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FOOD ENGINEERING DESIGN AND ECONOMICS CHAPTER IV Engineering Economics

FOOD ENGINEERING DESIGN AND ECONOMICS CHAPTER IV Engineering Economics

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FOOD ENGINEERING DESIGN AND ECONOMICS

CHAPTER IV

Engineering Economics

Economics is a social science dealing with Growth and population Productivity and growth Energy Unemployment and inflation Poverty, medical care, government policies, etc.

The science of economics, in fact, arise out of the use of scarce resources to satisfy unlimited human wants.

Economics is a part of our daily lives since it consists of the subjects as; getting income and goods, their use, inspection, budgeting, income taxes, future investments etc.

Engineering economics is a discipline, which evaluates systematically the cost and benefit of the technical and operational projects. In fact engineering economics is a “decision making” process.

Decision making is a quite broad topic for it is a major aspect of everyday human existence. Therefore, focus will be only those problems that are commonly faced by engineers and develop the tools to properly grasp, analyze and solve them.

Decision making and problem solving are synonymous with each other from the standpoint of a design engineer.

For those problems that we consider as “simple” the problem can be solved quickly by thinking without a need for analytical techniques to aid in their solution.

The “intermediate” level of problems are best suited for solution by engineering economic analysis. In this classification, the economics of the problem are the major component in decision making. There may well be a great many other aspects of the problem to consider before making a decision, but the economic aspects dominate the problem and are therefore dominant in determining its best solution.

The “complex” problems represent a mixture of economic political and humanistic elements. Engineering economic analysis is not expected to be helpful in solving complex problems.

The problems suitable for solution by engineering economic analysis have following properties;

1. The problem is sufficiently important that we are justified in giving it some serious thought and effort.

2. The problem can’t be worked in one’s head, that is, a careful analysis requires that we organize the problem and all the various consequences and this just too much to be done all at once.

3. The problem has economic aspects that are sufficiently important to be a significant component of the analysis leading to a decision.

Problem Solving (or Decision Making) in Engineering Economics

1. Determination of the Problem

The first stage is understanding the problem with all aspects. At this stage data are collected and the extend of the problem is determined.

2. Analysis of the Problem

After the determination and definition of the problem the next stage is the analysis of the problem. At this stage the necessary additional information is collected, other possibilities are determined and data are developed.

3. Investigation of Alternative Solutions to the Problem

The third stage is composed of investigating alternative solutions to problem. At this stage suitable solutions to the defined problem should be developed which requires engineering approach (ethics, efforts, believing).

Some of the factors which will be helpful for the design engineer are: asking questions related with all aspects of the problem using the results of similar projects to get new ideas being creative by considering human and economic

factors being concerned with the efficient use of limited

resources being aware of the fact that opportunities are not made

they are discovered placing the estimated future cash flows for all

alternatives on a comparable basis, considering time vaşue of money

not eliminating the ideas before reaching the definite results.

4. Determination of the Selected Alternative

At this stage alternatives are evaluated with engineering analysis methods and non-feasible ones are eliminated. For the feasible alternatives costs and benefits are determined and the selected alternative is transferred into form of engineering project.

In engineering economic analysis, the results are evaluated at the highest possible extend with money based items. However, there exist some other factors which cannot be based to money but still very important from economical standpoint. Before selecting the best alternative both money based items and others should be evaluated together.

In an economy or an industrial system the companies carry out their activities within some general rules. (To process certain raw materials, giving taxes, providing accommodation for the people etc.) The firm is successful if the goods or services produced are accepted by the customer and desired by the customer. However, this situation may change with time.

The factors which are not money based but important for a company to be successful are to accomplish consumer desires in an appropriate way getting consumer confidence production of high quality and defect free goods to accomplish the expectations of the workers to get a positive image from consumers to be capable of changing the production parallel to

consumer desires to regulate the fluctuations between consumption and

production occurring some times to improve the safety conditions within plant to reduce pollution made to nature to act in accordance with government policies.

Principles of Engineering Economics

1. rule: selection can be done among alternatives, therefore alternatives should be developed.

2. rule: while comparing and analyzing the alternatives the future out comings should be considered.

3. rule: the evaluation of possible alternatives should be based on realistic factors.

4. rule: for the analysis and comparison of the alternatives a common measurement unit should be used.

5. rule: for selection some criteria are needed. The factors which can be based on money and others should be evaluated together.

6. rule: the results of the alternatives are uncertain therefore contain risk.

7. rule: when implementation starts, the results observed should be continuously compared with the project.

INTEREST AND INVESTMENT COSTS

For a realistic evaluation of an engineering project the change of value of money with respected to time is as important as the cost of the project. When the estimation of capital cost has been finished the next stage is the preparation of the plans for financing the project and expending money. This plan brings about new definitions as “present value” or “future value” of money.

From the standpoint of engineers, interest is defined as; “the compensation paid for use of borrowed capital”. The rate at which interest will be paid is usually fixed at the time the capital is borrowed and guarantee is made to return the capital at some set time in the future or on an agreed upon pay-off schedule.

Interest, like taxes, is a very old concept; in 2000 BC it was used Babylon. There was a developed interest calculation method similar to modern methods. In some religions it’s forbidden like Islam. In some other systems as capitalism or some mixed systems interest is an item which supplies continuation by means of banks.

In the determination of interest rate, the following items should be considered: amount of money borrowed the period for which interest is to be calculated the economic conditions the probability of paying back the borrowed money

In economic terminology, the amount of capital on which interest is paid is designated as the “principal”. “Rate of interest” is defined as the amount of interest earned by a unit of principal in a unit of time.

If the period of time for the payment of the borrowed money is long, then the amount of interest should be much higher. By this way future value of an item will be higher than the present value.

Simple InterestSimple interest is the compensation of payment at a constant interest rate based only on the original principal. If,

P represents the principaln represents number of time units or

interest periodsi represents the interest rate based on

the length of one interest period, the amount of simple interest during n interest periods is;

niPI . .

Example: If $ 1000 were loaned for total time of four years at a constant interest rate of 10 percent / year what is the simple interest earned?

400

)4)(1.0)(1000(

..

I

I

niPI

The time unit used to determine the number of interest periods is usually one year and the interest rate is expressed on a yearly basis. When on interest period of less than one year is involved, the “ordinary” way to determine simple interest is to assume the year consists of 360 days. The “exact” method accounts for the fact that there are 365 days in a year. Thus,

P : principal i : interest rate, expressed on the regular yearly basisd : number of days in an interest period.Ordinary interest is commonly accepted in business

practices unless there is a particular reason to use the exact value ( banks ).

365interest

360interest

dPisimpleExact

dPisimpleOrdinary

The principal must be repaid eventually, therefore the entire amount S of principal plus simple interest due after n interest period is;

Example: Mr. A. Deposited a hundred million TL for a period of 5 years with 40 % simple interest rate. How much will be his Money at the end of 5th year and what is the amount of interest.

).1(.. niPniPPIPS

TL 000000200

)5)(4.0(000000100

..

TL 000000300

))5)(4.0(1(000000100

).1(

I

I

niPI

S

S

niPS

Compound InterestIn the payment of simple interest, it makes no difference whether the 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 ( promotion ) to pay the interest until the end of the total loan period. However, interest has a time value. If the interest were paid at the end of each time unit, the receiver could put this money to use for earning additional returns.In compound interest, determination of interest is done regularly at the end of each interest period. If the payment is not made, the amount due is added to the principal and interest is charged on this converted principal during the following time unit.

At the end of the first loan period;

By this time, the total amount is;

At the end of the second loan period;

)1(.1 niPI

)1(1

11

iPS

IPS

33

23

22

222

212

12

)1(

.)1(

)1(

)21(...).1(.

).1(.

iPS

iiPI

iPS

iiPiPiPiPPiiPiPPS

IIPS

iiPiSI

So at the end of n loan periods

If the future value of money is known and present value is to be calculated;

nii ifcompoundf

iPS

)1(factorinterest:

)1( 3

factordiscountf

iSP

iPS

d

n

n

)1(

)1(

Ex: Mr. A. Deposited a hundred TL for a period of 5 years with 40 % compound interest rate. How much will be his money at the end of the 5th year?

Loan period Amount at the beginning of the period (TL)

Amount of interest (TL)

Amount at the end of the period (TL)

1 100 40 140

2 140 56 196

3 196 78.40 274.40

4 274.40 109.76 384.160

5 384.16 153.66 537.82

NOMINAL AND EFFECTIVE INTEREST RATES (compound interest for short term )In common practice, the length of the interest period is assumed to be one year and the fixed interest rate is based on one year. However, there are cases where other time units are employed.- nominal interest rate per year, is the annual interest rate without considering the effect of any compounding i.e.; five percent interest compounded semi annually means that the bank pays 2 ½ % every six months. Five percent interest, compounded semi-annually means hat the bank pays 2 ½ % every six months. Thus the initial amount P = 100 $ will be,S1 = 100 + 100( 0,025 ) = 102.50 $This amount is left in the savings account. At the end of the second six months period,S2 = 102.50 + 102.50( 0.025 ) = 105.06 $

yearperperiodsmm

rPS m

interest:

)1(

- effective interest rate per year, is the annual interest rate taking into account the effect of any compounding during year.

for a principle of 1 $

)interest(11

$1ngsubstracti

11

yearperrateeffectivem

ri

sumthefrominitial

m

rS

m

eff

m

Example: It is desired to borrow 1000 $ to meet a financial obligation. This money can be borrowed from a loan agency at a monthly interest rate of 2 percent. Determine the following;

a. The total amount of principal plus simple interest due after 2 years if no intermediate payments are made.

b. The total amount of principal plus compounded interest due after 2 years if no intermediate payments are made

c. The nominal interest rate when the interest is compounded monthly

d. The effective interest rate when the interest is compounded monthly

a. length of one interest period: 1 month

number of interest periods in two years: 24

b.

1480 $ ))24)(02.0(1(1000

).1(

S

niPS

1608 $

)02.01(1000

)1(24

S

S

iPS n

c. Nominal interest rate per year compounded monthly.

60.1537 $

)24.01(1000

%2412*22

S

S

r

d. Effective interest rate;

)2interest( 80.1607 $

268.011000

)(24.1268 $

12

24.011000

2

12

yearsforcompoundedS

S

yearforrateeffectiveS

S

Continuous Interest

The type of interest considered up to this time have the common form of; payments are charged at periodic and discrete intervals. As can be seen from short term compound interest calculations with the increase of compounding periods the amount of interest increased. The extreme case, is when the time interval becomes infinitesimally small so that the interest is compounded continuously.

If the interest is compounded continuously;

nm

m

rPS

.

1

nr

nr

x

x

nm

m

ePS

xPS

nrr

mnm

m

rx

m

rPS

.

.

/1

0

.

.

1

...,

1

lim

lim

The previously defined “discount factor” for continuous interest is;

nrd ef .

nrd ef .

Example: For the case of a nominal annual interest rate of 20 %, determine total amount to which one dollar of initial principal would accumulate after one year with continuous compounding. Also find the effective annual interest rate.

nrePS ..

%14.222214,0

12214.1

1

224.1

)()0.1( )1)(20.0(

ori

i

ei

dollarS

eS

eff

eff

reff

Annuities An annuity is a series of equal payments occurring at equal

time intervals. Payments of this type can be used to accumulate a desired amount of capital pay off a debt receive a lump sum capital that is due in periodic

installments as in some life insurance plans.

The common type of annuity involves payments which occur at the end of each interest period. This is known as an ordinary annuity. Interest is paid on all accumulated and the interest is compounded each payment period.An annuity term is the time from the beginning of the first payment period to the end of the last payment period.The amount of an annuity is the sum of all the payments plus interest if allowed to accumulate at the definite rate of interest from the time of initial payment to the end of annuity term.

Let R (A.E.) represent the uniform periodic payment made during n discrete periods in an ordinary annuity.i ; is the interest rate based on the payment period.S ; is the amount of the annuity

The first payment of R is made at the end of the first period and will bear interest for ( n-1 ) periods. Thus, at the end of the annuity term, the first payment will have accumulated to an amount of R( 1 + i )n-1

The second payment of R is made at the end of second period and will bear interest for n-2 periods giving an accumulated aount of R*( 1 + i )n-2.

Similarly, each periodic payment will give an additional accumulated amount until the last payment of R is made at the end of the annuity term.

By definition, the amount of the annuity is the sum of all the accumulated amounts from each payment, therefore

In order to simplify the equation, both sides are multiplied with (1+i ) and written equation is subtracted from the resulting equation.

i

iRS

RiRSn

ni

1)1(

1

If there exist m interest periods per year, so i = r/m r representing the nominal interest rate. The total number of interest periods in n years is m.n.

mr

m

r

m

RS

nrr

m

11).)((

For the case of continuous cash flow:

r

eRS

nr 1.

Example: An engineer is paying 1000 $ at each year with four equal payments. If the nominal interest rate is 10 % calculate the total amount of money which will accumulate in 25 years.

0.1r

136118$S25n

4/1.0

40,11

250S4m

250$4

1000

11

)25)(4(

.

R

mrmr

RS

nm

InflationInflation is an important variable of economics, which is tried to kept constant and at a low value. The simplest definition of inflation is to increase in material and service prices at each year.

If the rate of inflation is e, the present price is Pe and the future price is Se;

In the inflational economies, the time value of money

decreases with time. Therefore,

nee ePS 1

e

ei

e

eei

e

i

e

ei

PS

e

iP

e

iPS

e

PS

n

n

n

n

n

11

1

1

1

1

1

1

1

1

1

1

1

Example: A researcher engineer receives $ 10 000 for a patent he developed. He decides to deposit this money with compound interest of % 8 for a period of fifteen years. If the inflation rate is 6 % for US Dollars, what will be the

total income of the engineer at the end of the period.

242.6113$

0189.0100010

1

0189.006.01

06.008.0

06.0

00010$08.0

15

F

F

PF

e

Pi

n

Cash Flow Diagrams

In accounting, cash flow should be fallowed with great attention. Cash flow is defined as the difference between in flows and outflows at a certain period of time. This period is usually one year and the term annual cash flow is used.

Use of cash flow diagrams is the easiest way of showing cash flow. In these diagrams each cash flow is represented by a vertical arrow on time axes. If it’s an inflow the arrow points up and if it’s an outflow the arrow points down. In addition, the length of the arrow is proportional with the amount of cash flow.