Present Worth Analysis

Chapter 4 and 5Application of Money-Time Relationship1Illustrate cash flow diagrams to show project costs and benefitsExplain and provide examples of the time value of moneyDiscuss and apply the present worth criteria (PW)Compare two competing alternatives choices using PWTopic Outcome (TO)2Discuss equivalent uniform annual cost (EUAC) and equivalent uniform annual benefits (EUAB)Analyze an engineering economic analysis problems into its annual cash flow equivalentApply future worth, benefit-cost ratio, payback period, breakeven and sensivity analysis methods in engineering economics calculationEvaluate project cash flows using the internal rate of return (IRR) methodologyApply EUAC and EUAB to compare alternatives with equal, common multiple or continuos lives, or over some fixed study period in engineering economics calculationTopic Outcome (TO)3 Topic to be covered-Minimum Attractive Rate of Return-Present Worth AnalysisPresent worth analysis for equal life alternativesPresent worth analysis for different life alternativesInfinite Analysis Period: Capitalized CostMultiple Alternatives

Present Worth Analysis4Two types of projectsMutually exclusive projectsOnly one of the viable projects can be selectedProjects compete with one anotherIndependent ProjectsMore than one viable project may be selected. A do-nothing (DN) option is considered one of the alternativesThese projects do not compete with each other. Each project is evaluated separately and comparison is between one or groups of projects and a do-nothing (DN) optionIf there are m independent projects, there will be 2m alternatives availableTypes of Projects5An 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 investmentnumber and purpose of good projects available for investmentamount of perceived risk of investment opportunities and estimated cost of administering projects over short and long runtype of organization involvedMARR is sometimes referred to as hurdle rateMinimum Attractive Rate Of Return ( MARR )6Based 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 investorsPresent Worth Analysis (PW)7Economic Criteria Restated Present Worth TechniqueSituationCriterionFixed InputAmount of capital available fixedMaximize present worth of benefitsFixed Output$ amount of benefit is fixedMaximize present worth of costsNeither FixedNeither capital nor $ benefits are fixedMaximize net present worth (NPV)8Economic Criteria Restated Present Worth TechniqueSituation ExampleCriterionAFixed Input$150,000 max.Maximum square feet of building for the priceBFixed Output20,000 ft2 building availableNegotiate for minimum cost/ft2CNeither Fixed$150,000 max 15-20,000 ft2Simultaneously negotiate for maximum building size and minimum cost/ft29PRESENT WORTH (PW) ANALYSIS FOR EQUAL LIFE ALTERNATIVES

If there is one alternative, calculate PW at MARR. If PW>0, accept projectIf there are two or more alternatives, calculate PW at MARR. Select project with highest numerical PWFor independent projects, select all projects with PW>0 at the MARRPresent Worth Analysis (PW)10Useful Lives Equal to the Analysis PeriodSituation ExampleCriterionFixed InputAmount of money or other input resources are fixedMaximize present worth of benefits or other outputsFixed OutputThere is a fixed task, benefit or other output to be accomplishedMinimize present worth of costs or other inputsNeither input nor output is fixedNeither amount of money, or other inputs, nor amount of benefits, or other output, is fixedMaximize (present worth of benefits minus present worth of costs), that is, maximize net present worth11PRESENT WORTH (PW) ANALYSIS FOR DIFFERENT LIFE ALTERNATIVESWhen projects have different lives (e.g. 3 years and 5 years), the PW of the alternatives must be compared over the same numbers of yearsThis can be achieved by:Comparing alternatives over a time period equal to the least common multiple (LCM) of their livesComparing alternatives using a study period of say, x years

Present Worth Analysis (PW)12Analysis period must be considered

Useful life of the alternative equals the analysis periodAlternatives have useful lives different from the analysis periodThe analysis period is infinite, n=Appling Present Worth Technique13A firm is considering which of two mechanical device to install to reduce costs in a particular situation. Both devices cost $1000 and have useful lives of 5 years and no salvage value. Device A can be expected to result in $300 savings annually. Device B will provide cost savings of $400 the first year but will decline $50 annually, making the second year saving $350, the third-year saving $300, and so fourth. With interest at 7%, which device should the firm purchase?Practice 114Wayne Country will build an aqueduct to bring water in from the upper part of the state. It can be built at a reduced size now for $300 million and be enlarged 25 years hence for an additional $350 million. An alternative is to construct the full-sized aqueduct now for $400 million. Both alternative would provide the needed capacity for the 50-year analysis period. Maintenance costs are small and maybe ignored. At 6% interest, which alternative should be selected?Practice 215Practice 3Perform a present worth analysis of equal-service machines with the cost shown below, if the MARR is 10% per year. The salvage life is 5 years and revenues for all three alternatives are expected to be the same.

Electric-PoweredGas-PoweredSolar-PoweredFirst Cost, $-2500-3500-$6000Annual Operating Cost, $-900-700-50Salvage Value, $20035010016Practice 4A purchasing agent is considering the purchase of some new equipment for the mailroom. Two different manufacturers have provided quotation as indicates the following table. Assuming MARR of 10%, which of the alternatives is more preferred.

AlternativesCostUniform Annual BenefitUseful Life (years)End-of-Useful-LifeSalvage ValueAltas Scale$2000$4505$100Tom Thumb Scale$3000$6005$70017Useful Lives Different from the Analysis PeriodPractice 5A diesel manufacturer is considering the two alternative production machines graphically depicted in Figure 5.1. Specific data are as follows:

The manufacturer uses an interest rate of 8% and wants to use the PW method to compare these alternatives over an analysis period of 10 years

Alternative 1Alternative 2Initial Cost$50000$75000Estimated salvage value at the end of useful life$10000$12000Useful life of equipment, in years713Alternative 1Alternative 2Estimated marker value, end of 10-year analysis period$20,000$1500018Practice 6A project engineer with EnviroCare is assigned to start a new office in a city where a 6-year contract has been finalized to take and to analyze ozone-level readings. Two lease options are available, each with a first cost, annual lease cost and deposit-return estimates shown below.

By taking MARR as 15% per yearEnviroCare has a standard practice of evaluating all projects over a 5-year period. If a study period of 5 year is used and the deposit returns are not expected to change, which location should be selected?Determine which lease option should be selected on the basis of present worth comparison if 5-year period is neglected.

Location ALocation BFirst Cost, $-15,000-18,000Annual lease cost, $-3,500-3,100Deposit return, $1,0002,000Lease term, years6919For:n = A = Pi

Then:Capitalized Cost, P = A/iThis requires first computing the future cost into an equivalent AInfinite Analysis Period: Capitalized Cost20How much should one set aside to pat $50 per year for maintenance on a gravesite if interest is assumed to be 4%? For perpetual maintenance, the principle sum must remain undiminished after making the annual disbursementPractice 621A city plants a pipeline to transport water from a distant watershed area to the city. The pipeline will cost $8 million and will have an expected life of 70 years. In addition, it will need $35,000 yearly maintenance. The city anticipates it will need to keep the water line in service indefinitely. Compute the capitalized cost, assuming 7% interest.Practice 722When more than one alternatives exists, we generally want to compare the alternatives in a single table

The goal is to develop a single table that automates as many of the calculations as possible.

Multiple Alternatives23Practice 8A purchasing agent is considering the purchase of new pipes. The analysis came up with different pipe size alternatives as shown below

The pipe and pump will have a salvage value at the end of 5 years, equal to the cost to remove them. The pump will operate 2000 hours per year. The lowest interest rate at which the contractor is willing to invest money is 7%. (the minimum required interest rate for invested money is called the minimum attractive rate of return, or MARR). Select the alternative with the least present worth of cost.

Pipe Sizes (in.)2345Installed cost of pipeline and pump$22000$23000$25000$30000Cost per hour for pumping$1.20$0.65$0.50$0.4024Practice 9An investor paid $8000 to a consulting firm to analyze what he might do with a small parcel of land on the edge of town that can be bought for $30,000. In their report, the consultants suggested four alternatives:

*includes the land and structures but does not include the $8000 fee to the consulting firmAssuming 10% is the minimum attractive rate of return, what should the investor do?

AlternativesTotal Investment Including Land*Uniform Net Annual BenefitTerminal Value at End of 20 yearsADo nothing $0$0$0BVegetable Market50000510030000CGas Station950001050030000D Small Motel3500003600015000025Practice 10Two pieces of construction equipment are being analyzed. Based on an 8% interest, which alternative should be selected?

CostAlternative AAlternative B0-$2000-$15001+1000+7002+850+3003+700+3004+550+3005+400+3006+400+4007+400+5008+400+60026