Engineering Economy Lecture b

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ENGINEERING ECONOMY

ENGINEERING ECONOMY

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Preview Problems

Preview Problems

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 What will be the equivalent amount at the end of What will be the equivalent amount at the end of five years of a five years of a uniform 5 yearly deposituniform 5 yearly deposit

of

of P5,000. P5,000. if tif the he nominal nominal annual intereannual interest st rate rate is 12% is 12% compounded compounded monthly?monthly?



 Determine the Determine the present present value of value of 10 sem10 semi-annual pai-annual payments of yments of P10,000 P10,000 , , the the first offirst of

which to be paid

which to be paid 2 years 2 years from nofrom noww. Mone. Money is wy is worth 12% pa compounded semi-orth 12% pa compounded semi-annually.

annually.



 A bond issue of P100,000 redeemable at par in 10A bond issue of P100,000 redeemable at par in 10-years, in P1,000 units paying10%-years, in P1,000 units paying10%

interest per annum payments,

interest per annum payments, must be retired by the use of sinking fund that earmust be retired by the use of sinking fund that earns 8%ns 8% pa.

pa. What is What is the total the total annual expense?annual expense?



 An economAn economy is experiencing inflation y is experiencing inflation at an annual rate of 7%. If at an annual rate of 7%. If the market interethe market interest ratest rate

is also

is also 5% per annum, wha5% per annum, what will be the real value of P500 two yet will be the real value of P500 two years from now ?ars from now ?



 A company purchased A company purchased an equipment for P 110,000. an equipment for P 110,000. It is estimated that it will have aIt is estimated that it will have a

useful life of 10 years.

useful life of 10 years. The scrap value of the equipment is P10,000,The scrap value of the equipment is P10,000, a.

a. Compute the book Compute the book value of the equipmvalue of the equipment at the start of the 6th year usingent at the start of the 6th year using declining-balance method.

declining-balance method.  b

 b. . Compute the book Compute the book value at the end of value at the end of the 5the 5thth _{year using Sinking Fundyear using Sinking Fund}

method, i=10% pa method, i=10% pa c.

c. Compute the Compute the book value book value at the at the start start of the of the 6th year 6th year using SYDMusing SYDM

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ENGINEERING ECONOMY

ENGINEERING ECONOMY





After completing this course,

After completing this cour

se, the student must be able to:

the student must be able to:

1.

1. Solve problems

Solve problems involving interest and

involving interest and the time

the time value of

value of

money;

money;

2.

2. Evalua

Evaluate project

te project alternatives b

alternatives by appl

y applying engineering

ying engineering

economic principles and methods and select the most

economic principles and methods and select the most

economically efficient one; and

economically efficient one; and

3.

3. Deal with ri

Deal with risk and uncer

sk and uncertainty in project outcomes by

tainty in project outcomes by

applying the basic economic

applying the basic economic decision making concepts.

decision making concepts.

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Course Objectives

Course Objectives

Course Outline

 1. Introduction

 2. Money-Time Relationship

 2.1. Time Value of Money

 2.2. Types of Interest  2.3. Inflation/Deflation  2.4. Annuity  3. Depreciation  4. Capital Investments  5. Operational Costs  6. Accounting Fundamentals

 7. Basic Methods of Profitability Analysis

 8. Methods of Financial Analysis

 9. Optimization

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 Plant Design and Economics for Chemical Engineers, Max S. Peters and Klaus D.

Timmerhaus, 4th _{Ed 2001}

 Process Engineering Economics, Schweyer

 Engineering Economy, de Garmo, Sullivan, Canada, 7th Ed  Engineering Economy, Arreola, 2nd Ed

 Engineering Economy Sta. Maria, 3rd Ed

 Engineering Economy, Sullivan, Bentadilli and Wichs, 12th Ed 2003  Contemporary Engineering Economics, Chan S. Park, 3rd Ed 2002

 Engineering Economy by L.T. Blank and A.J. Tarquin, 6th ed., McGraw Hill, 2005.  Engineering Economy by W.G. Sullivan, E.M. Wicks, and J.T. Luxhoj, 13th ed.,

Prentice Hall, 2006.

 Contemporary Engineering Economics by Chan S. Park, 4th ed., Prentice Hall,

2007.

 Excel for Engineering Economics by R.W. Larsen and Chan S. Park, Prentice Hall,

2003.

 Engineering Economy and the Decision Making Process by J.C. Hartman,

Prentice Hall, 2007.

 Engineering Economic Analysis by D. G. Newman, T. G. Eschenbach, and J. P.

Larelle, 9th ed., Oxford University Press, 2004.

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

Engineering Economy _ study of

economic theories

and

their applications to engineering problems with the concept

of obtaining maximum benefit at the least cost.

 “Economics

 _

is the study of scarcity. Resources are limited, and every society wants to figure out how to allocate its resources for maximum benefit.“_ 

 Jodie Beggs, PhD (Harvard U)

1. INTRODUCTION

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Reasons for Studying Engineering Economy:



Engineering designs and operations must be equated with

costs for practical applications



Engineers evolve into managers of their own or other

enterprises

Uses of Engineering Economy



Application on various fields of engineering



Determining of limiting factors



Tool in selection of alternates



Investment of capital



Tool in decision making

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Economists study topics such as:

 How prices and quantities of items are determined in market

economies

 How much value markets create for society

 How taxes and regulation affect economic value

 Why some goods and services are under-supplied in a market

economy

 How firms compete and maximize profit

 How households decide what to consume, how much to save, and

how much to work (or, more generally, how people respond to incentives)

 Why some economies grow faster than others

 What effect monetary and fiscal policy has on economic

well- being

 How interest rates are determined

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

It is NOT an economist’s job to tell

people what stocks and bonds they

should be investing in.

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SELECTION AND EQUIVALENCE IN PRESENT ECONOMY

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

Present economy involves the analysis of problems for

manufacturing a product or rendering a service upon the

 basis of present or immediate costs. It is highlighted when the

effects of time such as interest and depreciation are

negligible. Present economy is employed when the

alternatives to be compared will provide the same result and

the period involved in the study is relatively short.



When alternatives for accomplishing a specific task are being

compared over one year or least-and influence of time on

money can be ignored, engineering economic analyses are

referred to as present economy studies.

Present economy studies occur in the following situations:

 a. Selection of material_ In many cases, economic selection among materials cannot be

 based solely on the costs of materials. Frequently, a change in materials will affect the design and processing costs, and shipping costs may also be altered.

  b. Selection of method to be used_ In mechanical or chemical operations a product

may be made by two or more methods giving equivalent results. Some goods may be delivered by various methods such as using different capacity trucks, and the results would still be the same regardless of the truck used. These are but a few of the

examples that may be cited to show that certain operations are capable of being done  by two or more methods.

 c. Selection of design_ In the design of a machine to produce a certain product, the

engineer responsible for the work will usually make as many designs as possible and from which, by a process of elimination, he will select the design best fitted for the work to be done with particular care being given to the one which will do the work with most economy.

 d. Selection location or site for a project_ In the choice of a factory site many factors

are often considered such as the cost of the land, the cost of construction in the different sites, and the difference in transportation cost, and many other factors.

SELECTION AND EQUIVALENCE IN PRESENT ECONOMY

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 e. Comparison of proficiency among workers_ In industrial operations

where the efficiency of the workers is a factor affecting costs, it is usually observed that workers have varying efficiencies. In some occupations only those with better average proficiencies are acceptable. In teaching for example, other factors being equal, preference should be given to those teachers who did better than average in their studies.

 f. Economy of tool and equipment maintenance_ In many activities,

tools have to be sharpened from time to time, and equipment have to be kept in good operating condition all the time. In certain cases,

experience will show the best time to perform certain operations to maintain equipment at the optimum operating efficiency.

 g. Economy of number of laborers_ In certain industrial operations it is

observed that a certain number of workers cooperating on a certain

phase of work will lead to the highest efficiency. An increase beyond this number will usually cause the taking into effect of the law of diminishing returns. In certain cases, the excess of laborers will result in some

laborers not working at certain times while waiting for the work of other laborers to be finished.

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 A machine part to be machined may be made either from an alloy of 

aluminum or steel. There is an order for 8000 units. Steel costs P380/kg, while aluminum costs P870/kg. If steel is used, the steel per unit weighs 110 grams; for aluminum, 30 grams. When steel is used, 50 units can be produced per hour; for aluminum, 80 units per hour with the aid of a tool costing P64,000, which will be useless after 8,000 units are finished. The cost of the machine and the operator is P1080 per hour. If all other costs are identical, determine which material will be more economical.

 Solution:

Steel Aluminum

Material Cost 0.11kg (P380/kg)= P41.80 0.030 kg (P870) = P26.10 Labor and Machine P1,080/50 = P 21.60 P1,080/80 = P13.50

Tool P64,000/8,000 = P8.00

Cost per piece P63.40 P47.60

Answer: Aluminum is cheaper!  NTAldon

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Decrease in materials 1,000,000 (0.12) (P12) = P 1,440,000 Increase in processing

costs

1,000,000 (0.02) (P4) = P 80,000

 Net Savings P 1,360,000

• A company manufactures 1 million units of a product

annually. A new design of the product will reduce material cost by 12%, but will increase processing cost by 2%. If materials cost is P 12/unit and processing will cost P4/unit, how much can the company afford to pay for the preparation of the new design and making changes in equipment?

Solution:

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

_{Time affects the cost of money.}

A peso now is worth more than a peso a year from now because it can

earn interest during the year.



_{Interest represents two things}

_:

1.T

he compensation paid for the use of the borrowed capital 

2

. The risk taken in making the loan.



_{The rate at which interest will be paid is usually fixed at the time the}

capital is borrowed, and a guarantee is made to return the capital at

some set of time in the future or on an agreed-upon pay-off schedule.



_{Riskier loans require more interest to make them attractive.}

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Time Value of Money

 The concept of interest goes back to earliest recorded history.

 _{Babylon 2000 B.C. – money paid for use of grain that was borrowed.}  _{Typical rates were 6 to 25% per annum.}

 Usury is prohibited in the Law of Moses, and in Islamic cultures.

 In the middle ages, interest on loans was prohibited based on these

restrictions.

 In 1536, John Calvin adopted a theory of what constituted usury that

allowed interest.

 Islamic conventions developed to allow those with money to buy a stake in a

 business in return for an portion of the profit from that business.

 Interest and the cost of capital have become an essential part of doing

 business.

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INTEREST DEFINED

 Interest  __ the time value of money

 __ money paid for the use of money

 __ compensation paid for the use of borrowed capital.

 Elements of Interest

1. Principal, P = the sum of money lent or borrowed

2. Interest, I  = the price paid or charge made for the use

of money

3. Time, n = the period of time during which interest is charged, measured in some specific unit. The unit may be day, week, month, 3 months, 6 months, or a year.

4. Rate, i = the price paid for the use of money for a unit of time. It is given as a percentage of the original amount.

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CASH FLOW

Cash Flow _ is a systematic presentation of cash

receipts and disbursements for a given operating

period

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Cashflow Diagram: Investment Transaction

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Cashflow Diagram: Loan Transaction

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 A. Simple Interest:The interest is proportional to the original amount of the loan

Requires compensation payment at a constant interest rate based only on the orig inal amount

The principal Pmust be repaid eventually; therefore the entire amount F , of principal plus simple interest due after nperiods is;

In the payment of simple interest, it makesno difference whether interest is paid at the end of each time unit or after any number of time units. The same total amount of money is paid during a given length of time, no matter which method is used. Under these conditions, there is no incentive to pay the interest until the end of the total loan period.

1. Ordinary Simple Interest

The time unit used to determine the number of interest period is usually one year, and the interest rate is expressed on a yearly basis. When an interest period of less than one year is involved, the ordinary way to determine the simple interest is to assume the year consists of twelve 30-daymonths, or360 days.

2. Exact Simple Interest

The exact method accounts for the fact that there are 365 days in a nor mal year and 366 days in a leap year.

Types of Interest

366 ) )( )( ( ; 365 ) )( )( (  P  i d  or  d  i  P   Pin  I   

   

12 360 ) )( )( ( P  i d   P  i m  Pin  I     Pin  I   I   F   P   NTAldon 23

Example:

Determine the ordinary simple interest on P1000 for 8 months and 15 days if the rate of interest is 15%.

Solution:

For ordinary simple interest, it is assumed that 1 year = 12 months; n= 8.5 months P _{= 1000} i _{= 0.15/12}

  

106

.

25

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5

.

8

15

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0

1000

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Jan 1-31 31 August 31 February 29 September 30 March 31 October 15 April 30 May 31 June 30 July 31 213 76 289 days Example:

Determine the exact simple interest on P1000 for the period January 1 to October 15, 2012 if the rate of interest is 15% pa.

Solution: 2012 is a leap year = 366 days



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44

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118

366

289

15

.

0

1000

 P 

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366

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3. Discounted Interest,

i 

_d 

The interest for the money borrowed (discount) is deducted from the principal in advance.

   

 P 

i

n

 I 

_d 



_d 

P = amount loaned

P’= actual amount received

I _d = interest payment deducted in advance

Effective Interest Rate,i 

' 1  P   I  i i i d  d  d 







 P   I  i i i  I   P   P  d  d  d 







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1 ' NTAldon 26

Example:

Man borrowed P10,000 from a bank and agreed to pay the loan at the end of 1 year. The bank discounted the loan and gave him P8,000

1. What was the discounted rate?

2. What was the effective rate of interest?

 pa  P   I  i_d  d  20% 000 , 10 000 , 8 000 , 10









 pa  P   I  i d  25% 000 , 8 000 , 8 000 , 10 '



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 pa i i i d  d  _{0.25 25%} 2 . 0 1 20 . 0 1



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Compound Interest: The interest is proportional to the balance at any point in time

Stipulates that interest is due regularly at the end of each interest period. If

payment is not made, the amount due is added to the principal, and the interest is charged on this converted principal during the following unit time.

Thus an initial loan of P at an annual interest rate i would require payment of Pi

as interest at the end of the first year. If this payment were not made, the interest for the second year would be ( P + Pi ) i and the total amount due for interest after 2 years would be : i  Pi  P   Pi  I _T  ( ) 2







Therefore, the total amount of principal plus interest after 2 years equals

 

2 2  P   Pi ( P   Pi)i  P 1 i  F 

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1

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1           n n n n n i  P   P  i  P   P   F   I  i  P   F  Future Worth NTAldon 28

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1. Nominal Interest Rates, r 

Interests are compounded other than on annual basis. It always includes a qualifying statement indicating the compounding period. Example, 12% per annum compounded quarterly.

2. Effective Interest Rates ,

i 

Are always compounded on an annual basis.

Conversion of 

Nominal Interest rate ,

r 

to Effective Interest Rate,

i 

:

where: r  = nominal interest rate per year

m = number of interest periods per year

1 1



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 

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m m r  i NTAldon 30

1

4

1

4

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r 

i

Compounded Quarterly

(every three months)

Compounded Monthly

1

12

1

12

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r 

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1

2

1

2

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r 

i

Compounded Semi-annually (every six months)

Compounding Periods

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





i



e i  P   Pe  F  rn  P   F   Ln dn r   F  dF  rF  dn dF  r  n rn n n  F   P         





1 1 0

Continuous Interest

The concept of continuous interest is that the cost or income due to interest flows regularly.

If we let the change in the accumulated amount, F  over the change in the unit period, n a function of both the accumulated amount

F , and rate, r , then;

1





r 

e

i

Conversion of continuous interest rate to effective interest rate

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 

n ;

F 

F uture WorthF actor

= 1+i

(F /P,i%,n)

P



Example:

A person deposits P100,000 in the bank.

a. How much would be his money in the bank after 5 years if interest is 6% per annum compounded annually?





n





5

F =P 1+i =100,000 1.06 =P 133,822.56

 b. How much would be his money in the bank after 5 years if interest is 6% per annum compounded monthly?





   5 12 n   0.06 F =P 1+i =100,000 1+ =P 134,885.02 12













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CASHFLOW-Annual

1 year 2 years 3 years 4 years 5 years

74,726) x (1.06) 79,209.56 x (1.06) 83,962.13 x (1.06) 88,999.86 x (1.06) 94,339.85 x (1.06) = 79,209.56 = 83,962.13 = 88,999.86 = 94,339.85 = 100,000 SAMPLE PROBLEMS:

A person wishes to accumulate P100,000 in the bank after 5 years.

a. How much is he going to deposit now assuming that his money earns 6% per annum ?

   5 100,000 74,726 1 n 1 0.06  F   P P  i        1 ; ( / , %, ) 1   n  P   PresentWorth Factor P F i n  F  _i NTAldon 34

SAMPLE PROBLEMS:

A person wishes to accumulate P100,000 in the bank after 5 years.

    5 12 100,000 74,137.22 1   0.06 1 12 n  F   P P  i     _{ }     

Alternative solution (compute first,i )

12 0.06 1 1 0.061677 6.1677% 12 i    _ _          5 100,000 74,137.50 1 n 1 0.061677  F   P P  i      1 m r  i m    _{ }  

 b. How much is he going to deposit now assuming that his money earns 6% per annum ? compounded monthly

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Example:

Find the nominal interest compounded monthly which is

equivalent to 12% pa compounded quarterly?

Solution:

monthly compounded   pa r  r  m r  i m % 88 . 11 1 4 12 . 0 1 1 12 1 1 1 4 12                                         NTAldon 36

INFLATION or DEFLATION:

 f 

I nflation

is the increase in the price of goods and services from one year to the other, thus decreasing the purchasing power of money.

Deflation

_{involves a decrease in the average price of goods and services}

resulting to the increase in the purchasing power of money. It is usually associated with a prolonged erosion of economic activity and high

unemployment.

Some measures of price changes in our economy are the Consumer Price

Index (CPI) and Producer Price Index (PPI).

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Consumer Price Index

(CPI)

Consumer price index (CPI) measures changes in the price level of  consumer goods and services purchased by households.

The CPI is a statistical estimate constructed using the prices of a sample of  representative items whose prices are collected periodically. Sub-indexes and sub-sub-indexes are computed for different categories and sub-categories of goods and services, being combined to produce the overall index with weights reflecting their shares in the total of the consumer expenditures covered by the index. It is one of several price indices calculated by most national statistical agencies.

The annual percentage change in a CPI is used as a measure of inflation. A CPI can be used to index (i.e., adjust for the effect of inflation) the real value of  wages, salaries, pensions, for regulating prices and for deflating monetary magnitudes to show changes in real values.



100%



1 1  





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n n n  Index  Index  Index  f   Rate Change  Annual  NTAldon 38

Producer Price Index

(PPI)

Producer Price Index (PPI)_ is a family of indexes that measure the

average change over time in the prices received by domestic producers of goods and services. PPIs measure price change from the perspective of the seller.

The headline PPI (for finished goods) is a measure of the average price level for a fixed basket of capital and consumer goods for prices received  by producers. The producer price index for finished goods is a major

indicator of commodity prices in the manufacturing sector. These prices are more sensitive to supply and demand pressures than the more

comprehensive consumer price index. Changes in the producer price index are considered a leading indicator for consumer price changes, although only a small portion of the PPI is directly connected to less than half of the CPI.

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Example:

An economy is experiencing inflation at an annual rate of 5%. If this

continues, what will P500 be worth two years from now in terms of today’s

pesos? Solution:

  

1 0.05



453.51 500 1







2





_n  f   F   P  Example:

An item presently costs P500. If inflation is at the rate of 5% per year, what will be the cost of the item in 2 years?

Solution:



1







₂





1





2



500



1



0

.

05



2



551

.

25



 P 

 f 

 F 

 P 

f 

 F 

n NTAldon 40

Combination of Price Cost Index and Sixteenth Rule: 6 0.  A  B  K  n  A  B S  S   I   I  C  C  _                    

Example: Six years ago, an 80-kw diesel electric generator costs P400,000. The cost index for this class of equipment six years ago was 187 and is now 194. Determine the cost of a 120 kw unit now?

267 529 80 120 187 194 000 400 6 0  ,  P   , S  S   I   I  C  C  .  X   A  B  K  n  A  B                                            Sixteenth Rule:

The cost of the equipment at a new capacity can be computed if the cost of  the same equipment is known at a given capacity. The cost adjustment is the ratio of the two capacities raised to the power 0.6

6 0.  A  B  A  B S  S  C  C  _           NTAldon 41

'

i

   

_

_

 f  i i









1 1 ' 1 ₁ 1 1 '









 f  i i

The real interest rate, i' would be the REAL value of your money’s actual

earnings, i after considering the loss in its purchasing power due to inflation, f .

Example

IF you P100 deposit in the bank earning 8% pa, what would be the real interest rate if inflation is 5% pa.

Solution:

 After a year your P100 will become

But after a year your P108 is only worth

due to inflation. Therefore the real earnings is only  Therefore, the real interest rate is

 

1.08 108 ( ) 100  ACTUAL  F 





 

1.05 102.86 ( ) 108 '  REAL  F   86 . 2 100 86 . 102   

Real Interest Rate,

% 86 . 2 0286 . 0 05 . 0 08 . 0 '       f   f  i i % 86 . 2 % 100 100 86 . 2    x NTAldon 42

DIFFERENTIAL PRICE ESCALATION OR DEESCALATION RATE:

 _ is a real price change in good or service caused by various

 factors in the market . It is the increment (%) of price change above

or below the general inflation rate  f, during a period (normally a year). The increase or decrease in price is in REAL pesos.

For example, if the unit price of the goods is P100 and after a year its price will be escalated or increased by 10% ( e _j ) , then its unit price will

becomeP110 (  ACTUAL increase in pesos). But during that time if inflation rate (f) is 5%, therefore the inflated cost should have been

P110/1.05=104.76, then the differential price escalation or increase in

REAL pesos is 104.76-100= P 4.76; e’ _j= 4.76% '  j

e





1 05 1 0.0476 4.76% 10 . 1 1 1 1 '

















 f  e e _j  j



1



1 1 '       f   e e  _j j



  f  



e e  _j j     1 1 1 ' NTAldon 43

Total PRICE ESCALATION: _ The price escalation or de-escalation rate is total rate (%) of price change in the unit price, or cost for a fixed amount during the period (normally) a year for  good or service. It is the sum of the general price inflation rate

and the differential inflation rate plus their product. The increase or decrease in price is in ACTUAL pesos.

 j

e



 f 



e

e

 _j





 _j





1

1

1

'

For example, if the unit price of the goods is P100 and after a year its price will be escalated or increased by 10% ( e _j’ ) , then its unit price will becomeP110. But if inflation projected for the  year of 5% is considered, then the actual unit selling price of the goods would be:











1 0.10



1 0.05



115.50 100 1 1 1 '  P   f  e  PC  e  PC   FC   _{j  j}



















NTAldon 44



It is a series of equal PAYMENTS,

A

occurring at equal time

intervals. Interest is paid on all accumulated amounts, and the

interest is compounded each payment period. The amount of an

ANNUITY

F,

is the sum of all payments

A,

plus i

nterest

if

allowed to accumulate at a definite rate of interest from the

initial payment up to the end of the annuity term.



An

annuity term, n

is the time from the beginning of the first

payment period to the end of the last payment period.



It must be noted that the sum

F

, is expressed at the end of the

last payment period

NTAldon

45

P

0 1 2 3 4 n

 A A A A A

F 

_n

 A. Ordinary Annuity 

The most common type of annuity. It involves the payment of 

amount,

 A

at the end of each interest period.

















_

_



i

i

 A

 F 

n n 1 1











_









_

_





i

i

i

 A

 P 

n n 1 1 1

Present Worth of Ordinary Annuity Future Worth of Ordinary Annuity

NTAldon

 P 

0 1 2 3 4 n

 A A A A A

F 

_n

B. Annuity Due

The uniform payments,

 A

are made at the beginning of  

each interest period.

   













_

_

_





i

i

i

 A

 F 

n n 1 1 1





   



_









_

_

_







i

i

i

i

 A

 P 

n n 1 1 1 1

Future Worth of Annuity Due

Present Worth of Annuity Due

NTAldon

P

m

0

1

2

3

4

n

0

1

2

3

 A

A

A

A

A

F 

_n

C. Deferred Annuity 

It is also an ordinary annuity but the payment of the first amount is deferred a cer tain number of periods after the first. For example, the first annuity payment could be made after 3 annuity terms instead of after the first annuity term.

















_

_



i i  A  F  n n

1

1











_









_

_





_ i i i  A  P  n m n 1 1 1

Future Worth of Deferred Annuity

Present Worth of Deferred Annuity

NTAldon

D. Continuous Cash Flow Continuous Compounding

Continuous flow of funds means a series of cash flows occurring at infinitesimally short intervals of time, corresponding to an annuity having an infinite number of short periods.

m  A  A



















_

_



i

i

 A

 F 

n n

1

1

m

r 

i

 

A= cash flow in a period

= sum of cash flows in a year m = number of periods in a year n = total number of periods

 A

NTAldon

Continuous Cash Flow Continuous Compounding…..                                                                               r  m r   A m r  m r  m  A  F  rn r  m m n n 1 1 1 1 As m approaches infinity, _e m r  r  m

















 

 



 

 



1



 

 



 

 

_



r 

e

 A

 F 

rn n

1

Where  _A is the sum of continuous cash flows in one year 



 

 



 

 

_



r 

e

e

 A

 P 

rn rn

1

NTAldon 50

Discrete Cash Flow -Continuous Compounding

































r 

e

e

 A

 P 

r 

e

 A

 F 

rn rn rn n

1

;

1





































1

1

;

1

1

r  rn rn r  rn n

e

e

e

 A

 P 

e

e

 A

 F 

Continuous Cash Flow- Continuous Compounding

Ordinary Annuity

 

 

 



_









_

_

















_

_



i

i

i

 A

 P 

i

i

 A

 F 

n n n n

1

1

1

;

1

1

NTAldon 51

Interest Factors

1. Single-Payment Future-Worth Factor; F = P (F /P, i%, n)

  

i  F/P,i%,n



 P 

 F _{ } n _

1

2. Single-Payment Present-Worth Factor; F = P (F /P, i%, n)

 

i



 P/F,i%,n



 F   P  n    1 1

3. Future Worth Annuity Factor; F = A (F /A, i%, n)

 

_

_

n i%,  F/A, i i  A  F _ 1 n 1 _

 

 

i i



 P/A,i%,n



i  A  P  n n      1 1 1

4. Present Worth Annuity Factor; F = A (P/A, i%, n)

NTAldon

Example:

1. A loan of one million pesos is to paid in 10 years at an interest rate of 8.5% pa. a. How much should the 10 annual payments be?

Solution:        1.085 1 152,407.71 085 . 0 085 . 1 000 , 000 , 1 1 1 1 10 _{10 }                   n _n i i i  P   A P 0 1 2 3 4 5 6 7 8 9 10 A A A A A A A A A A F

b. What would the balance be after the 6th _{payment ?}

 

 





13 . 226 , 499 085 . 0 1 085 . 0 1 71 . 407 , 152 ) 085 . 1 ( 000 , 000 , 1 1 1 1 6 6        _{ }         _{ }    i i  A i  P   Balance n n

2. If one million pesos is to be accumulated in 10 years at an interest rate of 8.5% pa, how much should the 10 annual payments be?

  1.085  1 67,407.70 085 . 0 000 , 000 , 1 1 1 10 _                 _n i i  F   A NTAldon 53

GRADIENT

1. Uniform Arithmetic Gradient

It is also an ordinary annuity but disbursement or

payment A increases or decreases by a uniform amount G each period.

P

0 1 2 3 n

 A

A+G

A+2G A

+(n-1)

G

F_An+F_Gn

Future Worth of Arithmetic Gradient,

G

: F _Gn

























n i i i G  F  n G_n 1 1

















_

_



i i  A  F  n  A_n 1 1 G  A n

 F 

 F 

 F 





NTAldon 54

GRADIENT

2. Geometric Gradient

It is also an ordinary annuity but disbursement or payment A increases or decreases by a uniform rate

 g each period

P

0 1 2 3

_n

 A A(1+g) A(1+g)2  A(1+g)n-1

F 

_n

   



























n n n

 g 

i

 g 

i

 A

 F 

1

1

Future Worth of Geometric Gradient,

G

: F 

n

When g=i





1

1







n n

nA

i

 F 

NTAldon 55

CASH FLOW A A+G A+2G 0 1 2 3 P1,000 P1,200 P1,400 F₃      3  3 1 1 1 1 1 0.10 1 ₂₀₀ 1 0.10 1 1000 3 3,930 0.10 0.10 0.10    _{ }   _{ }                 _{ }   _{ }                n An Gn n n n n  F F F  i G i  F A n i i i  F P  SAMPLE PROBLEM

1.What is the accumulated amount of a series of payments when the initial payment of P1000 increases by P200 each period, until the third period if cost of money is 10% pa. Required: F_n Solution: A= P1000 ,G =P200, and i =10%   2   1000 1.1 1200 1.1 1400 3,930     3

 Another way of solving this is to project individually the  Future Worth of each payment.

 F P 

NTAldon

2. What is the accumulated amount of a series of payments when the initial  payment of P1,000 increases by 15% each period, until the third period if

cost of money is 10% pa. Required: F_n Solution: A= P1,000 , g =15%, and i =10%

   



 



3 1 1 3 3 1 0.10 1 0.15 1, 000 3, 798 0.10 0.15  _{  }                  _{ }      n n  gn i g   F A i g   F P   g  CASH FLOW A A(1+g) A(1+g)2 0 1 2 3 P1000 P 1000(1.15)=1,150 P1000(1.15)2_{= P1,725} F₃   1.1 1150 1.1 1725 3,798 1000 2 3  P   F   payment. each of  Worth  Future the ly individual   project  to is it   solving  of  way  Another   g      NTAldon 57

BONDS

 _is a certificate of  indebtedness of a corporation

usually for a period not less than ten years and

guaranteed by a mortgage on certain assets of the

corporation or its subsidiaries.

Bonds are issued when there is a need for more

capital such as for expansion of the plant or the services

rendered by the corporation.

The face or par value of a bond is the amount stated

on the bond.

When the face value has been repaid, the bond is

said to have been retired or redeemed.

The bond rate is the interest quoted on the bond

.

NTAldon

1. Registered Bonds _ The name of the owner of this bond is

recorded on the record books of the corporation and interest

payments are sent to the owner periodically without any action

on his part.

2. Coupon Bonds _ Have a coupon attached to the bond for each

interest payment that will come due during the life of the bond.

The owner of the bond can collect the interest due by

surrendering the coupon to the offices of the corporation or at

specified banks.

NTAldon

59

BONDS

…



Equipment obligation bonds _ refer primarily to bonds whose

guarantee is a lien on equipment.



Registered bonds _ the owner's name is recorded in the books of the

corporation, and the interest is paid periodically to the owner

without their asking for it.



 Joint bonds _ bonds which are issued by two or more corporations



Par value of the bond or face value_ is the amount stated on the

 bond.



Bond rate _ is the rate of interest quoted on the bond.



Redemption or disposal price

 — 

usually equal to par value.



Mortgage bonds _ bonds whose security is mortgaged on certain

specified assets of the corporation.

NTAldon

BONDS…

 Debenture bonds _ bonds without security behind them except a

promise to pay by the issuing corporation

 Callable bond _entitles the issuer to pay off the principal prior to the

stated maturity date. Similarly, the owner of a putable bond _can force the issuer to pay off the principal before the maturity date.

 Convertible bond _gives the bondholder the right to exchange the

 bond for shares of the issuer's common stock at a specified date.

 Municipal bonds _are issued by state and local governments and other

public entities, such as colleges and universities, hospitals, power authorities, resource recovery projects, toll roads, and gas and water utilities. Municipal bonds are often attractive to investors because the interest is exempt from federal income taxes and some local taxes. There are two types of municipal bonds: general obligation bonds and

revenue bonds.NTAldon

NTAldon



The corporation may issue another set of

 bonds equal to the amount of bonds due for

redemption.



The corporation may set up a sinking fund into

which periodic deposits of equal amount 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.

NTAldon

63

Bond Periodic Expense

 A

= periodic deposit to the sinking fund

I 

= interest on the bonds per period

 A

+

I 

= total periodic expense

F 

= accumulated amount, (par value of the bond)

needed to retire the bond

i =

rate of interest in the sinking fund

r =

 bond rate per period

NTAldon

 



_















1

1

i

n

i

 F 

 A

Interests

on the Bonds per Period,

I 

 I 



Fr 

+I 



i



Fr 

i

 F 

 I 

 A

_n

_





















1

1

i = rate of interest in the sinking fund

r = bond rate per period

n = number of periods

NTAldon

65

to the sinking fund

 A



The present worth of all future amounts that are expected to be

received through ownership of the bond.















_





















_n n n

i

i

i

 Fr 

i

C 

 P 

1

1

1

1

P

= value of the bond n periods before redemption

F

= amount needed to retire the bond

C 

= redemption price, usually equal to F (also known as Principal, Face Value, Par Value)

r 

=  bond rate per period

n

= number of periods before redemption

i

= investment rate or yield period

NTAldon

66

Sample Problems-BONDS

Sample Problems-BONDS

1.

1. A bond issue of P200,000, in 10-years, A bond issue of P200,000, in 10-years, in P1,000 units paying 16%in P1,000 units paying 16% nominal interest in

nominal interest in semi-annual payments, semi-annual payments, must be retired by the usmust be retired by the usee of si

of sinking fund nking fund that earthat earns 12% ns 12% pa compounded pa compounded semi-annuallysemi-annually. . What isWhat is the total semi-annual expense?

the total semi-annual expense?



 Solution:Solution:

F = P200,000 F = P200,000

r 

r = 16%/2= 8% per semi-annual= 16%/2= 8% per semi-annual ii = 12%/2 = 6% per semi-annual= 12%/2 = 6% per semi-annual



 Total semi-annual expenseTotal semi-annual expense

=

= A

A +

+ II

 

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 



2020

 



1 1 11 0.06 0.06 2 20000,, 00000 0 220000,, 00000 0 00..0088 1 1 00..006 6 11 21 21,, 437437

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n n ii T

Toottaall SSeemmii aannnnuuaall EExxppeennssee FF FFr  r   ii  P   P  NTAldon NTAldon 67 67

Sample Problems-Bonds

Sample Problems-Bonds

2.

2. Find the current price of a

Find the current price of a 10-year bond pa

10-year bond paying 6% per year

ying 6% per year

that is redeemable at par value, if bought by a purchaser to yield

that is redeemable at par value, if bought by a purchaser to yield

10% per

10% per year

year.

. The face

The face value of

value of the bo

the bond is

nd is P100,000

P100,000

Solution:

Solution:

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73 73 .. 42 4211 ,, 75 75 40 40 .. 86 8677 ,, 38 38 33 33 .. 55 5544 ,, 38 38 P P 10 10 .. 0 0 1 1 10 10 .. 0 0 1 1 1 1 .. 0 0 1 1 06 06 .. 0 0 00 0000 ,, 10 1000 1 1 .. 0 0 1 1 00 0000 ,, 10 1000 ii 1 1 ii 1 1 ii 1 1 F Fr r  ii 1 1 C C P P ₁₀₁₀ 10 10 10 10 n n n n n n

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NTAldon NTAldon 68 68

Sample Problems-Bonds

Sample Problems-Bonds

3.

3. Find the price of a

Find the price of a 10-year bond ,

10-year bond , two

two years before

years before

its redemption,

its redemption, paying

paying 6% per

6% per year

year that

that is

is

redeemab

redeemable at par value if bought by a purc

le at par value if bought by a purchaser to

haser to

yield

yield 10% p

10% per year

er year.

. The fa

The face valu

ce value of

e of the b

the bond i

ond iss

P100,000

P100,000

Solution:

Solution:

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05 05 .. 058 058 ,, 93 93 22 22 .. 413 413 ,, 10 10 83 83 .. 644 644 ,, 82 82 10 10 .. 00 11 10 10 .. 00 11 11 .. 00 11 06 06 .. 00 000 000 ,, 100 100 11 .. 00 11 000 000 ,, 100 100 11 11 11 11 22 22 22

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 P   P  ii ii ii  F  Fr r  ii C  C   P   P  _nn nn nn NTAldon NTAldon 69 69

Sample Problems-Bonds

Sample Problems-Bonds

4. A 10-year bond with a par value of P1,000 and with bond rate of

4. A 10-year bond with a par value of P1,000 and with bond rate of

10% payable annually is sold now for P1080. If the yield is to be

10% payable annually is sold now for P1080. If the yield is to be

12%,

12%, how

how much should the redemption price be at

much should the redemption price be at the end of

the end of 88

years?

years?

Solution:

Solution:

                      1,444.07  1,444.07  C  C  0.12 0.12 1 1 0.12 0.12 1 1 0.12 0.12 1 1 0.10 0.10 1,000 1,000 0.12 0.12 1 1 C  C  1080 1080 ii 1 1 ii 1 1 ii 1 1  Fr   Fr  ii 1 1 C  C   P   P  ₈₈ 8 8 8 8 n n n n n n                           NTAldon NTAldon 70 70

DISCOUNT FACTORS and EQUIVALENCE

NTAldon

DEPRECIATION

 The decrease in the value of equipment, building or other structures

due to the passage of time. The causes of depreciation may be physical or functional.

 Examples of physical depreciation are, wear and tear, corrosion, accident,

deterioration due to age or elements.

 The rest are functional depreciation and one good example is

obsolescence.This is caused by technological advances or developments which make an existing property obsolete. Even though the property has suffered no physical change, its economic serviceability is reduced  because it is inferior to improved types of similar assets that have been

made available through advancements in technology

 Depletion _ Another kind of depreciation is material loss due to

consumption or exploitation particularly applicable to natural resources._NTAldon

 To provide for the recovery of capital which has been invested in

physical property

 To enable the cost of depreciation to be charged to the cost of

producing products or services that results from the use of property.

 For engineers, depreciation is included as cost of production of

any product or the rendering of any service where equipment is used to provide for the replacement either at the end of its

physical or economic life or at the time when its operation no longer results in satisfactory profit or to provide for the

maintenance of capital to replace the decrease in the value of equipment

NTAldon

73

 Maintenance _conveys the idea of constantly keeping a property in good condition;  Repairs _connotes replacing or mending broken or worn parts of a property.

 Service life of the property _is the period during which the use of property is economically

feasible. Both physical and functional depreciation are taken into consideration in determining service life. The term is synonymous with economic or useful life. In estimating the probable service life, it is assumed that a reasonable amount of maintenance and repairs will be carried out at the expense of the property owner.

 Recovery Period _ The number of years over which the basis of the property is recovered through

the accounting process. For the classical methods of depreciation, this is normally the useful life. Under the MACRS, this period is the property class for the General Depreciation System (GDS), and it is the class life for the Alternative Depreciation System (ADS)

 PresentValue _The value of the asset in its condition at the time of valuation

 Salvage Value_ is the net amount of money obtainable from the sale of the used property over and

above any charges involved in removal and sale.

 _{If a property is capable of further service, its salvage value may be higher. This is not necessarily true, however, because}

other factors, such as location of the property, existing price levels, market supply and demand, and difficulty in dismantling, may have an effect. The term salvage value implies that the asset can give some type of further service and is worth more than merely its scrap or junk value.

NTAldon

 Scrap or junk value_ is the amount of money obtained when the property cannot

 be disposed as a useful unit but rather dismantled and sold as junk to be used again as a manufacturing raw material.

 Book value _ also known as depreciated value, is the worth of the property as

recorded in the books of account of the enterprise and is equal to the orig inal cost less the amounts which have been charged to depreciation. It is sometimes called the unamortized value.

 Market value _ The price which could be obtained for an asset if it were placed on

sale in the open market. Is the amount which a willing buyer will pay to a willing seller for the property when neither one is under compulsion to buy or sell.

 Fair Value _ The value is usually determined by a disinterested third party in order

to establish a price that is fair to both the seller and the buyer

 Replacement value _ The cost necessary to replace an existing property at any

given time with one at least equally capable of rendering the same service.

 Adjusted Cost basis _ The original cost of the asset, adjusted by allowable

increases or decrease , is used to compute depreciation and depletion deductions. For example, the cost of any improvement to a capital asset with a useful life greater than one year increases the original cost basis, an d a casualty or theft loss decreases it. If the basis is altered , the depreciation deduction may need to be adjusted.

 Basis, or cost basis _ The initial cost of acquiring an asset (purchase price plus tax)

, including transportation expenses and other normal costs of making an asset serviceable for its intended use. This amount is also called the unadjusted cost basisNTAldon.

Methods of Depreciation

 A. Uniform Depreciation

1. Straight Line Method

This is the simplest and most widely used method compared to any other method. It is  based on uniform annual charge. It doesn’t take into account the interest or profit earned

on accumulated depreciation fund. It is a standard accounting method acceptable by the Bureau of Internal Revenue.

where:

d = periodic depreciation

d_Tn= total depreciation after nth period

FC = First cost

SV = Salvage Value

BV _n= Book value after nth period

L = Service Life n = nth period

 L

SV 

 FC 

d 





  

n

d 

d 

_Tn



  

n d   FC   BV _n





NTAldon 76

2. Sinking Fund Method

It is based on uniform annual charge. It is assumed that a sinking fund is created to replace the original cost of equipment. All amounts in the sinking fund (including interest) earn interest. The company uses the amount accumulated in its operations, and therefore assumed to earn interest. It is generally used for economy-study purposes.

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i

i

d 

d 

n Tn

1

1

Tn n

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d 

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

NTAldon 77

1. Declining-Balance-Method

Also known as Matheson formula. The annual

depreciation cost is a constant percentage of the salvage value at

the beginning of the year. The annual depreciation cost differs

every year, and decreases in absolute value as time progresses.

The salvage value of the property can never depreciate to zero.

  f 

= fractional depreciation



 f 

  

 f 

 FC 

d 

_n

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1



n1

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f 

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1

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₁

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

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



n  L n

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SV 

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



1

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SV 

 f 

/ 1

1

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 



 

 





NTAldon 78

B. Non-Uniform Depreciation

2. Double-Declining Balance Method

This method is similar to the declining balance method except that the f is replaced by 2/L.

3. Sum-of-the-Years Digit Method

The annual depreciation cost differs each year and decreases as time progresses. It provides for a rapid depreciation during the early years of life of property, hence faster recovery of capital









1



/2 1  L  L n  L SV   FC  d _n







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



 L



 L n n  L SV   FC  d _Tn 1 1 2     

 BV 

_n



 FC 



d 

_Tn NTAldon 79

4. 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. This method has the advantage of making the unit cost of depreciation constant and giving low depreciation expense during periods of low production.

FC = first cost of equipment

SV = salvage value of equipment after its service life

Q  _T  = total units of output up to its service life

Q  _n = number of units of output during the nth year

Q  _Tn= total number of units of output on the nth year

d _n = annual depreciation during the nth year

d _Tn = total depreciation on the nth year





T  Q SV   FC  d  output  unit  on  Depreciati











n T  n Q Q SV   FC  d 









Tn T  Tn Q Q SV   FC  d 





Tn n  FC  d   BV 





NTAldon 80

5. Working Hours Method

This method assumes that the total depreciation that has taken place is directly proportional to the operating time of the equipment. This method has the advantage of making the unit cost of depreciation constant and giving low depreciation expense during low operating or utilization period.

FC = first cost of equipment

SV = salvage value of equipment after its service life

H _T  = total hours of operation up to its service life

H _n = hours of operation during the nth year

H _Tn = total hours of operation on the nth year

d _n = annual depreciation during the nth year

d _Tn = total depreciation on the nth year

  n T  n H   H  SV   FC  d      Tn T  Tn  H   H  SV   FC  d      T   H  SV   FC  d   Period  Operating  on  Depreciati    NTAldon 81