Chap 7 Eng'g Economy

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GENERAL ENGINEERING & APPLIED SCIENCES ENGINEERING ECONOMICS CHAPTER 7

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A cash - flow diagram is a graphical representation of cash flows drawn on a time scale. Arrow Convention: arrows directed upward

represents positive cash flow or cash inflow (receipts).

arrows directed downward represents negative cash flow or cash outflow (disbursement)

Single Payment Cash Flow - can occur at the beginning of

the time line ( t 0= ), at the end of the time line ( t n= ), or any time in between.

Uniform Series Cash Flow - consists of a series of equal

payments A starting at t 1= and ending at t n= .

Gradient Series Cash Flow - starts with a cash flow G at

t 2= , and increases by G each year until t n= , at which time the final cash flow is (n 1)G .

Exponential Gradient Cash

Flow

I. CASH FLOW DIAGRAMS

Uniform Series t 1= t n=

Borrowers Viewpoint

Lenders Viewpoint

t n= Single Payment

Gradient Series

G 2G

3G 4G

5G 6G

(n-1)G

t n= t 2=

Exponential Gradient t 1= t n=


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CHAPTER 7 Engineering Economics

II. SIMPLE INTEREST

Ordinary Simple Interest I Pin=" Future Worth, F:

F P I or F P(1 in)= + = +"

Where:

I = Interest earned ri

360= rate of interest per day

P = Present worth (capital) F = Future worth n = Total number of interest periods in days r = interest in one year " Note: For ordinary simple interest, the interest is computed based on one

bankers year.

1 banker' s year 12 months 360 days

Each month 30 days= ==

; Computation for n and i for ordinary simple interest:

Example: An interest rate of 10% for a period of 9 months:

0.10i360

= interest per day n 9(30) 270 days= =

An interest of 15% for 3 years:

0.15i360

= interest per day ( ) ( )n 3 12 30

1080 days==


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GEAS GENERAL ENGINEERING & APPLIED SCIENCES GEAS GEAS

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Exact Simple Interest I Pin=" Future Worth, F:

F P I P(1 in)= + = +"

Where: ri for ordinary year

365ri for a leap year

366

=

=

" Note: A year is a leap year if it is divisible by 4 and divisible by 400 for a

centennial year. (Centennial years are: 1800, 1900, 2000, etc) Example: Determine the exact simple interest on P5,000 for the period from January 1 to March 28, 2006 at 9% interest. Solution: Given:

P 5,000i 9%==

Solving for n:

January 31 daysFebruary 28days (leap year)March 28 days

_________n 87 days

=

Thus, solving for the interest:

I Pin0.09I 5,000 87365

I P107.26

= =

=


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III. COMPOUND INTEREST

Future Worth, F:

( ) mtn rF P 1 i or F P 1m

= + = + " o Present Worth, P:

( ) mtn rP F 1 i or P F 1m

= + = + "

Where: (for both cases) F Future worth= P = Present worth

i = Effective interest rate per interest period (per month, per quarter, per year, etc)

n = Total number of compounding periods m =mode of compounding r = specified nominal rate t number of years=

In compound interest formula, the quantity:

( )n1 i+ is known as the Single Payment Compound Amount Factor (SPCAF).

( ) n1 i + is known as the Single Payment Present Worth Factor (SPPWF) Continuous Compounding:

r t

r t

F Pe future worth

P Fe present worth

=

=

"

"


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; Values of n and i for different modes of compounding: 9 Annually (every 12 months)

m 1 ; i r ; n t= = =

9 Semi annually - (every 6 months) rm 2 ; i ; n 2t

2= = =

9 Quarterly - (every 3 months)

rm 4 ; i ; n 4t4

= = = 9 Bimonthly - (every 2 months)

rm 6 ; i ; n 6t6

= = =

9 Semi-quarterly - (every 1.5 months)

rm 8 ; i ; n 8t8

= = = 9 Monthly - (every month) rm 12 ; i ; n 12t

12= = =

9 Semi - monthly - (every 0.5 month)

rm 24 ; i ; n 24t24

= = =


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IV. RATES OF INTEREST

Nominal Rate of Interest (NRI): Nominal rate of interest specifies the rate of interest and the number of

interest periods per year. r im="

Where: r = nominal rate of interest i = interest rate per period m = number of periods

Thus, a nominal rate of interest of 6% compounded monthly simply means that there are 12 interest periods each year. The rate per interest period being:

r 6%i 0.5%m 12

= = =" " Note: For compound interest, the rate of interest usually quoted is the NRI. In

order to accurately reflect time - value considerations, NRI must be converted into ERI before applying the formulas for compound interest.

Effective Rate of Interest (ERI): mERI (1 i) 1= + " Where: ER = Effective Rate of Interest i = interest per interest period r

m=

m = number of periods p Effective Interest Rate for Continuous Compounding: rERI e 1= " Where: r = Nominal rate of interest


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To find the nominal rate when given the effective continuous rate: ( )r ln 1 i= +" Where: i = the effective continuous rate q Equivalent Nominal Rates For two nominal rates to be equal, their effective rates must be equal. Example: (Effective interest rate - Continuous compounding) Calculate the effective interest rate per month for an interest rate of 15%

in a continuously compounded account. Solution: The nominal monthly rate, r is:

15r 1.25%12

= =

0.0125i e 1 0.012578

i 1.2578%= ==

Example: (Equivalent Rates) What nominal rate, which if converted quarterly will have the same effect

as 12% compounded semi - annually? Solution: Let: r = the unknown nominal rate For two nominal rates to have the same effect, their corresponding

effective rates must be equal.

quarterly semi annualy

4 2

4

ERI ERI

r 0.121 1 1 14 2

r1 1.12364

r 0.11825 or 11.825%

= + = + + = =


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V. ESTIMATING DOUBLING AND TRIPLING TIME OF AN INVESTMENT Doubling Time: The time required for an initial single amount to double in value with

compound interest is:

log2nlog(1 i)

= +"

Approximate Formula: (Rule of 72) The time required for an initial single amount to double in value with

compound interest is approximately equal to:

72Estimated ni

=" Where:

i = effective interest in percent For example, at a rate of 2% per year, it would take 72 2 36= years for a

current amount to double in size. Tripling Time: The time required for an initial single amount to triple in value with

compound interest is:

log3nlog(1 i)

= +" . General Formula: The time required for an initial single amount to become k times in value

is:

logknlog(1 i)

= +" Where:

nk 2, 3,4,...k=


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VI. DISCOUNT ; Relationship Between Rate of Interest and Rate of Discount di

1 d= "

Where: i = rate of interest d = rate of discount ; Successive Discount Two successive discounts of p% and q% allowed on an item are

equivalent to a single discount of:

pqd p q %100

= + Example: Two discounts of 15% and 5% are equivalent to what single

discount? Solution:

( ) ( )15 5d 15 5 %

100

d 19.25%

= + =


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VII. ANNUITIES Annuity is a series of equal payments A made at equal intervals of

time. n Ordinary Annuity - the type of annuity where the payments are made at the end of

each period Future Worth of Ordinary Annuity:

( )n1 i 1F Ai

+ = "

Present Worth of Ordinary Annuity:

( )( )n

n

1 i 1P A

i 1 i

+ = + "

o Deferred annuity - is the type of annuity where the first payment is made later than the

first or is made several periods after the beginning of the annuity. Future Worth of Deferred Annuity:

( )n1 i 1F Ai

+ = "

Present Worth of Deferred Annuity:

( )( )n

m n

1 i 1P A

i 1 i + + = +

"

Where: A periodic equal payments= n number of periods (equal to the number of payments)= i = interest rate per payment m number of periods before the beginning of the first payment=

A A A A A A

F P

0 1 2 3 4 (n 1) n

Cash Flow Diagram (Ordinary Annuity)

Cash Flow Diagram (Deferred Annuity)

A A A A A

F P

0 1 2 m

m n+ periods

0 1 2 3 n

n periods


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p Annuity Due - is the type of annuity where the payment is made at the beginning

of each period Future Worth of Annuity Due:

( )n1 i 1F Ai

+ = "

Present Worth of Annuity Due:

( )( )n

n 1

1 i 1P A

i 1 i + = +

"

Where: (in both cases) A periodic equal payments= n number of periods (equal to the number of payments)= i = interest rate per payment q Perpetuity - is an annuity in which the periodic payments continue indefinitely. Present Worth of Perpetuity:

( For payments made at the end of each period)

APi

=" Where: A = periodic payment i = interest per payment

Cash Flow Diagram (Annuity Due)

A A A A A A A F

P

1 2 3 4 5 (n 1) n

n 1 periods

A A A A

P

0 1 2 3. n

Cash Flow Diagram (Perpetuity)


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VIII. CAPITALIZED COST Capitalized Cost refers to the present worth of a property that is

assumed to last forever. The capitalized cost of any property is the sum of the first cost and the present costs of perpetual replacement, operation and maintenance.

; CASE 1: No replacement, only maintenance.

ACC FCi

= +" Where: CC Capitalized Cost= FC = first cost or original cost A = Annual maintenance cost ; CASE 2: No maintenance, only replacement. ( )n

PCC FC1 i 1

= + + " Where:

CC Capitalized Cost= FC = first cost or original cost P = the amount needed to replace the property every n periods ; CASE 3: Replacement and maintenance every period.

( )nA PCC FCi 1 i 1

= + + + "

Where: CC Capitalized Cost= FC = first cost or original cost A = Annual maintenance cost P = the amount needed to replace the property every n periods


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IX. DEPRECIATION Depreciation is the decrease in the value of physical property due to

passage of time. Symbols used and their meaning: d = annual depreciation charge nd = depreciation charge during the nth year nD = total depreciation after n years FC = first cost /original cost SV = estimated salvage value after L years L = expected depreciable life of the property n = number of years before L ; METHODS OF COMPUTING DEPRECIATION n Straight Line Method Straight line method of depreciation assumes that the loss in value

of the property is directly proportional to the age of the property. Annual depreciation:

FC SVdL="

Depreciation after n years:

n

n

FC SVD nL

D d n

= =

"

"

Book value after n years: n nBV FC D= "


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o Sinking Fund Method Annual depreciation:

( )( )LFC SV i

d1 i 1

=+

"

Depreciation after n years:

( )n

n

d 1 i 1D

i

+ =" Book value after n years: n nBV FC D= " p Declining Balance Method Also called as the constant percentage method or the Mateson Formula: Depreciation during the nth year: ( ) ( )n 1nd k FC 1 k = " Salvage Value at the end of its useful life: ( )LSV FC 1 k= " Book Value at the end of n years:

( )n

nL

BV FC 1 k

SVBV FCFC

= =

"

"


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Rate of depreciation:

nn LBV SVk 1 1FC FC

= = q Double Declining Balance Method (DDBM) Also called as the constant percentage method or the Mateson Formula: Depreciation during the nth year:

( ) ( )n 1n 2 FC 1 kd L="

Salvage Value at the end of its useful life:

L2SV FC 1

L = "

Book Value at the end of n years:

n2BV FC 1

L = "

Note that the formulas for DDB method are obtained form the formulas for Declining Balance Method by simply replacing k with 2/L. r Sum - of - the - Years - Digits (SYD) Method Depreciation charge during the nth year:

( )n L n 1d FC SV SYD += "

Total depreciation after n years

( ) ( )( )nn 2L n 1

D FC SV2 SYD

+= "

WHERE:

( )SYD sum of the year's digit

nSYD n 12

== +


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TEST - 7

1. It is defined to be the capacity of a commodity to satisfy human want A. necessity B. utility C. luxuries D. discount 2. It is the stock that has prior right to dividends. It usually does not bring

voting right to the owners and the dividend is fixed and cannot be higher than the specified amount.

A. Common stock B. Voting stock C. Preferred stock D. Non par value stock 3. It is a amount which a willing buyer will pay to a willing seller for the

property where each has equal advantage and is under no compulsion to buy or sell.

A. Book value B. Market value C. Use value D. Fair value 4. _________ is the loss of value of the equipment with use over a period

of time. It could mean a difference in value between a new asset and the use asset currently in a service.

A. Loss B. Depreciation C. Extracted D. Gain


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5. An economic condition in which there are so few suppliers of a particular product that one suppliers actions significantly affect prices and supply.

A. Oligopoly * B. monopsony C. monopoly D. perfect competition 6. A market whereby there is only one buyer of an item for when there are

no goods substitute. A. Monopsony B. Monopoly C. Oligopoly D. Oligopsony 7. It is the worth of a property as recorded in the book of an enterprise. A. Salvage value B. Price C. Book value D. Scrap value 8. Reduction in the level of national income and output usually

accompanied by a fall in the general price level. A. Devaluation B. Deflation C. Inflation D. Depreciation 9. A formal organization of producers within industry forming a perfect

collusion purposely formed to increase profit and block new comers from the industry.

A. Cartel B. Monopoly C. Corporation D. Competitors


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10. A market situation where there is only one seller with many buyer. A. Monopoly* B. Monophony C. Oligopoly D. perfect competition 11. A market situation where there is one seller and one buyer. A. Bilateral monopoly* B. Monopoly C. Oligopoly D. Bilateral Monopoly 12. Reduction in the level of national income and output usually

accompanied by a fall in the general price level. A. Deflation B. Inflation C. Devaluation D. Depreciation 13. A series of equal payments made at equal interval of time.

A. Annuity* B. Amortization C. Depreciation D. Bonds 14. The money paid for the use of borrowed capital A. interest* B. amortization C. annuity D. bonds 15. The place where buyers and sellers come together. A. Market * B. Store C. Bargain center D. Port


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16. The value of the stock as stated on the stock certificate A. stock value B. par value* C. interest D. maturity value 17. An market situation in which two competing buyers exert controlling

influence over many sellers. A. bilateral monopoly B. oligopoly C. duopsony* D. duopoly 18. An market situation in which two powerful groups or organizations

dominate commerce in one business market or commodity. A. Oligopoly B. Duopoly* C. Bilateral oligopoly D. Bilateral Oligopsony 19. The type of annuity where the first payment is made after several

periods, after the beginning of the payment.

A. Perpetuity B. Ordinary annuity C. Annuity due D. Deferred annuity* 20. The condition in which the total income equals the total operating

expenses. A. tally B. par value C. Check and balance D. Break even *


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21. The amount which has been spend or capital invested which for some reasons cannot be retrieved.

A. sunk cost* B. fixed costs C. depletion cost D. construction cost 22. An obligation with no condition attached is called A. Personal B. Gratuitous * C. Concealed D. Private 23. The sum of all the costs necessary to prepare a construction project for

operation.

A. operation cost B. construction cost* C. depletion cost D. production cost 24. The amount received from the sale of an additional unit of a product. A. marginal cost B. marginal revenue* C. extra profit D. prime cost 25. The amount that the property would give if sold for junk.

A. junk value B. salvage value C. scrap value* D. book value


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26. The worth of the property which is equal to the original cost less the amount which has been charged to depreciation.

A. scrap value

B. salvage value C. book value* D. market value 27. The sum of the direct labor cost incurred in the factory and the direct

material costs of all materials that go into production is called A. net cost B. maintenance cost C. prime cost* D. operating cost 28. The difference between the present value and the worth of money at

some time in the future is called A. market value B. net value C. discount* D. interest 29. The additional cost of producing one more unit is A. prime cost B. marginal cost* C. differential cost D. sunk cost 30. A written contract by a debtor to pay final redemption value on an

indicated date or maturity date and to pay a certain sum periodically. A. annuity B. bond* C. amortization D. collateral


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31. Estimated value of the property at the end of the useful life. A. Market value B. Fair value C. Salvage value * D. Book value 32. Determination of the actual quantity of the materials on hand as of a

given date. A. physical inventory* B. counting principle C. stock assessment D. periodic material update 33. This consists of cash and account receivable during the next period or

any other material which will be sold. A. fixed assets B. deferred charges C. current asset* D. liability 34. A wrongful act that causes injury to a person or property and for which

the law allows a claim by the injured party to recover damages. A. fraud B. tort* C. libel D. scam 35. A series of uniform payment over an infinite period of time

A. depletion B. capitalized cost C. perpetuity * D. inflation


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36. These are products or services that are required to support human life

and activities that will be purchased in somewhat the same quantity event though the price varies considerably.

A. Commodities B. Necessities * C. Demands D. Luxury 37. The quantity of a certain commodity that is offered for sale at a certain

price at a given place and time. A. utility B. supply* C. stocks D. goods 38. It is sometimes called the second hand value A. Scrap value B. Salvage value* C. Book value D. Par value 39. Decreases in the value of a physical property due to the passage of time. A. Deflation B. Depletion C. Declination D. Depreciation* 40. An association of two or more individuals for the purpose of engaging

business for profit. A. Single proprietorship B. Party C. Corporation D. Partnership*


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41. The simplest form of business organization wherein the business is own entirely by one person.

A. partnership B. proprietorship* C. corporation D. joint venture 42. Parties whose consent or signature in a contract is not considered

intelligent. A. dummy person B. minors C. demented persons * D. convict 43. It is defined as the capacity of a commodity to satisfy human want. A. satisfaction B. luxury* C. necessity D. utility 44. This occurs in a situation where a commodity or service is supplied by a

number of vendors and there is nothing to prevent additional vendors entering the market .

A. perfect competition * B. monophony C. monopoly D. cartel 45. These are products or services that are desired by human and will be

purchased if money is available after the required necessities have been obtained.

A. Commodities B. Necessities C. Luxuries * D. Supplies


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46. Grand total of the assets and operational capability of a corporation. A. authorized capital* B. paid off capital C. subscribed capital D. investment 47. It is where the original record of a business transaction is recorded. A. ledger B. spreadsheet C. journal* D. logbook 48. The length of time which the property may be operated at a profit.

A. life span B. economic life* C. operating life D. profitable life 49. The right and privilege granted to an individual or corporation to do

business in a certain region. A. permit B. royalty C. license D. franchise* 50. The worth of an asset as shown in the accounting records of an

enterprise. A. fair value B. par value C. market value D. book value*


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1. A man expects to receive P20,000 in 10 years. How much is that money worth now considering interest at 6 %compounded quarterly? Solution: Given:

F 20,000t 10 yrsr 6%;compounded quarterly(m 4)

=== =

Formula: ( )nF 1 i= + Where: ri , n mt

m= =

Substitute the given values:

4(10).0620,000 P 14

P 11,025.25

= + =

2. An employee obtained a loan of P10,000 at the rate of 6% compounded annually in order to repair a house. How much must he pay monthly to amortize the loan within a period of 10 years? Solution: Given:

P 10,000r 6%m 1 (annuallyt 10 yrs

====

Formula: - (Annuity)

( )( )n

n1 i 1

P Ai 1 i

+ = +

Solved Problems In Economics


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Solving for the interest rate per month:

( ) ( )monthly annually

12 1

ERI ERI

1 i 1 1.06 1i 0.0048675

=+ =

=

Substitute to the formula:

( )( )

12(10)

1201.0048675 1

10,000 A0.0048675 (1.0048675)

10,000 90.72AA 110.22

= ==

3. What is the effective rate corresponding to 16% compounded daily? Take 1 year = 360 days. Solution: Given: r 16% ; m 360 (daily)= = Substitute:

3600.16ERI 1 1360

ERI 0.173517.35%

= + ==

4. What is the accumulated amount after 3 years of P6,500.00 invested at the rate of 12% per year compounded semi-annually? Solution: Given:

P 6,500 ; r 12%m 2 (semi annually)t 3 years

= == =

Substitute:

60.12F 6500 1 9,220.37

2 = + =

Formula - (Effective Rate)

( )m

m

ERI 1 i 1

r1 1m

= + = +

Where:r nominal ratem mode of compounding==

Formula - Future worth

( )n

mt

F P 1 i

rP 1m

= + = +

Where: r nominal ratem mode of compoundingt no. of years

===


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5. Fifteen percent (15%) when compounded semi-annually will have an effective rate of Solution: Given:

r 15%m 2==

Substitute:

20.15ERI 1 12

ERI 0.155615.56%

= + ==

6. What rate of interest compounded annually is the same as the rate of interest of 8% compounded quarterly? Solution: Given:

r 8%m 4==

Let: x = unknown rate

( )annually quarterly

41

ERI ERI

0.081 x 1 1 14

x 0.0824x 8.24%

= + = +

==

Formula - (Effective Rate)

( )m

m

ERI 1 i 1

r1 1m

= + = +

Where: r nominal ratem mode of compounding==

" Note: For two or more rates to be equivalent, their corresponding effective rates must be equal.


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7. How long will it take the money to triple itself if invested at 10% compounded semi-annually? Solution: Let: P = present worth F 3P= Substitute to the formula:

( )

2t

2t

0.103P P 12

3 1.05ln3 2t(ln1.05)

t 11.3 years

= + ===

8. A man wishes his son to receive P500,000.00 ten years from now. What amount should he invest now if it will earn interest of 12% compounded annually during the first 5 years and 15% compounded quarterly during the next 5 years? Solution: Given:

F 500,000r 12%m 1 (annually)

===

From: ( )nF 1 i= + For the first 5 years: r 12% ; m 1 (annually)= =

( )555

F P 1.12F 1.76P

==

Formula - Future worth (CI)

( )n

mt

F P 1 i

rP 1m

= + = +


===

0 5 10

P 5F

10F


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For the next 5 years,:

( ) ( )

10

20

10 5

5 20

F 500,000r 15% ; m 4 (quarterly)

0.15F F 14

500,000 1.12 P 1.0375P 135,868.19

== =

= + =

=

9. What interest rate compounded monthly is equivalent to 10% effective rate? Solution: Given:

r 8%m 4==

Let: x = unknown rate

12

12

x0.10 1 112

x1.10 112

x 0.0957x 9.57%

= + = +

==

Formula - Effective rate

mrER 1 1

m = +


===


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10. A machine costs P8,000.00 and an estimated life of 10 years with a salvage value of P500.00. What is its book value after 8 years using straight-line method? Solution: Given:

oC 8,000 ; n 8 yearsL 10 years(life)

SV 500

= ===

Solving for the total depreciation:

n

o

D d n (d annual depreciation)C SV n

L8000 500 8 6000

10

= = = = =

Thus, the Book Value (BV) is:

o nBV C D8000 6000 2000

= = =

11. By the condition of a will, the sum of P20,000 is left to a girl to be held in trust fund by her guardian until it amounts to P50,000. When will the girl receive the money if the fund is invested at 8% compounded quarterly? Solution: Given:

P 20,000 F 50,000r 8%; m 4 (quarterly)= == =

Substitute given values:

( )

4t

4t

.0850,000 20,000 14

2.5 1.02

= + =

Take ln both sides:

( )ln2.5 4t ln1.02

t 11.57 years==

Formula - (Compound Interest)

( ) mtn rF P 1 i P 1m

= + = +


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12. The amount of P12,800 in 4 years at 5% compounded quarterly is Solution: Given:

P 12,800t 4 yearsr 5%m 4 (quarterly)

====

Substitute the given values to the formula:

4(4)0.05F 12,800 14

F 15,614.59

= + =

13. How much money must you invest today in order to withdraw P2000 annually for 10 years if the interest rate is 9%? Solution: Given:

A 2000t 10 yearsr 9%m 1 (annually)P ?

=====

Substitute the given values:

( )10

10

1.09 1P 2000

0.09(1.09)

P 12,835.32

= =



= + = +

Formula: - (Annuity)

( )( )n

n1 i 1

P Ai 1 i

+ = +

Where: A annual payment / widrawalP present worthn number of payment / widrawal

===


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14. Money Borrowed today is to be paid in 6 equal payments at the end of 6 quarters. If the interest is 12% compounded quarterly, how much was initially borrowed if quarterly payments is P2,000.00? Solution: Given:

A 2000r 12%m 4 (quarterly)

r 0.12i 0.03m 4

===

= = =

Substitute:

( )

( ) ( )6

6

1.03 1P 2000

0.03 1.03

P 10,834.38

= =

15. The effective rate of 14% compounded semi-annually is Solution: Given:

r 14%m 2 (semi annually)ER ?

== =

From the formula:

20.14ER 1 12

0.144914.49%

= + ==

Formula: (Present worth -Annuity)

( )( )n

n1 i 1

P Ai 1 i

+ = +

Where:

A periodic paymentP present worthn number of payments

ri int erest per periodm

====


mrER 1 1

m = +


===


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16. A man expects to receive P25,000 in 8 years. How much is that money worth now considering interest at 8% compounded quarterly? Solution: Given:

F 25,000r 8%m 4 (quarterly)t 8 yearsP ?

=====

From the formula:

4(8)0.0825,000 P 14

P 13,262.83

= + =

17. What is the accumulated amount of the five-year annuity paying P6,000 at the end of each year with interest at 15% compounded annually? Solution: Given:

A 6000r 15%m 1 (annually)t 5 yearsF ?

=====

Substitute:

( )51.15 1F 6000

0.15

F 40,454.29

= =



= + = + Where; F future worthP present worthr nominal rate of int erestm mode of compoundingt no. of years

=====

Formula: (Future worth -Annuity)

( )n1 i 1F Ai

+ =

Where:



====


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18. ABC Corporation makes its policy that for every new equipment purchased, the annual depreciation cost should not exceed 20% of the first cost at any time without salvage value. Determine the length of service life necessary if the depreciation used is the SYD method. Solution : Using SYD method:

( )

( ) ( )

n o L

o o

reversed digitd C CSYD

n0.2C C 0 n n 12

20.20n 1

n 9 years

= = +

= +=

19. At an interest rate of 10% compounded annually, how much will a deposit of P1500 in 15 years? Solution: Given:

r 10%m 1 (annually)P 1500t 15 yearsF ?

=====

From:

mtrF P 1

m = +

Substitute:

( )15F 1500 1.10F 6,265.87==


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20. A debt of P10,000 with 10% interest compounded semi-annually is to be amortized by semi-annual payments over the next 5 years. The first due is 6 months. Determine the semi-annual payments. Solution: Given:

P 10,000r 10%m 2 (semi annually)

r 0.10i 0.05m 2

n mt 2(5) 10A ?

===

= = == = ==

Substitute:

( )10

10

1.05 110,000 A

0.05(1.05)

A 1,295.05

= =

21. If you borrowed money from your friend with simple interest of 12%, find the present worth of P50,000.00 which is due at the end of 7 months. Solution: Given:

r 12%0.12i360

F 50,000n 7(30) 210 days

==== =

Substitute to the formula:

0.1250,000 P 1 210360

P 46,728.97

= + =


( )( )n

n1 i 1

P Ai 1 i

+ = +

Where:



====

Formula: (Ordinary Simple Interest) ( )F P 1 in= + 1month 30 days= Where:

P present worthi rate of int erest / dayn no. of int erest periods

===


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22. The amount of P50,000.00 was deposited in the bank earning an interest of 7.5% per annum. Determine the total amount at the end of 5years if the principal and interest were not withdrawn during the period. Solution: Given:

P 50,000r 7.5%m 1 (per annum)t 5 yearsF ?

=====

Substitute:

( )5F 50,000 1 0.075F 71,781.47= +=

23. What is the corresponding effective rate of 18% compounded semi- quarterly? Solution: Given:

r 18%m 8 (semi quarterly)==

Substitute:

80.18ER 1 18

ER 0.1948ER 19.48%

= + ==




=====


mrER 1 1

m = +

Where:

r nominal ratem mode of compoundingt no. of years

===


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24. A telephone company purchased a microwave radio equipment for P6 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 depreciated cost during the 5th year using SYD. Solution:

oo

C 6M 0.04(6M)C 6.24M

= +=

oSV 0.08(C )

SV 0.08(6.24M)SV 499,200

===

( )( )

nSYD n 1210SYD 10 1 552

= +

= + =

Solving for the depreciation during the 5th year:

( ) ( )55

10 5 1d 6,240,000 499,200

55d 626,269.10

+= =

25. In how many years is required for P2,000 to increase by P3,000 if interest at 12% is compounded semi-annually? Solution: Given:

P 2000 ; F 5000r 12% ; m 2 (semi annually)= == =

Substitute:

( )

2t

2t

0.125000 2000 12

2.5 1.06ln2.5 2t(ln1.06)t 7.86 (say 8 years)

= + ==

=

Formula :Depreciation (SYD Method)

( )m n m 1d FC SV SYD +=

Where: F = First Cost SV = Salvage Value n = life of the property in years m = number of years used before n




=====


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26. A VOM has a current selling price of P400. If its selling price is expected to decline at a rate of 10% per annum due to obsolence, what will be its selling price after 5 years? Solution: Given:

oC 400 ; k 10%m 5 years

= ==

Substitute:

( )5mm

C 400 1 0.1C 236.20

= =

27. Find the nominal rate which if converted quarterly could be used instead of 12% compounded semi-annually. Solution: Given: r 12% ; m 2 (semi annually)= = Let: x = the unknown nominal rate xm 4= (mode of compounding of x) Equate Effective rates of interest:

quarterly semi annually

4 2

ER ER

x 0.121 1 1 14 2

x 0.1183x 11.83%

= + = +

==

28. Find the present worth of a future payment of a P100,000 to be made in 10 years with an interest of 12% compounded quarterly. Solution: See Formula in Problem 25:

4(10)0.12100,000 P 14

P 30,655.68

= = + =

Formula - Matesons Formula

( )mm oC C 1 k= Where: mC =Book Value at the

end of m years k = rate of depreciation oC = first cost



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29. What nominal rate, compounded semi-annually, yields the same amount as 16% compounded quarterly? Solution: Given:

r 16%m 4 (quarterly)==

Let: x = the unknown nominal rate xm = (mode of compounding of x) Equate Effective rates of interest:

semi annually quarterly

2 4

ER ER

x 0.161 1 1 12 4

x 0.1632x 16.32%

= + = +

==

30. What rate of interest compounded annually is the same as the rate of interest of 8% compounded quarterly? Solution: Equate Effective rates of interest:

( )annually quarterly

4

ER ER

0.081 x 1 1 14

x 0.0824x 8.24%

= + = +

==



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31. You loan from a loan firm an amount of P100,000 with the rate of simple interest of 20%, but interest was deducted from the loan at the time the money was borrowed. If at the end of one year you have to pay the full amount of P100,000, what is the actual rate of interest? Solution: Given: n 1 year= P 100,000 20%= advance interest

P 100,000 0.20(100,000)P 80,000F 100,000 (to be paid at the end of 1 year)

= ==

From: ( )F P 1 in= + Substitute:

[ ]100,000 80,000 1 i(1)

i 0.25i 25%

= +==

Alternate Solution: Think of the 20% advance interest as a rate of discount: Relationship between rate of interest i and rate of discount d:

di1 d

= Then,

0.20i1 0.20

i 0.25i 25%

= ==


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32. 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 a loan period? Solution: Given:

P 5000n 15(30) 450days

0.15i360

== ==

Substitute:

0.15F 5000 1 450360

F 5,937.5

= + =

33. Mr. J. Reyes borrowed money from a bank. He received from the bank P1,842 and promise to repay P2,000 at the end of 10 months. Determine the simple interest. Solution:

Given:

( )P 1842F 2000n 10 30 300daysi ?

=== ==

See formula in problem 32:

i2000 1842 1 300360

i 0.1029i 10.29%

= + ==



===


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34. What is the effective rate corresponding to 18% compounded daily?

Take 1 year is equal to 360 days. Solution:

Given:

r 18%m 360 (daily)==

Substitute:

3600.18ER 1 1360

ER 0.1972ER 19.72%

= + ==

35. Find the annual payment to extinguish a debt of P 10,000 payable for 6

years at 12% interest annually. Solution:

Given:

P 10,000t 6 yearsr 12%m 1 (annually)n mt (1)(6) 6

r 0.12i 0.12m 1

===== = == = =

Substitute:

( )

( )6

6

1 0.12 110,000 A

0.12 1 0.12

A 2,432.257

+ = + =


mrER 1 1

m = +

Where:

r nominal ratem mode of compoundingt no. of years

===


( )( )n

n1 i 1

P Ai 1 i

+ = +

Where:



====


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36. What annuity is required over 12 years to equate with a future amount of P 20,000? Assume i 6%= annually.

Solution:

Given:

F 20,000t 12 yearsr 6%m 1 (annually)

r 0.06i 0.06m 1

n mt 12A ?

====

= = == ==

Substitute:

( )121 0.06 120,000 A

0.06

A 1,185.54

+ = =

37. Find the present worth of a future payment of 80,000 to be made in six

years with an interest of 12% compounded annually. Solution:

Given:

F 80,000t 6 yearsr 12%m 1 (annually)P ?

=====

Substitute:

( )680,000 P 1 0.12P 40,530.48= +=


( )n1 i 1F Ai

+ =

Where:

A periodic paymentF Future worthn number of payments


====




=====


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38. The amount of P 20,000 was deposited in a bank earning an interest 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.

Solution:

Given:

P 20,000r 6.5%m 1 (per annum)t 7 yearsF ?

=====

Substitute:

( )7F 20,000 1 0.065F 31,079.7= +=

39. Today a businessman borrowed money to be paid in 10 equal payments

for 10 quarters. If the interest rate is 10% compounded quarterly and the quarterly payment is 2,000 pesos, how much did he borrowed?

Solution: Given:

r 10%m 4 (quarterly)

r 0.10i 0.025m 4

n 10 equal paymentsA 2000P ?

==

= = ====

Substitute:

( )

( )10

10

1 0.025 1P 2000

0.025 1 0.025

P 17,504.13

+ = + =




=====


( )( )n

n1 i 1

P Ai 1 i

+ = +

Where:



====


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40. What is the present worth of a P 500 annuity starting at the end of the third year and continuing to the end of fourth year, if the annual interest rate is 10%.

Solution:

Given:

A 500m 2n 2r 10% (annually)

====

Substitute:

( )

( )2

2 2

1 0.10 1P 500

0.10 1 0.10

P 717.16

+ + = +

=

Alternate Solution:

( )( )( )

1 n

1 3

2 4

FP1 i500P 375.66

1.10500P 341.51

1.10P 375.66 341.51

717.17

= += =

= =

= +=

Formula: (Present worth of deferred Annuity)

( )( )n

m n

1 i 1P A

i 1 i + + = +

Where:

A periodic paymentP present worthn number of paymentsm=number of periods before

the beginning of the firstpayment

===

0 1 2 3 4

500 500 1P

2P


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41. A copying machine has a useful life of 3 years and a salvage value of P

20,000 was bought for P 135,000. If the owner decides to sell after using it for 2 years, how much should the selling price be so that he will not lose or gain if the interest is 5%. (Hint: apply the Sinking Fund method).

Solution: Selling Price = Book Value after n years

From Sinking Fund Formula:

( )( )o

L

C SVd

1 i 1

= + annual depreciation cost Where: d = annual depreciation cost oC = first cost or original cost SV = salvage value at the end of useful life L = life Solving for d:

( ) ( )

( )3135,000 20,000 0.05

d1.05 1

d 36,479

= =

Total depreciation at the end of 2 years:

( )

( )

n

2

2

2

2

1 i 1D d

i

1.05 1D 36,479

0.05

D 74,782

+ = =

=

Solving for the Book Value after 2 years:

2 o 2BV C D

135,000 74,78260,218 selling price

= = =


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42. A loan for P 50,000 is to be paid in 3 years at the amount of P65, 000. What is the effective rate of money?

Solution: Given:

P 50,000F 65,000n 3 years

===

From:

I Pin65,000 50,000 50,000i(3)

i 0.10i 10%

= =

==

43. Today an investor withdraws P50,000.00 representing the accrued

amount of his investment that matured. If he invested at 10% compounded semi-annually for 10 years, how much did he invest in pesos?

Solution:

Given:

F 50,000r 10%m 2 (semi annually)t 10 years

=== =

Substitute:

2(10)0.1050,000 P 12

P 18,844.47

= + =




=====


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44. A man invested part of P 20,000 at 18% and the rest at 16%. The annual

income from 16% investment was P 620 less than three times the annual income from 18% investment. How much did he invest at 18%?

Solution:

Let: x = amount invested at 18% interest 20,000 x = the amount invested at 16% interest Then,

( ) ( )0.16 20,000 x 3 0.18x 620

3200 0.16x 0.54x 620x 5,457.14

= =

=

45. What is the accumulated amount after three (3) years of P 6,500

invested at the rate of 12% per year compounded semi-annually? Solution:

Given:

P 6,500t 3 yearsr 12%m 2 (semi annually)F ?

==== =

Substitute:

2(3)0.12F 6,500 12

F 9220.37

= + =




=====


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46. If the authorize capital stock of corporation is P 2,000,000 how much must the paid-up capital be?

Solution:

Formula:

Authorized Capital Stock 16(Paid upCapital)= Hence,

2,000,000Paid up Capital

16125,000

==

47. In how many years will the amount of P 10,000 triple if invested at an

interest rate of 10% compounded per year? Solution:

Given:

P 10,000F 3P 30000r 10%m 1 (per year)t ?

== ====

Substitute:

( )( )( )

t

t

30,000 10,000 1 0.10

3 1.10

ln3 t ln1.10 taking ln both sidest 11.53

= +== =




=====


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48. What is the amount of an annuity of P5000 per year at the end of each

year for 7 years at 5% interest compounded annually? Solution:

( ) ( )n 71 i 1 1 0.05 1F A 5000

i 0.05

F 40,710

+ + = = =

50. What is the book value of an equipment purchased three years ago for

P15,000 if it is depreciated using the sum of the years digit method? The expected life is 5 years.

Solution:

( ) ( ) ( )

( )( )( )( )

1 o n

1

2

3

n

Using SYD method :n n 1 5 6

year 152 2

nd C Cyear

5d 15,000 0 5000154d 15,000 0 4000

153d 15,000 0 3000

15Total depreciation :D 5000 4000 3000

12,000Book Value 15,000 12,000Book Value 3,000

+= = = =

= =

= =

= =

= + +=

= =


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51. The parents planned for their son to receive P50,000 ten years from now. What amount in pesos should they invest now if it will earn interest of 12% compounded annually for the first five years and 15% compounded quarterly during the next five years?

Solution: For the first 5 years:

( )

( )5

5

55

F P 1 0.12

F 1.12 P

= +=

For the next 5 years:

( )

mt

10 5

4(5)5

rF F 1m

0.1550,000 1.12 P 14

P 13,586.82

= + = +

=

52. A customer buys an electric fan from a store that charges P1,500 at the

end of 90 days. The customer wishes to pay cash. What is the cash price if the money is worth 10% simple interest?

Solution:

F 1,500i 10%P ?

===

0.101500 P 1 90360

P 1,463.41

= + =

10F 50,000=

5

5F

P

10 0



===


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53. In how many years will P1000 double if interest is 10% compounded

quarterly? Solution: Given:

P 1000F 2P 2000r 10%m 4 (quarterly)t ?

== ====

Substitute:

( )

4t

4t

0.102000 1000 14

2 1.025ln2 4t(ln1.025)

t 7 years

= + ===

54. Mr. Reyes borrowed money from a bank. He received from the bank

P1842 and promised to pay P2,000 at the end of 10 months. Determine the simple interest rate.

Solution:

Given:

P 1842F 2000n 10(30) 300 days

=== =

Substitute:

( )i2000 1842 1 300360

i 0.1029i 10.29%

= + ==




=====



===


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55. A man borrows P15,000 from Hong Kong Bank. The rate of simple interest is 12%, but the interest is to be deducted from the loan at the time the money is borrowed. At the end of one year he has to pay back P15,000. What is the actual rate of interest?

Solution:

Given:

P 15,000 12%(15,000)P 13,200F 15,000n 360 days

= ===

Substitute:

i15,000 13,200 1 360360

i 0.1364i 13.64%

= + ==

56. In how many years will it take money to quadruple if invested at 8%

compounded semi-annually? Solution:

Let: P = present worth

F 3P (quadruple)r 8% ; m 2 (semi annually)== =

Substitute:

( )

2t

2t

0.084P P 12

4 1.04ln4 2t(ln1.04)t 17.67 years

= + ==

=



===




=====


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57. A brand new machine is estimated to have a salvage value of P 10,000

after ten years and a book value of P 30,000 after 5 years, what is the initial cost of the machine? (Assume straight line depreciation).

Solution: Given:

SV 10,000BV 30,000L 10 yearsn 5 years

==

==

Substitute:

( )oo

oo

oo

o

o

C SVBV C n

L

C 10,00030,000 C 510

C30,000 C 5,0002

C25,0002

C 50,000

= =

= +

==

58. A machine was brought for P 50,000 with an estimated useful life of 10

years and a salvage value of P 5,000. What is its annual depreciation assuming straight line trend?

Solution:

Substitute:

50,000 5,000d

10d 4,500

==

Formula: Depreciation (Straight Line Method)-Book Value

oo n nC SVBV C D ; D n

L = =

Where:

o

n

BV Book ValueSV Salvage ValueC first cos tD total depreciation after n yearsL lifen no. of years before L

====

==

Formula: Depreciation (Straight Line Method)-Book Value

oC SVdL=

Where:

o

d annual depreciationSV Salvage ValueC first cos tL life

===

=


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59. What annuity is required over 10 years to equate with a future amount of P20,000. Assume i = 8% per year.

Solution:

Given:

F 20,000r 8%m 1A ?

====

Substitute:

( )101 .08 120,000 A

0.08

A 1,380.59

+ = =

60. A merchant loaned P 500,000 payable in 10 years at an interest rate of

12 percent compounded annually. What is the monthly amortization for 10 years?

Solution: Solving for the interest per month:

( ) ( )

( )( ) ( )

12 1

120

120

1 i 1 1.12 1i 0.009489 or 0.9489%

A 1.009489 1500,000

0.009489 1.009489A 6997.43

+ = =

=

=


( )n1 i 1F Ai

+ =

Where:

A periodic paymentF Future worthn number of payments

===


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61. 10 years ago a businessman purchased a machine for P50,000 with an

expected life of 20 years based on a straight line depreciation. Today, he decided to replace it with a modern one that costs P120,000. If the salvage value of the old unit is P20,000, how much more will he raise to buy the new machine?

Solution:

Given:

oC 50,000

SV 20,000L 20n 10

==

==

Let: x = amount needed to buy the new equipment Solving for the Book Value:

( )( )

oo

C SVBV C n

L

50,000 20,000BV 50,000 10

20BV 45,000

= =

=

Solving for x:

ox C BV

x 120,000 45,000x 75,000

= = =


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62. A heavy duty copying machine was procured for P100, 000 with an estimated salvage value of P10, 000 after 10 years. What is the book value after 5 years? (Assume straight line depreciation).

Solution:

Given:

oC 100,000SV 10,000L 10 yearsn 5 years

==

==

Substitute:

( )( )

o5 o

5

5

C SVBV C n

L

100,000 10,000BV 100,000 5

10BV 55,000

= =

=

63. Based on its purchased price a machine is expected to depreciate at a

uniform rate of 18 percent annually until it has zero salvage value. Approximate the useful life of the machine in years using the SYD method?

Solution:

Given:

( ) ( )0 o

o 0

n0.18C C 0 n n 12

20.18C Cn 1

0.18n 0.18 2n 10

= + = +

+ ==


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64. A man deposited P5, 000 on the date his son celebrated his 1st birthday.

If the money is worth 10% compounded semi-annually, what is the maximum amount the son can withdraw on his 21st birthday?

Solution:

P 5000t 20 yearsr 10%, semi annually

===

Solving for the future worth, F:

( )n2(20)

F P 1 i

0.10F 5000 12

F P35,200

= + = +

=

65. What will be the value of P 6,000 four and one-half years from now if

invested at 15% compounded quarterly?

Solution:

P 6,000r 15%, quarterlyt 4.5 yearsF ?

====

Solving for F:

4.5(4)0.15F 6000 14

F P11,640

= + =

Documents

Chap 7 Eng'g Economy