OUTLINE Questions? News? New homework due Wednesday Present Worth Analysis, Annual Equivalent Internal Rate of Return

Example

CAPITALIZED EQUIVALENT METHOD APPLIES TO LONG HORIZONS OR PERPETUAL INVESTMENTS COMPUTING THE PRESENT WORTH OF AN INFINITE SERIES IS CALLED THE CAPITALIZATION OF THE PROJECT’S COST APPLYING L’HOPITAL’S RULE TO THE (P/A,i,N), WE GET A/i SO THE PRESENT WORTH OF A SERIES OF CONSTANT PAYMENT STRETCHING OUT TO INFINITE IS A/i

Capitalized Equivalent Method

MUTUALLY EXCLUSIVE ALTERNATIVES IF ONE ALTERNATIVE IS SELECTED, ALL OTHERS ARE REJECTED “DO NOTHING’ IS AN ALTERNATIVE REVENUE PROJECT -- INCOME DEPENDS ON THE CHOICE OF THE ALTERNATIVE SERVICE PROJECT -- REVENUES ARE THE SAME REGARDLESS OF THE CHOICE OF ALTERNATIVE

PROJECT LIVES ANALYSIS PERIOD, STUDY PERIOD OR PLANNING HORIZON: TIME PERIOD OVER WHICH THE IMPACT OF THE PROJECT WILL BE EVALUATED WE HAVE TO CONSIDER: REQUIRED SERVICE PERIOD USEFUL LIFE OF THE PROJECT FOR MULTIPLE PROJECTS WE MUST COMPARE THEM OVER AN EQUAL TIME SPAN

PROJECT LIVES (CONTINUED) EXAMPLES PROJECT LIVES LONGER THAN ANALYSIS PERIOD (5.11) PROJECT LIVES SHORTER THAN ANALYSIS PERIOD (5.10) ANALYSIS PERIOD = LONGEST LIFE (5.12) UNEQUAL LIVES - LOWEST COMMON MULTIPLE METHOD (5.13)

Example 5.11 – Project lives longer than the analysis period THIS ONE IS VERY STRAIGHT FORWARD: WE ANALYZE FOR THE LENGTH OF TIME THAT WE ARE INTERESTED IN. WHATEVER THE EQUIPMENT IS WORTH AT THE END OF THE ANALYSIS PERIOD, WE CONSIDER AS SALVAGE VALUE

Example 5.10 Project lives shorter than analysis period THIS EXAMPLE HAS TWO OPTIONS, ONE THAT WILL LAST 3 YEARS, THE OTHER 4. UNFORTUNATELY WE HAVE TO PLAN FOR 5 YEARS. WE HAVE TO CHOOSE AN APPROACH THAT WILL FILL IN THE LAST TWO YEARS OF THE FIRST OPTION, AND THE LAST YEAR OF THE SECOND OPTION

Example 5.12 Analysis period = longest life

Example 5.13 Unequal lives - lowest common multiple method Option A has a life of 3 years, Option B 4 years. We use a lowest common denominator (12) as the analysis period For each option we calculate NPV and the end of its life and then repeat the investment until we have reached the common denominator

A FUN NOTE TO END ON (GARY LARSEN) “Notice all the computation, theoretical scribbling, and lab equipment, Norm … Yes, curiosity killed these cats

Annual Equivalent CALCULATE THE NPW, THEN CONVERT TO THE ANNUAL EQUIVALENT Examples 6.1 and 6.2

Contemporary Engineering Economics Incremental Analysis Lecture No. 13 Chapter 7 Contemporary Engineering Economics Copyright © 2006 Contemporary Engineering Economics, 4th edition, © 2007

Comparing Mutually Exclusive Alternatives Based on IRR Issue: Can we rank the mutually exclusive projects by the magnitude of its IRR? n A1 A2 -$1,000 -$5,000 $2,000 $7,000 100% > 40% $818 < $1,364 1 IRR PW (10%)

Contemporary Engineering Economics, 4th edition, © 2007 Who Got More Pay Raise? Bill Hillary 10% 5% Contemporary Engineering Economics, 4th edition, © 2007

Can’t Compare without Knowing Their Base Salaries Bill Hillary Base Salary $50,000 $200,000 Pay Raise (%) 10% 5% Pay Raise ($) $5,000 $10,000 For the same reason, we can’t compare mutually exclusive projects based on the magnitude of its IRR. We need to know the size of investment and its timing of when to occur.

Incremental Investment Project A1 Project A2 Incremental Investment (A2 – A1) 1 -$1,000 $2,000 -$5,000 $7,000 -$4,000 $5,000 ROR PW(10%) 100% $818 40% $1,364 25% $546 Assuming a MARR of 10%, you can always earn that rate from other investment source, i.e., $4,400 at the end of one year for $4,000 investment. By investing the additional $4,000 in A2, you would make additional $5,000, which is equivalent to earning at the rate of 25%. Therefore, the incremental investment in A2 is justified.

Incremental Analysis (Procedure) Step 1: Compute the cash flow for the difference between the projects (A,B) by subtracting the cash flow of the lower investment cost project (A) from that of the higher investment cost project (B). Step 2: Compute the IRR on this incremental investment (IRR ). Step 3: Accept the investment B if and only if IRR B-A > MARR B-A NOTE: Make sure that both IRRA and IRRB are greater than MARR.

Example 7.10 - Incremental Rate of Return B1 B2 B2 - B1 1 2 3 -$3,000 1,350 1,800 1,500 -$12,000 4,200 6,225 6,330 -$9,000 2,850 4,425 4,830 IRR 25% 17.43% 15% Given MARR = 10%, which project is a better choice? Since IRRB2-B1=15% > 10%, and also IRRB2 > 10%, select B2.

IRR on Increment Investment: Three Alternatives Step 1: Examine the IRR for each project to eliminate any project that fails to meet the MARR. Step 2: Compare D1 and D2 in pairs. IRRD1-D2=27.61% > 15%, so select D1. D1 becomes the current best. Step 3: Compare D1 and D3. IRRD3-D1= 8.8% < 15%, so select D1 again. Here, we conclude that D1 is the best alternative. n D1 D2 D3 -$2,000 -$1,000 -$3,000 1 1,500 800 2 1,000 500 2,000 3 IRR 34.37% 40.76% 24.81%

Practice Problem You are considering four types of engineering designs. The project lasts 10 years with the following estimated cash flows. The interest rate (MARR) is 15%. Which of the four is more attractive? Project A B C D Initial cost $150 $220 $300 $340 Revenues/Year $115 $125 $160 $185 Expenses/Year $70 $65 $60 $80 IRR (%) 27.32 24.13 31.11 28.33

Example 7.13 Incremental Analysis for Cost-Only Projects Items CMS Option FMS Option Annual O&M costs: Annual labor cost $1,169,600 $707,200 Annual material cost 832,320 598,400 Annual overhead cost 3,150,000 1,950,000 Annual tooling cost 470,000 300,000 Annual inventory cost 141,000 31,500 Annual income taxes 1,650,000 1,917,000 Total annual costs $7,412,920 $5,504,100 Investment $4,500,000 $12,500,000 Net salvage value $500,000 $1,000,000

Incremental Cash Flow (FMS – CMS) CMS Option FMS Option Incremental (FMS-CMS) -$4,500,000 -$12,500,000 -$8,000,000 1 -7,412,920 -5,504,100 1,908,820 2 3 4 5 6 $2,408,820 Salvage + $500,000 + $1,000,000

Solution:

Summary Rate of return (ROR) is the interest rate earned on unrecovered project balances such that an investment’s cash receipts make the terminal project balance equal to zero. Rate of return is an intuitively familiar and understandable measure of project profitability that many managers prefer to NPW or other equivalence measures. Mathematically we can determine the rate of return for a given project cash flow series by locating an interest rate that equates the net present worth of its cash flows to zero. This break-even interest rate is denoted by the symbol i*.

Internal rate of return (IRR) is another term for ROR that stresses the fact that we are concerned with the interest earned on the portion of the project that is internally invested, not those portions that are released by (borrowed from) the project. To apply rate of return analysis correctly, we need to classify an investment into either a simple or a nonsimple investment. A simple investment is defined as one in which the initial cash flows are negative and only one sign change occurs in the net cash flow, whereas a nonsimple investment is one for which more than one sign change occurs in the net cash flow series. Multiple i*s occur only in nonsimple investments. However, not all nonsimple investments will have multiple i*s either.

For a pure investment, the solving rate of return (i For a pure investment, the solving rate of return (i*) is the rate of return internal to the project; so the decision rule is: If IRR > MARR, accept the project. If IRR = MARR, remain indifferent. If IRR < MARR, reject the project. IRR analysis yields results consistent with NPW and other equivalence methods. For a mixed investment, we need to calculate the true IRR, or known as the “return on invested capital.” However, if your objective is simply to make an accept or reject decision, it is recommended that either the NPW or AE analysis be used to make an accept/reject decision. To compare mutually exclusive alternatives by the IRR analysis, the incremental analysis must be adopted.