ENGINEERING ECONOMY
Engineering Economy – is the analysis and evaluation of the factors
that will affect the economic success of engineering projects to the
end that a recommendation can be made which will ensure the
best use of capital.
SET 1A: INTEREST AND MONEY-TIME RELATIONSHIPS
Interest – is the amount of money paid for the use of borrowed
capital (borrower’s viewpoint) or the income produced by money
which has been loaned (lender’s viewpoint).
F=P+I
Where:
I = interest
P = principal or present worth
F = accumulated amount or future worth
Cash-Flow Diagrams
Cash-Flow Diagram – is a graphical representation of cash flows
drawn on a time scale.
↑ - receipt (positive cash flow or cash inflow)
↓ - disbursement (negative cash flow or cash outflow)
Example: A loan of P100 at simple interest will becomeP150 after
5 years.
Cash flow diagram on the viewpoint of the lender
Example 2: Determine the (a) ordinary and (b) exact simple
interests on P 100,000 for the period January 15 to June 20 2012
if interest is 15%.
Ans. (a) P 6,541.67; (b) P 6434.43
Example 3: Calculated the exact interest on an investment of P
2,000.00 for a period from January 30 to September 15, 2001 if the
rate of interest is 10%.
Ans. P124.93
Example 4: If P 4000 is borrowed for 75 days at 16% per annum.
How much will be due at the end of 75 days?
Ans. P 4,133.33
Example 5: How long will it take for a deposit of P 1, 500.00 to earn
P 186 if invested at the simple interest rate of 7 1/3%?
Ans. 1.6909 years
Example 6: If you borrow money from your friend with simple
interest of 12%, find the present worth of P 20,000 at the end of 9
months.
Ans. P 18,348.60
Example 7: (CE Board) A deposit of P 110,000 was made for 31
days. The net interest after deducting 20% withholding tax is P
890.36. Find the rate of return annually.
Ans. 11.75%
Example 8: A man buys an electric fan from a merchant that
charges P1500.00 at the end of 90 days. The man wishes to pay
cash. What is the cash price if money is worth 10% simple
interest?
Ans. P 1,463.41
Example 9: What amount will be available in eight months if P
15,000.00 is invested now at 10% simple interest per year?
Ans. P 16,000.00
Example 10: P 1000.00 becomes P 1500.00 in three years. Find the
simple interest rate.
Ans. 16.67%
Example 11: An engineer borrowed a sum of money under the
following terms: P 650,000.00 if paid in 90 days, or P 600,000.00
if paid in 30 days. What is the equivalent annual rate of simple
interest?
Ans. 50%
Compound Interest
Compound Interest – the interest for an interest period is
calculated on the principal plus total amount of interest
accumulated in previous period.
Cash flow diagram on the viewpoint of the borrower
Simple Interest
Simple Interest – is calculated using the principal only, ignoring
any interest that has been accrued in preceding periods.
I = Pni
F = P(1 + in)
Where:
I = interest
P = principal or present worth
n = number of interest periods
i = rate of interest per period
F = accumulated amount or future worth
For Ordinary Simple Interest:
Interest period = 1 year = 360 days
For Exact Simple Interest:
Interest period = 1 year = 365 days (ordinary year)
= 366 days (leap year)
SAMPLE PROBLEMS
Example 1: Determine the ordinary simple interest on P 20,000
for 9 months and 10 days if the rate of interest is 12%.
Ans. P 1,866.67
1
Principal at
Beginning
of Period
P
2
P(1+i)
P(1+i)i
P(1+i)2
3
P(1+i)2
P(1+i)2i
P(1+i)3
…
…
…
…
n
P(1+i)n-1
P(1+i)n-1i
P(1+i)n
Interest
Period
Interest Earned
During Period
Amount at End
of Period
Pi
P(1+i)
F = P(1 + i)n
F⁄ = (1 + i)n = (F⁄ , i%, n)
P
P
P = F(1 + i)−n
P⁄ = (1 + i)−n = (P⁄ , i%, n)
F
F
Where:
F = accumulated amount or future worth
P = principal or present worth
i = rate of interest per interest period
n = number of compounding periods
F/P = single payment compound amount factor
P/F = single payment present worth factor
Nominal Rate of Interest – specifies the rate of interest and the
number of interest periods in one year.
r
i=
m
n = my
r my
F = P (1 + )
m
Where:
i = rate of interest per interest period
n = number of compounding periods
r = nominal rate of interest
m = number of compounding periods per year
y = number of years
Compounding Period
Compounded Quarterly
Compounded Semi-annually
Compounded Monthly
Compounded Bi-monthly
m
4
2
12
6
Effective Rate of Interest – is the actual or exact rate of interest on
the principal during 1 year, or simply the ratio of accumulated
interest in one year to the principal amount.
F−P
ER =
P
r m
ER = (1 + i)m − 1 = (1 + ) − 1
m
SAMPLE PROBLEMS
Example 1: The amount of P 20,000 was deposited in a bank
earning an interest rate of 6.5% per annum. Determine the total
amount at the end of 7 years if the principal and interest were not
withdrawn during this period.
Ans. P 31,079.73
Example 2: A man expects to receive P 25,000 in 8 years. How
much is that money worth now considering interest at 8%
compounded quarterly?
Ans. P 13,265.83
Example 3: How many years will P 100,000 earn a compounded
interest of P 50,000 if interest is 9% compounded quarterly?
Ans. 4.56 years
Example 4: Find the effective rate of interest corresponding to 8%
compounded quarterly.
Ans. 8.24%
Example 5: Find the nominal rate, which if converted quarterly
could be used instead of 12% compounded semiannually?
Ans. 11.825%
Example 6: If money is worth 5% compounded quarterly, find the
equated time for paying a loan of P 150,000 due in one year and P
280,000 in 2 years.
Ans. 1.6455 years
Example 7: Five years ago, you paid P 340,000 for a lot. Today you
sold it at P 500,000. What is the annual rate of appreciation?
Ans. 8%
Example 8: John borrowed P50, 000.00 from the bank at 25%
compounded semi-annually. What is the equivalent effective rate
of interest?
Ans. 26.56%
Example 9: Find the present worth of a future payment of P
300,000 to be made 5 years with an interest rate of 8% per annum.
Ans. P 204,174.96
Example 10: How long will it take money to double itself if
invested at 5% compounded annually?
Ans. 14.2 years
Example 11: The amount of P 50,000 was deposit in the bank
earning an interest of 7.5% per annum. Determine the total
amount at the end of 5 years, if the principal and interest were not
withdrawn during the period?
Ans. P 71,781.47
Example 12: Compute the equivalent rate of 6% compounded
semi-annually to a rate compounded quarterly.
Ans. 5.96% compounded quarterly
Example 13: If P5, 000.00 shall accumulate for 10 years at 8%
compounded quarterly. Find the compounded interest at the end
of 10 years.
Ans. P 6,040.20
Example 14: A sum of P 1,000.00 is invested now and left for eight
years, at which time the principal is withdrawn. The interest has
accrued is left for another eight years. If the effective annual
interest rate is 5%, what will be the withdrawal amount at the end
of the 16th year?
Ans. P 705.42
Example 15: By the condition of a will, the sum of P 2,000 is left to
a girl to be held in trust fund by her guardian until it amounts to P
5,000, when will the girl received the money if the fund is invested
at 8% compounded quarterly?
Ans. 11.57 years
Example 16: A student plan to deposit P1, 500 in the bank now
and another P3, 000 for the next 2 years. If he plans to withdraw
P5, 000 3 years after his last deposit for the purpose of buying
shoes, what will be the amount of money left in the bank after one
year of his withdrawal? Effective annual interest rate is 10%.
Ans. P 1,549.64
Example 17: If the interest rate of a certain account is 6.5%,
compute the (a) single payment present worth factor; and (b)
single payment compound amount factor at the end of 18 years.
Ans. (a) 0.322; (b) 3.107
Continuous Compounding Interest
From the compound interest formula for m periods per year:
r my
F = P (1 + )
m
m
Let = k, then m = rk, as m increases, so must k:
r
(1 +
r my
1 rky
1 k
) = (1 + ) = [(1 + ) ]
m
k
k
ry
1 k
The limit of (1 + ) as k approaches infinity is e, thus:
k
F = Pery
The effective rate of interest for continuous compounding is:
ER = er − 1
Where:
F = accumulated amount or future worth
P = principal or present worth
r = nominal rate of interest
y = number of years
e = Euler’s number
ery = continuous compound amount factor
1⁄ ry = present worth of continuous compounding factor
e
SAMPLE PROBLEMS
Example 1: P 100,000 is deposited in a bank that earns 5%
compounded continuously. What will be the amount after 10
years?
Ans. P 164,872.13
Example 2: Money is deposited in a certain account for which
interest is compounded continuously. If the balance doubles in 6
years, what is the annual percentage rate?
Ans. 11.55%
Example 3: A man wishes to have P 40,000 in a certain fund at the
end of 8 years. How much should he invest in a fund that will pay
6% compounded continuously?
Ans. P 24, 751.34
Example 4: If the effective annual interest rate is 4%, compute the
equivalent nominal interest compounded continuously.
Ans. 3.922%
Example 5: What is the nominal rate of interest compounded
continuously for 10 years if the compound amount factor is
1.34986?
Ans. 3%
Example 6: Deposits of P35,000.00, P48,000.00 and P25,000.00
were made in a savings account eight years, five years, and two
years ago, respectively. Determine the accumulate amount in the
account today if a withdrawal of P55,000.00 was made four years
ago. The applied interest rate is 11% compounded continuously.
Ans. P 113,330.66
Discount
Discount – is the difference between the future worth of a certain
commodity and its present worth.
2 Types of Discount:
Trade Discount – discount offered by the seller to induce trading.
Cash Discount – is the reduction on the selling price offered to a
buyer to induce him to pay promptly.
D = F−P
Where:
D = amount of discount
F = accumulated amount or future worth
P = principal or present worth
Discount Rate – is the discount on one unit of principal per unit of
time.
F−P
d=
= 1 − (1 + i)−1
F
If the commodity is discounted in a certain period of time:
Fd = F − P
P = F(1 − d) For 1 year
P = F(1 − nd) For n years
The relationship between discount rate and interest rate
becomes:
d
i=
1−d
and
i
d=
1+i
Where:
d = discount rate for the period involved
i = rate of interest for the same period
SAMPLE PROBLEMS
Example 1: Mr. T borrowed money from the bank. He receives
from the bank P 1,340 and promised to pay P 1,500 at the end of
9 months. Compute: (a) Simple interest rate; and (b) Discount
Rate.
Ans. (a) 15.92%; (b) 13.73%
Example 2: Find the discount if P 2,000 is discounted for 6 months
at 8% simple discount.
Ans. P 80
Example 3: Discount 1650 for 4 months at 6% simple interest.
What is the discount?
Ans. P 32.35
Inflation
Inflation – is the increase in the prices for goods and services from
one year to another, thus decreasing the purchasing power of
money.
FC = PC(1 + f)n
Where:
FC = future cost of a commodity
PC = present cost of a commodity
f = annual inflation rate
n = number of years
In an inflationary economy, the buying power of money decreases
as cost increases:
P
F=
(1 + f)n
If interest is computed as the same time that inflation is occurring:
1+i n
F = P(
)
1+f
Where:
F = future worth of today’s present amount P
f = annual inflation rate
n = number of years
i = rate of interest
If the uninflated present worth is to be determined:
F
F
P=
=
(1 + i)n (1 + f)n (1 + icf )n
icf = i + f + if
SAMPLE PROBLEMS
Example 1: A man invested P 130,000 at an interest rate of 10%
compounded annually. What will be the final amount of his
investment, in terms of today’s peso, after 5 years, if inflation
remains the same at the rate of 8% per year?
Ans. P 142,491
Example 2: What is the uninflated present worth of a P 200,000
future value in two years if the average inflation rate is 6% and
interest rate is 10%.
Ans. P 147,107
SET 1B: ANNUITIES
Annuity – is a series of equal payments occurring at equal periods
of time.
Ordinary Annuity
Ordinary Annuity – a type of annuity were equal payments are
made at the end of each period.
P = A(1 + i)−1 + A(1 + i)−2 + A(1 + i)−3 + ⋯ + A(1 + i)−(n−1)
+ A(1 + i)−n → Eq. 1
Multiplying this equation by (1 + i), the equation becomes:
P + Pi = A + A(1 + i)−1 + A(1 + i)−2 + ⋯ + A(1 + i)−n+2
+ A(1 + i)−n+1 → Eq. 2
Subtracting Eq. 1 from Eq. 2:
Pi = A − A(1 + i)−n
A
A (1 + i)n − 1
P = [1 − (1 + i)−n ] = [
]
(1 + i)n
i
i
n
P⁄ = [(1 + i) − 1] = (P⁄ , i%, n)
A
A
i(1 + i)n
The functional symbol (P/A, i%, n) is called the “uniform series
present worth factor”.
n
A⁄ = [ i(1 + i) ] = (A⁄ , i%, n)
n
P
P
(1 + i) − 1
The functional symbol (A/P, i%, n) is called the “capital recovery
factor”.
Where:
P = value or sum of money at present
A = series of periodic equal amount of payments
i = interest rate per interest period
n = number of interest periods/number of equal payments
P/A = uniform series present worth factor
A/P = capital recovery factor
Substituting P = F(1 + i)−n from the equation of P, it becomes:
A
F = [(1 + i)n − 1]
i
n
F⁄ = [(1 + i) − 1] = (F⁄ , i%, n)
A
A
i
The functional symbol (F/A, i%, n) is called the “uniform series
compound amount factor”.
i
A⁄ = [
] = (A⁄F , i%, n)
F
(1 + i)n − 1
The functional symbol (A/F, i%, n) is called the “sinking fund
factor”.
Where:
F = value or sum of money at some future time
A = series of periodic equal amount of payments
i = interest rate per interest period
n = number of interest periods/number of equal payments
F/A = uniform series compound amount factor
A/F = sinking fund factor
SAMPLE PROBLEMS
Example 1: Find the annual payment to extinguish a debt of P
100,000 payable for 6 years at 12% interest annually.
Ans. P 24,322.57
Example 2: What annuity is required over 12 years to equate to a
future amount of P 200,000? i = 8%.
Ans. P 10,539.00
Example 3: A man paid 10% downpayment of P 200,000 for a
house and lot and agreed to pay the 90% balance on monthly
installment for 60 months at an interest rate of 15% compounded
monthly. Compute the amount of monthly payment.
Ans. P 42,821.87
Example 4: Mr. Y bought a house and lot for $ 2,800,000 with a
downpayment of $ 300,000. Interest is 5% to be paid for 30 years
on a monthly basis. Compute the amount of monthly payment.
Ans. $ 13,420.54
Example 5: A piece of machinery can be bought for P 10,000 cash,
or for P 2,000 downpayment and payments of P 750 per year for
15 years. What is the annual interest rate of the time payments?
Ans. 4.6%
Example 6: A man inherited a regular endowment of P 100, 000
every of 3 months for 10 years. However, he may choose to get a
single lump sum payment at the end of 4 years. How much is this
lump sum if the cost of money is 14% compounded quarterly?
Ans. P 3,702,939.73
Example 7: A service car whose cash price was P 540,000 was
bought with a down payment of P 162,000 and monthly
installment of P 10,874.29 for 5 years. What was the rate of
interest if compounded monthly?
Ans. 24% compounded monthly
Example 8: If P500.00 is invested at the end of each year for 6
years, at an annual interest rate of 7%, what is the total peso
amount available upon the deposit of the sixth payment?
Ans. P 3,576.65
Example 9: A man purchased a car with a cash price of P 350,000.
He was able to negotiate with the seller to allow him to pay only a
down payment of 20% and the balance payable in equal 48 end of
the month installment at 1.5% interest per month. On the day he
paid the 20th installment, he decided to pay the remaining
balance. How much is the monthly payment and what is the
remaining balance that he paid?
Ans. P 8,224.99; P 186,927.02
Example 10: For having been loyal, trustworthy and efficient, the
company has offered a superior a yearly gratuity pay of P
20,000.00 for 10 years with the first payment to be made one year
after his retirement. The supervisor, instead, requested that he be
paid a lump sum on the date of his retirement less interest that the
company would have earned if the gratuity is to be paid on yearly
basis. If interest is 15%, what is the equivalent lump sum that he
could get?
Ans. P 100,375.37
Example 11: In anticipation of a much bigger volume of business
after 10 years, a fabrication company purchased an adjacent lot
for its expansion program where it hopes to put up a building
projected to cost P 4,000,000.00 when it will be constructed 10
years after. To provide for the required capital expense, it plans to
put up a sinking fund for the purpose. How much must the
company deposit each year if interest to be earned is computed at
15%?
Ans. P 197,008.25 annual deposits
Example 12: A new office building was constructed 5 years ago by
a consulting engineering firm. At that time the firm obtained the
bank loan for P 10,000,000 with a 20% annual interest rate,
compounded quarterly. The terms of the loan called for equal
quarterly payments for a 10-year period with the right of
prepayment any time without penalty. Due to internal changes in
the firm, it is now proposed to refinance the loan through an
insurance company. The new loan is planned for a 20- year term
with an interest rate of 24% per annum, compounded quarterly.
The insurance company has a onetime service charge 5% of the
balance. This new loan also calls for equal quarterly payments.
a.) What is the balance due on the original mortgage (principal) if
all payments have been made through a full five years?
b.) What will be the difference between the equal quarterly
payments in the existing arrangement and the revised proposal?
Ans. (a) P 7,262,747.03; (b) P 120,862
Example 13: An annual payment is made for 10 years with an
annual interest rate of 8%. Compute the following:
(a) Uniform series present worth factor;
(b) Capital recovery factor;
(c) Uniform series compound amount factor;
(d) Sinking fund factor
Ans. (a) 6.710; (b) 0.149; (c) 14.487; (d) 0.069
Annuity Due
Annuity Due – a type of annuity were equal payments are made at
the beginning of each period.
P =A+
F=
beginning of each year. How much should he deposit if the fund is
invested at 5% compounded annually?
Ans. P 6,057.49
Example 4: Determine the present worth and the accumulated
amount of an annuity consisting of 6 payments of P120, 000 each,
the payment are made at the beginning of each year. Money is
worth 15% compounded annually.
Ans. P = P 522,259; F = P 1,208,016
Example 5: A farmer bought a tractor costing P 25,000 payable in
10 semi-annual payments, each installment payable at the
beginning of each period. If the rate of interest is 26%
compounded semi-annually, determine the amount of each
installment.
Ans. P 4,077.20
Example 6: A certain manufacturing plant is being sold and was
submitted for bidding. Two bids were submitted by interested
buyers. The first bid offered to pay P 200,000 each year for 5 years,
each payment being made at the beginning of each year. The
second bid offered to pay P 120,000 the first year, P 180,000 the
second year, and P 270,000 each year for the next 3 years, all
payments being made at the beginning of each year. If money is
worth 12% compounded annually, which bid should the owner of
the plant accept?
Ans. second bid, Present worth = P 859,727.18
Deferred Annuity
Deferred Annuity – a type of annuity were the first payment is
made several periods after the beginning of annuity.
A (1 + i)n − 1
P= [
] (1 + i)−m
(1 + i)n
i
A (1 + i)n−1 − 1
[
]
(1 + i)n−1
i
A
[(1 + i)n+1 − 1] − A
i
Where:
P = value or sum of money at present
F = value or sum of money at some future time
A = series of periodic equal amount of payments
i = interest rate per interest period
n = number of interest periods/number of equal payments
SAMPLE PROBLEMS
Example 1: If money is worth 4% compounded semiannually, find
the present amount of an annuity due paying P 5,000
semiannually for a term of 3.5 years.
Ans. P 33,007.15
Example 2: A man agrees to make equal payments at the beginning
of each 6 months for 10 years to discharge a debt of P 50,000 due
now. If money is worth 8% compounded semiannually, find the
semiannual payment.
Ans. P 3,537.58
Example 3: To accumulate a fund of P 80,000 at the end of 10
years, a man will make equal annual deposits in the fund at the
Where:
P = value or sum of money at present
F = value or sum of money at some future time
A = series of periodic equal amount of payments
i = interest rate per interest period
n = number of interest periods/number of equal payments
m = number of interest periods when there is no payment made
SAMPLE PROBLEMS
Example 1: The present value of an annuity of R pesos payable
annually for 8 years, with the 1st payment at the end of 10 years is
P 187,481.25. Find the value of R if money if money is worth 5%.
Ans. P 45,000
Example 2: A parent on the day that child is born wishes to
determine what lump sum would have to be paid into an account
bearing interest of 5% compounded annually, in order to
withdraw P 20,000 each on the child’s 18th, 19th , 20th and 21th
birthdays?
Ans. P 30,941.73
Example 3: An asphalt road requires no upkeep until the end of 2
years when P60, 000 will be needed for repairs. After this P90, 000
will be needed for repairs at the end of each year for the next 5
years, then P120, 000 at the end of each year for the next 5 years.
If money is worth 14% compounded annually, what was the
equivalent uniform annual cost for the 12-year period?
Ans. P 79,245.82
Example 4: A man wishes to provide a fund for his retirement such
that from his 60th to 70th birthdays he will be able to withdraw
equal sums of P18, 000 for his yearly expenses. He invests equal
amount for his 41st to 59th birthdays in a fund earning 10%
compounded annually. How much should each of these amounts
be?
Ans. P 2,285.25
Example 5: A lathe for a machine shop costs P 60,000, if paid in
cash. On the installment plan, a purchaser should pay P 20,000
downpayment and 10 quarterly installments, the first due at the
end of the first year after purchase. If money is worth 15%
compounded quarterly, determine the quarterly installment.
Ans. P 5,439.18
Example 6: A man invests P 10,000 now for the college education
of his 2 year old son. If the fund earns 14% effective interest rate,
how much will his son get each year starting from his 18 th to the
22nd birthday?
Ans. P 20,791.64
Continuous Compounding for Discrete Payments
For an annuity compounded continuously, replace interest rate
with the effective rate for compounded continuously. Recall that:
ER = er − 1
Replacing the interest rate for the formula of ordinary annuity
with ER, the formula becomes:
A
ern − 1
P= r
( rn )
e −1
e
A
(ern − 1)
F= r
e −1
Perpetuity
Perpetuity – a type of annuity in which payments continue
indefinitely.
Capitalized Cost
Capitalized Cost – is the sum of the first cost and the present worth
of all costs of replacement, operation and maintenance for a long
period of time of any property.
Capitalized Cost = First Cost + Present Worth of Perpetual
Operations and Maintenance + Present Worth of Perpetual
Replacement by Sinking Fund Method
R
Present Worth of Perpetual Replacement =
(1 + i)L − 1
R = FC − SV
Where:
R = replacement cost
FC = first cost
SV = salvage value
i = interest rate per interest period
L = useful life in years
P=
A
i
Where:
P = value or sum of money at present
A = series of periodic equal amount of payments
i = interest rate per interest period
SAMPLE PROBLEMS
Example 1: Find the present worth of perpetuity of P 5,200
payable monthly if the interest is 16% compounded monthly.
Ans. P 390,000
Example 2: Find the present value of a perpetuity of P 15,000
payable semiannually if money is worth 8% compounded
quarterly.
Ans. P 371,287
Example 3: If money is worth 8%, determine the present value of
a perpetuity of P 1,000 payable annually with the 1st payment due
at the end of 5 years.
Ans. P 9,187.87
Example 4: If money is worth 8% compounded quarterly,
calculate the present worth of the following:
(a) An annuity of P 1,000 payable quarterly for 50 years
(b) An annuity of P 1,000 payable quarterly for 100 years
(c) A perpetuity of P 1,000 payable quarterly
Ans. (a) P 49,047.35; (b) P 49,981.85; (c) P 50,000
Example 5: It costs P 50,000 at the end of each year to maintain a
section of Kennon road in Baguio City. If money is worth 10%, how
much would it pay to spend immediately to reduce the annual cost
by P 10,000?
Ans. P 400,000
SAMPLE PROBLEMS
Example 1: Determine the accumulated amount to an account
paying P 5,000 annually (payments are made at the beginning of
each period) for 18 years if money is worth 8% compounded
continuously. Also determine the present worth.
Ans. P 209,452.57; P 49,625.13
SAMPLE PROBLEMS
Example 1: The first cost of a certain equipment is P 324,000 and
a salvage value of P 50,000 at the end of its life of 4 years. If money
is worth 6% compounded annually, find the capitalized cost.
Ans. P 1,367,901.15
Example 2: Find the capitalized cost of a bridge whose cost is P
250M and life is 20 years. If the bridge must be partially rebuilt at
a cost of P 100M at the end of each 20 years. i = 6%.
Ans. P 295.3076M
Example 3: A machine cost P 150,000 and will have a scrap value
of P 10,000 when retired at the end of 15 years. If money is worth
4%, find the annual investment and the capitalized cost of the
machine.
Ans. P 324,793.85
Example 4: A bridge that was constructed at a cost of P 7.5M is
expected to last 30 years at the end of which time its renewal cost
will be P 4M. Annual repairs and maintenance is P 300,000. What
is the capitalized cost of the bridge at an interest of 6%?
Ans. P 13,343,260.77
Example 5: Calculate the capitalized cost of a project that has an
initial cost of P 3,000,000 and an additional cost of P 1,000,000 at
the end of every 10 yrs. The annual operating costs will be P100,
000 at the end of every year for the first 4 years and P160, 000
thereafter. In addition, there is expected to be recurring major
rework cost of P 300,000 every 13 yrs. Assume i =15%.
Ans. P 4,281,936
Uniform Arithmetic Gradient
Uniform Arithmetic Gradient – is the increase by a relatively
constant amount each period.
The cash flow above is equal to the sum of the two cash flows
below:
A = P 1000, n = 5
We denote the difference between two preceding amount
(increase per period) as G, which is also known as uniform
gradient amount, in this case:
G = 500, n = 5
The formulas that may be used in this type of cash flow may be
analyzed using this formulas:
P = PA + PG
A (1 + i)n − 1
PA = [
]
(1 + i)n
i
G (1 + i)n − 1
PG = [
− n] (1 + i)−n
i
i
1 (1 + i)n − 1
PG⁄
=
− n] (1 + i)−n
G i[
i
Where:
PG/G = Gradient to present worth factor
SAMPLE PROBLEMS
Example 1: An individual makes 5 deposits that increase
uniformly by P 300 every month in a savings account that earns
12% compounded monthly. If the initial deposit is P 4,500,
determine the accumulated amount in the account just after the
last deposit.
Ans. P 25,984.67
Example 2: An amortization of a debt is in a form of a gradient
series. (a) What is the equivalent present worth of the debt if
interest is 5%. (b) Determine also the future amount of
amortization as well as the equivalent uniform periodic payment.
(c) What is the equivalent uniform annual cost?
Ans. P 15,178.34; P 18,449.37; P 4,280.47
A contract has been signed to lease a building at
P20,000 per year with an annual increase of P1,500 for
8 years. Payments are to be made at the end of each
year, starting one year from now. The prevailing
interest rate is 7%. What lump sum paid today would
be equivalent to the 8-year lease-payment plan?
SET 2: DEPRECIATION
Depreciation – is the decrease in value of physical property with
the passage of time.
Book Value – is the worth of property as shown on the accounting
records of an enterprise
Salvage/Resale Value – is the price that can be obtained from the
sale of the property after it has been used.
Scrap Value – the amount of property would sell if disposed of as
junk.
BVm = FC − Dm
DL = FC − SV
Where:
BVm = book value of a property at any time m
Dm = total depreciation of a property at any time m
DL = total depreciation at the end of its useful life
FC = first cost
SV = salvage or scrap value
Straight Line Method
Straight Line Method –a method which assumes that the loss in
value is directly proportional to the age of the property.
d1 = d2 = ⋯ = dm = dL = d
FC − SV
d=
L
Dm = md
DL = Ld
Where:
d = depreciation at any year
Dm = total depreciation of a property at any time m
DL = total depreciation at the end of its useful life
L = useful life in years
FC = first cost
SV = salvage or scrap value
SAMPLE PROBLEMS
Example 1: A machine has an initial cost of P 50,000 and a salvage
value of P 10,000 after 10 years. Using Straight Line Method of
Depreciation:
(a) What is the annual depreciation?
(b) What is the book value after 5 years?
(c) What is the total depreciation after 3 years?
Ans. (a) P 4,000; (b) P 30,000; (c) P 12,000
Example 2: An Engineer bought an equipment for P 500,000. He
spent an additional amount of P 30,000 for installation and other
expenses. The salvage value is 10% of the first cost. If the book
value at the end of 5 years is P 291,500 using straight line
depreciation, compute the life of the equipment in years.
Ans. 10 years
Example 3: A machine which cost P 10,000 was sold as scrap after
being used for 10 years. The scrap value is P 500. Determine the
total depreciation at the end of 5 years.
Ans. P 4750
Example 4: An engineer bought an equipment for P 500,000.00.
He spent an additional amount of P 30,000 for installation and
other expenses. The salvage value is 10% of the initial first cost.
Life = 15 years. Compute the following:
(a) Annual Depreciation.
(b) Book Value after 6 years.
(c) Total depreciation after 10 years.
Ans. (a) P 31,800; (b) P 339,200; (c) P 318,000
Example 5: A machine cost P 73,500 and has a life of 8 years with
a salvage value of P 3500 at the end of 8 years. Determine the
book value at the end of 4 years using straight line method.
Ans. P 38,500
Example 6: What is the book value of electronic test equipment
after 8 years of use if it depreciates from its original value of P
120,000.00 to its salvage value of 13% in 12 years? Use straightline method.
Ans. P 50,400
Example 7: The initial cost of paint sand mill, including its
installation is, P800 000.00. The BIR approved life of this machine
is 10 years for depreciation. The estimated salvage value of the
mill is P 50,000.00 and the cost of dismantling is estimated to be P
15,000.00. Using straight line depreciation, what is the annual
depreciation charge and what is the book value of the machine at
the end of six years?
Ans. P 76,500; P 341,000
Example 8: An equipment has a salvage value of P1M at the end of
50 years. The straight line depreciation charge is P2M.
(a) What is the first cost of the machine?
(b) What is the book value after 25 years?
(c) At what year will its total depreciation be P30M?
Ans. P 101M; P 51M; 15th year
Sinking Fund Method
Sinking Fund Method – a method which assumes that the sinking
fund established in which funds will accumulate for replacement.
The total depreciation that has been taken place up to any given
time is assumed to be equal to the accumulated amount in the
sinking fund at any time.
d1 = d2 = ⋯ = dm = dL = d
(FC − SV)i
d=
(1 + i)L − 1
d
Dm = [(1 + i)m − 1]
i
d
DL = [(1 + i)L − 1]
i
Where:
d = depreciation at any year
Dm = total depreciation of a property at any time m
DL = total depreciation at the end of its useful life
L = useful life in years
FC = first cost
SV = salvage or scrap value
SAMPLE PROBLEMS
Example 1: Given FC = 100,000, SV = 10,000, L = 10 years, i = 5%.
(a) Annual Depreciation, d.
(b) Book Value after 3 years.
(c) Book Value after 8 years.
Ans. P 7,155.41; P 77,442.56; P 31,672.21
Example 2: An equipment cost P 100,000 with a salvage value of P
5,000 at the end of 10 years.
Using Sinking Fund Method with interest rate= 4%.
(a) Compute the annual depreciation cost.
(b) Find the book values at years 1 to 4.
Ans. (a) P 7,912.64; (b) P 92,087.36; P 83,858.21; P 75,299.90; P
66,399.26
Example 3: A plant erected to manufacture socks with a first cost
of P 10,000,000 with an estimated salvage value of P 100,000 at
the end of 25 years. Find the appraised value to the nearest 100
by sinking fund method at 6% interest rate at the end of
a. 10 years
b. 20 years
Ans. P 7,621,600; P 3,362,200
Example 4: A factory is constructed at a 1st cost of P 8,000,000 and
with an estimated salvage value of P 200,000 at the end of 25
years. Find its appraised value to the nearest 100 at the end of 10
years by using sinking fund of depreciation assuming an interest
of 5%.
Ans. P 5,944,400
Example 5: A four-stroke motorbike costs P75 000.00. It will have
a salvage value of P10 000.00 when worn out at the end of eight
years. Determine the annual replacement deposit using the SFM
at 5%.
Ans. P 6,806.92
Example 6: A machine that costs P75 000.00 five years ago now
cost P45 864.31, when 7% interest is applied using the sinking
fund formula. Determine the salvage value of the machine for an
estimated useful life of 10 years.
Ans. P 5,000
Declining Balance Method (Matheson’s Method)
Declining Balance Method – a method which assumes that the
annual cost of depreciation is a fixed percentage (k) of the salvage
value at the beginning of the year.
dm = FC(1 − k)m−1 k
BVm = FC(1 − k)m
SV = FC(1 − k)L
L SV
k=1− √
FC
Dm = FC − BVm
Note: This method is not applicable if there is no salvage value.
Where:
dm = depreciation at any time m
BVm = book value of a property at any time m
Dm = total depreciation of a property at any time m
L = useful life in years
FC = first cost
SV = salvage or scrap value
k = rate of depreciation
SAMPLE PROBLEMS
Example 1: A machine costing P 720,000 is estimated to have a
book value of P 40,545.73 when retired at the end of 10 years.
Depreciation cost is computed using a constant percentage of the
declining value.
(a) What is the annual rate of depreciation?
(b) What is the book value after 3 years?
(c) What is the depreciation charge at the 4th year?
(d) What is the total depreciation after 6 years?
Ans. (a) 0.25; (b) P 303,750; (c) P 75,937.50; (d) P 591,855.47
Example 2: A machine having a certain 1st cost has a life of 10 years
and a salvage value of 6.633% of the first cost of 10 years. If it has
a book value of P 58,914 after 6 years, how much is the first cost
of the machine using Matheson’s Method?
Ans. P 300,049.23
Example 3: A machine has a current price of P 400,000. If its selling
price is expected to decline at the rate of 10% per annum, what
will be the selling price after 5 years?
Ans. P 236,196.00
Example 4: A radio service panel truck initially costs P 56,000. I
resale value at the end of the fifth year is estimated at P 15,000. By
means of the Declining Balance Method, determine the yearly
depreciation charge for the first and second years.
Ans. P 12,969.60; P 9,965.84
Example 5: An engineer bought an equipment for P 800,000. Other
expenses, including installation, amounted to P 50,000. At the end
of its estimated useful life of 10 years, the salvage value will be
10% of the first cost. Using the constant percentage method of
depreciation, what is the book value after 5 years?
Ans. P 268,793.20
Double Declining Balance Method
Double Declining Balance Method – a method which is similar to
declining balance method except that the rate of depreciation k is
replaced by 2/L.
2 m−1 2
dm = FC (1 − )
L
L
2 m
BVm = FC (1 − )
L
Dm = FC − BVm
Where:
dm = depreciation at any time m
BVm = book value of a property at any time m
Dm = total depreciation of a property at any time m
L = useful life in years
FC = first cost
SAMPLE PROBLEMS
Example 1: A machine has a first cost of P 140,000 and a life of 8
years with a salvage value of P 10,000 at the end of its useful life.
Using double declining balance method:
(a) What is the Book Value on the 3rd year?
(b) What is the depreciation charge on the 4 th year?
Ans. P 59,062.50; P 14,765.63
Example 2: An equipment costs P 500,000 and has a salvage value
of P 25,000 after its 25 years of useful life. Using Double Declining
Balance Method, what will be the book value after 8 years?
Ans. P 256,609.44
Example 3: XYZ Company has an equipment that cost P 90,000.
After 8 years, it will have a salvage value of P 18,000.00. Using
Double declining balance method, find the book value at the end
of 5 years.
Ans. P 21,357
Example 4: Given the following data for a construction equipment:
Initial cost = P 1,200,000.00; Economic Life = 12 years; Estimated
salvage value = P 320,000.00.
(a) What is the book value after seven years?
(b) What is the depreciation charge on the 4th year?
(c) What is the total depreciation charge at the end of the 10 th
year?
Ans. P 334,898; P 115,740.74; P 1,006,193.30
Example 5: A machine costing P 550,000 has an estimated scrap
value of P 85,000 at the end of its economic life of 8 years. Using
DDBM of depreciation:
(a) What is the book value after 4 years of service?
(b) What is the book value at the end of its life?
Ans. P 174,023; P 55,062.10
Sum of the Years Digit Method
(FC − SV)
dm =
(reverse digit)
sum of the digits
(FC − SV)
Dm =
(sum of reverse digits)
sum of the digits
L
sum of the digits = (L + 1)
2
reverse digit = L − m + 1
m
sum of reverse digits = (2L − m + 1)
2
Where:
dm = depreciation at any time m
Dm = total depreciation of a property at any time m
L = useful life in years
FC = first cost
SV = salvage or scrap value
SAMPLE PROBLEMS
Example 1: An asset is purchased for P 9000. Its estimated life is
10 years, after which is will be sold for P 1,000. Using SOYD
(a) Find the book value during the 3rd year.
(b) Find the depreciation during the 2nd year.
(c) Find the total depreciation after 4 years.
Ans. P 5,072.72; P 1,309.09; P 4,945.45
Example 2: Mr. Q purchased a Bulk Milk Cooler for P 480,000.00.
Shipping, tax, and installation costs amounted to P 25,000.00, P
20,000.00 and P 15,000.00. The machine has a useful life of 7 years
and salvage value of P 40,000.
(a) Determine the book value after four years.
(b) Determine the depreciation charge on its last year of service.
(c) Determine the total depreciation after 3 years.
Ans. P 147,142.86; P 17,857.14; P 321,428.57
Example 3: A telephone company purchased microwave radio
equipment for P 6 million, freight and installation charges
amounted to 4% of the purchased price. If the equipment will be
depreciated over a period of 10 years with a salvage value of 8%,
determine the depreciation cost during 5th year using SYD.
Ans. P 626,269.10
Example 4: A company purchases an asset for P 10,000.00 and
plans to keep it for 20 years. If the salvage value is zero at the end
of the 20th year:
(a) What is the depreciation in the third year?
(b) What is the total depreciation at the end of 14 years?
(c) What is the book value of the asset at the end of 8 years?
Use sum-of-the-year’s digits depreciation.
Ans. P 857.14; P 9,000; P 3,714.29
Example 5: An equipment costing P 500,000.00 has a life
expectancy of 5 years. Using some-of-the-year’s digit method of
depreciation, what must be its salvage value such that its
depreciation charge for the first year is P 100,000.00?
Ans. P 200,000.00
Service-Output Method
Service-Output Method – a method which assumes that the total
depreciation that has taken place is directly proportional to the
quantity of output of the property up to that time.
(FC − SV)
Depreciation per unit output =
T
(FC − SV)
(Q)
Dm =
T
Where:
Dm = total depreciation of a property at any time
FC = first cost
SV = salvage or scrap value
T = total units of output up to the end of its life
Q = total number of units of output at any time
SAMPLE PROBLEMS
Example 1: A television company purchased machinery for P
100,000 on July 1, 1979. It is estimated that it will have a useful of
10 years, scrap value of P 4,000, production of 400,000 units and
working hours of 120,000. The company uses the machinery for
14,000 hours in 1979 and 18,000 hours is 1980. The machinery
produces 36,000 units in 1979 and 44,000 units in 1980. Compute
the depreciation charge for 1980 using each method given below:
(a) Straight Line Method
(b) Working Hours Method
(c) Output method
Also compute the total depreciation at the end of 1980 using:
(d) Working Hours Method
(e) Service Output Method
Ans. P 9,600; P 14,400; P 10,560; P 25,600; P 19,200
Example 2: An asphalt and aggregate mixing plant having a
capacity of 50 cu.m. every hour costs P 2,500,000. It is estimated
to process 800,000 cu.m. during its life. During a certain year it
processed 60,000 cu.m. If its scrap value is P 100,000, determine
the total depreciation during the year and the depreciation cost
chargeable to each batch of 50 cu.m. using the service output
method.
Ans. P 180,000.00; P 150.00
SUPPLEMENTARY PROBLEMS
Example 1: A machine costs P 7,000 which last for 8 years with a
salvage value at the end of its life of P 350. Determine the
depreciation charge during the 4th year and the book value at the
end of 4 years by:
(a) Straight Line Method;
(b) Declining Balance Method;
(c) SOYD Method;
(d) Sinking Fund Method with interest of 12%;
(e) Double Declining Balance Method
Ans. (a) P 831.25, P 3,675; (b) P 710.96, P 1,565.25; (c) P 923.61,
P 2,197.22; (d) P 540.66, P 4,416.00; (e) P 738.28; P 2,214.84
Example 2: A P 110,000 chemical plant had an estimated life of 6
years and a projected scrap value of P 10,000. After 3 years of
operation, an explosion made it a total loss. How much money
would have to be raised to put up a new plant costing P 150,000,
if the depreciation reserve was maintained during its 3 years of
operation by:
(a) Straight Line Method;
(b) Sinking Fund Method at 6% interest
Ans. P 100,000; P 104,359.08
Example 3: A contractor imported a bulldozer for his job, paying P
250,000 to the manufacturer. Freight and insurance charges
amounted to P 18,000; customs’, broker’s fees and arresters
services amounted to P 8,500; taxes, permits, and other expenses
which is 10% of the purchasing cost. If the contractor estimates
the life of the bulldozer to be 10 years with a salvage value of P
20,000, determine the book value at the end of 6 years using the:
(a) Straight Line Method;
(b) Sinking Fund Method with interest at 8%;
(c) Matheson’s Formula;
(d) SOYD Method
Ans. P 132,600; P 158,949.69; P 59,201.53; P 71,181.82
Example 4: An equipment costs P 10,000 with a salvage value of P
500 at the end of 10 years. Calculate the annual depreciation cost
by:
(a) Straight Line Method;
(b) Sinking Fund Method at 4% interest
Ans. P 950; P 791.26
Example 5: A radio service panel truck initially costs P 56,000. Its
resale value at the end of the 5th year is estimated at P 15,000.
(a) Determine the annual depreciation charge by SLM.
(b) By means of the Matheson’s Formula, determine the yearly
depreciation charge for the first, second and third year.
Ans. P 8,200; P 12,969.60, P 9,965.84, P 7,657.75
SET 3: BASIC METHODS FOR ECONOMY STUDIES
Rate of Return Method (ROR)
Rate of Return – is a measure of the effectiveness of an investment
of capital and its financial efficiency. When this method is used, it
is necessary to decide whether the computed rate of return is
sufficient to justify the investment.
net annual profit
ROR =
capital invested
net annual profit = annual revenue − annual cost
Annual cost includes depreciation, labor and material cost,
overhead, rental, tax and insurances, etc.
SAMPLE PROBLEMS
Example 1: An investment of P 270,000 can be made in a project
that will produce a uniform annual revenue of P 185,400 for 5
years and having a salvage value of 10% of the investment. Out of
pocket costs for operation and maintenance will be P 81,000 per
year. Taxes and insurance will be 4% of the first cost per year. The
company expects capital to earn not less than 25% before income
taxes. Using ROR method, determine the rate of return of the
investment. Is this a desirable investment?
Ans. ROR = 23.70%; It is not a desirable investment
Example 2: A young mechanical engineer is considering
establishing his own small company. An investment of P 800,000
will be required which will be recovered in 15 years. It is
estimated that sales will be P 800,000 per year and that operating
expenses will be as follows.
Materials
P 160,000 per year
Labor
P 280,000 per year
Overhead
P 40,000 +10% of sales per year
Selling expense
P 60,000
The man will give up his regular job paying P 216,000 per year and
devote full time to the operation of the business; this will result in
decreasing labor cost by P40,000 per year, material cost by P
28,000 per year and overhead cost by P32,000 per year. If the man
expects to earn at least 20% of his capital, should he invest?
Compute for the actual rate of return.
Ans. The man should not invest; ROR = 6.6118%
Example 3: The ABC Company is considering constructing a plant
to manufacture a proposed new product. The land costs P
15,000,000, the building costs P 30,000,000, the equipment costs
P 12,500,000, and P 5,000,000 working capital is required. At the
end of 12 years, the land can be sold for P 25,000,000, the building
for P 12,000,000, the equipment for P 250,000 and all of the
working capital recovered. The annual disbursements for labor,
materials, and all other expenses are estimated to cost P
23,750,000. If the company requires a minimum return of 25%,
what should be the minimum annual sales for 12 years to justify
the investment?
Ans. P 39,748,563.43
Example 4: A man formerly employed as chief mechanic of an
automobile repair shop has saved P 1,000,000.00 which are now
invested in certain securities giving him an annual dividend of
15%. He now plans to invest this amount in his own repair shop.
In his resent job, he is earning P 25,000.00 a month, but he has to
resign to run his own business. He will need the services of the
following: 2 mechanics each earning P400.00 a day, and 8 helpers
each earning P200.00 a day. These men will work on the average
300 days per year. His other expenses are the following:
Rental
P30,000.00 a month
Miscellaneous
P25,000.00 a month
Sales tax
3% of gross income
Insurance
2%
The length of his lease is 5 years. If the average charge for each car
repaired by his shop is P 1,000.00. Determine the number of cars
he must service in one year so that he will obtain a profit of at least
20% on his investment?
Ans. 2112 cars
Annual Worth Method
Annual Worth Method – in this method, interest on the original
investment (sometimes called minimum required profit) is
included as cost. If the excess of annual cash inflows over annual
cash outflows is not less than zero, the proposed investment is
justified.
Annual cash inflow − Annual cash outflow ≥ 0
SAMPLE PROBLEMS
Example 1: An investment of P 270,000 can be made in a project
that will produce a uniform annual revenue of P 185,400 for 5
years and having a salvage value of 10% of the investment. Out of
pocket costs for operation and maintenance will be P 81,000 per
year. Taxes and insurance will be 4% of the first cost per year. The
company expects capital to earn not less than 25% before income
taxes. Using annual worth method, determine the annual cash
flow. Is this a desirable investment?
Ans. –P 3,509; It is not a desirable investment
Example 2: A man is considering investing P 500,000 to open a
semi-automatic auto washing business in a city of 400,000
population. The equipment can wash, on the average, 12 cars per
hour, using two men to operate it and to do small amount of hand
work. The man plans to hire two men, in addition to himself, and
operate the station on an 8-hour basis, 6 days per week, and 50
weeks per year. He will pay his employees P 25.00 per hour. He
expects to charge P 25.00 for a car wash. Out-of-pocket
miscellaneous cost would be P 8,500 per month. He would pay his
employees for 2 weeks vacation each year. Because of the length
of his lease, he must write off his investment within 5 years. His
capital now is earning 15%, and he is employed at a steady job that
pays P 25,000 per month. He desires a rate of return of at least
20% of his investment. Using annual worth method, determine the
excess of annual revenue over the annual cost. Would you
recommend the investment?
Ans. P 19,040; the man should invest
Present Worth Method
Present Worth Method – this method is based on the concept of
present worth. If the present worth of all net cash flows is equal to
or greater than zero, the project is justified. Cash inflow includes
annual revenue and salvage or scrap value. Depreciation is
excluded in cash outflow.
Present Worth of all net cash flows ≥ 0
SAMPLE PROBLEMS
Example 1: An investment of P 270,000 can be made in a project
that will produce a uniform annual revenue of P 185,400 for 5
years and having a salvage value of 10% of the investment. Out of
pocket costs for operation and maintenance will be P 81,000 per
year. Taxes and insurance will be 4% of the first cost per year. The
company expects capital to earn not less than 25% before income
taxes. Using present worth method, determine the present worth
of all net cash flows. Is this a desirable investment?
Ans. –P 9,436.00; It is not a desirable investment
Future Worth Method
Future Worth Method – this method is exactly comparable to the
present worth method except that all cash inflows and outflows
are compounded forward to a reference point in time called the
future.
Future Worth of all net cash flows ≥ 0
SAMPLE PROBLEMS
Example 1: An investment of P 270,000 can be made in a project
that will produce a uniform annual revenue of P 185,400 for 5
years and having a salvage value of 10% of the investment. Out of
pocket costs for operation and maintenance will be P 81,000 per
year. Taxes and insurance will be 4% of the first cost per year. The
company expects capital to earn not less than 25% before income
taxes. Using future worth method, determine the future worth of
all net cash flows. Is this a desirable investment?
Ans. –P 28,796.50; It is not a desirable investment
Payback (Payout) Period Method
Payback Period – is defined as the length of time required to
recover the first cost of an investment from the net cash flow
produced by the investment for an interest rate of zero.
Depreciation is not included in cash outflow.
Investment − Salvage Value
Payback Period =
Net annual cash flow
SAMPLE PROBLEMS
Example 1: An investment of P 270,000 can be made in a project
that will produce a uniform annual revenue of P 185,400 for 5
years and having a salvage value of 10% of the investment. Out of
pocket costs for operation and maintenance will be P 81,000 per
year. Taxes and insurance will be 4% of the first cost per year. The
company expects capital to earn not less than 25% before income
taxes. Determine the payback period.
Ans. 2.6 years
Example 2: A fixed capital investment of P 10,000,000.00 is
required for a proposed manufacturing plant and an estimated
working capital of P 2,000,000.00. Annual depreciation is
estimated to be 10% of the fixed capital investment. Determine
the rate of return on the total investment and the payout period is
the annual profit is P 2,500,000.00.
Ans. ROR = 12.5%; 4.8 years
SET 4: COMPARING ALTERNATIVES
Rate of Return on Additional Investment Method
The formula for the ROR on additional investment is:
annual net savings
ROR on additional investment =
additional investment
annual net savings = difference in annual cost
additional investment = difference in capital invested
If the rate of return on additional investment is satisfactory, then,
the alternative requiring a bigger investment is more economical
and should be chosen.
SAMPLE PROBLEMS
Example 1: A company is considering two types of equipment for
its manufacturing plant. Pertinent data are as follows:
Type A
Type B
First cost
P 200,000 P 300,000
Annual operating cost
P 32,000
P 24,000
Annual labor cost
P 50,000
P 32,000
Insurance and property taxes
3%
3%
Payroll taxes
4%
4%
Estimated Life
10
10
If the minimum required rate of return is 15%, which equipment
should be selected? What is the rate of return of return on
additional investment?
Ans. Type B; 18.79%
Annual Cost Method
To apply this method, the annual cost of alternatives including
interest on investment is determined. The alternative with the
least annual cost is chosen.
SAMPLE PROBLEMS
Example 1: Example 1: A company is considering two types of
equipment for its manufacturing plant. Pertinent data are as
follows:
Type A
Type B
First cost
P 200,000 P 300,000
Annual operating cost
P 32,000
P 24,000
Annual labor cost
P 50,000
P 32,000
Insurance and property taxes
3%
3%
Payroll taxes
4%
4%
Estimated Life
10
10
If the minimum required rate of return is 15%. What is the annual
cost of Type A? Type B? Which equipment should be selected?
Ans. P 129,850; P 126,056; Type B
Equivalent Uniform Annual Cost Method
In this method, all cash flows (irregular or uniform) must be
converted to an equivalent uniform annual cost, that is, a year-end
amount which is the same each year. The alternative with the least
equivalent uniform annual cost is preferred. When the EUAC
method is used, the equivalent uniform annual cost of the
alternatives must be calculated for one life cycle only.
SAMPLE PROBLEMS
Example 1: Example 1: A company is considering two types of
equipment for its manufacturing plant. Pertinent data are as
follows:
Type A
Type B
First cost
P 200,000 P 300,000
Annual operating cost
P 32,000
P 24,000
Annual labor cost
P 50,000
P 32,000
Insurance and property taxes
3%
3%
Payroll taxes
4%
4%
Estimated Life
10
10
If the minimum required rate of return is 15%. What is the EUAC
of Type A? Type B? Which equipment should be selected?
Ans. P 129,850; P 126,060; Type B
Present Worth Cost Method
In comparing alternatives by this method, determine the present
worth of the net cash outflows for each alternative for the same
period of time. The alternative with the least present worth of cost
is selected.
SAMPLE PROBLEMS
Example 1: Example 1: A company is considering two types of
equipment for its manufacturing plant. Pertinent data are as
follows:
Type A
Type B
First cost
P 200,000 P 300,000
Annual operating cost
P 32,000
P 24,000
Annual labor cost
P 50,000
P 32,000
Insurance and property taxes
3%
3%
Payroll taxes
4%
4%
Estimated Life
10
10
If the minimum required rate of return is 15%. What is the present
worth of Type A? Type B? Which equipment should be selected?
Ans. P 651,689; P 632,643; Type B
Break-Even Analysis
SAMPLE PROBLEMS
Example 1: The cost of producing a small transistor radio set
consists of P 23.00 for labor and P 37.00 for materials. The fixed
charges in operating the plant are P 100,000 per month. The
variable cost is P 1.00 per set. The radio set can be sold for P 75.00
each. Determine how many sets must be produced per month to
break-even.
Ans. 7,143 sets
Example 2: A company has a production capacity of 500 units per
month and its fixed costs are P 250,000 a month. The variable
costs per unit are P 1,150 and each unit can be sold for P 2,000.
Economy measures are instituted to reduce the fixed costs by 10%
and the variable cost be 20%. (a) Determine the old break-even
point.
(b) Determine the new break-even point.
(c) What is the old monthly profit at 100% capacity?
(d) What is the new monthly profit at 100% capacity?
Ans. 294 units per month; 208 units per month; P 175,000; P 315,
000
Example 3: Two machines are being considered for the production
of a particular part for which there is a long term demand.
Machine A costs P 50,000 and is expected to last 3 years and have
a P 10,000 salvage value. Machine B costs P 75,000 and is expected
to last 6 years and have zero salvage value. Machine A can produce
a part in 18 seconds; Machine B requires only 12 seconds per part.
The out-of-pocket hourly cost of operation is P 38 for A and P 30
for B. Monthly maintenance costs are P 200 for A and P 220 for B.
If interest on invested capital is 25%, determine the number of
parts per year at which the machines are equally economical.
Ans. 29, 544 parts
Benefit-Cost Ratio
SAMPLE PROBLEMS
Example 1: A non-profit educational research organization, is
contemplating an investment of P 1,500,000 in grants to develop
new ways to teach people the rudiments of profession. The grants
would extend over a ten-year period and would achieve an
estimated savings of P 500,000 per year in professors’ salaries,
student tuition, and other expenses. The program would be an
addition to ongoing and planned activities, thus an estimated P
100,000 a year would have to be released from the other program
to support the educational research. A rate of return of 15% is
expected, calculate the B/C ratio. Is this a good program?
Ans. B/C = 1.34; It is a good program
Example 2: The National Government intends to build a dam and
hydroelectric project in the Cagayan Valley at a total cost of P
455,500,000. The project will be financed by soft foreign load with
a rate of interest of 5% per year. The annual cost for operation and
maintenance, distribution facilities and others would total P
15,100,000. Annual revenues and benefits are estimated to be P
56,500,000. If the structures are expected to last for 50 years, with
no salvage value, determine the B/C ratio of the project.
Ans. B/C = 1.41