2.0 engineering costs and cost estimating rev 2

2.0 Engineering Costs and Estimating*ECONOMIC EQUIVALENCE FOR FOUR REPAYMENT PLANS OF AN $8,000 LOAN

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2.0 Engineering Costs and Estimating*CASH FLOW DIAGRAM NOTATION

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2.0 Engineering Costs and Estimating*CASH FLOW DIAGRAM NOTATION

MSc AMM UTP

Asset Life Study 28th Sept to 1st Oct 2012

*Break

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2.0 Engineering Costs and Estimating*Cost TerminologyFixed cost is the cost that are unaffected by changes in activity level. Eg- Insurance, interest, rentalVariable cost is the cost that vary with output or activity level. Eg- Material cost, labor costSunk Cost- Occurred in the past and has no relation to the future cost or revenue and normally disregarded from engineering analysis.Opportunity Cost-The benefit that are foregone by engaging a business resource in a chosen activity instead of engaging that same resource in the foregone activity.

2.0 Engineering Costs and Estimating


2.0 Engineering Costs and Estimating*Fixed, variable and total costCostVolumeTotal cost = Total Fixed cost + Total Variable Cost



2.0 Engineering Costs and Estimating*Cost TerminologyRecurring cost- Cost that are repetitive and occurs when the organization produce similar product or services.Eg-Space rental, annual maintenanceNon-recurring cost- Cost that are not repetitive or one time deal charge. Eg- new machine installation, emergency maintenanceDirect cost- Cost that can be measured and allocated to specific activities. Eg- labor cost, material costIndirect cost- Cost other than direct cost that related to the product. eg.- Equipment maintenance cost, building maintenance



2.0 Engineering Costs and Estimating*LCCs TerminologySalvage values : the cost of the equipment/asset at the end of the study period. Values can be positive or negative (cost)Study period : Over which the operating expenses and ownership are evaluated.Discount rate : Rate of interest reflecting investors time value of money.



2.0 Engineering Costs and Estimating*Cost EstimatingThe basic of engineering economic analysis focuses on future consequences of current decisionsAs the exact future results are not known with certainty, the consequences must be estimated.The estimate depends largely on the accuracy of data inputExamples of estimates normally used in economic analysis:-Purchase CostAnnual revenueAnnual MaintenanceEquipment salvage value



2.0 Engineering Costs and Estimating*Types of Estimates

Types of EstimatesUsageAccuracyRough estimates High level planning Macro feasibility-30% to +60%Semi-Detailed estimatesBudgeting purposesPreliminary design decision-15% to +20%Detailed estimatesProjects detailed designContract bidding-3% to +5%



2.0 Engineering Costs and Estimating*Estimating ModelsPer unit modelUses per unit factor such as cost per square foot (meter)No allowance for economies of scale(higher quantity cost less on per unit basis)Applicable for order of magnitude type of estimatesExamples:Service cost per customerUtility cost per equipmentSegmenting model (bottom up approach)An estimate is decomposed into individual componentsEstimates are then made on those individual componentsThose individual estimates are aggregated(added) back together)



2.0 Engineering Costs and Estimating*Cost estimate-bottom up approachMaterial cost estimates for a PumpTotal Material cost estimates = $173.45

Cost ItemsUnit material cost estimatesCost ItemUnit material cost estimatesA. ChassisC. ControlsA.1 Deck7.40C.1 Handle Assembly3.85A.2 Wheels10.20C.2 Engine linkage8.55A.3 Axles4.85C.3 Blade linkage4.7022.45C.4 Speed control21.50B. Drive TrainC.5 Drive control6.70B.1 Engine38.5C.6 Cutting height7.40B.2 Starter Assembly5.9052.70B.3 Transmission5.45D. Cutting systemB.4 Drive disc assembly10.00D.1 Blade assembly10.80B5. Clutch Linkage5.15D.2 Side Chute7.05B6. Belt assembly7.70D.3 Grass bag7.7572.7025.60

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2.0 Engineering Costs and Estimating*Estimating ModelCost IndexesDimensionless number to reflect historical change in costs. Eg CPI or consumer price index.Indexes can be used to update historical costs with basic ratio relationship given in the equation below:

The equation states that the ratio of the cost index numbers at two points in time is equal to the dollar cost ratio of the item at the same times.



2.0 Engineering Costs and Estimating*Cost indexes -ExampleEstimate the annual labor cost for a new production facility given the following data:Labor cost index value was 124 ten years ago and is 188 todayAnnual labor cost for a similar facility were $575K ten years ago.Cost at Time B = $871,800



2.0 Engineering Costs and Estimating*Estimating ModelPower Sizing ModelUsed to estimate the cost of industrial plants and equipmentThe model scale up or scale down of previously known costs by taking into account the economies of scale.The model uses the following equation for the estimate:-

Where:x is the power sizing exponentCost of A and B are at the same point in timeSize or capacity of A and B are in the same physical units



2.0 Engineering Costs and Estimating*Power Sizing Exponent ValuesNote : Other values can be obtained from Plant design and economics for chemical engineers, Chemical Engineering Handbooks, other design handbooks and equipment companies.



2.0 Engineering Costs and Estimating*Power sizing ExampleA preliminary estimate is needed for the cost of building a 600MW fossil fuel plant. It is known that a 200MW plant cost $100 Million 20 year ago when the approximate cost index was 400. The cost index is now at 1200 and the power sizing exponent value for a fossil fuel plant is 0.79. 1. Use cost index to calculate the current cost of 200MW plant.Cost at Time B = $300 Mil2. Use power sizing exponent equation to calculate the cost for 600MW plant 300Cost of 600 MW =2006000.79= Cost of 600MW plant =$714mil



2.0 Engineering Costs and Estimating*Cash Flow Diagram (CFD)CFD illustrates the size, sign and timing of individual cash flows



2.0 Engineering Costs and Estimating*Cash Flow CategoriesThe expense and revenue due to engineering projects can be divided into the following categories:-First Cost Expense to build or to buy and installOperation and Maintenance (O&M) Annual expense such as repairs, utilities, laborSalvage value Receipt at project termination Revenues Annual receipt due sale of product and servicesOverhaul Major capital expenditure that occurs during the assets life.



2.0 Engineering Costs and Estimating*Cash Flow Diagram ExampleA piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost is $25000 and the equipment will have a market value of $5000 at the end 5 years. The annual O&M is $2500 and an overhaul needs to be done at the end of 3rd year at a cost of $12000. Draw the cash flow diagram.

YearCapital CostO&MOverhaulNet Cash Flow0-25000-250001-2500-25002-2500-25003-2500-12000-145004-2500-250055000-25002500



2.0 Engineering Costs and Estimating*Cash Flow Diagram ExerciseA rental company spent $2500 on a new air compressor 7 years ago. The annual rental income from the compressor has been $750. Additionally, the $100 spent on maintenance during the first year has increased each year by $25. The company plans to sell the compressor at the end of next year for $150. Construct the cash flow diagram from the companys perspective.

YearIncomeCostNet Cash Flow-70-2500-2500-6750-100650-5750-125625-4750-150600-3750-175575-2750-200550-1750-2255250750 -2505001750 + 150-275625



MINIMUM ATTRACTIVE RATE OF RETURN ( MARR )An interest rate used to convert cash flows into equivalent worth at some point(s) in timeUsually a policy issue based on: - amount, source and cost of money available for investment - number and purpose of good projects available for investment - amount of perceived risk of investment opportunities and estimated cost of administering projects over short and long run - type of organization involved

MARR is sometimes referred to as hurdle rate


CAPITAL RATIONINGEstablishing MARR involves opportunity cost viewpoint results from phenomena of CAPITAL RATIONINGExists when management decides to restrict the total amount of capital invested, by desire or limit of available capitalSelect only those projects which provide annual rate of return in excess of MARRAs amount of investment capital and opportunities available change over time, a firms MARR will also change


PRESENT WORTH METHOD ( PW )Based on concept of equivalent worth of all cash flows relative to the present as a baseAll cash inflows and outflows discounted to present at interest -- generally MARRPW is a measure of how much money can be afforded for investment in excess of costPW is positive if dollar amount received for investment exceeds minimum required by investors


FINDING PRESENT WORTHDiscount future amounts to the present by using the interest rate over the appropriate study period

PW = Fk ( 1 + i ) - k

i = effective interest rate, or MARR per compounding periodk = index for each compounding periodFk = future cash flow at the end of period kN = number of compounding periods in study periodinterest rate is assumed constant through projectThe higher the interest rate and further into future a cash flow occurs, the lower its PWk = 0N


FUTURE WORTH METHOD (FW )FW is based on the equivalent worth of all cash inflows and outflows at the end of the planning horizon at an interest rate that is generally MARR

The FW of a project is equivalent to PWFW = PW ( F / P, i%, N )

If FW > 0, it is economically justifiedFW ( i % ) = Fk ( 1 + i ) N - kk = 0Ni = effective interest ratek = index for each compounding periodFk = future cash flow at the end of period kN = number of compounding periods in study period


ANNUAL WORTH METHOD ( AW )AW is an equal annual series of dollar amounts, over a stated period ( N ), equivalent to the cash inflows and outflows at interest rate that is generally MARRAW is annual equivalent revenues ( R ) minus annual equivalent expenses ( E ), less the annual equivalent capital recovery (CR)AW ( i % ) = R - E - CR ( i % )AW = PW ( A / P, i %, N ) AW = FW ( A / F, i %, N )If AW > 0, project is economically attractiveAW = 0 : annual return = MARR earned


CAPITAL RECOVERY ( CR )CR is the equivalent uniform annual cost of the capital investedCR is an annual amount that covers:Loss in value of the assetInterest on invested capital ( i.e., at the MARR )CR ( i % ) = I ( A / P, i %, N ) - S ( A / F, i %, N ) I = initial investment for the project S = salvage ( market ) value at the end of the study period N = project study period


Example of capital recovery calculation Consider a machine that cost $10,000, last five years, and have a salvage (market) value $2,000. Thus, the loss in value of this asset over five years is $8,000. Additionally, the MARR is 10% per year. Thus

CR (i%)=I(A/P,i%,N) S(A/F,i%,N) Solution:

CR (10%) = $10,000(A/P,10%,5) - $2,000(A/F,10%,5) = $10,000 (0.2638) - $2,000 (0.1638) = $2,310.40


CAPITAL RECOVERY ( CR)CR is also calculated by adding sinking fund amount (i.e., deposit) to interest on original investmentCR ( i % ) = ( I - S ) ( A / F, i %, N ) + I ( i % )

CR is also calculated by adding the equivalent annual cost of the uniform loss in value of the investment to the interest on the salvage valueCR ( i % ) = ( I - S ) ( A / P, i %, N ) + S ( i % )


Example of Annual Worth Method:A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost $25,000, and the equipment will have a market value $5,000 at the end of a study period of five years. Increased productivity attributable to this equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the firm MARR is 20% per year, is this proposal a sound one? Use the AW method. Solution: AW (i%) = R - E CR (i%) AW (20%) = $8,000 [$25,000(A/P,20%,5) - $5,000(A/F,20%,5)] = $8,000 [$8360 - $672] = $312 Conclusion: because AW (20%) is positive, the equipment more than pays for itself over the planning horizon.


Problem A certain service can be performed satisfactorily by process X, which has a capital investment cost of $8,000, an estimated life of 10 years, no salvage value, an annual net receipts (revenue expenses) of $2,400. Assuming a MARR of 18% before income taxes, find the AW of this process and specify whether you would recommend it.

Solution: AW (i%) = R - E CR (i%) AW (18%) = $2,400 [$8,000(A/P,18%,10) - $0(A/F,18%,10)] = $2,400 [$1,780 - $0] = $620

Conclusion: because AW (18%) is positive, the process X is recommended.


INTERNAL RATE OF RETURN METHOD ( IRR )IRR solves for the interest rate that equates the equivalent worth of an alternatives cash inflows (receipts or savings) to the equivalent worth of cash outflows (expenditures)Also referred to as:investors methoddiscounted cash flow methodprofitability indexIRR is positive for a single alternative only if:both receipts and expenses are present in the cash flow patternthe sum of receipts exceeds sum of cash outflows


INTERNAL RATE OF RETURN METHOD ( IRR )IRR is i %, using the following PW formula: R k ( P / F, i %, k ) = E k ( P / F, i %, k ) R k = net revenues or savings for the kth yearE k = net expenditures including investment costs for the kth yearN = project life ( or study period )If i > MARR, the alternative is acceptableTo compute IRR for alternative, set net PW = 0

PW = R k ( P / F, i %, k ) - E k ( P / F, i %, k ) = 0

i is calculated on the beginning-of-year unrecovered investment through the life of a project Nk = 0Nk = 0Nk = 0Nk = 0


INTERNAL RATE OF RETURN PROBLEMSThe IRR method assumes recovered funds, if not consumed each time period, are reinvested at i %, rather than at MARRThe computation of IRR may be unmanageableMultiple IRRs may be calculated for the same problemThe IRR method must be carefully applied and interpreted in the analysis of two or more alternatives, where only one is acceptable


Example of Internal Rate of Return method A capital investment of $10,000 can be made in a project that will produce a uniform annual income revenue of $5,310 for five years and then have a salvage value of $2,000. Annual expenses will be $ 3,000. The company is willing to accept any project that will earn at least 10% per year on all invested capital. Determine whether it is acceptable by using IRR method.

Solution: Set net PW = 0 PW = 0 = -$10,000 + ($5,310 $3,000)(P/A,i%,5) + $2,000(P/F,i%,5) ; i% = ? By trial-and-error process; at i = 5%; PW = $1,568 at i = 15%; PW = -$1,262 By linear interpolation; i = 10.5%. Therefore, the project is minimally acceptable (i% > MARR)


Problem Determine the IRR of the following engineering project when the MARR is 15% per year. Investment cost $10,000 Expected life 5 years Salvage value -$ 1,000 Annual receipts $ 8,000 Annual expenses $ 4,000

Solution: PW = 0 = -$10,000 $1,000(P/F,i%,5) + (8,000-$4,000)(P/A,i%,5) By trial and error process, at i = 20%, PW = $1560.50 at i = 25%, PW = $ 429.50 By linear extrapolation, i = 27%, hence the project is acceptable; > MARR


Example of IRR Method: A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost $25,000, and the equipment will have a market value $5,000 at the end of a study period of five years. Increased productivity attributable to this equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the firm MARR is 20% per year, is this proposal a sound one? Use the IRR method. Solution: PW = 0 = -$25,000 + $8,000(P/A,i%,5) + $5,000(P/F,i%,5) By trial and error process, at i = 20%, PW = $ 934.30 at i = 25%, PW = -$1847.10 By linear interpolation, i = 22 %. Therefore, the project is acceptable (i% > MARR)


Example of IRR method Barron Chemical uses a thermoplastic polymer to enhance the appearance of certain RV panels. The first cost of one process was $126,000 with annual costs of $49,000 and revenues of $88,000. A salvage value of $33,000 was realized when the process was discontinued after 8 years. What rate of return did the company make on the process ?

Solution: Set FW = 0 FW = 0 = -$126,000 (F/P,i%,8) + ($88,000 - 49,000) (F/A,i%,8) + $33,000 By trial and error process, at i = 25%, FW = -$55,811 at i = 30%, FW = $64,370 By linear interpolation, i = 27.3%


LCC Selection of HVAC System Location: XDiscount Factor: 10% real for constant dollar analysisEnergy prices: Fuel type: Electricity at 30 sen /kWhUseful life of systems: 20 yearsStudy period: 20 yearsBase Date: Jan 2010. Base Case is a constant volume HVAC system with a reciprocal chiller , without night-time setback and economiser cycle. Relevant cash flow:Initial investment: RM400,000, assumed to occur in a lump sump.Replacementcost for fan at the end of year 12: RM36,000Salvage value at the end of 20-year period: RM14,000Annual electricity cost (250,000 kWh at 30 sen/kWh): RM75,000Annual OM&R costs: RM30,000Cash flow Diagram:Total LCC:


LCC Selection of HVAC System Location: XDiscount Factor: 10% real for constant dollar analysisEnergy prices: Fuel type: Electricity at 30 sen /kWhUseful life of systems: 20 yearsStudy period: 20 yearsBase Date: Jan 2010.

Alternative: Energy Saving DesignRelevant Cash Flow:Initial investment: RM440,000, assumed to occur in a lump sump.Replacementcost for fan at the end of year 12: RM38,000Salvage value at the end of 20-year period: RM15,000Annual electricity cost (162,500 kWh at 30 sen/kWh): RM48,750Annual OM&R costs: RM32,000Cash flow Diagram:Total LCC:


LCC Conventional HVAC System Location: XDiscount Factor: 10% real for constant dollar analysisEnergy prices: Fuel type: Electricity at 30 sen /kWhUseful life of systems: 20 yearsStudy period: 22 yearsBase Date: Jan 2010.

Cost ItemsBase date Cost (RM)Year of occur-renceDiscount factor @ i=10%Present valueInitial investment cost1st installment2nd installment206,000206,00012Capital replacement (fan)48,00014Capital replacement (Plant)240,00017Salvage value80,00022Electricity125,000 kWh at 0.3/kWh37500annualNatural gas1700 GJ at RM24/GJ40800annualOM&R28000annualTotal LCC


LCC Energy saving HVAC System Location: XDiscount Factor: 10% real for constant dollar analysisEnergy prices: Fuel type: Electricity at 30 sen /kWhUseful life of systems: 20 yearsStudy period: 22 yearsBase Date: Jan 2010.

Cost ItemsBase date Cost (RM)Year of occur-renceDiscount factor @ i=10%Present valueInitial investment cost1st installment2nd installment220,000220,00012Capital replacement (fan)50,00014Salvage value80,00022Electricity100,000 kWh at 0.3/kWh37500annualNatural gas1180 GJ at RM24/GJ40800annualOM&R32000annualTotal LCC


*Back Up

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THE EXTERNAL RATE OF RETURN METHOD ( ERR )ERR directly takes into account the interest rate ( ) external to a project at which net cash flows generated over the project life can be reinvested (or borrowed ).

If the external reinvestment rate, usually the firms MARR, equals the IRR, then ERR method produces same results as IRR method


CALCULATING EXTERNAL RATE OF RETURN ( ERR )1. All net cash outflows are discounted to the present (time 0) at % per compounding period.2. All net cash inflows are discounted to period N at %.3. ERR -- the equivalence between the discounted cash inflows and cash outflows -- is determined. The absolute value of the present equivalent worth of the net cash outflows at % is used in step 3.A project is acceptable when i % of the ERR method is greater than or equal to the firms MARR


CALCULATING EXTERNAL RATE OF RETURN ( ERR ) Ek ( P / F, %, k )( F / P, i %, N ) = Rk ( F / P, %, N - k )

Rk = excess of receipts over expenses in period kEk = excess of expenses over receipts in period kN = project life or period of study = external reinvestment rate per periodNk = 0Nk = 0i %= ?TimeN0 Rk ( F / P, %, N - k )Nk = 0 Ek ( P / F, %, k )( F / P, i %, N )Nk = 0


ERR ADVANTAGESERR has two advantages over IRR:1. It can usually be solved for directly, rather than by trial and error.2. It is not subject to multiple rates of return.


Example of ERR Method: A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost $25,000, and the equipment will have a market value $5,000 at the end of a study period of five years. Increased productivity attributable to this equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the external investment rate = MARR = 20% per year, what is the alternatives ERR, and is this proposal a sound one?

Solution: $25,000(F/P,i%,5) = $8,000(F/A,20%,5) + $5,000 (F/P,i%,5) = $64,532.80/$25,000 = 2.5813 i = 20.88%

Since i > MARR, the alternative is minimally justified.


Problem 4.32 Summary of the projected costs and annual receipts for a new product line is presented as follows:End of Year Net Cash Flow 0 - $450,000 1 - 42,500 2 92,800 3 386,000 4 614,600 5 - 202,200 The companys external reinvestment rate per year = MARR = 10% per yearSolution:[$450,000 + $42,500(P/F,10%,1) + $202,200(P/F,10%,5)](F/P,i%,5) = $92,800(F/P,10%,3) + $386,000(F/P,10%,2) + $614,600(F/P,10%,1)$614,182.73(F/P,i%,5) = $1,265,544 (F/P,i%,5) = 2.0622By interpolation, i% , ERR = 15.6%


Problem 4.34 Given a cash flow as stated below:

End of Year Net Cash Flow 0 - $10,000 1 7 1,400 7 10,000 The external rate of reinvestment for the company = 8% per year.

Solution: $10,000(F/P.i%,7) = $1,400(F/A,8%,7) + $10,000 (F/P,i%,7) = $22,491.92 / 10,000 = 2.2492

By interpolation, ERR = i% = 12.3%


Problem A certain project has a net receipts equaling $1,000 now, has costs of $5,000 at the end of the first year, and earns $6,000 at the end of the second year. If the external reinvestment rate of 10% is available, what is the rate of return for this project using the ERR method ?

Solution:

$5,000(P/F,10%,1)(F/P,i%,2) = $1,000(F/P,10%,2) + $6,000 (F/P,i%,2) = $7,210 / $4,545.50 = 1.5862

By interpolation, ERR = i% = 25.9%


PAYBACK PERIOD METHODSometimes referred to as simple payout methodIndicates liquidity (riskiness) rather than profitabilityCalculates smallest number of years ( ) needed for cash inflows to equal cash outflows -- break-even life ignores the time value of money and all cash flows which occur after ( Rk -Ek) - I > 0If is calculated to include some fraction of a year, it is rounded to the next highest yeark = 1


PAYBACK PERIOD METHODThe payback period can produce misleading results, and should only be used with one of the other methods of determining profitabilityA discounted payback period ( where < N ) may be calculated so that the time value of money is considered

i is the MARR I is the capital investment made at the present time ( k = 0 ) is the present time is the smallest value that satisfies the equation ( Rk - Ek) ( P / F, i %, k ) - I > 0k = 1


Simple Payback Period vs. Discounted Payback Period

EOY Net c / flow Cum PW @ i=0% PW @ i=20% Cum PW @ MARR 20% 0 - $25,000 - $25,0000 - $25,000 - $25,000 1 8,000 - 17,000 6,667 - 18,333 2 8,000 - 9,000 5,556 - 12,777 3 8,000 - 1,000 4,630 - 8,147 4 8,000 7,000 3,858 - 4,289 5 13,000 20,000 5,223 934

The payback period is at 4th years, because the cumulative balance turns positive at EOY 4[time value of money is not considered]The payback period is at 5th year, because the cumulative discounted balance turns positive at EOY 5[time value of money is considered]

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2.0 engineering costs and cost estimating rev 2

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