SOLVED PROBLEMS IN
ENGINEERING
ECONONOMY
Charlie A. Marquez, PIE
AUGUST 2014
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
ENGINEERING ECONOMY – 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.
FORMULAS IN ENGINEERING ECONOMY
SIMPLE INTEREST – the interest on a loan that is based only on the principal. Usually
used for short-term loans where the period is measured in days rather than years.
I = Pni
(1)
F = P + I = P + Pni
F = P(1+ni)
where:
(2)
I = interest
P = principal or present worth
n = number of interest periods
i = rate of interest per interest period
F = accumulated amount or future worth
TYPES OF SIMPLE INTEREST
ORDINARY SIMPLE INTEREST - interest is computed on the basis of 12 months of 30 days
each which is equivalent to 360 days a year. In this case, the value of n that is used in the
preceding formulas may be computed as:
where d is the number of days the principal was invested
EXACT SIMPLE INTEREST – interest is computed based on the exact number of days in a
given year which is 365 days for a normal year and 366 days during a leap year (which occurs
every 4 years, or if it is a century year, it must be divided by 400). Note that during leap years,
February has 29 days and 28 days only during a normal year. In this case, the value of n that is
used in the preceding formulas may be computed as:
for a normal year
for a leap year
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
DISCOUNT – discount in simple terms is the interest deducted in advance. It is the
difference between the amount a borrower receives in cash (present worth) and the amount he
pays in the future (future worth).
Discount = Future Worth – Present Worth
D=F–P
(3)
Rate of discount is the discount on one unit of principal for one unit of time.
d = 1 – (1 + i)-1
(4)
i=
(5)
Where: d = rate of discount
i = rate of interest for the same period
COMPOUND INTEREST – interest which is based on the principal plus the previous
accumulated interest. It may also be defined as ‘interest on top of interest.” This is usually
used in commercial practice especially for longer periods.
CASH FLOW DIAGRAMS – a graphical representation of cash flows drawn on a time scale.
↑ = receipts (positive cash flow or cash inflow)
↓ = disbursements (negative cash flow or cash outflow)
F = P(1+i)n
(6)
0 •
1
P = F(1+i)-n
(7)
P
•
2
Where: F = future amount of money
P = present worth or principal
i = rate of interest per interest period
n = number of interest periods
(1+i)n = single payment compound amount factor
(1+i)-n = single payment present worth factor
F
•
3
•
n
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
RATE OF INTEREST – the cost of borrowing money or the amount earned by a unit
principal per unit time.
TYPES OF RATES OF INTEREST
NOMINAL RATE OF INTEREST – is the basic annual rate of interest. It specifies the rate of
interest and the number of interest periods in one year.
i=
(8)
Where: i = rate of interest per interest period
r = nominal rate of interest
m = number of compounding periods per year
EFFECTIVE RATE OF INTEREST – is the actual or the exact rate of interest earned on the
principal during a one-year period.
ERi = (1+i)m – 1 (9)
Where: ERi = effective rate of interest
CONTINUOUS COMPOUNDING – based on the assumption that cash payments occur once
per year but compounding is continuous throughout the year.
nm
F=P
xnm
Let
x=
F=P
But
Therefore,
=e
F=P
(10)
EQUATION OF VALUE – this is obtained by setting the sum of the values on a certain
comparison or focal date of one set of obligations to the sum of the values on the same
date of another set of obligations.
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
ANNUITIES – a series of equal payments occurring at equal interval of time.
TYPES OF ANNUITIES
ORDINARY ANNUITY – this type of annuity is one where the payments are made at the
end of each period beginning from the first period.
P0
•
0
Fn
1
•
2 3..
•
•
• n
A A A A
Finding F when A is given:
F = A{[(1+i)n – 1] / i}
(11)
Finding P when A is given:
P = A{[1-(1+i)-n ] / i}
(12)
Where: F = future worth of an annuity
A = a series of periodic, equal amounts of money
P = present worth of an annuity
i = interest rate per interest period
n = number of interest periods
DEFERED ANNUITY – this type of annuity is one where the first payment is made
several periods after the beginning of the annuity.
P0
•
0
Fn
1
•
2
•
3
•
4
•
5…
• •
n
A A A A A
Finding F when A is given:
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
F = A{[(1+i)n – 1] / i}(1+i)n
(13)
Finding P when A is given:
P = A{[1-(1+i)-n ] / i}(1+i)-n
(14)
PERPETUITY – is an annuity wherein the payments continue indefinitely.
P
•
•
•
n ∞
P = A{[1-(1+i)-n ] / i} = A{[1-(1+i)- ∞ ] / i}
P=
(15)
CAPITALIZED COST – this is one of the most important applications of perpetuity.
The capitalized cost of any property is the sum of its first cost and the present worth of all
costs for replacement, operation, and maintenance for a long period or forever.
Case 1: No Replacement, maintenance and/or operation every period.
CC = FC + P
(16)
Where: CC = capitalized cost
FC = first cost
P = present worth of perpetual operation and maintenance
Case 2: Replacement only, no operation and maintenance
CC = FC + X
(17)
X = S / (1+i)k-1
(18)
Where: X = present worth of perpetual replacement
S = amount needed to replace the property every k period
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
k = periodic replacement
GRADIENT – A series of disbursements or receipts that increases or decreases in each
succeeding period by a constant amount
=
P
PA
PG
P = P A + PG
P = A(P/A, i%,n) + G(P/G, i%, n)
PG = G(P/G, i%,n) =
(19)
[(1+i)n-1/i]-n}(1+i)-n
Where: PA = present worth of an annuity
PG = present worth of gradient
CAPITAL FINANCING WITH BONDS
BONDS – a financial security note issued by businesses or corporations and by the
government as a means of borrowing long-term fund. It may also be defined as a long-term
note issued by the lender to the borrower stipulating the terms of repayment and other
conditions.
BOND VALUE – the value of a bond is the present worth of all future amounts that
are expected to be received through ownership of the bond.
METHODS OF BOND RETIREMENT
1. The corporation may issue another set of bonds equal to the amount of
bonds due for redemption.
2. The corporation may set up a sinking fund into which periodic deposits
of equal amounts are made. The accumulated amount in the sinking fund
is equal to the amount needed to retire the bonds at the time they are due.
C
Fr Fr Fr
F
Fr
0
1
2
n-1
n
0
1
2
3
n
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
A A
A = F/(F/A, i%, n)
A
A
(20)
P = Fr [(1-(1+i)-n)/i] + C(1+i) -n
Where:
P
(21)
A = periodic deposit into the sinking fund
F = amount needed to retire the bonds, face / par value
C = redemption price (often equal to F)
r = bond rate
i = investment rate or yield per period
P = purchase price of the bond / value of the bond n periods before redemption.
DEPRECIATION – the decrease in the value of a physical property with the passage of
time.
TYPES OF DEPRECIATION
1. Physical depreciation – this is due to the reduction of the physical ability of an
equipment or asset to produce results.
2. Functional depreciation – this is due to the lessening in the demand for the
function which the property was designed to render.
METHODS OF DEPRECIATION
1. Straight Line Method – this method assumes that the loss in the value is
directly proportional to the age of the equipment or asset.
d=
(22)
Dn = (n){
Cn = Co – Dn
Where:
(23)
(24)
d = annual depreciation charge
Co = original cost of the property
CL = scrap value / salvage value
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
L = useful life of the property
Dn = accumulated depreciation up to n years.
Cn = book value at the end of n years.
2. Sinking Fund Method – in this method, it is assumed that a sinking fund is
established in which funds will accumulate for replacement purposes.
d = (Co – CL)(i) / (1+i)L-1
(25)
Dn = d(1+i)n-1
(26)
Cn = Co – Dn
(27)
Note: all parameter definitions are the same with SLM.
3. Declining Balance Method – in this method, it is assumed that the annual cost
of depreciation is a fixed percentage of the book value at the beginning of the
year. This method is also called the constant percentage method or the
Matheson Formula.
L
k=1-
or
k=1-
dn = Co(1-k)n-1(k)
(29)
Cn = Co(1-k)n
(30)
Cn = Co(CL/Co)n/L
(31)
(28)
Where: k = decline rate, whose value must always be < 1 and the salvage
value must not be zero.
4. Double Declining Balance Method – this method is very similar to DBM but
the decline rate, k, is replaced by 2/L.
dn = Co[1-(2/L)]n-1(2/L)
(32)
Cn = Co[1-(2/L)]n
(33)
5. SUM-OF-THE-YEARS’ DIGIT METHOD (SYD)
dn = (Co – CL)(reverse digit/ digits)
 digits =
(34)
(35)
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
6. SERVICE-OUTPUT METHOD – this method assumes that the total
depreciation that has taken place is directly proportional to the quantity of
output of the property up to that time.
dn = (Co-CL / T)(Qn)
(36)
where: T = total units of output up to the end of life
Qn = total number of units of output during the nth year
DEPLETION – depletion cost is the reduction of the value of a certain natural
resource such as mines, oil, quarries, etc. due to the gradual extraction of its contents.
METHODS OF COMPUTING DEPLETION CHARGE
1. UNIT OR FACTOR METHOD – this method is dependent on the initial cost of the
property and the number of units in the property.
Depletion cost in any year =
(37)
(units sold during the year)
2. PERCENTAGE OR DEPLETION ALLOWANCE METHOD
Depletion charge = Fixed percentage of Gross Income
(38)
Depletion charge = 50% of the Net Taxable Income
(39)
INVESTMENT OF CAPITAL
METHODS OF MAKING ECONOMY STUDIES
1. RATE OF RETURN (ROR) METHOD – this method is a measure of the
effectiveness of an investment of capital. When this method is used, it is necessary to
decide whether the computed rate of return is sufficient to justify the investment. If
the computer ROR is
RORreq’d, the proposed investment is justified.
Conditions:
1. A single investment of capital is made at the beginning of the first year.
2. The capital invested is the total amount of capital investment required to finance
the project.
3. There is identical revenue and cost date for each year.
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
ROR =
(40)
2. ANNUAL WORTH METHOD – in this method, interest on the original investment is
included as a cost. If the excess if annual cash inflows over annual cash outflows is
0, the proposed investment is justified. Same conditions apply as the ROR
method.
3. PRESENT WORTH (PW) METHOD – this method is based on the concept of
present worth. If the present worth of net cash flows is
than 0, the proposed
project is economically justified.
4. PAYBACK PERIOD METHOD – this is the length of time required to recover the
first cost of an investment from the net cash flow.
Payback Period (years) =
(41)
COMPARING ALTERNATIVES METHODS OF COMPARING ALTERNATIVES
1. ROR ON ADDITIONAL INVESTMENT – in this method, if the ROR on additional
investment is
than the ROR required, then the alternative requiring an additional
investment is more economical and therefore, should be chosen.
ROR on additional investment =
(42)
2. ANNUAL COST (AC) METHOD – to use this method, the annual cost of the
alternatives including interest on capital is determined. The alternative with the least
annual cost is chosen. This applies only to alternatives which has a uniform cost data
for each year and a single investment of capital at the beginning of the project.
3. EQUIVALENT UNIFORM ANNUAL COST (EUAC) METHOD – in this method,
all cash flows must be converted to an equivalent uniform annual cost. The
alternative with the least EUAC should be selected. This method is flexible and can
be used for any type of alternative selection problems.
4. PRESENT WORTH COST (PWC) METHOD – in 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 PW should be selected.
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
5. PAYBACK PERIOD METHOD – in this method, the payback period for each
alternative is computed. The alternative with the shortest payback period is adopted.
BREAK-EVEN ANALYSIS – this is used in situations where the cost of two or more
alternatives may be affected by a common variable.
BREAK-EVEN POINT – is the value of the variable for which the costs of the
alternatives will be equal.
C1 = f1(x) and C2 = f2(x)
(43)
To Break-Even,
Income = fixed costs + variables cost (x)
(44)
Income = fixed costs + variables cost (x) + profit/loss
(45)
Income = fixed costs + variables cost (x) + dividends + profit/loss (46)
Where: C1 = certain specified total cost applicable to alternative 1
C2 = certain specified total cost applicable to alternative 2
x = a common independent variable affecting alternatives 1 & 2
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
SIMPLE INTEREST AND DISCOUNT
1. Determine the ordinary simple interest on P10,000 for 9 months and 15 days if
the rate of interest is 6%.
a. P475
c. P467.21
b. P468
d.NOTA
SOLUTION:
I = Pni
Solving for the number of days:
n = (9 X 30) + 15 = 285 days
I = (10,000)(
)(0.06) = P475
2. Determine the exact simple interest on P20,000 for the period from January 15 to
November 15, 2012 if the rate of simple interest is 5%.
a. P847.22
b. P835.16
c.P833.33
d.NOTA
SOLUTION:
I = Pni
Solving for the number of days:
January 15-31
= 16 days
February
= 29 (2012 is a leap year)
March
= 31
April
= 30
May
= 31
June
= 30
July
= 31
August
= 31
September
= 30
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
October
= 31
November 1-15
= 15 days
Total n = 305 days
I = Pni
I = (20,000)(
)(0.05) = P833.33
3. You loan from a bank the amount of P100,000 with a rate of simple interest of
20% but the interest was deducted from the loan at the time the money was
borrowed. If at the end of 1 year, you have to pay the full amount of P100,000,
what is the actual rate of interest?
a. P80,000
b. i = 25%
c. i = 20%
d. NOTA
SOLUTION:
Interest = (0.20)(100,000) = 20,000
Amt. received = 100,000 - 20,000 = 80,000
I = Pni
20,000 = 80,000(1)i
i = 25%
ANOTHER SOLUTION:
F=P(1+ni)
100,000 = 80,000[1+(1)i]
1.25 – 1 = i
i = 0.25 = 25%
4. Mr. Dee Jay borrowed money from the bank and he received the amount of
P18,420 and promised to repay P20,000 at the end of 10 months. Determine the
rate of simple interest.
a. i = 10.29%
c. i = 8.58%
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
b. P = 1,580
d. NOTA
SOLUTION:
F=P(1+ni)
20,000 = 18,420[1+(
)i]
i = 10.29%
5. If P50,000 is borrowed for 90 days at 8% per annum simple interest. How much
will be due at the end of 90 days?
a. P50,986
b. i = 8%
SOLUTION:
c. P51,000
d. NOTA
F = P(1+ni)
F = 50,000[1+
(0.08)]
F = P51,000
6. On his recent birthday on April 18, 2012, Cai was given by his mother a certain
amount of money as birthday present. He decided to invest the said amount on
a 12% exact simple interest. If the investment will mature on Christmas day at
an amount of P15,000, how much did Cai receive from his mother on his
birthday?
a. P13,859.37
b. P13,841.47
c. P15,000
d. NOTA
SOLUTION:
Solving for the number of days, n
April 18-30
= 12 days
May
= 31
June
= 30
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
July
= 31
August
= 31
September
= 30
October
= 31
November
= 30
December 1-25= 25
Total = 251 days
F = P(1+ni)
15,000 = P[1+
(0.12)]
P = P13,859.37
7. Carle borrowed money from Cy with simple interest of 12%. Find the present
worth of P50,000 which is due at the end of 6 months.
a. P74,691.18
b. P50,000
c. P47,169.81
d. NOTA
SOLUTION:
F = P(1+ni)
50,000 = P[1+(6/12)(0.12)]
P = 47,169.81
8. A loan of P5,000 is made for a period of 15 months at a simple interest rate of
15%. What future amount is due at the end of the loan period?
a. P5,750.00
b. P5,937.50
SOLUTION:
F = P(1+ni)
c. P5,600.00
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
F = 5,000[1+(15/12)(0.15)]
F = P5,937.50
9. Ciel borrowed P20,000 from the bank and promised to pay the amount after one
year. He received the amount of P19,200 after the bank collected an advance
interest of P800. What was the rate of discount and the rate of interest that the
bank collected in advance?
a. d=4%; i=4.17%
b. i=4%; d=4.17
c. P19,200; P20,000
d. NOTA
SOLUTION:
Solving for the rate of discount,
d = Interest (I) / Future Amount
d = 800 / 20,000 = 4%
Solving for the rate of interest,
i=
i=
i = 4.17%
10. You pawned your PSP for P5,000 and agreed to pay the amount at the end of 9
months. The pawnshop gave you P4,000 cash after deducting the interest in
advance. What was the rate of discount?
a. i = 20%
b. d = 25%
c. d = 20%
d. NOTA
SOLUTION:
d = Interest (I) / Future Amount
d=
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
d = 20%
11. What was the rate of interest for the problem in # 10?
a. i=20%
b. i=25%
c. d=25%
d. NOTA
SOLUTION:
i=
i=
i = 25%
COMPOUND INTEREST / EQUATION OF VALUE
12. If the sum of P15,000 is deposited in an account earning interest at the rate of
8% compounded quarterly, what will it become at the end of 5 years?
a. P22,039.92
b. P32,866.85
c. P22,289.21
d. NOTA
SOLUTION:
F = P(1+i)n
F = 15,000[1+(0.08/4)]4*5
F = P22,289.21
13. An employee has a promissory note, due 5 years hence, whose maturity value is
P70,000. If the rate of interest is 12% compounded semi-annually, what is the
value of this note now?
a. P39,720.82
b. P39,087.63
SOLUTION:
c. P96,675.79
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
P = F(1+i)-n
P = 70,000[1+(0.12/2)]-2*5
P = P39,087.63
14. IEdeas is planning to expand its operations and plans to purchase a parcel of
land on which to construct a building for their review center, which they will need
5 years hence. The current costs are: land: P2M; building: P3.5M. Since these
are not needed immediately, the company plans to defer the purchase of the land
and the construction of the building until they are needed. If the value of the land
and the cost of the building are expected to appreciate at the rates of 10% and
8% per annum, respectively. What will be the total cost of the investment after 5
years?
a. P8,580,275.74
b. P7,244,668.37
c. P8,363,668.27
d. NOTA
SOLUTION:
Solving for the future worth of the land,
F = P(1+i)n
F = 2,000,000(1.10)5 = 3,221,020
Solving for the future worth of the building,
F = P(1+i)n
F = 3,500,000(1.08)5 = 5,142,648.27
Total cost of investment after 5 years = 3,221,020 + 5,142,648.27 =
Total cost of investment after 5 years = P8,363,668.27
15. A debt of P150,000 was paid for as follows: P40,000 at the end of 3months,
P50,000 at the end of 12 months, P30,000 at the end of 15 months, and a final
payment F at the end of 21 months. If the rate of interest is 12% compounded
quarterly, find the final payment F.
a. P15,774.56
b. P157,745.58
c. P12,359.52
d. NOTA
150,000
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
SOLUTION:
3
12
15
21
0
40,000
30,000
F
50,000
Using “21” as focal point
i = r / m = 0.12/4 = 0.03/quarter
F + 30,000(1.03)2 + 50,000(1.03)3 + 40,000(1.03)6 – 150,000(1.03)7
F = 15,774.56
16. Adriste Philippines, Inc. expects to retire an existing machine at the end of 2013
and will replace it with a new machine for the same task at an estimated cost of
P600,000. The old machine can be sold for P50,000 when it is replaced. To
provide for replacement, the company intends to deposit the following amounts in
an account earning interest at 8% compounded quarterly:
P200,000 at the end of 2010
P150,000 at the end of 2011
P100,000 at the end of 2012
What additional amount is needed at the end of 2013 to purchase the new machine?
a. P15,774.56
b. P35,952.12
c. P12,359.52
d. NOTA
600,000-50,000
SOLUTION:
2010
i = 0.08/4 = 0.02/quarter
2011 2012
150,000 100,000
200,000
F
F + 100,000(1.02)4 + 150,000(1.02)8 + 200,000(1.02)12 = 600,000 – 50,000
F = P12,539.52
17. If P500,000 is invested at 8% interest compounded quarterly, how many years
will it take for this amount to accumulate to P900,000?
a. 29.68
b. 7.42
SOLUTION:
c. 14.84
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
F = P(1+i)n
900,000 = 500,000[1+(0.08/2)]n
1.8 = 1.02n
n = ln 1.8 / ln 1.02
n = 29.68 quarters / 4
n = 7.42 years
18. Charlie wishes to bequeath his youngest son P200,000 ten years from now.
What amount should he invest now if it will earn interest of 6% compounded
annually during the first 5 years and 8% compounded quarterly during the next 5
years?
a. P200,000
b. P134,594.27
c. P100,576.67
d. NOTA
SOLUTION:
P = F (1+i)-n
P = 200,000[1+(0.08/4)]-5*4 (1.06)-5
P = P100,576.67
19. Find the nominal rate compounded monthly which is equivalent to 16%
compounded quarterly. What is the corresponding effective rate?
a. r=15.79% ; Eri = 16.98%
b. i=15.97% ; r=16.98%
c. r=16.98% ; Eri=15.97%
d. NOTA
SOLUTION:
Note: For two nominal rates to be equal, their corresponding effective rates must
also be equal.
Nominal Rate
(1 +
)12 – 1
Effective Rate
=
(1 +
)4 – 1
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
1+
= 1.04(4/12)
r = 15.79% compounded quarterly
Solving for ERi,
ERi = (1 +
)4 – 1 = 16.98%
20. What payment X 10 years from now is equivalent to a payment of P500,000 7
years from now, if interest rate is 12% compounded semi-annually?
a. P221,150.05
b. P155,902.36
c. P709,259.56
d. NOTA
7
SOLUTION:
10
500,000
X
NOTE: For two future amounts to be equal, their corresponding present values
must also be equal.
500,000[1+(0.12/2)]-2*7 = X[1+(0.12/2)]-2*10
X = P709,259.56
21. At a certain interest rate compounded annually, P100,000 will become P450,000
in 15 years. What is the amount at the end of 10 years?
a. P272,638.68
b. P450,000.00
c. P100,000.00
d. NOTA
F10
SOLUTION:
450,000
0
F = P(1+i)n
450,000 = 100,000(1+i)
15
100,000
15
450,000/100,000 = (1+i)15
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
i = 1 – (4.5)1/15
i = 10.55%
solving for F10,
F10 = 100,000(1.1055)10
F10 = P272,638.68
22. Compare the accumulated values at the end of 5 years if P100,000 is invested at
the rate of 12% per year compounded annually, semi-annually, quarterly,
monthly, daily, and continuously. Which gives the highest return interest.
a. ANNUALLY
b. DAILY
c. CONTINUOUSLY
d. NOTA
a) Fannual = 100,000(1.12)5 = P176,234.17
b) Fsemi-annual = 100,000(1.06)5*2 = P179,084.77
c) Fquarterly = 100,000(1.03)5*4 = P180,611.24
d) Fmonthly = 100,000(1.01)5*12 = P181,669.67
e) Fdaily = 100,000[1+(0.12/365)]5*365 = P182,193.91
f) Fcontinuously = 100,000e0.12*5 = P182,211.88
ANNUITIES
23. In preparation for the college education of his son, DJ plans to deposit equal
amounts money for five consecutive years starting one year from now. If he
wants his son to be able to withdraw P1,000,000 ten years from now at 8%
interest compounded annually, what is the amount of each deposit?
a. P190,651.28
b. P161,010.09
c. P116,010.09
d. NOTA
F10 = 1M
SOLUTION:
0
1
5
10
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
A A A A A
Using today as focal point, take the present worth of the annuity…
P = A{[1-(1+i)-n ] / i}
A{[1-(1.08)-5]/0.08} = 1,000,000(1.08)-10
A = P116,010
Using year “5” as focal point
A{[(1.08)5 – 1]/0.08} = 1,000,000(1.08)-5
A = P116,010
24. A farmer bought a farm and he paid P100,000 cash and agreed to pay P20,000
at the end of each 6 months for 5 years. He failed to pay the first 5 payments. At
the end of 3 years, he is required to pay the seller the entire debt consisting of
his accumulated and future liabilities, otherwise the farm will be foreclosed. What
must he pay if money is worth 14% compounded semi-annually?
a. P200,863.70
b. P197,018.73
c. P68,661.62
d. NOTA
SOLUTION:
Let X – amount the farmer must pay to settle his entire indebtedness.
X = F5 + P5
X = 20,000[(1.075 – 1)/0.07] + 20,000[1-(1.07)-5/0.07]
X = 115,014.7802 + 82,003.9487
X = P197,018.7289
25. IEDeas constructed its building 6 years ago financed by a 10-year bank loan of
P10M with 12% annual interest rate compounded monthly. The company has
paid all the monthly instalments for the past 6 years. However, to reduce the
amount of instalment payments, it was decided to refinance the balance of the
loan through an insurance company. The new loan is for 10 years with interest
at 8% compounded quarterly. A service charge of 5% will be added to the loan
amount. Determine the balance on the original mortgage and the amount of the
quarterly payments to the insurance company.
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
a. P209,119.55; P5,448,161.75
b. P5,448,161.75; P209,119.55
c. P5,448,161,75
d. NOTA
10M
SOLUTION:
120
1
0
A A A A A
A
Solving for the balance on the original loan,
10M = A (P/A,12%,120)
A = P143,470.9938
Balance on the original loan = present worth of all future payments from period
73 to period 120.
P72 = 143,470.9939[(1-1.01-48)/0.1)] = P5,448,161.7
New Loan = P5,448,161.7 + service charge
New loan = P5,448,161.7 + (0.05 X P5,448,161.7) = P5,720,569.84
Solving for the quarterly payments to the Insurance Company,
P5,720,569.84
SOLUTION:
1
40
A A A A A
A
0
Solving for the quarterly payments to the Insurance Company,
5,720,569.84 = A(P/A,2%,40)
A = P209,119.5496
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
26. Bright Futures Inc. will make the following disbursements for their retiring
employees:
P500,000 on Dec. 31, 2011
P1,000,000 on Dec. 31, 2012
P2,000,000 on Dec. 31, 2013
To accumulate these sums, a sinking fund was established by making equal
year-end deposits starting Dec. 31, 2006 up to the end of 2013. If the fund earns
8% compounded annually, what is the required amount of the annual deposits?
a. P344,395.77
b. P3,633,200
c. P106,366
d. NOTA
2M
1M
SOLUTION:
500k
2006
0
2013
A A A A A =?
A
Using “2013” as focal point
A[(1.088-1)/0.08] = 2M + 1M(1.08)1 + 500,000(1.08)2
A = 3,363,200 / 10.6366 = P344,395.7656
27. Ceila invests P100,000 for the college education of her 3-year old son, Ciel. If
the fund earns 12% effective interest rate, how much will Ciel get each year
starting from his 18th to his 22nd birthday?
a. P324,486.82
b. P135,574.26
c. P324,483.37
d. NOTA
SOLUTION:
Using “22” as focal point
A[(1.125-1)/0.12] = 100,000(1.12)19
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
A = P135,574.2616
28. Carle buys a piece of property for P100,000 down payment and 10 deferred
semi-annual payments of P10,000 each starting three years from now. What is
the present value of the investment if the rate of interest is 8% compounded
semi-annually?
a. P166,665.65
b. P691,665.56
c. P60,165.45
d. NOTA
SOLUTION:
P = 100,000 + 10,000[(1-1.04-10)/0.04][1.04-5]
P = P166,665.65
29. Sir DJ spent the following amounts for the maintenance of the machine he
bought: P4,000 each year for the 1st five years, P6,000 each year for the next five
years. In addition, he spent P15,000 for overhauling at the end of the 4 th year
and another P20,000 for overhauling at the end of the 8th year. If money is worth
6% compounded annually, what was the equivalent uniform annual cost for the
10-year period?
a. P8,174.54
b. P16,849.46
c. P60,165.45
d. NOTA
SOLUTION:
Solve for the present worth of all future payments,
P = 4,000[(1-1.06-5)/0.06] + 6,000[(1-1.06-5)/0.06][1.06-5] + 15,000(1.06)-4 +
20,000(1.06)-8
P = P60,165.4471
Solving for the equivalent uniform annual cost,
A = (60,165.4471)(0.06) / (1-1.06-10)
A = P8,174.5421
30. UST Hospital purchased a CT Scan machine on the basis of guaranteed
performance. However, initial tests indicate that the operating cost will be
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
P5,000 more per year than guaranteed. If the expected life of the machine is 20
years and money is worth 15%, what deduction from the purchase price would
compensate UST for the additional operating cost?
a. P13,269.99
b. P84,590.92
c. P31,296.66
d. NOTA
SOLUTION:
P = 5,000[(1-1.15-20)/0.15]
P = P31,296.6574
31. A brand new car can be purchased with a down payment of P150,000 and equal
instalments of P18,000 each paid at the end of every month for the next 36
months. If money is worth 12% compounded monthly, determine the equivalent
cash price of the car.
a. P150,000
c. 168,000
b. P691,935.09
d. NOTA
SOLUTION:
Cash Price = 150,000 + 18,000[(1-1.01-36)/0.01]
Cash Price = P691,935.0907
32. A brand new house and lot costs P5M if paid in cash. On the instalment plan,
the buyer should pay 30% down and 20 quarterly instalments. The first payment
due at the end of the 1st year. If money is worth 12% compounded quarterly,
determine the amount of quarterly instalments.
a. P257,069.41
b. P4.85M
c. P3.5M
d. NOTA
SOLUTION:
5M = 1.5M + A[(1-1.03-20)/0.03][1.03-3]
A = P257,069.41
33. You have a building for rent and you were offered two options by a prospective
client:
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
A) P1.2M a year for 6 years, the rent being paid at the beginning of each year.
B) P300,000 for the 1st year, P400,000 for the 2nd year, P500,000 for the 3rd year,
and P600,000 a year for the next three years. All rentals are paid at the
beginning of each year.
If money is worth 12%, which is the better offer?
a. OFFER A
b. OFFER B
c. SAME OFFER
d. NOTA
SOLUTION:
OFFER A:
POffer A = 1.2M + 1.2M[(1-1.12-5)/0.12]
POffer A = P5,525,731.443
OFFER B:
POffer B = 300,000 + 400,000(1.12-1) + 500,000(1.12-2) +
600,000[(1-1.12-3)/0.12](1.12-2)
POffer B = P2,204,574.91
Therefore, choose Offer A!
PERPETUITY & CAPITALIZED COST
34. It costs P500,000 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 to P100,000?
a. P4M
c. P1M
b. P5M
d. NOTA
SOLUTION:
Savings = P500,000-100,000 = 400,000
P = A / I = 400,000 / 0.10
P = P4M
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
35. For its maintenance, a bridge in NLEX requires P250,000 at the end of 3 years
and annually thereafter. If money is worth 8%, determine the capitalized cost of
all future maintenance.
a. P214,334.71
c. P3,125,000
b. P2,679,183,81
d. NOTA
SOLUTION:
Capitalized Cost = (250,000/0.08)(1.08) -2
Capitalized Cost = PP2,679,183.81
36. API installed a new steam boiler at a total cost of P1.5M and is estimated to have
a useful life of 10 years. It is estimated to have a scrap value of P50,000 at the
end of its life. If interest is 8% compounded annually, determine its capitalized
cost.
a.P1.5M
c. P1.45M
b. P2,751,159.48
d. NOTA
SOLUTION:
Capitalized Cost = FC + X (present worth of perpetual replacements)
Solving for X
X = (FC – CL)/[(1+i)k-1]
X = (1.5M – 50,000) / [(1.08)10-1] = P1,251,159.48
Capitalized Cost = 1.5M + 1,251,159.48
Capitalized Cost = P2,751,159.48
37. The capitalized cost of a certain piece of equipment was found to be P1.42M.
The rate of interest is 12% with a salvage value of P100,000 at the end of its
useful life of 10 years. Assuming that the cost of perpetual replacement remains
constant, determine the original cost of the equipment.
a. P994,973.76
b. P47,487.89
SOLUTION:
CC = FC + X
c. 1.42M
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
X = (FC – CL)/[(1+i)k-1]
1.42M = FC + (FC – 100,000) / (1.1210-1)
1.42M = FC + 0.4749FC – 47,486.8
FC = P994,973.76
38. DJ & Charlie Foundation wishes to set-up a trust fund to provide the following
expenditures: P15M for the lot and the building and P5M for the initial equipment
of a Methods and Ergonomics Laboratory; P1M for the annual operating costs
every year; and P3M for the purchase of new equipment and replacement of
some equipment every 5 years beginning 5 years from now. If money is worth
10% compounded annually, how much money should be put into the fund?
a. P34,913,924.42
b. P30M
c. P24M
d. NOTA
SOLUTION:
CC = FC + P + X
Solving for P
P = 1M / 0.1 = P10M
Solving for the present worth of perpetual replacement, X
X = 3M / (1.105 – 1) = 4,913,924.42
CC = FC + P + X
CC = (15M + 5M) + 10M + 4,913,924.42)
Amount of money that should be put into the fund = CC = P34,913,924.42
39. A fund is to be donated by APIEMS Foundation to provide annual scholarships to
deserving Industrial Engineering students. The fund will grant P500,000 each for
the first 5 years, P800,000 for each of the next 5 years, and P1M each year
thereafter. The scholarship will start one year after the fund is established. If the
fund earns 8% interest, what is the amount of the donation?
a. P1,996,355.02
b. P9.96M
c. P2,173,897.09
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
SOLUTION:
Solving for the amount of the donation, X
X = 500,000(P/A,8%,5) + 800,000(P/A,8%,5)(P/F,8%,5) + (1M/0.08)(1.08) -10
X = 500,000(3.9927) + 800,000(3.9927)(0.6806) + (1M/0.08)(1.08) -10
X = P9,960,170.71
GRADIENT
40. IEdeas signed a contract to lease a building at P60,000 a year with an annual
increase of P1,000 a year for 5 years. Payments are to be made at the end of
each year, starting one year from now. If money is worth 8%, what lump sum
paid today is equivalent to the 5-year lease-payment plan?
a. P246,935.03
b. P239,562.60
c. P7,372.43
d. NOTA
SOLUTION:
P = PA + PG
P = 60,000[(1-1.08-5)/0.08] +
{[(1.085-1)/0.08]-5}(1.08)-5
P = P246,935.03
41. Find the equivalent annual payment of the following obligations at 12% interest:
End of Year
1
2
3
4
5
a. EUAC = 127,940.32
b. EUAC = 360,477.62
SOLUTION:
Payment (Pesos)
100,000
80,000
60,000
40,000
20,000
c. P64,507.68
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
P = PA + PG
{[(1.125-1)/0.12]-5}(1.12)-5
P = 100,000(P/A,12%,5) -
P = 360,477.62 – 127,940.32 = 232,537.2994
EUAC = 232,537.2994(0.12) / (1-1.12-5)
EUAC = P64,507.68
42. Two brothers, Cy and Ciel decided to save money in a fund that earns 8%
interest compounded annually but on different ways. Cy decided to save money
by making an end of year deposit of P10,000 on the 1st year, P11,000 on the 2nd
year, 12,000 on the 3rd year, and so on increasing the next year’s deposit by
P1,000 until the end of 10 years. Ciel decided to save by just making annual
deposits of P14,000 annually for 10 years. Who has more savings at the end of
10 years?
a. CY
b. CIEL
c. SAME
d. NOTA
SOLUTION:
Solving for the saving of Cy after 10 years,
Pcy = PA + PG
Pcy = 10,000(P/A,8%,10) +
{[(1.0810-1)/0.08]-10}(1.08)-10
Pcy = 67,100.81 + 25,976.83 = 93,077.64
Savings of Cy after 10 years, F10
F10 = 93,077.64(1.08)10 = P200,947.65
Solving for the saving of Ciel after 10 years,
PCIEL = PA + PG
PCIEL = 14,000(P/A,8%,10)
PCIEL = P93,941.14
Savings of Ciel after 10 years, F10
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
F10 = 93,941.14(1.08)10 = P202,811.87
Therefore, Ciel has more savings at the end of 10 years!
43. As an IE consultant for the World Bank you are earning an average annual salary
of $5M for 10 years. A private company would like to acquire your services and
offers you an initial salary of $3M but is increasing at the rate of $400,000
annually. If money is worth 10%, are your going to accept the offer?
a. YES, ACCEPT THE OFFER
b. NO, Don’t accept the offer
c. P93,941.14
d. NOTA
SOLUTION:
Solving for the present worth of your salary as a consultant for the WB,
PWB = 5M(P/A,10%,10) = 30,722,835.53
Solving for the present worth of the private company’s offer,
Pcompany = PA + PG
Pcompany = 3M (P/A,10%,10) = 20,130,244.20 +
{[(1.1010-1)/0.10]-10}(1.10)-10
Pcompany = 30,520,976.79
PWB > Pcompany; Therefore, do not accept the offer!
44. An Industrial Engineer wrote a best selling book in IE and was offered two
alternatives by a publishing company for the rights to publish and sell his book.
The 1st offer was a single lump sum payment of P3.5M. The 2nd offer was an
initial payment of P2.5M plus 4% of the gross receipts for the next 5 years
estimated to be as follows:
End Of Year
Gross Receipts (Pesos)
1
2
3
4
10M
8M
6M
4M
4% of Gross
Receipts(Pesos)
400,000
320,000
240,000
160,000
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
5
2M
80,000
After the 5th year, the author will not receive further royalties. If money is worth
10%, which alternative should he select?
a. Choose Offer 1
b. Choose Offer 2
c. Get them both
d. NOTA
SOLUTION:
OFFER 1: P3.5M
OFFER 2: 2.5M + 4% of the gross receipts for the next 5 years
P = PA + PG
P = 2.5 + 400,000(P/A,10%,5) -
{[(1.105-1)/0.10]-5}(1.10)-5
P = 4,565,258.12
POFFER 2 > POFFER 1 , THEREFORE, choose Offer 2!
CAPITAL FINANCING WITH BONDS
45. APIEMS Corporation is establishing a sinking fund for the purpose of
accumulating a sufficient capital to retire its outstanding bonds at maturity. The
bonds are redeemable in 10 years and their maturity value is P10M. How much
should be deposited each year if the fund pays interest at the rate of 4%.
a. P10M
b. P400,000
c. P832,909.44
d. NOTA
SOLUTION:
Maturity Value = 10M
F10 = A(A/F,4%,10)
10,000,000 = A[(1.0410-1)/0.04]
A = 832,909.44
46. PIIE Corporation issued 5,000 bonds of P1,000 face value each, redeemable at
par at the end of 10 years. To accumulate the funds required for redemption, the
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
company established a sinking fund consisting of annual deposits, the interest of
the fund being 5%. What was the principal in the fund at the end of 7 years?
a. P397,522.64
b. P3,236,632.68
c. P5M
d. NOTA
SOLUTION:
C = (5,000X1,000) = 5M
Solving for the accumulated amount after 7 years, F 7
F10 = A(A/F,5%,10)
5,000,000 = A[(1.0510-1)/0.05]
A = 397,522.64
F7 = 397,522.64[(1.057-1)/0.05]
F7 = P3,236,632.68
47. You were offered a Land Bank certificate with a face value of P1M. The bond is
bearing an interest of 6% per year payable semi-annually and due in 5 years. If
you want to earn 8% semi-annually, how much are you willing to pay for the
certificate?
a. P1M
b. P918,891.04
c. P30,000
d. NOTA
SOLUTION:
P = Fr(P/A,i%,n) + C(1+i)-n
Fr = (1M)(0.06/2) = 30,000
P = 30,000[(1-1.04-2*5)/0.04] + 1M(1.04)-2*5
P = P918,891.04
48. A P100,000, 6% bond pays dividend semi-annually and will be redeemed at
120% on December 31, 2015. If you bought this bond on December 31, 2011 to
yield 5% interest compounded semi-annually, find the price of the bond.
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
a. P120,000
b. P98,489.59
c. P21,510.41
d. NOTA
SOLUTION:
P = Fr(P/A,i%,n) + C(1+i)-n
Fr = (100,000)(0.06/2) = 3,000
C = 100,000(1.2) = 120,000
P = 3,000[(1-1.025-2*4)/0.025] + 120,000(1.025)-2*4
P = P120,000
49. You want to make 14% nominal interest compounded quarterly on a bond
investment. How much are you willing to pay now for a 12%, P100,000 bond that
will mature in 5 years and pays interest quarterly?
a. P100,000
b. P50,256.59
c. P92,893.80
d. NOTA
SOLUTION:
P = Fr(P/A,i%,n) + C(1+i)-n
Fr = (100,000)(0.12/4) = 3,000
P = 3,000[(1-1.035-5*4)/0.035] + 100,000(1.035)-5*4
P = P92,893.80
50. You bought a bond having a face value of P10,000 for only P9,000. The bond
rate was 14% nominal and interest payments were made to you semi-annually
for a total of 7 years. At the end of this period, you sold the bond at a price that
yielded 16% nominal on your investment. For how much did you sell the bond?
a. P9,484.30
b. P26,434.75
SOLUTION:
c. 410.453.66
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
P = Fr(P/A,i%,n) + C(1+i)-n
Using “14” as focal point
9,000(1.08)2*7 = (10,000)(0.14/2)[(1.0814-1)/0.08] + C
C = P9,484.30
DEPRECIATION AND DEPLETION
A. Straight Line Method
51. A machine costsP80,000 and an estimated life of 10 years with a salvage value
of P5,000. What is the book value after 5 years using the straight line method?
a. P42,500
b. P75,000
c. P7,500
d. NOTA
SOLUTION:
d = (Co – CL) / L = (80,000-5,000) / 10 = 7,500
D5 = nd = (5)(7,500) = 37,500
C5 = Co - D5
C5 = 80,000-37,500
C5 = P42,500
52. A knitting machine is purchased for P500,000. The salvage value in 10 years is
P100,000. What is the total depreciation in the first 6 years using the SLM?
a. P40,000
b. P240,000
c. P400,000
d. NOTA
SOLUTION:
d = (Co – CL) / L = (500,000-100,000) / 10 = 40,000
D6 = nd = (6)(40,000)
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
D6 = 240,000
53. The cost of a Tajima embroidery machine is P1.2M and the cost of installation is
P50,000. If the salvage value is 10% of the cost of the machine at the end of 10
years, determine the book value at the end of the 3 rd year. Use SLM.
a. P339,000
b. P911,000
c. P1.25M
d. NOTA
SOLUTION:
Co = 1.2M + 50,000 = 1,250,000
CL = (0.10)(1.2M) = 120,000
d = (Co – CL) / L = (1,250,000-120,000) / 10 = 113,000
D3 = 3(113,000) = 339,000
C3 = 1,250,000 – 339,000
C3 = P911,000
B. Sum-of-the-Year’s Digit Method
54. A linking machine is purchased for P20,000. It is estimated to have a useful life
of 10 years and a salvage value of P10,000. Find the depreciation for the 1 st
year using the SYD method.
a. P10,000
b. P18,812
c. P1,818.18
d. NOTA
SOLUTION:
dn =
(Co – CL)
dn = {10 / [10(10+1)/2]}(20,000-10,000)
dn = P1,818.18
55. Zach’s Corporation purchased a Jacquard machine for P6M, freight and
installation charges amounted to 3% of the purchase price. If the machine will be
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
depreciated over a period of 10 years with a salvage value of 6%, determine the
depreciation cost during the 5th year using the SYD method.
a. P633,730.91
b. P6.18M
c. 360,000
d. NOTA
SOLUTION:
(Co – CL)
dn =
Co = 6M + (0.03)(6M) = 6,180,000
CL = (0.06)(6.18M) = 370,800
= n(n+1)/2 = 10(10+1)/2 = 55
Reverse digit:
1
2
3
4
5
6
7
8
9
10
10
9
8
7
6
5
4
3
2
1
d5 = (6/55)( 6,180,000-370,800)
d5 = P633,730.91
56. Kaizen Corporation makes it a policy that for any new equipment purchased, the
annual depreciation cost should not exceed 20% of the 1st cost at any time with
no salvage value. Determine the service life necessary if the depreciation
method used is the SYD.
a. 7 YEARS
b. 8 YEARS
c. 9 YEARS
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
SOLUTION:
dn = 0.20Co
dn =
(Co – CL)
0.20Co = 2Co / n+1
n +1 = 2Co/0.20Co
n = 9 yrs.
C. DECLINING BALANCE METHOD (MATHESON FORMULA)
57. A stop watch used for TMS has a selling price of P1,500. If its selling price is
expected do decline at a rate of 10% per year due to obsolescence, what will be
its selling price after 3 years?
a. P1,093.5
b. P1,500
c. P150.50
d. NOTA
SOLUTION:
k=1-
0.10 = 1 -
(1-0.10)3 = (
)3
C3 = (0.90)3(1,500)
C3 = P1,093.5
58. A circular linking machine costing P45,000 has a book value of P4,500 when
retired at the end of 10 years. Depreciation cost is computed using a constant
percentage of the declining book value. What is the annual rate of depreciation
in %.
a. P4,500
c. K=20.57%
b. P45,000
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
SOLUTION:
k=1-
k=1k = 20.57%
59. A generator costing P750,000 depreciates in value each year by 20% of its
value. Find its book value when the generator is 5 years old.
a. P245,760
b. P150,000
c. 750,000
d. NOTA
SOLUTION:
C5 = CO(1-K)5
C5 = 750,000(1-0.20)5
C5 = P245,760
D. SINKING FUND METHOD
60. An electric sewing machine costs P10,000 with a salvage value of P500 at the
end of 10 years. Calculate the annual depreciation cost by the sinking fund
method at 5% interest.
a. P755.29
b. P755,293.5
c. P500
d. NOTA
SOLUTION:
d = Co – CL / (F/A,i%,L)
d = (10,000-500)(0.05) / (1.0510 – 1)
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
d = P755.29
61. A set of Wire Bond machine costs $500,000. This amount includes freight and
installation charges estimated at 10% of the original price. If the machine shall
be depreciated over a period of 10 years with a salvage value of $5,000, and
money is worth 6% per year, what is the annual depreciation charge?
a.41,348.04
b. P37,554.64
c. P545,000
d. NOTA
SOLUTION:
d = Co – CL / (F/A,i%,L)
d = (500,000 – 5,000)(0.06) / (1.0610 - 1)
d = P37,554.64
62. A steam boiler was bought for P300,000. Other expenses including freight and
installation amounted to P20,000. The boiler is expected to have a life of 12
years with a salvage value of 10% of the purchase price. Determine the book
value at the end of 10 years using the Matheson formula at 8% annual interest.
a. P15,281.55
b. P98,622.80
c. P320,000
d. NOTA
SOLUTION:
Co = 300,000 + 20,000 = 320,000
CL = (0.10)(300,000) = 30,000
d = (320,000-30,000)(0.08) / (1.0812-1) = 15,281.5549
D10 = d[(1.0810-1)/0.08] = 221,377.1996
C10 = 320,000 – 221,377.1996
C10 = P98,622.80
E. DOUBLE DECLINING BALANCE METHOD
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
63. A Die Attach machine was bought for $200,000 and used it for 10 years. What is
the book value of the machine after 5 years if the salvage value is $20,000 using
the double declining balance method?
a. P13,107.2
b. P200,000
c. P65,536
d. NOTA
SOLUTION:
Cn = Co[1-(2/L)]n
C5 = 200,000[1-(2/10)]5
C5 = P65,536
64. Determine the total depreciation up to the end of 6 years for a large dryer that
costs P50,000 when new and has an estimated salvage value of P2,000 at the
end of 10 years by the double declining balance method.
a. P36,892.8
b. P56,308.12
c. P46,892.18
d. NOTA
SOLUTION:
D6 = Co – [Co(1-2/L)n]
D6 = 50,000 – [50,000(1-2/10)6]
D6 = P36,892.80
F. SERVICE-OUTPUT METHOD
65. Julie’s Bakeshop bought a loaf bread machine for P500,000 on June 1, 2010. It
is estimated that it will have a useful life of 10 years, a scrap value of P10,000,
production of 400,000 loaves of bread and working hours of 150,000. The
company uses the machine for 15,000 hours in 2010 and 20,000 hours in 2011.
The machine produces 40,000 loaves in 2010 and 50,000 loaves in 2011.
Compute the depreciation in 2011 using (a) service-output method (b) working
hours method.
a. P61,250; P65,333.33
c. P56,366.66;P61.520
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
b. P25,000; 65,000
d. NOTA
SOLUTION:
a) dn = [(Co-CL)/T](Qn)
d2011 = [(500,000-10,000)/400,000] X (50,000)
d2011 = P61,250
b) dn = [(Co-CL)/H](Qn)
d2011 = [(500,000-10,000)/150,000] X (20,000)
d2011 = P65,333.33
G. DEPLETION
66. Balatok Mines has a total gross income for a particular year of P100M. The
taxable income after taking all deductions except for depletion is P20M. What is
the allowable depletion allowance for that particular year? Use percentage of
gross income for iron ore of 15%.
a. P10M
b. P15M
c. P25M
d. NOTA
SOLUTION:
Using depletion allowance method.
d = Fixed percentage of Gross income OR 50% of Net Taxable Income
d = (0.15)(100M) = 15M
OR
d = (0.50)(20M) = 10M
Therefore, d = P10M
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
67. Philex Oil Refinery developed an oil well which is estimated to contain 7M barrels
of oil at an initial cost of $75M. What is the depletion charge during the year
where it produces 1M barrels of oil? Use the Unit or Factor Method in computing
depletion.
a. P1M
b. P7.5M
c. P10.7143M
d. NOTA
SOLUTION:
Using Unit or Factor method,
d = (initial cost of property / total units in property)(units sold during the year)
d = (75M / 7M)(1M)
d = P10.7143M
INVESTMENT OF CAPITAL
A. ROR METHOD
68. A company estimates that insulation of steam pipes in the factory will reduce the
fuel bill by as much as 25%. The cost of the insulation is P100,000 installed and
the annual cost of taxes and insurance is 6% of the first cost. Without the
insulation, the annual fuel bill is P200,000. If the insulation is worthless after 5
years of use, and a minimum return on investment of 10% is desired, would it be
worthwhile to invest in the insulation? Solve using the ROR method.
a. ROR = 10%, yes invest
b. ROR = 27.62%, yes invest
c. ROR = 7.62%, do not invest
d. NOTA
SOLUTION:
Annual Savings = (0.25)(200,000) = P50,000
Annual Costs:
Dep’n = (100,000-0)(0.10)/(1015-1) = 16,379.75
T&I = (0.06)(100,000) = 6,000
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
TAC = 22,379.75
ROR = net annual savings / capital invested
ROR = (50,000 - 22,379.75) / 100,000
ROR = 27.62% > 10%
Therefore, it is worthwhile to invest in the insulation!
69. An investor buys a land for P1M and constructs a three-door apartment for
students worth P5M. He estimates that the annual receipts from the apartment
to be P1.2M and annual expenses to cover taxes, insurance, and maintenance of
the apartment to be P100,000. He also estimates that the land can be sold for
P3M at the end of 10 years. If his money is now earning 8% interest before
taxes, will this property earn enough for the investment to be justified? Use ROR
method.
a. ROR = 1.6%, not justified
b. ROR = 16.03%, justified
c. ROR = 13.06%, justified
d. NOTA
SOLUTION:
Annual Income = 1.2M
Annual Costs:
Dep’nland = (1M-3M)(0.08)/(1.0810-1) = -138,058.98
Dep’nbldg = (5M-1M)(0.08)/(1.0810-1) = 276,117.95
O&M = 100,000
TAC = 238,058.97
ROR = (1.2M – 238,058.97) / 6M
ROR = 16.03% > 8%; therefore, investment is justified!
B. ANNUAL COST METHOD
70. An investment of P3M can be made in a business that will produce a uniform
annual revenue of P1.2M for five years and then have a salvage value of 10% of
the investment. Operations and maintenance will be P100,000 per year. Taxes
and insurance will be 5% of the first cost per year. The investor expects to earn
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
not less than 20% before income taxes. Is this a good investment? Use the
annual cost method.
a. P2,112,825.2; good investment!
b. P1.2M; good investment
c. -P912,825.199; do not invest!
d. NOTA
SOLUTION:
Annual Income: P1.2M
Annual Costs:
Dep’n = [3M-(0.10*3M)](0.20) / (1.205-1) = 362,825.20
O&M = 100,000
T&I = (0.05*3M)
Min Req’d Profit = (0.20*3M) = 600,000
TAC = P2,112,825.199
Difference = -P912,825.199; therefore, not a good investment!
71. You are considering building a 25-unit apartment near a Call Center. If you want
a return of 25%, will this be a good investment? Other related data are shown
below:
Land
P5,000,000
Building
P10,000,000
Study period
25 years
Cost of land after 25 years
P20,000,000
Cost of building after 25 years
P1,000,000
Rent per unit per month
P8,000
Maintenance per unit per year
P1,000
Property taxes
2%
Insurance
1%
a. P2.4M; good investment
b. –P1.918M; do not invest
c. P4.219M; good investment
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
SOLUTION:
Annual Income: 8,000*25*12 = 2.4M
Annual Costs:
Dep’nland = (5M-20M)(0.25) /1.2525-1 = -14,220.8242
Dep’nbldg = (10M-1M)(0.25) / 1.2525-1 = 8,532.4945
O&M = 1,000*25 = 25,000
Taxes = 0.02*15M = 300,000
Insurance = 0.01*15M = 150,000
MRP = 0.25*15M = 3.75M
Total Annual Cost = P4,219,311.67
Difference = 2.4M - 4,219,311.67 = -P1,819,311.67
therefore, not a good investment!
C. PRESENT / FUTURE WORTH METHOD
72. A project capitalized for P5,000,000 is expected to earn a uniform annual
revenue of P1,000,000 in 10 years. The cost of operation and maintenance is
P100,000 a year, taxes and insurance will cost 3% of the first cost per year. If
the company expects its capital to earn 12% income before taxes, would you
recommend investing in the project? Use the PW method.
a. PW income < PW costs; don’t
c. PW income = PW costs ; invest
invest
b. PW income > PW costs; invest
d. NOTA
SOLUTION:
PW income = 1M(P/A,12%,10) = 1M[(1-1.12-10)/0.12] = 5,650,223.03
PW costs = 5M + [(100,000+(0.03*5M)][(1-1.12-10)/0.12] = 6,412,555.76
Since PW income < PW costs ; therefore; do not invest!
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
73. A proposed project will require the immediate investment of P500,000 and is
estimated to have year-end revenues and costs as follows:
Year
1
2
3
4
5
Revenue (P)
750,000
900,000
1,000,000
950,000
600,000
Costs
600,000
770,000
750,000
800,000
470,000
An additional investment of P200,000 will be required at the end of the 2 nd year.
The project will terminate at the end of the 5 th year, and the assets are estimated
to have a salvage value of P250,000 at that time. If interest is 12% per year, is
this a good investment? Use the PWC method.
a. PW income < PW costs; don’t
c. PW income = PW costs ; invest
invest
b. PW income > PW costs; invest
d. NOTA
SOLUTION:
PW income = 150,000(1.12)-1 + 250,000(1.12)-3 + 150,000(1.12)-4 + 803,000(1.12)-5
PW income = 862,845.11
PW costs = 500,000 + 70,000(1.12)-2
PW costs = 555,803.57
Since PW income > PW costs ; therefore; it is a good investment!
D. PAYBACK PERIOD METHOD
74. A proposed manufacturing plant will require a fixed capital investment of P6M
and an estimated working capital of P1M. Annual depreciation is estimated to be
5% of the fixed capital investment. If the annual profit is P3M, determine the rate
of return on the total investment and the minimum payback period.
a. ROR = 38.57%; Payback
Period = 2.33 years
b. ROR = 28.75%; Payback
Period = 3.23 years
c. ROR = 35.57%; Payback Period =
3.33 years
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
SOLUTION:
A) ROR = net annual profit / capital invested
ROR = [3M – (0.05*6M)] / (6M+1M)
ROR = 38.57%
B) Payback period = investment – salvage value / net annual cashflow
Payback period = 7M / 3M = 2.33 years.
COMPARING ALTERNATIVES
A.
B.
C.
D.
E.
ROR ON ADDITIONAL INVESTMENT
ANNUAL COST METHOD
EUAC METHOD
PWC METHOD
PAYBACK PERIOD METHOD
75. A company is considering the purchase of a generator. Three offers were
received and the basis of selection have been tabulated as follows:
Price of Generator
Economic Life
Salvage Value
Annual O&M
OFFER A
P600,000
5
P50,000
P15,000
OFFER B
P800,000
7
P100,000
P12,000
OFFER C
P1M
10
P120,000
P10,000
If the cost of money is 8%, what offer would you recommend to be purchased?
Using ROR on additional investment?
a. OFFER A
b. OFFER B
c. OFFER C
d. NOTA
OFFER A
OFFER B
Dep’n (600,000-50,000)(0.08) (800,000-100,000)(0.08)
(1.08)5-1
(1.08)7-1
O&M
15,000
12,000
TAC
108,751.05
90,450.68
OFFER A VS OFFER B
OFFER C
(1M-120,000)(0.08)
(1.08)10-1
10,000
70,745.95
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
RORon addl investment = (108,751.05-90,450.68) / 800,000-600,000 = 9.51%
Choose OFFER B!
OFFER B VS OFFER C
RORon addl investment = (90,450.68-70,745.95) / 1M-800,000 = 9.85%
Since RORC > RORB > RORA
Therefore, Choose OFFER C!
Using Annual Cost Method?
a. OFFER A
b. OFFER B
Dep’n
O&M
MRP
TAC
c. OFFER C
d. NOTA
OFFER A
(600,000-50,000)(0.08)
(1.08)5-1
15,000
0.08*600,000
156,751.05
OFFER C
OFFER B
(800,000-100,000)(0.08) (1M-120,000)(0.08)
(1.08)7-1
(1.08)10-1
12,000
10,000
0.08*800,000
0.08*1M
150,745.95
154,450.68
Since TACC < TACB < TACA ;
Therefore, choose Offer C!
76. To remedy the traffic situation at a busy intersection in Quezon City, two plans
are being considered. Plan A is to build a complete clover-leaf costing P100M
which would provide for all the needs during the next 30 years. Maintenance
costs are estimated to be P200,000 a year for the first 15 years, and P300,000 a
year for the next 15 years.
Plan B is to build a partial clover-leaf at a cost of P70M which would be sufficient
for the next 15 years. At the end of 15 years, the clover-leaf will be completed at
an estimated cost of P50M. Maintenance would cost P120,000 a year during the
first 15 years and P220,000 a year during the next 15 years.
If money is worth 10%, which of the two plans would you recommend?
a. Recommend Plan A
b. Recommend Plan B
SOLUTION:
c. Recommend neither of the 2 Plans
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
Solve for the PWCA :
PWCA = 100M + 200,000[(1-1.10-15)/0.10] + 300,000[(1-1.10-15)/0.10][1.10-15]
PWCA = 102,067,466.4
Solve for the PWCB :
PWCB = 70M + 50M(1.10)-15 + 120,000[(1-1.10-15)/0.10] +
220,000[(1-1.10-15)/0.10][1.10-15]
PWCB = 82,882,332.01
Since PWCA > PWCB ; therefore, choose Plan B!
77. You are considering two offers of an equipment for your manufacturing plant and
pertinent data are as follows:
First Cost
Annual O&M Cost
Annual Labor Cost
Insurance and Taxes
Payroll Taxes
Estimated Life
Offer A
P500,000
P128,000
P125,000
3%
4%
10
Offer B
P850,000
P96,000
P80,000
3%
4%
10
If the minimum required rate of return is 12%, which equipment should be
selected?
Using ROR on additional investment?
a. Offer A
b. Offer B
Dep’n
O&M
LABOR
INSURANCE
TAXES
TAC
c. both offers
d. NOTA
OFFER A
(500,000-0)(0.12)
(1.12)10-1
128,000
125,000
0.03*500,000
0.04*500,000
316,492.08
OFFER B
(850,000-0)(0.12)
(1.12)10-1
96,000
80,000
0.03*850,000
0.04*850,000
283,936.54
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
ROR = (316,492.08 – 283,936.54) / (850,000 – 500,000)
ROR = 9.30% < 12%; therefore, choose Offer A!
REPLACEMENT STUDIES
78. Shutdowns costs a company an average of P5,000 per day. Based on past
data, there are 5 average shutdowns per year. To prevent future shutdowns, it is
proposed to purchase a standby machine for P30,000 which is expected to have
a service life of 10 years and a salvage value of P6,000. Annual O&M would
cost 2,000. Would it pay to buy the standby machine if the minimum required
return is 12%?
Using ROR on additional investment?
a. do not buy the machine
b. buy the machine
c. buy a 2nd hand machine
d. NOTA
Annual costs = 5,000 * 5 = 25,000
Standby Machine:
Dep’n = (30,000-6,000)(0.12) / (1.1210-1) = 1,367.62
O&M = 2,000
TAC = 3,367.62
ROR = (25,000 – 3,367.62) / 30,000
ROR = 72.10% > 12% ; therefore, but the standby machine!
Using Annual Cost Method?
a. do not buy the machine
b. buy the machine
TAC = 3,367.62 +
(0.12*30,000) = 6,967.62
c. buy a 2nd hand machine
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
Since TAC (new machine) < 25,000; therefore, buy the standby machine!
79. An existing machine of a company requires P20,000 for O&M. If the company
decides to sell the machine now, it can be sold for P15,000. If the company
decides to use it, the machine will last for 10 more years but the salvage value at
that time will be zero.
Another plan is to overhaul the existing machine at a cost of P10,000 which is
expected to reduce the annual O&M to P15,000. Its economic life will also be
extended to 10 years but with a salvage value of P5,000.
Alternatively, a new machine can be purchased for P50,000 which is expected to
reduce annual costs for O&M to P10,000 for a period of 10 years, at which time
the salvage value is expected to be P10,000. If money is worth 12% to the
company, which alternative would you recommend?
Using Annual Cost Method?
a. Use existing machine
b. Overhaul old machine
Dep’n
O&M
MRP
TAC
c. Buy new machine
d. NOTA
EXISTING MACHINE
(15,000-0)(0.12)
(1.12)10-1
20,000
0.12*15,000
22,654.76
OVERHAUL MACHINE
(10,000-5,000)(0.12)
(1.12)10-1
15,000
0.12*10,000
16,484.92
NEW MACHINE
(50,000-10,000)(0.12)
(1.12)10-1
10,000
0.12*50,000
18,279.37
Since TAC of overhauling the machine is the lowest, therefore, recommend to overhaul
the machine.
BREAK-EVEN ANALYSIS
80. A leather gloves manufacturer produces certain items at a labor cost of P500,
materials cost per unit is P150, variable cost of P50.00 each. If the gloves has a
selling price of P1,500 per pair, how many units must be manufactured each
month for the company to break-even if the monthly overhead is P500,000?
a. 525
b. 600
c. 625
d. NOTA
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
SOLUTION:
Let X – number of units to be produced per month to break-even
Income = 1,500X
Expenses:
Labor cost = 500X
Material cost = 150X
Variable cost = 50X
Overhead = 500,000
To Break-Even:
Income = Expenses
1,500X = 500X + 150X + 50X + 500,000
1,500X = 700X + 500,000
800X = 500,000
X = 625 pairs
81. Baysports Manufacturing produces m&m chocolate dispensers at a labor cost of
P280 each, material cost of P100 each and variable cost of P4.50 each. If the
item has a unit price of P900, how many units must be manufactured each month
for the company to break-even if the monthly overhead is P450,000.
a. 872.9
b. 873
c. 297
d. NOTA
SOLUTION:
Let X – number of units to be produced per month to break-even
Income = 900X
Expenses:
Labor cost = 280X
Material cost = 100X
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
Variable cost = 4.50X
Overhead = 450,000
To Break-Even:
Income = Expenses
900X = 280X + 100X + 4.50X + 450,000
900X = 384.5X + 450,000
515.5X = 450,000
X = 873 dispensers
82. A pharmaceutical firm manufactures a whitening pill at a labor cost of P500 and
material cost of P300. The fixed charges on the company are P2M a month and
the variable costs are P200 per pill. Royalty to the movie actress promoting the
product is P600 per pill sold. If the whitening pill can be sold for P2,500 each,
how many pills must be produced each month for the firm to break-even?
a. 2,223
b. 2,222.22
c. 2,222
d. NOTA
SOLUTION:
Let X – number of whitening pills to be produced per month to break-even
Income = 2,500X
Expenses:
Labor cost = 500X
Material cost = 300X
Variable cost = 200X
Royalty = 600X
Overhead = 2M
To Break-Even:
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
Income = Expenses
2,500X = 500X + 300X + 200X + 600X + 2M
2,500X = 1,600X + 2M
900X = 2M
X = 2,223 whitening pills
83. A small company manufacturing car stereos with usb has a capacity of 250 units
a month. The variable costs are P1,200 per unit. The average selling price of
the stereos is P2,700. Fixed costs of the company amount to P175,000 a month
including all taxes. The company pays annual dividend of P15.00 per share on
each of the 25,000 shares of common stocks. (A) Determine the number of
stereos that must be produced and sold each month to break-even and the sales
volume corresponding to the unhealthy point. (B) What is the profit or loss if 200
units were produced and sold per month?
BEP:
a. 117
b. 116
c. 116.67
d. NOTA
SOLUTION:
Let X – number of car stereos to be produced per month to break-even
Income = 2,700X
Expenses:
Variable cost = 1,200X
Fixed costs = 175,000
To Break-Even:
Income = Expenses
2,700X = 1,200X + 175,000
1,500X = 175,000
X = 117 stereos
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
Unhealthy Point:
a. 137.5
b. 138
c. 137
d. NOTA
Let Y – the unhealthy point
Income = 2,700Y
Expenses:
Variable cost = 1,200Y
Fixed costs = 175,000
Dividends = (15*25,000) / 12 = 31,250
To Break-Even:
Income = Expenses + dividends
2,700Y = 1,200Y + 175,000 + 31,250
1,500Y = 206,250
X = 138 car stereos
Profit / Loss
a. 93,750 – loss
b. 93,750 – profit
c. 540,000
d. NOTA
SOLUTION:
If 200 stereos were produced and sold:
Income = 2,700(200) = 540,000
Expenses:
Variable cost = 1,200(200) = 240,000
Fixed costs = 175,000
Dividends = (15)(25,000) / 12 = 31,250
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
To Break-Even:
Income = Expenses + dividends + P/L
540,000 = 175,000 + 240,000 + 31,250 + P/L
P/L = P93,750
84. A local factory making watches produces 500 units per month and sells them at
P2,000 each. Dividends are 6% on the 10,000 shares with par value of P300
each. The fixed operating cost per month is P30,000. Other costs are P1,200
per watch. Determine the break-even point. If only 30 units were produced and
sold per month, determine the profit or loss.
Break-Even
a. 38
b. 37.5
c. 37
d. NOTA
SOLUTION:
Let X – number of watches to be produced per month to break-even
Income = 2,000X
Expenses:
Variable cost = 1,200X
Fixed costs = 30,000
To Break-Even:
Income = Expenses
2,000X = 1,200X + 30,000
800X = 30,000
X = 38 watches
Profit / Loss
a. -21,000 – loss
c. 60,000
SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014
b. 21,000 – profit
d. NOTA
SOLUTION:
If 30 watches were produced and sold:
Income = 2,000(30) = 60,000
Expenses:
Variable cost = 1,200(30) = 36,000
Fixed costs = 30,000
Dividends = (0.06)(10,000)(300) / 12 = 15,000
To Break-Even:
Income = Expenses + dividends + P/L
60,000 = 36,000 + 30,000 + 15,000 + P/L
P/L = -21,000
“Whether you think you can or you can’t, either
way you are right.”
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