5 6 7 8 9 10 11 12 13
2 yrs
Ad = Ran+d ¬i−Rad ¬ i
Ad = ₱ 5,500 a20 +5¬ 5 −₱ 5,500 a5 ¬5
Ad = ₱ 5,500 a25 ¬5 −₱ 5,500 a 5¬ 5
Ad = ₱5,500 [ 1−(1.05)−25
.05 ] - ₱5,500 [ 1−(1.05)−5
.05 ]
Ad = ₱77,516.70 – ₱23,812.12
Ad = ₱73,704.58
Given:
Ad = ₱100,00
J = 12%
m =2
n = 10
d = 10
12
i = 2 = 6%
Solution:
Ad
R= ad +n ¬i −ad ¬ i
P 100,000
R= a20+6 ¬ i−a 10¬ 6
P100,00
[ ][ ]
−20
R= 1−(1.06) 1−10
−
.06 .06
P 100,000
R= 11,46992122−7.360087051
P 100,000
R= 4.109834169
R = ₱24,331.88
Exercise 3.5
For each problem, draw a line diagram illustrating the data and the solution
Find the present values of the deferred annuity which is described, at the
specific interest rate.
1. A farm costs ₱750,000 cash. Paolo will pay ₱150,000 cash and a sequence of
15 annual payments, the first due at the end of 5 years. If money is worth
12%, find the annual payment.
2. A house and lot is offered for sale ₱1,500,000 cash, or ₱500,000 cash and a
sequence of 20 semi-annual payments of 120,000 each, the first is due at
the end of 3 years, if money is worth 14% compounded semi-annually, find
the cash value of the house and lot.
3. Find the present value of a series of quantity payments of ₱2,500 each, the
first payment is due at the end of 4 years and 6 months, and the last at the
end of 10 years and 3 months, if money is worth 16% compounded quarterly.
4. Find the cash equivalent of a computer set that sells for ₱10,000 cash and a
sequence of 10 semi-annual payments of ₱3,500 each, the first is due at the
1
end of 3 years and 6 months, if money is worth 10 2 % compounded semi-
annually.
5. Jason borrows ₱50,000 and agrees to pay his obligation by making 15 equal
annual payments, the first is due at the end of 3 years . Find the annual
payment, if money is worth 12%.
7. Find the present value of the pension of a man, now 60 years old, and who
will receive a pension of ₱10,000 per month for 15 years, with the first
payment to occur one month after his 65th birthday, if money is worth 18%
compounded monthly.
8. An investment in a stock market will yield no operating profit until the end of
5 years, when investor will receive P100,000 after that, he will receive
₱100,000 at the end of each quarter for 10 more years. Find the present
1
value of this income if money is worth12 2 % compounded quarterly,
9. On November 20, 2002 a house and lot was bought for ₱500,000 down and
12 quarterly payments of ₱50,000 each, the first is due on November 20,
2003. Find the cash value of the house and lot, if money is worth 15%
compounded quarterly.
10. Aling Nelie won ₱25,000,000 in Mega Lotto. She invest winnings at
16% compounded semi-annually with the conditions that she receive 25
semi-annual withdrawals starting at the end of 2 years. Find the size of the
withdrawals.
3.8.2 Value of an Annuity in an Arbitrary Date
Example 1
A sequence of 5 semi-annual payments of ₱2,500 each will start with a
1
payment at the end of 2 2 years. If money is worth 12% compounded semi-
annually, find the sum of the values of these payment at the end of a)2 years,
1
b)4 2 years, c) 6 years, d)the actual present value of the payments.
Given;
R = ₱2, 500
j = 12%
m=2
12
i= 2 = 6%
n=5
Time Diagram:
R R R R R
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 yr. 2 yrs. 3 yrs. 4 yrs. 5 yrs. 6 yrs. 7 yrs.
Solution;
1
a) At the end of 2 years b) At the end of 4 years
2
A₂ = ₱ 2, 500 a5 ¬ 6 S = ₱2,500 a5 ¬ 6
c) To find the value of the annuity at the end of 6 yrs, accumulate a) for 4 years, or
1
the result of b) for 1 2 years.
Exercise
For each problem in the exercise, draw a time diagram showing the payment dates,
and the beginning and the end of the term of any ordinary annuity involved.
1. A sequence of 16 quarterly payments of ₱5,000 each will start with a payment at the
1
end of 5 2 years. If money is worth 18% compounded quarterly, find the sum of
1
the values of these payments at the end of a) 4 years b) 8 2 years c) 10 years d)
3. If the fund accumulation at 15% compounded semi-annually, how much will be in the
fund at the end of 30 years if ₱5,000 is deposited in it at the end of each 6 months
for the first 25 years?
5. The first payment of 10,000 by Paul is due in 2 years. Payment continues every 3
months until the 15th year. If money is worth 20% compounded quarterly, find the
annuity value at the beginning of the first year.
CHAPTER 4
Amortization and Sinking Fund
Amortization
Amortization is a means of repaying a debt by a series of equal time interval.
the periodic payments form an annuity in which the present value is the principal
of an interest-bearing debt. Hence we use the following annuity formulas;
A=R [ 1−(1+i) ⁻ ⁿ
i ]
R= [ Ai
1−(1+i) ⁻ ⁿ ]
Where: A = principal
R = periodic payment
i = interest per period
n = total number of payment periods
Examples:
1. An obligation of ₱21,000 with interest of 8% compounded semi-annually
must be paid at end of every 6 months for 4 years. a) Find the size of periodic
payment. b) Find the remaining liability just after making the 5th payment. c)
Prepare the amortization table.
Solution:
A = 21, 000 m=2 i = 0.04
j = 8% t = 4 years n=8
Ai
a) R = 1−(1+i )⁻ ⁿ
21,000 (0.04)
= 1−(1.04)⁻ ⁸
= ₱3,119.08
b)
R R R R R R R R
yrs
0 1 2 3 4
The remaining liability after the 5th payments is the present value of the
3,119.09
remaining periodic payments.
3,119.09
3,119.09
1−(1+i)−n
A=R ⁻ i
⁻
−3
1−(1.04)
A = 3,119.08 ⁻ 0.04
⁻
= ₱8,655.73
c) Amortization schedule
Payment for
Period Balance Payment Interest Paid Principal
1. ₱21, 000.00 ₱3,119.08 ₱840.00 ₱2,279.08
2. 18, 720.92 3,119.08 748.84 2,370.24
3. 16, 350.68 3,119.08 654.03 2, 465.05
4. 13, 885.63 3, 119.08 555.43 2,563.65
5. 11, 321.98 3, 119.08 452.88 2, 666.20
6. 8, 655.78 3, 119.08 346.23 2, 772.85
7. 5, 882.93 3, 119.08 235.31 2, 883.76
8. 2, 999.16 3, 119.08 119.97 2, 999.11
Total ₱24, 952.64 ₱3, 953.10 ₱21, 000˟
˟The actual value is ₱20, 999.94. This is due to rounding error.
Exercise 4.1
1. A debt of ₱70, 000.00 with 8% converted quarterly must be paid at the end
of every 3 months for 3 years.
a) Find the size of each payment.
b) Find the remaining liability after the 4th payment.
c) Construct the amortization schedule.
3. A loan of ₱100, 000 with interest at 10% converted quarterly will be settled
by paying ₱12,000 at the end of every 3 months.
a) How many payments amounting ₱12, 000 will be made?
b) What final smaller payment is needed?
c) Construct the amortization schedule.
7. Aster borrows ₱360, 000. She plans to amortize her debt with equal quarterly
payments for 2 years. Interest is allowed at 11% converted quarterly.
a) Find the quarterly cost of his debt.
b) By how much is the debt reduced by the 3rd payment?
c) Construct the amortization schedule.
8. The cash value of motorcycle is ₱32, 500. Andro brought the item by
installment. He pays 7,000 down payment and promised to settle the balance
with equal payments due at the end of every month for 1 year. Interest is
converted monthly at 12%
a) How large should each installment payment be?
b) After 1 year, how much of his balanced is reduced?
c) Construct the amortization schedule.
9. Prepare the amortization table for the extinction of a debt of ₱12,000 with
interest at 8% compounded bi-monthly in equal bi-monthly installment for 1
year.
R [ ( 1+i ) ⁿ −1 ]
S= i
S (i)
R= 1−(1+i)⁻ⁿ
Illustrative Examples
1. A fund is created by making equal monthly deposits of ₱3,000 at 9%
converted monthly.
a. Determine the sum after half year.
b. What is the amount in the fund after the 4th deposit?
c. Construct the sinking fund schedule for a 6-month period.
Solution:
[ (1.0075 )6 −1 ]
a) S=3,000
.0075
S=3,000(6.113631847)
S=₱18, 340.89
b) n = 4
[ (1.0075) ⁴−1 ]
S=3,000 .0075
S=₱12,135.68
2. Three years from now, Mr. T needs ₱30,000 to liquidate a certain debt, at 6%
converted semi-annually.
a. How much must he deposit at the end of every 6 months to provide for the
payment of the debt?
b. Prepare a sinking fund table showing the growth of the fund for 3 years.
a. Given:
S = ₱30,000
j = 6% = .06
m=4
30,000(.05)
R= [ (1.015)6−1 ]
450
R= .093443263
R = ₱4,815.75
2. Mr. Imperial will deposit ₱10,000 at the QUAK Bank at the end of each
quarter for 2 years. If the banks gives out 9% compounded quarterly, find
the amount to his credit just after the last deposit.
5. Prepare a sinking fund table that shows the growth of fund in 5 months if
₱1,300 is to be paid at the rate of 12% converted monthly.
Given:
S = ₱45,000
n=6
Solution:
S (i)
R= [ (1+i)n −1 ]
40,000(.02)
R= [ (1.02)n6 −1 ]
R = ₱6,341.03