INTRODUCTION TO ENGINEERING

FUNDAMENTALS OF ENGINEERING PROJECT

ECONOMICS - 2

Prepared by Prof T.M.Lewis

Location of Notes – to be updated

•http://www.eng.uwi.tt/depts/faculty/ugrad/courses/engr0300/engr0300.html

• Interest is the amount that has to be paid for the use of borrowed money, and it is a rate (usually a percentage) for a period of time (usually a year), and so interest is a measure of the time value of money.

• Within the level of accuracy normally required for engineering projects, interest rates can be assumed proportional to the length of time period:

• 12% per annum = 1% per month

• 1% per month = (1/30)% per day

Effect of Time – Interest Rates

Time Value of Money• Cash inflows to a project are normally

treated as positive (+) whilst cash outflows are negative (-).

• The net cash flow during any time period is simply the difference between the cash inflows and the outflows, and it may be either positive or negative

• For an engineering project, receipts (benefits) and expenditures (costs) are generally compared at the start of the project - this is ‘the end of year zero' or ‘the start of year 1’ – so they are all usually converted to their equivalent Present Value.

Time Value of Money• Which is better?

Now -$1,000 -$5,000 -$10,000

End yr 1

+$500 +$1,000

+$9,000

End yr 2

+$500 +$2,400

+$6,500

End yr 3

+$500 +$8,000

+$500

Cash Total

+$500 +$6,500

+$6,000

• How can you tell which gives the best return?• By converting them all to equivalent values.

Present Value• All cash sums can be converted to an

equivalent value by working out what sum would have to be 'invested' now (start of year 1 = end year 0) at the going interest rate to give that specific sum at that specific time in the future.

• The equivalent sums at the start of year 1, are Present Values (P)

• The process of converting an amount to its equivalent at an earlier date is known as discounting, and the interest rate involved is called the discount rate.

SIMPLE INTEREST• Simple Interest is where the interest on

some capital (present) sum or principal is not reinvested to earn more interest on itself.

• Where: i = interest rate; I = Interest earned; P = Present sum; F = Future sum

• Interest earned in one year I1 = P.i

• Interest earned in 2 years I2 = Pi + Pi = P.2i

• Interest earned in n years In = P.ni

SIMPLE INTEREST• At end of n years the future sum will be: • F = P + In = P + P.ni = P(1+ni) • F = P(1+ni) • Alternatively, by rearranging the equation,

we can get an expression for the present value given the future sum.

• P = F/(1+ni) • Simple interest is rarely used in

practice• Normally it is compound interest that

is required.

COMPOUND INTEREST• Where the interest each year is reinvested to earn

more interest on itself. • Again where: i = interest rate; P = Principal (or

Present Sum); F = Future sum • Year Amount at Interest earned

Amount at end start of year during year of year (a) (b) (a+b)

• 1 P Pi P(1+i)• 2 P(1+i) P(1+i).i P(1+i)2

• 3 P(1+i)2 P(1+i)2.i P(1+i)3

• n P(1+i)n-1 P(1+i)n-1.i P(1+i)n

+ =+ =+ =+ =

Present Values• Thus, the future value $F, of the principal

$P, after n years at compound interest rate i, is given by the last entry in this table – this is the equation we arrived at by simple logic last week:

F = P(1+i)n • Rearranging the equation also tells us that:

P = F (1 + i)n

Simple or Compound Interest?• Why do we rarely use simple interest?• In 1626 the Government of the Dutch

settlement in America – New Amsterdam – bought Manhattan Island from the Indians for beads and trinkets worth $24. If that money had been invested to earn an average of 8% annually – what would it be worth today?

• Simple Interest – F = 24(1+380(0.08))

= $753.60• Compound Interest – F = 24(1+.08)380

= $120,569,740,656,495

Present & Future Value Factors

• If you have to do a number of calculations, it is possible to take the money out of the equation and simply calculate a PV Factor for any period and any interest rate. When you have done it once you can just use the factor afterwards :

Hence the Future Value factors can be calculated for all values of I and n so that the

PV Factor = 1/(1+i)n

And the FV Factor = (1 + i)n

Present Value Factors•PV Factor = 1/(1+i)n

After 1 year, with 10% per annum interest ratePV Factor = 1 = 0.9091 (1.1) After 2 years, with 10% per annum interest ratePV Factor = 1 = 0.8264 (1.1)2 After 3 years, with 10% per annum interest rate PV Factor = 1 = 0.7513 (1.1) 3

Going back to the table earlier

Option 1

Option 2

Option 3

Now -$1,000 -$5,000 -$10,000End 1st year +$500 +

$1,000+$9,000

End 2nd year +$500 +$2,400

+$6,500

End 3rd year +$500 +$8,000

+$500

CASH TOTAL +$500 +$6,400

+$6,000

Which is best? •Option 1 requires lowest investment - so risk is low•Option 2 gives the highest cash value.•Option 3 gives earliest payback

10% factor

Option 1 Option 2 Option 3

Now 1.0000

-$1,000 -$5,000 -$10,000

For 1 year

0.9091

+$500 +$1,000

+$9,000

For 2 year

0.8264

+$500 +$2,400

+$6,500

For 3 year

0.7513

+$500 +$8,000

+$500

To Convert to Present Values at an Interest Rate of 10% per year apply Factors (P=F(1/(1+i)n)

Present valuesPV Factor Project 1 Project 2 Project 31.0000 -$1,000 -$5,000 -$10,0000.9091 +$455 +$909 +$8,1820.8624 +$431 +$1,983 $5,372 0.7513 +$376 +$6,010 $376 TOTALS +$261 +$3,903 $3,929

• Which one has the highest net present value? • Project 3. • But Project 2 gives the highest return per $

invested.• Different choices may be appropriate at different

times, but usually people want to know the highest returns

Uniform Cash Flows - Annuities• Often cash flows are uniform – with

the same amount being paid or received during each time period – as in a mortgage, pension or insurance premium.

• These can be discounted individually but doing the 240 calculations for a monthly payment on a 20 year mortgage is tiresome so they are usually discounted as an ‘annuity’

Annuities• An annuity is a periodic payment of a fixed

amount, over a given period of time, whose accumulated value, whether present or future, is required (the payments do not have to be yearly to be called an annuity).

• If a sum of $A is invested at the end of each year for n years, the total final amount will obviously be the sum of the compounded amounts for the individual investments.

• Remember that the Future Value of a Present Sum is given by the expression F = P(1+i)n

ANNUITIES

Payment $A

Payment $A

Payment $A

Payment $A

Payment $A

Payment $A

Payment $A

Annuities• The present sum of money $A invested at the

end of the first year will earn interest for (n-1) years, so it will mount to F = A(1+i)n-1.

• The second year's payment of $A will be compounded up to A(1+i)n-2, and so on until the (n-1)th year, when that payment will earn interest for 1 year

• The last payment on the last day will earn no interest.

• The total future amount $F is, thus, given by the expression:

• F = A[1 + (1+i) + (1+i)2.....(1+i)n-1]

Annuities• F = A[1 + (1+i) + (1+i)2.....(1+i)n-1]• This is not a very convenient expression

but it can be put into a simpler form by multiplying both sides by (1+i) to get

• F(1+i) = A[(1+i) + (1+i)2.......(1+i)n-1 + (1+i)n]

• Now subtract the first equation from the second, and you will get:

• Fi = A[(1+i)n – 1]

Annuities• F = A{[(1+i)n - 1]/i}• Thus, A = F[ i ]• (1+i)n - 1• Previous calculations showed that F = P(1+i)n ,

so P can be obtained in terms of A, i and n from these equations by substitution, i.e.

• P(1+i)n = A{[(1+i)n - 1]/i}• giving P = A [ (1+i)n - 1 ]• i(1+i)n

• and from this it is easy to see that:• A = P [ i(1+i)n ]• (1+i)n - 1

The Formulae• To Find the Future Value given

the Present Value• F = P(1+i)n (1)

• To Find the Present Value given the Future Value

• P = F (2) (1 + i)n

The Formulae• To Find the Future Value given the

Annual Value • F = A[(1+i)n - 1] (3) i• To Find the Annual Value given the

Future Value • A = F[ i ] (4) (1+i)n – 1)

The Formulae• To Find the Present Value given the

Annual Value • P = A[(1+i)n - 1] (5) i(1+i)n

• To Find the Annual Value given the Present Value

• A = P[ i(1+i)n ] (6) ((1+i)n – 1)

Uniform Periodic Cash Flows• When it is often necessary to consider

payments over a significant period of time in equal amounts.

• This is the way that mortgages on properties are normally amortised.

• Likewise, in providing capital for future use (e.g. pensions), payments are often made monthly or annually into a fund which is built up over a period of time while accumulating interest.

Tables• The factors from these equations can

be used to create look-up tables that can be used to calculate the equivalent values of cash flows.

• These tables are widely published (occasionally in diaries) and are built into spreadsheet packages like MS Excel

• Typical tables are shown below

To Find a Future Value given a Present Value

To Find a Present Value given a Future Value

To Find a Future Value given a Periodic (Annual) Value

To Find a Present Value given aPeriodic (Annual) Value

Example• If you set aside a sum of $500 each month

for 50 months (just over 4 years) at an interest rate of 12% per annum, how much will you have at the end of that period?

• Use table "To Find a Future Value given a Periodic (Annual) Value” using the 1% interest rate (per month) and 50 periods, you get a factor 64.463. Multiply by the sum of $500.

• And the answer is $32,231

Internal Rate of Return• When it is not clear what interest rate should

be used to discount the cash flows on a project. • The problem becomes one of determining what

interest rate the cash flows from the project are actually earning - it is the interest rate that would make the discounted costs and benefits exactly equal.

• In other words, it involves finding which discount rate would result in a zero NPV.

• This interest rate is referred to as the Internal Rate of Return.

Example 7

• Find the internal rate of return for the investment with a capital outlay of $100,000 now, which earns receipts of $40,000, $50,000, and $40,000 at the end of the first, second and third years

Solution• In order to find the internal rate of return, it is necessary to try a discount rate and see what net present value (NPV) it gives. • If the resulting NPV is positive then a higher interest rate must be used for the discounting (to reduce the value of the discounted future flows). • If the NPV is negative, a lower interest rate must be used. • Hence, for this example, using the expression • NPV = -100,000 + 40,000 + 50,000 + 40,000 • 1+i (1+i)2 (1+i)3

• Solving for i = 10% NPV = +$7,738.54

• i = 15% NPV = -$1,109.56

Internal Rate of Return• To find the value at which the NPV is zero, we

can interpolate by using the rule of simple proportions, i.e. the internal rate of return i is

• i = {10 + 5(7,738.54/8,848.10)} = 14.3%

• The criterion usually applied under this method is that a project is acceptable if it produces an internal rate of return greater than the minimum return required on capital.

• Thus, in this example, if the cost of capital is 10%, the project should be accepted, whereas if it is 15%, it should be rejected.

Using Excel

Excel Functions

Description

Excel Functions

FormulaCell ID Result

IRR Function

Problem

• THE ANNUAL GROSS DOMESTIC PRODUCT PER CAPITA OF TRINIDAD AND TOBAGONIANS IS US$9,000 AND IS EXPECTED TO GROW AT 4% PER ANNUM FOR THE NEXT 15 YEARS (UNTIL 2026), WHAT WILL IT BE THEN?

Solution

• F = P (1+i)n

• F = 9,000 (1.04)15

• F = $16,208

Problem

• DUE TO IMPROVEMENTS IN RICE SEED VARIETIES, FERTILISATION AND IRRIGATION, THE PRODUCTIVITY OF RICE PADDY HAS INCREASED BY A STEADY 3% PER YEAR SINCE 1960. WHAT WAS THE PRODUCTIVITY OF PADDY IN 2010 COMPARED WITH 1960?

Solution

• PRODUCTION IN 1960 = X TONS/HECTARE• N = 50 YEARS• F = X(1.03)50

• F = 4.4X• HENCE RICE PADDY PRODUCTIVITY • HAS GROWN BY 440%

Problem

• POPULATION OF TRINIDAD & TOBAGO IS 1.3 MILLION NOW WHAT WILL IT BE IN 2025 IF IT CONTINUES TO GROW AT 1.7% PER YEAR?

Solution

• F = 1.3 (1.017)14

• F = 1.65 • POPULATION IN 2024 = 1.65

MILLION

Problem• THE GOVERNMENT WANTS TO SPEND

$10 MILLION A YEAR ON HOUSING BETWEEN 2013 AND 2023.

• IT WANTS TO BUDGET FOR THIS NOW BY SETTING ASIDE A FUND TO PAY FOR THE SCHEME.

• HOW MUCH DOES IT NEED TO BUDGET NOW TO COVER THE COST OF THIS? INTEREST = 5% PER ANNUM.

Solution – Part 1• AT THE START OF 2013 THE GOVT.

NEEDS • P2013 = A[(1+i)n –1]/[i(1+i)n)• P2013 = 10m *[(1.05)10-1]/[0.05(1.05)10]• P2013 = 10m *[1.629-1]/[0.05(1.629)]• P2013 = 10m (0.629)/(0.0814)• P2013 = 10m * 7.722• P2013 = $77.22 million

Solution – Part 2• Need to set aside $77.22 million at the start

of 2013 to pay out $10m per year for next decade

• How much do you need to put aside at the start of 2012 to have $77.22m at the start of 2013?

• Need P = F/(1+i)n

• P2012 = 77.22/(1.05)1

• P2012 = 77.22/1.05• P2012 = $73.54 million

Problem• If you want to have a million

dollars by the time you are 50 how much do you need to put away each month for 30 years when the interest rate is 8% per annum?

Solution• A = F [i/((1+i)n – 1)]• A= 1000000x(8%/12)

(1+(8%/12))360 – 1)• A= 1000000x(.00667) = $671 approx

(1.00667)360 – 1• If you put away $671 every month for the next

30 years and earn 8% interest you will have $1m at the end

• If you could get interest of 12%, then you would only have to put aside approx $286 each month

• If you could get interest of 2%, then you would have to put aside approx $2,029 each month

Problem• THE AVERAGE LIFE EXPECTANCY

FOR WOMEN IN THE CARIBBEAN HAS INCREASED BY APPROXIMATELY 4 YEARS, IN EACH DECADE SINCE 1960, WHEN IT WAS 64. TO WHAT AGE WILL WOMEN LIVE (ON AVERAGE) IN THE YEAR 2020 IF THIS RATE OF INCREASE CONTINUES?

Solution• RATE OF INCREASE = (4/64)x100%

PER 10 YEARS

• PV = 64 (IN 1960)

• FV = TO FIND

• N = 6 DECADES

• FV = 64 (1.0625)6

• = 92

Problem•What MONTHLY periodic

payment would be required to pay off a loan of $300,000 over a period of 20 years at interest rate of 13% per year

Solution

• A = P {i(1+i)n/[(1+i)n - 1]}

• A = 300,000[0.0108(1.0108)240](1.0108)240 - 1

• = 300,000[0.1438/12.2768]

• = 300,000 [0.0117]

• = $3,514.73

Problem• Find the present and future values of $1,000 per

year, over a period of 20 years at 10% compound interest

• Solution The equations given above can be used to carry out this calculation. Hence:

• P = 1,000[(1.1)20 -1] = 1,000[6.7275-1] 0.1(1.1)20 0.67275

• P = 1,000[8.5136] = $8,514

• This means that if you had $8514 in an account earning 10% per annum compound interest, you could take out $1000 every year for 20 years.

Problem• Having obtained the present value, the

future value can easily be obtained from • F = 8,514(1.1)20 • = 8,510(6.7275) • = $57,275

• This means that if you put $1000, every year for 20 years, into an account earning 10% per annum compound interest, then at the end you would have $57,275.

Problem• A building can be insulated at a cost of

$165,000 in order to reduce cooling costs by $26,000 a year.

• If the insulation has a useful life of 10 years, calculate whether the proposal is worthwhile, assuming the required return on capital is 8% per annum.

• The task here is to express the initial capital outlay as an equivalent annual expenditure, to compare with the annual saving achieved using the insulation.

Solution• The initial cost of the building insulation can be expressed as

the equivalent of a constant amount payable at the end of each of 10 years.

• Let this amount be A• A = 165,000[0.08(1.08)10] = 165,000 [ 0.1727 ] (1.08)10 - 1 2.159-1 • = 165,000 x 0.149 = $24,588• Now, considering the problem as a whole:• Annual savings on cooling +

$26,000.00• Annual equivalent of initial cost -$24,589.87• Net annual value +$ 1,410.13

• This is positive and so the project is worthwhile.

• Would you rather have $1 million in 50 years time or $35,000 in 5 years time when the interest available is 9.6% interest?

• P1 = F/(1+i)n = 1000000/(1.096)50

= $10,221

• P2 = 35000/(1.096)5

= $22,132

Annual Percentage Rate (APR) or Annual Equivalent Rate (AER)• The annual percentage rate (APR) is the

percentage cost of credit on a yearly basis. • Suppose you borrow $100 for one year and pay

a finance charge (interest plus other costs) of $10. If you can keep the entire $100 for the whole year and then pay back $110 at the end of the year, you are paying an APR of 10 percent.

• But, if you repay the $100 and finance charge (a total of $110) in twelve equal monthly installments, you don't really get to use $100 for the whole year. In fact, you get to use less and less of that $100 each month. In this case, the $10 charge amounts to an APR of 18 percent.

Competition for Scarce Resources• There are never sufficient resources to

undertake everything that one wants to.• There is a need for economy in the use of

resources because there are competing uses for them - if a resource is used for one purpose it is denied to another.

• Thus there is competition for scarce resources: 1. between alternative projects designed to serve

the same purpose, 2. between projects which serve different purposes,

and 3. between engineering projects and other potential

modes of investment.

Competition for Scarce Resources• The decision on how to employ

resources often lies with politicians, administrators, bankers and others.

• If they are to make a wise choice they must be properly informed not only of the comparative technical merits of alternatives but also of their relative costs and benefits.

• This responsibility rests with professional engineers amongst others.

Economic Evaluation• The aim of economic evaluation is to secure

the greatest benefit from the resources available.

• More specifically, economic evaluation is an assessment of the costs and benefits of carrying out a project as compared with doing nothing.

• This is to establish if any of the proposed schemes are economic, and if they are, to establish a priority ranking for the competing proposals.

Economic Evaluation• Where projects require related

investments to be made in order to be effective, these must be taken into account.

• For example, the evaluation of proposals for the provision of electrical power must take into account not only the cost of building and operating the power stations but also the costs of developing their respective distribution systems.

Financial Planning• Financial planning is concerned with the

movement of money both required for and arising from the project.

• It is directed at assessing the cash inflows and outflows during the life of the project and ensuring that the financial implications of these cash flows are accounted for – i.e. where the money will come from when it is needed, and how you will service the debt interest and repayments as they come due.

Need for Judgement• Planning relates to the future. • As a result, even the most sophisticated

forecasting techniques have fundamental uncertainties associated with them.

• One major uncertainty is the way technology will change, and this can have the most profound effect on the future course of events.

• When it is necessary to make assumptions about the technological capabilities and needs of the future, personal judgement and experience are the best guides.

Project Life• In the economic analysis of a project, a limited

period is assigned as the useful life of the project. • This may mean that the project components will

have perished, or that the project will have become obsolete, perhaps due to advances in technology.

• The useful life assumed for the project determines the period over which its economic viability has to be established.

• The useful life of a private sector project depends heavily on predictions made about technological change; for public works projects the useful life is normally not taken to be more than 50 years.