ECLT 5930/SEEM 5740: Engineering Economics Second Term

Transcription

1 ECLT 5930/SEEM 5740: Engineering Economics Second Term Master of Science in ECLT & SEEM Instructors: Dr. Anthony Man Cho So Department of Systems Engineering & Engineering Management The Chinese University of Hong Kong March 4, 2016

2 Recap: Evaluating a Single Project 1. Convert cash flows of a given alternative to a particular (and fixed) point of time using the Minimum Attractive Rate of Return (MARR) and evaluate the alternative s economic merit. Present/Future Worth Annual Worth Internal/External Rate of Return 2. Two powerful tools the fundamental formula, from which all other formulae follow: F = P(1+i) N cash flow diagrams for visualizing the calculations ECLT 5930/SEEM 5740 ( Second Term) March 4,

3 Equivalence among Decision Rules Recall that an alternative is economically justifiable if the cash flows, discounted at the MARR, is non negative: PW(MARR) 0, FW(MARR) 0, AW(MARR) 0, the internal/external rate of return exceeds the MARR: IRR MARR, ERR MARR. As shown in the last lecture, the PW, FW and AW decision rules are all equivalent. This follows from the fundamental formula. Question: How about the IRR/ERR? ECLT 5930/SEEM 5740 ( Second Term) March 4,

4 Non Equivalence of the IRR and PW Decision Rules We claim that the IRR decision rule and the PW decision rule do not imply each other. Indeed, consider the following cash flows: $100, $20, 000 $90, 000 Figure 1: Illustrating the IRR and PW Decision Rules Questions If IRR 100% (which is a reasonable assumption), what is the IRR of the above cash flows? Suppose that MARR = 20%. What is the PW of the above cash flows? ECLT 5930/SEEM 5740 ( Second Term) March 4,

5 Equivalence of the IRR and ERR Decision Rules Recall that IRR satisfies the equation N R k (P/F,IRR%,k) = k=0 N E k (P/F,IRR%,k), k=0 while ERR satisfies the equation N E k (P/F,u%,k) (F/P,ERR%,N) = k=0 N R k (F/P,u%,N k), k=0 where u is the external reinvestment interest rate per period. We claim that when u = IRR, then the IRR decision rule is equivalent to the ERR decision rule. In fact, we shall show that IRR = ERR in this case. ECLT 5930/SEEM 5740 ( Second Term) March 4,

6 Equivalence of the IRR and ERR Decision Rules (Cont d) Suppose that u = IRR. Then, from the equation defining ERR, we have N E k (P/F,IRR%,k) (F/P,ERR%,N) = k=0 N R k (F/P,IRR%,N k). k=0 Dividing both sides by (F/P,ERR%,N), we get N E k (P/F,IRR%,k) = k=0 N R k k=0 (F/P,IRR%,N k). (*) (F/P,ERR%,N) Now, note that ERR = IRR is a solution to the above equation. Indeed, since (F/P,IRR%,N k) (F/P,IRR%,N) = (1+IRR)N k (1+IRR) N = (1+IRR) k = (P/F,IRR%,k), we see that (*) reduces to the equation defining IRR. ECLT 5930/SEEM 5740 ( Second Term) March 4,

7 This Lecture: Comparison and Selection among Alternatives So far we have developed various decision rules to evaluate the economic merit of a single alternative. Typically, there are many different alternatives for a project. These alternatives can differ wildly: investment of capital annual revenues and costs duration Fundamental Question: Do the added benefits from a more expensive alternative bring a positive return relative to the added costs? ECLT 5930/SEEM 5740 ( Second Term) March 4,

8 Example: An Investment Project Suppose that there are two projects: A and B. Project A requires an investment of $10,000 and will return $12,000 in one year. Project B requires an investment of $100,000 investment and will return $115,000 in one year. Suppose that your MARR is 10%, that the projects are mutually exclusive, and that you can take up either project (i.e., there are no budget concerns). Question: What do you do? First of all, are the projects economically justified? ECLT 5930/SEEM 5740 ( Second Term) March 4,

9 Example: An Investment Project (Cont d) By the PW decision rule, we have PW A (10%) = $10,000+ $12,000 = $909.09, 1.1 PW B (10%) = $100,000+ $115,000 = $ By the IRR decision rule, we have $10,000 = $12,000 1+IRR A = IRR A = 20%, $100,000 = $115,000 1+IRR B = IRR B = 15%. Thus, both projects are economically justified. However, it is not clear which is better: PW A (10%) PW B (10%) but IRR A IRR B. ECLT 5930/SEEM 5740 ( Second Term) March 4,

10 Basic Concepts in Comparing Alternatives To compare the alternatives, remember one of the engineering economic analysis rule: focus on the difference. In our current setting, we take it to mean the following: The alternative that requires the minimum investment of capital and produces satisfactory functional results will be chosen, unless the incremental capital associated with an alternative having a larger investment can be justified with respect to its incremental benefits. ECLT 5930/SEEM 5740 ( Second Term) March 4,

11 Example: An Investment Project (Cont d) Continuing the previous example, Project B requires an additional $90,000 in investment and produces an additional return of $103,000 in one year over Project A. (Why?) The present worth of the incremental cash flows at MARR = 10% is $90,000+ $103, = $ Alternatively, we can compute the IRR of the incremental cash flows: $90,000 = $103,000 1+IRR = IRR = 14.44% MARR. Conclusion: The use of the additional $90,000 for Project B is justified, and hence Project B is preferred over Project A. From this example, we see that using the highest IRR can lead to the wrong conclusion. One reason is that the additional investment may not earn a return at the IRR. ECLT 5930/SEEM 5740 ( Second Term) March 4,

12 Cash Flow Diagram Representation The evaluation of two mutually exclusive alternatives can be conveniently carried out by cash flow diagrams. Consider the following two projects, both of which will run for 4 years: Project A Project B (B A) Capital Investment $60, 000 $73, 000 $13, 000 Annual net revenue $22, 000 $26, 225 $4, 225 They can be represented by the following cash flow diagrams: $22, 000 $22, 000 $22, 000 $22, 000 $26, 225 $26, 225 $26, 225 $26, $60, 000 Project A $73, 000 Project B $4, 225 $4, 225 $4, 225 $4, $13, 000 (B A) Figure 2: Cash Flow Diagrams of the Two Projects ECLT 5930/SEEM 5740 ( Second Term) March 4,

13 Cash Flow Diagram Representation (Cont d) Our analysis of the alternatives then reduces to applying your favorite decision rules to each of the diagrams. Continuing the above example, if MARR = 10% and we use the PW decision rule, then PW A (10%) = $60,000+$22,000 (P/A,10%,4) = $9,738, PW B (10%) = $73,000+$26,225 (P/A,10%,4) = $10,131, PW (B A) (10%) = $13,000+$4,225 (P/A,10%,4) = $393. The first two equations imply that both A and B are economically justified projects. The last equation shows that Project B is preferred over Project A. Question: Note that PW (B A) (10%) = PW B (10%) PW A (10%). Is this a coincidence? ECLT 5930/SEEM 5740 ( Second Term) March 4,

14 Comparing and Selecting an Alternative: A Summary Basic assumptions The alternatives are mutually exclusive. The durations of the alternatives are the same. There is no budget concern. Incremental analysis procedure based on the PW Calculate the PWs of each project under the MARR. Delete any project whose PW is negative. Arrange the remaining projects in increasing order of their initial capital requirement. Compute the PW of the difference of the first two remaining projects: Delete the first project if the PW is non negative; otherwise, delete the second project. Repeat until there is only one project left. ECLT 5930/SEEM 5740 ( Second Term) March 4,

15 Remarks on the Incremental Analysis Procedure Instead of PW, one can also use FW, AW or IRR. In all cases, the incremental cash flows must be properly identified. In essence, we have the following rules for selecting between two projects A and B: PW Selection Rule: If PW (A B) (MARR) 0, then A is preferred over B; otherwise, B is preferred over A. IRR Selection Rule: If IRR (A B) MARR, then A is preferred over B; otherwise, B is preferred over A. ECLT 5930/SEEM 5740 ( Second Term) March 4,

16 Illustration of the Incremental Analysis Procedure Consider the following three projects, which last for 10 years. The MARR is 10%. Project A Project B Project C Capital Investment $390, 000 $920, 000 $660, 000 Annual net revenue $69, 000 $167, 000 $133, 500 We compute their PWs: PW A (10%) = $390,000+$69,000 (P/A,10%,10) = $33975, PW B (10%) = $920,000+$167,000 (P/A,10%,10) = $106143, PW C (10%) = $660,000+$133,500 (P/A,10%,10) = $ In particular, they are all qualified for further considerations. ECLT 5930/SEEM 5740 ( Second Term) March 4,

17 Illustration of the Incremental Analysis Procedure (Cont d) We consider the projects in the order of increasing capital requirement, i.e., A,C,B. Consider the difference (C A): We then compute Project A Project C (C A) Capital Investment $390, 000 $660, 000 $270, 000 Annual net revenue $69, 000 $133, 500 $64, 500 PW (C A) (10%) = $270,000+$64,500 (P/A,10%,10) = $126,325. Thus, we remove Project A from consideration. Now, we are left with Projects B and C. ECLT 5930/SEEM 5740 ( Second Term) March 4,

18 Illustration of the Incremental Analysis Procedure (Cont d) Consider the difference (B C): Then, we have Project B Project C (B C) Capital Investment $920, 000 $660, 000 $260, 000 Annual net revenue $167, 000 $133, 500 $33, 500 PW (B C) (10%) = $260,000+$33,500 (P/A,10%,10) = $54,157. Thus, we remove Project B from consideration. Since Project C is the only one left, we choose it as our preferred alternative. ECLT 5930/SEEM 5740 ( Second Term) March 4,

19 Consistency of the Incremental Analysis Procedure We use A B to denote the fact that A is more preferred over B. By the PW Selection Rule, A B if and only if PW (A B) (MARR) 0. Suppose that we have A B and B C. Question: Does it follow that A C? ECLT 5930/SEEM 5740 ( Second Term) March 4,

20 Consistency of the Incremental Analysis Procedure (Cont d) In general, the incremental analysis procedure produces a consistent global ranking of all the alternatives. To see this, observe that for any two alternatives A and B, PW (A B) (MARR) = PW A (MARR) PW B (MARR). Moreover, we have A B if and only if PW A (MARR) PW B (MARR). Hence, if PW (A B) (MARR) 0 and PW (B C) (MARR) 0, then PW (A C) (MARR) = PW A (MARR) PW C (MARR) = PW A (MARR) PW B (MARR) + PW B (MARR) PW C (MARR) 0; i.e., if A B and B C, then we necessarily have A C. ECLT 5930/SEEM 5740 ( Second Term) March 4,

21 Consistency of the Incremental Analysis Procedure (Cont d) From the above argument, it follows that the most preferred alternative returned by the incremental analysis procedure is the one that has the highest PW. However, as we have seen, it is not correct to choose the alternative that has the highest IRR. Question: Do the PW Selection Rule and the IRR Selection Rule give the same conclusion? It turns out that if the cash flows of (A B) are given by where I,R 1,...,R N 0, then I,R 1,...,R N, (*) PW (A B) (MARR) 0 IRR (A B) MARR. In other words, under condition (*), the PW Selection Rule and the IRR Selection Rule are equivalent; i.e., they give the same conclusion. ECLT 5930/SEEM 5740 ( Second Term) March 4,

22 The PW and IRR Selection Rules Recall that PW (A B) (x) = I + N n=1 R n (1+x) n. Question: What is PW (A B) (IRR (A B) )? Now, observe that by our assumption, PW (A B) (x) is a decreasing function of x. Hence, we conclude that PW (A B) (MARR) PW (A B) (IRR (A B) ) = 0 if and only if IRR (A B) MARR. ECLT 5930/SEEM 5740 ( Second Term) March 4,

23 Further Illustration of the Incremental Analysis Procedure Consider 6 projects A, B, C, D, E and F. The durations for all projects are all equal to 10 years, and the MARR is 10%. Suppose that their cash flows and IRRs are given as follows: Project Capital Investment Annual Profit IRR A $900 $ % B $1, 500 $ % C $2, 500 $ % D $4, 000 $ % E $5, 000 $1, % F $7, 000 $1, % Note that if MARR = 19%, then Project D is the only acceptable choice. However, if the MARR is lower, then we may waste some investment opportunity by investing in D. ECLT 5930/SEEM 5740 ( Second Term) March 4,

24 Further Illustration of the Incremental Analysis Procedure By the incremental analysis procedure, we can get the following ordering: E F D B A C. In other words, if the projects are mutually exclusive and we have no budget concerns, then we should pick E. Questions Suppose that we only have $4,000 to invest. Which project should we pick? Suppose that the projects are not mutually exclusive. What should be the preference of the projects? ECLT 5930/SEEM 5740 ( Second Term) March 4,

25 Application: Choosing Air Compressor Suppose that one must install an air compressor among 4 mutually exclusive choices for 5 years. The MARR is 20%, and the expenses are given as follows: Questions Alt. Price Annual Cost Resale Value D 1 $100,000 $29,000 $10,000 D 2 $140,000 $16,900 $14,000 D 3 $148,200 $14,800 $25,600 D 4 $122,000 $22,100 $14,000 What are the PWs of the alternatives? Which is the most preferred alternative? ECLT 5930/SEEM 5740 ( Second Term) March 4,

26 Application: Parking Lot Project A downtown parking center was out of capacity. You are called to perform a project evaluation. The alternatives are P: keep and improve existing parking lot B 1 : construct one-story building B 2 : construct two-story building B 3 : construct three-story building Suppose that the MARR is 10%. The cash flows are given as follows: Alt. Investment Net Annual Income P $200, 000 $22, 000 B 1 $4,000,000 $600,000 B 2 $5,550,000 $720,000 B 3 $7,500,000 $960,000 In 15 years time, the residual value of each alternative is estimated as the same as its construction cost today. ECLT 5930/SEEM 5740 ( Second Term) March 4,

27 Application: Parking Lot Project (Cont d) The IRR of each alternative can be computed as follows: Project IRR P 11% B 1 15% B 2 13% B % The management team of the parking lot is inclined to choose B 1, since it has the highest IRR. Question: Is this a good choice? Why? ECLT 5930/SEEM 5740 ( Second Term) March 4,

28 Application: Parking Lot Project (Cont d) The PWs of the choices are given by PW P (10%) = $200,000+$22,000 (P/A,10%,15) + $200,000 (P/F,10%,15) = $15,212, PW B1 (10%) = $1,521,216, PW B2 (10%) = $1,255,003, PW B3 (10%) = $1,597,277. Question: What course of action would you recommend? ECLT 5930/SEEM 5740 ( Second Term) March 4,

29 Beyond the Basic Assumptions So far it is assumed that the alternatives are mutually exclusive, and that we have no budget concerns. We have also discussed the case when one of those assumptions are relaxed. Not mutually exclusive: Choose the projects in decreasing order of PW/FW/AW. Budget constraints: Choose the most preferred project that is within budget according to the incremental analysis. How about the case where the alternatives are not mutually exclusive, but we have a budget constraint? This leads to capital budgeting. ECLT 5930/SEEM 5740 ( Second Term) March 4,

30 Capital Budgeting Suppose that we have n alternatives, say, A 1,...,A n. The capital investment and PW of A i are given by c i and w i, respectively. Let C be the budget for investment. Then, to choose the best alternatives subject to the budget constraint, we need to solve the following optimization problem: maximize subject to n w i x i i=1 n c i x i C, i=1 x i = 0 or 1 for i = 1,...,n. ECLT 5930/SEEM 5740 ( Second Term) March 4,

31 Capital Budgeting: Example Suppose that we have a budget of $500,000. The following alternatives are available to us: Alt. Investment PW 1 $100,000 $300,000 2 $20,000 $50,000 3 $150,000 $350,000 4 $50,000 $110,000 5 $50,000 $100,000 6 $150,000 $250,000 7 $150,000 $200,000 It can be shown that it is optimal to choose alternatives 1,3,4,5,6. ECLT 5930/SEEM 5740 ( Second Term) March 4,

32 What s Next? Assignment: Read Chapter 6 of the course textbook. Next: Continuing Chapter 6 ECLT 5930/SEEM 5740 ( Second Term) March 4,