ENGINEERING ECONOMICS

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ENGINEERING ECONOMICS Ref. Grat, Ireso & Leaveworth, "Priciples of Egieerig Ecoomy'','- Roald Press, 6th ed., New York, 1976. INTRODUCTION Choice Amogst Alteratives 1) Why do it at all? 2) Why do it ow? 3) Why do it this way? Aswers to 2) ad 3) require a uderstadig of the priciples of egieerig ecoomy - i particular a uderstadig of the time- value of moey. Recogisig ad Defiig Alteratives decisios alteratives Multiple objectives merits Ecoomic objectives Eve the most careful estimates of the moetary cosequeces of choosig differet alteratives almost certaily will tur out to be icorrect. It ofte is helpful to a decisio maker to make use of secodary criteria that reflect i some way the lack of certaity associated with all estimates of the future. The Need for a "Systems Viewpoit' Ofte there are side-effects that ted to be disregarded whe idividual decisios are made. To cosider such side-effects adequately, it may be ecessary to examie the iterrelatioships amog a umber of decisios before ay of the idividual decisios ca be made.

Ecoomics Most problems i ecoomy ivolve determiig what is ecoomical i the log ru, i.e., over a cosiderable period of time. I such problems, it is ecessary to recogise the time value of moey; because of the existece of iterest, a dollar ow is worth more tha the prospect of a dollar ext year or at some later date. Iterest may be defied as moey paid for the use of borrowed moey. Broadly speakig, iterest may be thought of as the retur obtaiable by the productive ivestmet of capital. The rate of iterest is the ratio betwee the iterest chargeable or payable at the ed of a period of time, usually a year or less, ad the moey owed at the begiig of that period. eg. $6 iterest aually i debt of $100 iterest rate -- 6/100 =.06 per aum = 6% (p.a., uderstood) Iterest Formulae Symbols i = iterest rate per iterest period (usually per year, writte as pa from the Lati per aum) = umber of iterest periods (usually years) P = the preset sum of moey F = the sum of moey at the ed of periods, which is equivalet to P at i. A= ed of period paymet i a uiform series for periods, etire series equivalet to P at i. F = P(1 + i) (1.1) P = F/(1 + i) (1.2) (1 + i) is called the sigle paymet compoud amout factor 1/(1 + i) is called the sigle paymet preset worth factor There are a umber of other formulae which are frequetly used which result from summatio of a series of the same paymets at regular itervals i the same way as you make a regular loa repaymet to a ledig authority such as a bak. You might for example borrow a sum to buy a stereo system, which you pay off by a mothly

repaymet of $27, each moth for three years. The coverse is that you receive paymets i a series like this ad ivest each paymet at iterest rate i whe it has bee received. Because the iterest accumulates i a compoud fashio, with decreasig as we move through the series, the total value is a geometric series summatio. The paymet received at the ed of the th year attracts o iterest, while 1 the paymet received at the ed of the first year accumulates to A(1 i) +. F A i i i 2 1 = [1 + (1 + ) + (1 + ) +... + (1 + ) ] (1.3) The formula for this sum of a GP is derived by multiplyig both sides of the equatio by (1 + i) which results i : + if= A + i + + i + + + i (1.4) 2 (1 ) [(1 ) (1 )... (1 ) ] If (1.3) is subtracted from (1.4) we get: if = A[(1 + i) 1] (1.5) ad hece A[(1 + i) 1] F = (1.6) i The coverse of this relatioship is easily obtaied by rearragig (1.6): Fi A = (1.7) [(1 + i) 1] A series of paymets desiged to produce a future sum F at the ed of years i this way is called a sikig fud. There are a couple of other useful formulae which ca easily be derived by maipulatig Equatios (1.1), (1.2), (1.6) ad (1.7). By substitutig for F from (1.1) ito (1.7), we get: i(1 + i) A= P (1 + i) 1 (1.8) I (1.8), the bracketed expressio is called the capital recovery factor because it idicates the paymets required to sum to the value of a curret capital sum. The coverse of this expressio is called the series preset worth factor ad is give by: (1 + i) 1 P = A i (1 + i ) (1.9)

Examples E.g. 1 A local govermet egieer desigs a reiforced cocrete culvert with a aticipated life of fifty years ad cost of $4000. He is approached by a steel maufacturer who offers him a steel culvert costig $3000. If the egieer estimates a desig life of the steel culvert of twety- five years, for a discout rate of 5%, which should he choose? R.C. P.W. = $4000 Steel $3000 P.W. = $3000 + = $3885.9 25 (1.05) Steel Culvert has lowest preset worth, choose it. E.g. 2 A dam is to be built for irrigatio purposes i a major irrigatio area. Two sizes of dam are cosidered, the larger of which would cost $20 millio ad the smaller $10 millio. It is estimated that the aual irrigatio returs from the larger dam would be $1.5 millio ad from the smaller dam $0.8 millio. At a discout rate of 6% p.a., which dam should be chose? What is the beefit cost ratio if the duratio of the project is forty years? Large Preset Worth Small P. Worth Capital Cost $20millio $10millio Returs (1.06) 1 (1.06) 1 1.5 = 22.57 0.06(1.06) 0.8 = 12.04 0.06(1.06) B-C $2.57millio $2.04millio B/C 1.13 1.20

Comparisos Usig Aual Paymets E.g. 3 A materials hadlig operatio i a cocrete plat is at preset carried out by had. The total aual cost for this labour is $19,500. Machiery could be istalled which would reduce the labour costs substatially. The equipmet costs $45000 ad would reduce labour costs to $7,000 but additioal isurace ad tax paymets of about $6,300 per year would have to be made. If the life of the plat is 10 years ad iterest rates are about 8.5%, assumig the ew machiery would have a salvage value of $15,000 at the ed of te years, should it be istalled? Preset Costs With New Machiery $19,500 CRF (8.5%, 10 years) =.1524 45,000 x.1524 = 6858 Labour 7000 Tax 6300 20158 Salvage (Sikig fud 8.5%, 10 yrs. =.0674) -1011 19,147 E.g. 4 Solve example I by comparig equivalet series of uiform paymets R.C. Steel $4000 x.05478 = $219.12 $3000 x.07095 = $212.85

Homework examples: A dam is to be built for irrigatio purposes i a major irrigatio area. Two sizes of dam are cosidered, the larger of which would cost $20 millio ad the smaller $10 millio. It is estimated that the aual irrigatio returs from the larger dam would be $1.5 millio ad from the smaller dam $0.8 millio. At a discout rate of 6% p.a., which dam should be chose? What is the beefit cost ratio if the duratio of the project is forty years? Use aualized values. Large Aual Small Aual Cost 1.329231 0.664615 Returs (B) 1.5 0.8 B/C 1.128472 1.203704 E.g. 3 A materials hadlig operatio i a cocrete plat is at preset carried out by had. The total aual cost for this labour is $19,500. Machiery could be istalled which would reduce the labour costs substatially. The equipmet costs $45000 ad would reduce labour costs to $7,000 but additioal isurace ad tax paymets of about $6,300 per year would have to be made. If the life of the plat is 10 years ad iterest rates are about 8.5%, assumig the ew machiery would have a salvage value of $15,000 at the ed of te years, should it be istalled? Use preset worth. Preset worth of labour cost is : $127946.3 Preset worth of machiery labour ad tax is: $45929.4 +$ 41336.49 = $87265.93 Preset worth of machiery salvage is: $6634.281 Total preset worth costs of machiery+labour + tax - salvage: $125631.6 Machiery is cheaper ote that you ca covert this figure back to $19,147 directly usig A=f(P), where f is a capital recovery factor.