Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. CHAPTER V PRESENT WORTH ANALYSIS McMc Graw Hill ENGINEERING ECONOMY Fifth Edition Blank and Tarquin

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2 CHAPTER TOPICS Formulating Alternatives PW of Equal-Life Alternatives PW of Different-Life alternatives Future Worth Analysis Payback Period PW of Bonds Spreadsheet Applications

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. CHAPTER V 5.1 FORMULATING ALTERNATIVES McMc Graw Hill ENGINEERING ECONOMY Fifth Edition Blank and Tarquin

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display FORMULATING MUTUALLY EXCLUSIVE ALTERNATIVES Viable firms/organizations have the capability to generate potential beneficial projects for potential investment Two types of investment categories Mutually Exclusive Set Independent Project Set

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display FORMULATING MUTUALLY EXCLUSIVE ALTERNATIVES Mutually Exclusive set is where a candidate set of alternatives exist (more than one) Objective: Pick one and only one from the set. Once selected, the remaining alternatives are excluded.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display INDEPENDENT PROJECT SET Given a set of alternatives (more than one) The objective is to: Select the best possible combination of projects from the set that will optimize a given criteria. Subjects to constraints

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Evaluating Alternatives Part of Engineering Economy is the selection and execution of the best alternative from among a set of feasible alternatives Alternatives must be generated from within the organization One of the roles of engineers!

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Alternatives Problem Do Nothing Alt. 1 Alt. 2 Alt. m Analysis Selection Execution

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Type of Alternatives Revenue/Cost – the alternatives consist of cash inflow and cash outflows Select the alternative with the maximum economic value Service – the alternatives consist mainly of cost elements Select the alternative with the minimum cost value

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. CHAPTER V 5.2 Present Worth Approach McMc Graw Hill ENGINEERING ECONOMY Fifth Edition Blank and Tarquin

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Present Worth Approach A process of transforming all of the current and future estimated cash flow back to a point in time (time t = 0) – called the Present Worth P(i%) = P( + cash flows) +P( - cash flows)

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display THE PRESENT WORTH METHOD If P(i%) > 0 then the project is deemed acceptable. If P(i%) < 0 – the project is usually rejected. If P(i%) = 0 Present worth of costs = Present worth of revenues – Indifferent!

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display THE PRESENT WORTH METHOD – Depends upon the Discount Rate Used The present worth is purely a function of the discount rate. If one changes the discount rate, a different present worth will result.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Equal Lives – Straightforward! Given two or more alternatives with equal lives…. Alt. 1 Alt. 2 Alt. N N = for all alternatives Find PW(i%) for each alternative then compare Smallest if cost and largest if profit.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display PRESENT WORTH: Example Consider:Machine AMachine B First Cost$2,500$3,500 Annual Operating Cost Salvage Value Life5 years5 years i = 10% per year Which alternative should we select? Note: Where costs dominate a problem it is customary to assign a positive value to cost and negative to inflows

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display PRESENT WORTH: Cash Flow Diagram $2,500 A = $900 F 5 =$200 MAMA $3,500 F 5 =$200 A = $700 MBMB

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display PRESENT WORTH: Solving P A = 2, (P|A,.10, 5) – 200 (P|F,.01, 5) = 2, (3.7908) (.6209) = 2, , = $5,788 P B = 3, (P|A,.10, 5) – 350 (P|F,.10, 5) = 3, , = $5,936 SELECT MACHINE A: Lower PW cost!

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. CHAPTER V 5.3 PRESENT WORTH: Different Lives McMc Graw Hill ENGINEERING ECONOMY Fifth Edition Blank and Tarquin

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display PRESENT WORTH: Different Lives Comparison must be made over equal time periods Compare over the least common multiple, LCM, for their lives Example: {3,4, and 6} years.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display PRESENT WORTH: Example Unequal Lives EXAMPLE Machine AMachine B First Cost$11,000$18,000 Annual Operating Cost 3,500 3,100 Salvage Value 1,000 2,000 Life6 years9 years i = 15% per year Which alternative should we select?

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display PRESENT WORTH: Example Unequal Lives A common mistake is to compute the present worth of the 6-year project and compare it to the present worth of the 9-year project. NO! NO! NO!

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display PRESENT WORTH: Unequal Lives i = 15% per year $11,000 F 6 =$1,000 A 1-6 =$3,500 Machine A F 6 =$2,000 A 1-9 =$3,100 $18,000 Machine B LCM(6,9) = 18 year study period will apply for present worth

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Unequal Lives: 2 Alternatives i = 15% per year Machine A Cycle 1 for ACycle 2 for ACycle 3 for A 6 years Cycle 1 for B Cycle 2 for B 18 years 9 years Machine B

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Example: Unequal Lives Solving LCM = 18 years Calculate the present worth of a 6-year cycle for A P A = 11, ,500 (P|A,.15, 6) – 1,000 (P|F,.15, 6) = 11, ,500 (3.7845) – 1,000 (.4323) = $23,813, which occurs at time 0, 6 and 12

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Example: Unequal Lives P A = 23,813+23,813 (P|F,.15, 6)+ 23,813 (P|F,.15, 12) = 23, , ,451 = 38, $23,813 Machine A

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Unequal Lives Example: Machine B Calculate the Present Worth of a 9-year cycle for B F 6 =$2,000 A 1-9 =$3,100 $18,000

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Year Cycle for B Calculate the Present Worth of a 9-year cycle for B P B = 18,000+3,100(P|A,.15, 9) – 1,000(P|F,.15, 9) = 18, ,100(4.7716) - 1,000(.2843) = $32,508 which occurs at time 0 and 9

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Alternative B – 2 Cycles P B = 32, ,508 (P|F,.15, 9) = 32, ,508(.2843) P B = $41,750 Choose Machine A $32,508 Machine A: PW =$38,558

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Unequal Lives – Assumed Study Period Study Period Approach Assume alternative: 1 with a 5-year life Alternative: 2 with a 7-year life Alt-1: N = 5 yrs Alt-2: N= 7 yrs LCM = 35 yrs Could assume a study period of, say, 5 years.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Unequal Lives – Assumed Study Period Assume a 5-yr. Study period Estimate a salvage value for the 7-year project at the end of t = 5 Truncate the 7-yr project to 5 years Alt-1: N = 5 yrs Alt-2: N= 7 yrs Now, evaluate both over 5 years using the PW method!

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. CHAPTER V 5.4 Future Worth Approach McMc Graw Hill ENGINEERING ECONOMY Fifth Edition Blank and Tarquin

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display FUTURE WORTH APPROACH FW(i%) is an extension of the present worth method Compound all cash flows forward to some specified time using (F/P), (F/A),… factors or, Given P, find F = P(1+i) N

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Future Worth Example (Figure 5.3) Example 5.3(P172) Calculate the Future Worth of determining the selling price in order to earn exactly 25% on the investment Draw the cash-flow diagram!!

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. CHAPTER V 5.6 Payback Period Analysis McMc Graw Hill ENGINEERING ECONOMY Fifth Edition Blank and Tarquin

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Payback Period Analysis Payback is the period of time it takes for the cash flows to recover the initial investment. Two forms for this method Discounted Payback Period (uses an interest rate) Conventional Payback Period (does not use an interest rate)

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Payback Period Analysis Discounted Payback Approach Find the value of n p such that:

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Payback Period Analysis - Example Example 5.8 Machine 1: N=7 Machine 2: N=14 i = 15%

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Payback Period Analysis - Example 5.8 Tabular Format: Machine 1

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Payback Period Analysis- Machine A Payback is between 6 and 7 yeas (6.57 yrs) PW(15%)= +$481.00

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Non-Discounted Analysis – Machine A At a “0” interest rate the PB time is seen to equal 4 years!

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Payback for Machine B at 15%, N = 14 yrs Payback for B is between 9 and 10 years! Longer time period to recover the investment years

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Payback at “0”% for Machine B Payback for B at 0% is 6 years!

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Payback for Example 5.8 Discounted Machine A: 6.57 years Machine B: 9.52 years Undiscounted Machine A: 4.0 years Machine B: 6.0 years Go with Machine A – lower time period payback to recover the original investment

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Payback Method Summarized Payback is only a rough estimator Use as an initial screening method Avoid using this method as a primary analysis technique for selection projects

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. CHAPTER V 5.8 Present Worth of Bonds McMc Graw Hill ENGINEERING ECONOMY Fifth Edition Blank and Tarquin

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Present Worth of Bonds Bonds represent a source of funds for the firm. Bonds are sold (floated) by investment banks for firms in order to raise additional debt capital Bonds are evidence of Debt

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Present Worth of Bonds – Overview The Firm Investment Bankers Commissions/Fees Proceeds from The sale Sell the Bonds to The lending public Bondholders

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Bonds – Notation V – The face value of the bond ( 债券面值） b – Bond coupon rate( 债券票面利率） I – Bond interest c – number of payment periods per year I=(face value) (bond coupon rate)/c = Vb /c

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Bonds – Notation and Example Example: V = $5,000 (face value) b= 4.5% per year paid semiannually N = 10 years What is the interest the firm would pay to the current bondholder semiannually?

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Bonds – Example – Continued The interest is calculated as: The bondholder, buys the bond and will receive $ every 6 months for the life of the bond

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Buy the bonds or not PW the present worth of a bond PW ≥ bond purchase price, buy the bond. PW ＜ bond purchase price, do not buy the bond. Bonds are bought at sold in a bond market. Thus the price of the bond is subject to the pressures of the bond market.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Bonds – Example 5.11 Given V = $5,000 (Face value of the bond) r = 4.5% paid semiannually $I/6 months = $5,000(0.045/2) = $ paid to the current bondholder N = 10 years or 20 interest periods Determine the purchase price you should be willing to pay now.Assume your MARR is 8%/yr c.q.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Example 5.1 – Continued 8% c.q is the same as 0.08/4 = 0.02 = 2% per quarter. Bond interest flows every 6 months Need an effective 6-month rate The effective 6-month rate is then  (1.02) 2 – 1 = = 4.04%/6 months  This is the potential buyer’s required interest rate

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Example 5.11 – Cash-Flow Diagram A = /6 months …. … P=?? $5,000 i=4.04%/6 months Find the PW(4.04%) of the future cash flows to the potential bond buyer

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display Example 5.11 – Solving P = $112.50(P/A,4.04%,20) + $5,000(P/F,4.04%,20) P = $3,788 IF the buyer can buy this bond for $3,788 or less, he/she will earn at least the 8% c.q. rate.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 56 End of Chapter 5 Lecture Set