The document discusses different types of annuities including ordinary annuities, annuity due, deferred annuities, and perpetuities. It provides formulas and examples for calculating present and future worth for these annuity types. It also covers capitalized cost, which represents the present worth of the total cost associated with an asset over an infinite time period.

1. CE 213 – ENGINEERING ECONOMICS

2. II. ANNUITIES
An annuity is a series of equal payments occurring at equal
periods of time.
TYPES OF ANNUITY:
1. Ordinary Annuity
2. Annuity Due
3. Deferred Annuity
4. Perpetuity

3. ORDINARY ANNUITY
is one where the payments are made at the end of each period.
Present Worth
• 𝑃 = 𝐴
[(1+𝑖)𝑛−1
(1+𝑖)𝑛𝑖
Future Worth
• 𝐹 = 𝐴
[(1+𝑖)𝑛−1
𝑖

4. SAMPLE PROBLEMS
1. A man paid a 10% down payment of P200,000 for a house and lot and
agreed to pay the 90% balance on monthly installments for 60 months at
an interest rate of 15 % compounded monthly. Compute the amount of
the monthly payment.
2. An instructor plans to retire in one year and wants an account that will
pay him P25,000.00 a year for the next 15 years. Assuming a 6% annual
effective interest rate, what is the amount he would need to deposit now?
(the fund will be depleted after 15 years).
3. How much money must you invest today in order to withdraw P2000
annually for 10 years if the interest rate is 9%?
4. What is the accumulated amount of the five-year annuity paying P6000 at
the end of each year, with interest at 15% compounded annually?
5. The maintenance cost of equipment is P20,000 for 2 years, P40,000 at the
end of 4 years, and P80,000 at the end of 8 years. Compute the semi-

5. ASSIGNMENT
1. A piece of machinery can be bought for P10,000.00 cash, or for P2,000.00 down and
payments of P750.00 per year for 15 years. What is the annual interest rate for the time
payments?
2. A manufacturing firm wishes to give each 80 employees a holiday bonus. How much is
needed to invest monthly for a year at a 12% nominal interest rate, compounded monthly,
so that each employee will receive a P2,000.00 bonus?
3. Money borrowed today is to be paid in 6 equal payments at the end of 6 quarters. If the
interest is 12% compounded quarterly, how much was initially borrowed if the quarterly
payment is P2,000.00?
4. A debt of P10,000 with 10% interest compounded semi-annually is to be amortized by semi-
annual payments over the next 5 years. The first due is 6 months. Determine the semi-
annual payments.
5. A man purchased on a monthly installment of P100,000 worth of land. The interest rate is
12% nominal and payable in twenty years. What is the monthly amortization?
6. The maintenance cost of equipment is P23,000 for 2 years, P40,000 at the end of 4 years, and

6. ASSIGNMENT
7. An employee is about to receive the sum of P300.00 at the end of each year for 5 years. One
year prior to the receipt of the first sum, he decides to discount all 5 sums. If the interest
rate is 6%, what proceeds will he obtain?
8. The president of a growing engineering firm wishes to give each of 50 employees a holiday
bonus. How much is needed to invest monthly for a year at a 12% nominal interest rate
compounded monthly, so that each employee will receive a P1,000 bonus?
9. Mr. Ramirez borrowed P15,000 two years ago. The terms of the loan are 10% interest for 10
years with uniform payments. He just made his second annual payment how much principal
does he still owe?
10. A machine is under construction for investment. The cost of the machine is P25,000.00. Each
year it operates, the machine will generate a savings of P15,000.00. Given an effective
annual interest rate of 18%, what is the discounted payback period, in years, on the
investment in the machine.

7. ANNUITY DUE
is one where the payments are made at the beginning of each period.
• Present Worth
• 𝑃 =
𝐴[(1+𝑖)𝑛−1−1]
(1+𝑖)𝑛−1𝑖
+ 𝐴
• Future Worth
• 𝐹 =
𝐴 [(1+𝑖)𝑛+1−1
𝑖
− 𝐴

8. SAMPLE PROBLEMS
1. Engr. Sison borrows P100,000.00 at 10% effective annual interest. He must pay back the
loan over 30 years with uniform monthly payments due on the first day of each month.
What does Engr. Sison pays each month?
2. A man wishes to have P35,000 when he retires 15 years from now. If he can expect to
receive 4% annual interest, how much did he set aside in each of the 15 equal annual
beginning-of-year deposits?
3. A man owes P12,000 today and agrees to discharge the debt by equal payments at the
beginning of each 3 months for 8 years, where these payments include all interest at 8%
payable quarterly. Find the quarterly payment.
4. At what interest rate is payable quarterly will payments of P500 at the beginning of each 3
months for 7 years discharge a debt of P12500 due immediately?

9. ASSIGNMENT
1. A man owes P10,000.00 with interest at 6% payable semi-annually. What equal payments
at the beginning of each 6 months for 8 years will discharge his debt?
2. On retirement, a workman finds that his company pension calls for payment of P300 to
him or to his estate if he dies at the beginning of each month for 20 years. Find the
present value of this pension at 5% compounded monthly.
3. Under a factory saving plan, a workman deposits P25 at the beginning of each month for
4 years, and the management guarantees accumulation at 6% compounded monthly. How
much stands to the work man’s credit at the end of 4 years.
4. A man will deposit P200 with a savings and loan association at the beginning of each 3
months for 9 years. If the association pays interest at the rate of 5.5% quarterly, find the
sum to his credit just after the last deposit
5. Engr. Aventajado loan an amount of P100,000 at a local commercial bank at 10%
compounded annually. How much is his monthly payment if he is required to pay at the
beginning of the first day of the month for a period of 30 years?

10. DEFERRED ANNUITY
is also an ordinary annuity but the payment of the first amount is deferred for a certain
number of periods after the first.
Present Worth
• 𝑃 =
𝐴 [(1+𝑖)6−1
(1+𝑖)6𝑖
• 𝑃 =
𝑃
(1+𝑖)4
Future Worth
• 𝐹 =
𝐴 [(1+𝑖)6−1
𝑖
• 𝑃1 =
𝐹
(1+𝑖)10

11. SAMPLE PROBLEMS
1. A man loans P187,400 from a bank with interest at 5% compounded annually. He agrees to
pay his obligations by paying 8 equal annual payments the first being due at the end of 10 yrs.
Find the annual payments.
2. A father wishes to provide P4,000 for his son on his deposit 21st birthday. How much should he
deposit every 6 months in a saving bank which pays 3% converted semi-annually, if the first
deposit is made when the son is 3 ½ year old?
3. XYZ Inc. plans to construct and additional building at the end of 10 years for an estimated cost
of P5,000,000.00. To accumulate this amount it will have equal year end deposits in a fund
earning 13%. However, at the end of the 5th year, it was decided to have a larger building that
originally intended to an estimated cost of P8,000,000.00 what should be the annual deposit
for the last 5 years?
4. Two years ago the rental for the use of equipment and facilities as paid 5 years in advance, with
option to renew the rent for another years by payment P15,000 annually at the start of each
year for the renewal period. Now, the lessor asks the lessee if it could be possible to prepay the
rental that will be paid annually in the renewed 5 years period. If the lessee will consider the
request, what would the fair prepayment to be made to the lessor if the interest is now figured
at 8%?

12. ASSIGNMENT
1. A house and lot can be acquired with a down payment of P500,000.00 and a yearly payment
of P100,000.00 at the end of each year for a period of 10 years, starting at the end of 5 years
from the date of purchase. If money is worth 14% compounded annually, what is the cash
price of the property?
2. A parent on the day the child is born wishes to determine what lump sum would have to be
paid into an account annually, in order to withdraw P20,000.00 each on the child’s 18th, 19th,
20th, and 21st birthdays
3. In five years, P18,000 will be needed to pay for a building renovation. In order to generate this
sum, a sinking fund consisting of three annual payments is established now. For tax purposes,
no further payments will be made after three years. What payments are necessary if money is
worth 15% per annum?
4. Annual maintenance costs for a rotary kiln are P2,000.00. The placement of a new refractory
would reduce the annual maintenance cost to P500 a year for the first 5 years and to P1000.00
a year for the next 5 years. The annual maintenance cost after 10 years would again be
P2,000.00 if the maintenance cost is the only savings, what maximum investment can be

13. ASSIGNMENT
5. The old boiler costs P2,400 a year to maintain. What expenditures for a new boiler are
justified if no maintenance will be required for the first 3 years, P600 per year for the next 7
years, and P2,400 a year thereafter? Assume money to cost 4% compounded annually and
no other cost to be considered.
6. A shirt factory has just installed a boiler. It is expected that there will be no maintenance
expenses until the end of the 11th year when P400 will be spent at the end of each
successive year until the boiler is scrapped at the age of 35 years. What sum of money was
set aside at his time at 6% interest will take care of all maintenance expenses for the boiler?
7. A new generator has just been installed. It is expected that there will be no maintenance
charges until the end of the 6th year when P300 will be spent at the end of each successive
year until the generator is scrapped at the end of the fourteenth year of service. What sum
of money set aside at the time of installation of the generator at 6% will take care of all
maintenance expenses for the generator?

14. ASSIGNMENT
8. A fund for the replacement of machinery in a plant must contain P30,000 at the end of 9
years. If the fund is invested at 3.5% compounded semi-annually, what equal deposits
should be placed in the fund at the end of each 6 months just for the first four years?
9. A house costs P400,000 cash. A purchaser will pay P90,000 cash, P60,000 at the end of 2
years, and a sequence of 6 equal annual payments starting with one at the end of 4 years, to
discharge all his liability as to the principal and interest at 7% compounded annually. Find
the annual payments which must be made for 6 years.
10. A businessman borrowed P300,000 and agrees to discharge his obligation by paying a series
of equal payments of P57,434.78 the first being due at the end of 5 ½ years. Find the rate of
interest he is paying if it is compounded semi-annually.

15. PERPETUITY
is an annuity whose payments continue forever
Formulas:
• 𝑃 =
𝐴
𝑖
Perpetuity payable annually beginning of each year
• 𝑃 = 𝐴 +
𝐴
𝑖
Perpetuity payable annually with the first payment due at the end of “nth” year
• 𝑃 =
𝐴
𝑖
−
𝐴 [(1+𝑖)𝑛−1]
(1+𝑖)𝑛𝑖
Where: P = Present value of perpetuity
A = annuity or amount of perpetuity
i = rate of interest

16. SAMPLE PROBLEM
1. What present sum would be needed for annual end of year payments of P15,000 each,
forever if money is worth 8%.
2. What amount of money deposited 20 years ago at 4% interest would provide a perpetual
payment of P2,000 per year?
3. Find the present value of a perpetuity of P100 payable semi-annually if money is worth
4% compounded quarterly.
4. If money is worth 8%, obtain the present value of a perpetuity of P1000 payable annually
when the first payment is due at the end of 5 years.

17. ASSIGNMENT
1. P45,000 is deposited in a savings account that pays 5% interest compounded semi-
annually. Equal annual withdrawals are to be made from the account, beginning one year
from now and continuing forever. Compute the maximum amount of the equal annual
withdrawal.
2. If money is worth 4% find the present value of a perpetuity of P100 payable at the
beginning of each year.
3. Find the present value, in pesos, of the perpetuity of P15000 payable semi-annually if
money is worth 8% compounded quarterly.
4. What is the present value in pesos of the perpetuity of P16000 payable every end of 3
months if money is worth 10%, compounded monthly?
5. Find the present value of a perpetuity of P18000 payable at the end if every 6 months if
money is worth 10% compounded quarterly.

18. CAPITALIZED COST
• One of the most important applications of perpetuity.
• is defined as the sum of money such that the annual interest on it at the rate
“i ” % is equal to the annual investment cost of the asset.
• the present worth of cost associated with an asset for an infinite period of
time
• From the annual cost equipment (amortization method)
• Annual Cost =
𝐹.𝐶−𝑆.𝑉 𝑖
(1+𝑖)𝑛−1
+ 𝐹. 𝐶 𝑖 + 𝑂. 𝐶
Dividing the equation by i:
• Capitalized cost =
(𝐹.𝐶−𝑆.𝑉)
(1+𝑖)𝑛−1
+ 𝐹. 𝐶 +
𝑂.𝐶
𝑖
From the annual cost equation (Capital recovery method)

19. CAPITALIZED COST
• Annual cost =
(𝐹.𝐶−𝑆.𝑉)(1+𝑖)𝑛𝑖
(1+𝑖)𝒏−1
+ 𝑆. 𝑉 +
𝑂.𝐶
𝑖
Dividing the equation by i:
• Capitalized cost =
(𝐹.𝐶−𝑆.𝑉)(1+𝑖)𝑛
1+𝑖 𝑛−1
+ 𝑆. 𝑉 +
𝑂.𝐶
𝑖
Capitalized cost for perpetual life:
• Capitalized cost = 𝐹. 𝐶 +
𝑂.𝐶
𝑖
Annual cost for perpetual life
• Annual cost = 𝐹. 𝐶 𝑖 + 𝑂. 𝐶
• Annual Investment Cost
• =
𝐹.𝐶−𝑆.𝑉 𝑖
(1+𝑖)𝑛−1
+ 𝐹. 𝐶 𝑖 + 𝑂. 𝐶
• Capitalized Cost
• =
𝐹.𝐶−𝑆.𝑉
(1+𝑖)𝑛−1
+ 𝐹. 𝐶 +
𝑂.𝐶
𝑖

20. SAMPLE PROBLEMS
1. At 6%, find the capitalized cost of a bridge whose cost is P250M and life is 20 years if the
bridge must be partially rebuilt at a cost of P100M at the end of each 20 years.
2. A new boiler was installed by a textile plant at a total cost of P300,000 and is projected
to have a useful life, it is estimated to have a salvage value of P30,000. Determine its
capitalized cost if interest is 18%compounded annually.
3. A machine costs P150,000 and will have a scrap value of P10,000 when retired at the
end of 15 years. If money is worth 4% find the annual investment and the capitalized
cost of the machine.
4. If a machine costs P3,000 with life of 15 years and a final salvage value of P500, compute
the capitalized cost if money is worth 5%

21. ASSIGNMENT
1. A suspension bridge was constructed for P24M. The annual maintenance cost is P500,000. if
the rate of interest is 6%. Compute the capitalized cost of the bridge including maintenance.
2. A concrete pavement on a street would cost P10000 and would last for 5 years with negligible
repairs. At the end of each 5 years. P1000 would be spent again to lay a new surface. Find the
capitalized cost of the pavement at 5%.
3. Find the capitalized cost of an asset whose cost is P100,000, salvage value is P10000, and life
is 15 years at 5%.
4. A bridge that was constructed at a cost of P75,000 is expected to last 30 years, at the end of
which time its renewal cost will be P40,000. Annual repairs and maintenance are P3000. What
is the capitalized cost of the bridge at an interest of 6%?
5. An item is purchased for P100,000. The annual cost is P18,000. using an interest rate of 8%,
what is the capitalized cost of perpetual service?