Engineering Project Economics Lecture 6 PDF

ENGINEERING PROJECT ECONOMICS Lecture 6 Present Worth Analysis Sherif H. El-Gohary , Phd Sherif ElGohary, Ph.D. [email protected] Recap of Main Points in Last Lecture Must understand: interest period, compounding period, compounding...

ENGINEERING PROJECT ECONOMICS Lecture 6 Present Worth Analysis Sherif H. El-Gohary , Phd Sherif ElGohary, Ph.D. [email protected] Recap of Main Points in Last Lecture Must understand: interest period, compounding period, compounding frequency, and payment period Always use effective rates in interest formulas i = (1 + r / m)m – 1 Interest rates are stated different ways; must know how to get effective rates For continuous compounding, use i = er – 1 to get effective rate Sherif ElGohary, Ph.D.  Introduction The evaluation and selection of economic proposals require: – Cash flow estimates over a stated period of time, – Mathematical techniques to calculate the measure of worth, – A guideline for selecting the best proposal. Sherif ElGohary, Ph.D.  The Single Project Called “The Unconstrained Project Selection Problem; No comparison to competing or alternative projects; Acceptance or Rejection is based upon: – Comparison with the firm’s opportunity cost; – Opportunity Cost is always present; Sherif ElGohary, Ph.D.  Projects to Alternatives Creation of Alternatives. Ideas, Data, Experience, Plans, And Estimates Generation of Proposals P1 P2 Pn 5 Sherif ElGohary, Ph.D.  Projects to Alternatives Proposal Assessment. P1 Pj Economic Analysis P2 And Assessment. Pk. Infeasible or Pn Rejected! Viable POK POK 6 Sherif ElGohary, Ph.D.  Assessing Alternatives Feasible Alternatives Economic Analysis And Assessment.. POK Feasible Set POK Mutually Exclusive Set OR Independent Set 7 Sherif ElGohary, Ph.D.  Formulating Alternatives Two types of economic proposals Mutually Exclusive (ME) Alternatives: Only one can be selected; Compete against each other Independent Projects: More than one can be selected; Compete only against DN Do Nothing (DN) – An ME alternative or independent project to maintain the current approach; no new costs, revenues or savings Sherif ElGohary, Ph.D.  Mutually Exclusive Set  Only one of the feasible (viable) projects can be selected from the set.  Once selected, the others in the set are “excluded”.  Each of the identified feasible (viable) projects is (are) considered an “alternative”.  It is assumed the set is comprised of “do-able”, feasible alternatives.  ME alternatives compete with each other! Sherif ElGohary, Ph.D.  Mutually Exclusive Set Do Nothing Alt. 1 Analysis Problem Alt. 2 Selection Alt. Execute! Sherif ElGohary, Ph.D.  The Independent Project Set Independent Set – Given the alternatives in the set: – More than one can be selected; – Deal with budget limitations; – Project Dependencies and relationships. More Involved Analysis – – Often formulated as a 0-1 Linear Programming model – With constraints and an objective function. May or may not compete with each other – depends upon the conditions and constraints that define the set! Sherif ElGohary, Ph.D.  THE PRESENT WORTH METHOD A process of obtaining the equivalent worth of future cash flows BACK to some point in time. At an interest rate usually equal to or greater than the Organization’s established MARR. Sherif ElGohary, Ph.D.  Minimum Attractive Rate of Return MARR is a reasonable rate of return (percent) established for evaluating and selecting alternatives  An investment is justified economically if it is expected to return at least the MARR  Also termed hurdle rate, benchmark rate and cutoff rate Sherif ElGohary, Ph.D.  Net – Present Worth Criterion Evaluation of a single project Step 1: Determine the interest rate that the firm wishes to earn on its investments. Firm always invest the money in its investment pool. This interest rate often referred to required rate of return or a minimum attractive rate of return (MARR). Usually this selection is a policy decision made by the top management. Step 2: Estimate the service life of a project. Step 3: Estimate the cash inflow for each period over the service life. Step 4: Estimate the cash outflow for each period over the service life. Step 5: Determine the net cash flow for each period net cash flow = cash inflow – cash outflow Sherif ElGohary, Ph.D. Basic parameters needed to be considered An engineering economic analysis evaluates cash flow estimates for parameters such as: Initial cost Annual costs and revenues Nonrecurring costs Possible salvage value over an estimated useful life of the product; process, or service. Sherif ElGohary, Ph.D.  Net – Present Worth Criterion Step 6: Find the present worth of each net cash flow at the MARR. Add up these present worth figures; their sum is identified as the project’s NPW. Step 7: In this context, a positive NPW means that the equivalent worth of the inflows is greater than the equivalent worth of outflows so that the project makes a profit. If PW( i ) > 0, accept the investment. If PW( i ) = 0, remain indifferent to the investment If PW( i ) < 0, reject the investment Sherif ElGohary, Ph.D.  Net Present Worth (NPW) Measure  Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i.  Decision Rule: Accept the project if the net surplus is positive. Inflow 0 1 2 3 4 5 Outflow Net surplus PW(i ) inflow 0 PW (i ) > 0 PW(i )outflow Sherif ElGohary, Ph.D.  Selection of Alternatives byPW For the alternatives shown below, which should be selected if they are (a) mutually exclusive; (b) independent? Project ID Present Worth A $30,000 B $12,500 C $-4,000 D $ 2,000 Solution: (a) Select numerically largest PW; alternative A (b) Select all with PW > 0; projects A, B & D Sherif ElGohary, Ph.D.  Example: Three Alternatives Assume i = 10% per year A1 A2 A3 Electric Power Gas Power Solar Power First Cost: -2500 First Cost: -3500 First Cost: -6000 Ann. Op. Cost: -900 Ann. Op. Cost: -700 Ann. Op. Cost: -50 Sal. Value: +200 Sal. Value: +350 Sal. Value: +100 Life: 5 years Life: 5 years Life: 5 years Which Alternative – if any, Should be selected based upon a present worth analysis? Sherif ElGohary, Ph.D.  Example: Cash Flow A1: Electric Diagrams FSV = 200 0 1 2 3 4 5 -2500 A = -900/Yr. FSV = 350 A2: Gas 0 1 2 3 4 5 A = -700/Yr. -3500 FSV = 100 A3:Solar 0 1 2 3 4 5 A = -50/Yr. -6000 i = 10%/yr and n = 5 Sherif ElGohary, Ph.D.  Calculate the Present Worth's Present Worth's are: 1. PWElec. = -2500 - 900(P/A,10%,5) + 200(P/F,10%,5) = $-5788 2. PWGas = -3500 - 700(P/A,10%,5) + 350(P/F,10%,5) = $-5936 3. PWSolar = -6000 - 50(P/A,10%,5) + 100(P/F,10%,5) = $-6127 Select “Electric” which has the min. PW Cost! Sherif ElGohary, Ph.D.  Present Worth – A Function of the assumed interest rate. Present Worth transforms the future cash flows into: – Equivalent Dollars NOW! – One then COMPARES alternatives using the present dollars of each alternative. Problem: – Present Worth requires that the lives of all alternatives be EQUAL. Sherif ElGohary, Ph.D.  Life-Cycle Cost Analysis LCC analysis includes all costs for entire life span, from concept to disposal Best when large percentage of costs are M&O Includes phases of acquisition, operation, & phaseout Use PW analysis if there are revenues and other benefits considered Sherif ElGohary, Ph.D.  Phases and stages of systems engineering Acquisition phase: all activities prior to the delivery of products and services. Requirements definition stage—Includes determination of user/customer needs, assessing them relative to the anticipated system, and preparation of the system requirements documentation. Preliminary design stage—Includes feasibility study, conceptual, and early-stage plans; final go–no go decision is probably made here. Detailed design stage—Includes detailed plans for resources— capital, human, facilities, information systems, marketing, etc.; there is some acquisition of assets, if economically justifiable. Sherif ElGohary, Ph.D. Phases and stages of systems engineering Operation phase: all activities are functioning, products and services are available. Construction and implementation stage—Includes purchases, construction, and implementation of system components; testing; preparation, etc. Usage stage—Uses the system to generate products and services; the largest portion of the life cycle. Phase-out and disposal phase: covers all activities to transition to a new system; removal/recycling/disposal of old system. Sherif ElGohary, Ph.D. Typical Life-Cycle Cost Distribution by Phase Sherif ElGohary, Ph.D.  Different Lives With alternatives with Un-equal lives the rule is: The PW of the alternatives must be compared over the same number of years. Called “The Equal Service” requirement Sherif ElGohary, Ph.D.  Two Approaches for Unequal Lives IF present worth is to be applied, there are two approaches one can take to the unequal life situation: 1. Least common multiple (LCM) of their lives: Compare the alternatives over a period of time equal to the least common multiple (LCM) of their lives. 2. The planning horizon approach: Compare the alternatives using a study period of length n years, which does not necessarily take into consideration the useful lives of the alternatives! Sherif ElGohary, Ph.D. (The LCM procedure is used unless otherwise specified)  LCM Approach The assumptions of a PW analysis of different- life alternatives are a follows: 1.The service provided by the alternatives will be needed for the LCM of years or more. 2.The selected alternative will be repeated over each life cycle of the LCM in exactly the same manner. 3.The cash flow estimates will be the same in every life cycle. Sherif ElGohary, Ph.D. Example: Different-Life Alternatives Compare the machines below using present worth analysis at i = 10% per year Machine A Machine B First cost, $ 20,000 30,000 Annual cost, $/year 9000 7000 Salvage value, $ 4000 6000 Life, years 3 6 Solution: LCM = 6 years; repurchase A after 3 years PWA = -20,000 – 9000(P/A,10%,6) – 16,000(P/F,10%,3) + 4000(P/F,10%,6) = $-68,961 20,000 – 4,000 in PWB = -30,000 – 7000(P/A,10%,6) + 6000(P/F,10%,6) year 3 = $-57,100 Select alternative B Sherif ElGohary, Ph.D.  LCM Observations For the LCM method: – Becomes tedious; – Numerous calculations to perform; – The assumptions of repeatability of future cost/revenue patterns may be unrealistic. However, in the absence of additional information regarding future cash flows, this is an acceptable analysis approach for the PW method. Sherif ElGohary, Ph.D. PW Evaluation Using a Study Period Once a study period is specified, all cash flows after this time are ignored Salvage value is the estimated market value at the end of study period Short study periods are often defined by management when business goals are short-term Study periods are commonly used in equipment replacement analysis Sherif ElGohary, Ph.D.  Example: Study Period PW Evaluation Compare the alternatives below using present worth analysis at i = 10% per year and a 3-year study period Machine A Machine B First cost, $ -20,000 -30,000 Annual cost, $/year -9,000 -7,000 Salvage/market value, $ 4,000 6,000 (after 6 years) 10,000 (after 3 years) Life, years 3 6 Solution: Study period = 3 years; disregard all estimates after 3 years PWA = -20,000 – 9000(P/A,10%,3) + 4000(P/F,10%,3) = $-39,376 PWB = -30,000 – 7000(P/A,10%,3) + 10,000(P/F,10%,3) = $-39,895 Marginally, select A; different selection than for LCM = 6 years Sherif ElGohary, Ph.D.  Practice Problem Sherif ElGohary, Ph.D.  Solution-a Sherif ElGohary, Ph.D.  Solution-a Since the equipment has different lives, compare them over the LCM of 18 years. For life cycles after the first, the first cost is repeated in year 0 of each new cycle, which is the last year of the previous cycle. These are years 6 and 12 for vendor A and year 9 for B. Vendor B is selected, since it costs less in PW terms; that is, the PWB value is numerically larger than PWA Sherif ElGohary, Ph.D.  Solution-b Sherif ElGohary, Ph.D. Future Worth Analysis FW exactly like PW analysis, except calculate FW Must compare alternatives for equal service (i.e. alternatives must end at the same time) Two ways to compare equal service: Least common multiple (LCM) of lives Specified study period (The LCM procedure is used unless otherwise specified) Sherif ElGohary, Ph.D. FW of Different-Life Alternatives Compare the machines below using future worth analysis at i = 10% per year Machine A Machine B First cost, $ -20,000 -30,000 Annual cost, $/year -9000 -7000 Salvage value, $ 4000 6000 Life, years 3 6 Solution: LCM = 6 years; repurchase A after 3 years FWA = -20,000(F/P,10%,6) – 9000(F/A,10%,6) – 16,000(F/P,10%,3) + 4000 = $-122,168 FWB = -30,000(F/P,10%.6) – 7000(F/A,10%,6) + 6000 = $-101,157 Select B (Note: PW and FW methods will always result in same selection) Sherif ElGohary, Ph.D.  Capitalized Cost Analysis A perpetual or infinite life is the effective planning horizon. Permanent endowments for charitable organizations and universities also have perpetual lives. The economic worth of these types of projects or endowments is evaluated using the present worth of the cash flows. Sherif ElGohary, Ph.D.  Capitalized Cost (CC) Analysis CC refers to the present worth of a project with a very long life, that is, PW as n becomes infinite Basic equation is: CC = P = Aw i “Aw” essentially represents the interest on a perpetual investment For example, in order to be able to withdraw $50,000 per year forever at i = 10% per year, the amount of capital required is 50,000/0.10 = $500,000 For finite life alternatives, convert all cash flows into an A value over one life cycle and then divide by i Sherif ElGohary, Ph.D.  Example: Capitalized Cost Compare the machines shown below on the basis of their capitalized cost. Use i = 10% per year Machine 1 Machine 2 First cost,$ -20,000 -100,000 Annual cost,$/year -9000 -7000 Salvage value, $ 4000 ----- Life, years 3 ∞ Solution: Convert machine 1 cash flows into A and then divide by i A1 = -20,000(A/P,10%,3) – 9000 + 4000(A/F,10%,3) = $-15,834 CC1 = -15,834 / 0.10 = $-158,340 CC2 = -100,000 – 7000/ 0.10 = $-170,000 Select machine 1 Sherif ElGohary, Ph.D. Procedure to determine the CC for an infinite sequence of cash flows is as follows: 1. Draw a cash flow diagram showing all nonrecurring (one-time) cash flows and at least two cycles of all recurring (periodic) cash flows. 2. Find the present worth of all nonrecurring amounts. This is their CC value. 3. Find the A value through one life cycle of all recurring amounts. Add this to all other uniform amounts (A) occurring in years 1 through infinity. The result is the total equivalent uniform annual worth (AW). 4. Divide the AW obtained in step 3 by the interest rate i to obtain a CC value. 5. Add the CC values obtained in steps 2 and 4. Sherif ElGohary, Ph.D. Practice Problem Sherif ElGohary, Ph.D. 5-44  Nonrecurring cache flow Sherif ElGohary, Ph.D. 5-45  Recurring cache flow Sherif ElGohary, Ph.D. 5-46  Total Capitalized cost: Sherif ElGohary, Ph.D. 5-47 Summary of Important Points PW method converts all cash flows to present value at MARR Alternatives can be mutually exclusive or independent Cash flow estimates can be for revenue or cost alternatives PW comparison must always be made for equal service Equal service is achieved by using LCM or study period Capitalized cost is PW of project with infinite life; CC = P = A/i Sherif ElGohary, Ph.D. Sherif ElGohary, Ph.D.