The objective of chapter 6 isto evaluate correctly capitalinvestment alternatives whenthe time value of money is akey influence.
Making decisions meanscomparing alternatives. In this chapter we examine feasible design alternatives. The decisions considered are those selecting from among a set of mutually exclusive alternatives (when selecting one excludes the choice of any of the others).
Principle 2 from Chapter 1.The alternative that requires the minimum investment ofcapital and produces satisfactory functional results will bechosen(the base alternative)unless the incremental capital associated with an alternativehaving a larger investment can be justified with respect to itsincremental benefits.The alternative requiring the least investment is the basealternative.
For alternatives that have a largerinvestment than the base…If the extra benefits obtained by investing additionalcapital are better than those that could be obtained frominvestment of the same capital elsewhere in thecompany at the MARR, the investment should be made.(Please note that there are some cautions whenconsidering more than two alternatives, which will beexamined later.)
Types of Alternatives Investment alternatives (capital & Revenue) Cost alternatives (Negative & salvage value) same expected revenue
Ensuring Comparable Basis Rule 1. When revenues and other economic benefits are present(Investment alternatives ), select alternative that has greatest positive equivalent worth at i %= MARR% and satisfies project requirements. Rule 2. When revenues and economic benefits are not present(Cost alternatives ), select alternative that minimizes cost.
In Other Words:Select the alternative that gives youthe most money! For investment alternatives the PW of all cash flows must be positive, at the MARR, to be attractive. Select the alternative with the largest PW. For cost alternatives the PW of all cash flows will be negative. Select the alternative with the largest (smallest in absolute value) PW.
Example: Investment AlternativeUse a MARR of 10% and useful life of 5 years to selectbetween the investment alternatives below. Alternative A B Capital investment -$100,000 -$125,000 Annual revenues less expenses $34,000 $41,000 Both alternatives are attractive, but Alternative B provides a greater present worth, so is better economically.
Cost alternative exampleUse a MARR of 12% and useful life of 4 years to selectbetween the cost alternatives below. Alternative C D Capital investment -$80,000 -$60,000 Annual expenses -$25,000 -$30,000 Alternative D costs less than Alternative C, it has a greater PW, so is better economically.
Cost Alternative with selvage valuePW(10%)C= -477,077$PW(10%)D= -463,607$PW(10%)D-C = 13,470$
Determining the study period. A study period (or planning horizon) is the time period over which mutually exclusive alternatives are compared, and it must be appropriate for the decision situation. mutually exclusive alternatives can have equal lives (in which case the study period used is these equal lives), or they can have unequal lives, and at least one does not match the study period. The equal life case is straightforward.
Useful Lives of all Alternativesis Equal to the Study Period1. Equivalent worth methods2. Rate of return methods
1.Equivalent Worth Methods If : PWA (i) < PWB (i) Then AWA (i) < AWB (i) and FWA (i) < FWB (i) Select Alternative B
Example:When lives are equal adjustments to cash flows are notrequired. The MEAs can be compared by directly comparingtheir equivalent worth (PW, FW, or AW) calculated using theMARR. The decision will be the same regardless of theequivalent worth method you use. For a MARR of 12%, selectfrom among the MEAs below. Alternatives A B C DCapital investment -$150,000 -$85,000 -$75,000 -$120,000Annual revenues $28,000 $16,000 $15,000 $22,000Annual expenses -$1,000 -$550 -$500 -$700Market Value $20,000 $10,000 $6,000 $11,000(EOL)Life (years) 10 10 10 10
Selecting the best alternative.Present worth analysis select Alternative A (but C is close).Annual worth analysis—the decision is the same.
Using rates of return isanother way to comparealternatives.
The return on investment (rate of return) is a popular measure of investment performance. Selecting the alternative with the largest rate of return can lead to incorrect decisions do not compare the IRR of one alternative to the IRR of another alternative. Remember, the base alternative must be attractive (rate of return greater than the MARR), and the additional investment in other alternatives must itself make a satisfactory rate of return on that increment.
The Incremental Investment Analysis Procedure. Arrange (rank order) the feasible alternatives based on increasing capital investment. Establish a base alternative. Cost alternatives the first alternative is the base. Investment alternatives the first acceptable alternative (IRR>MARR) is the base.
Iteratively evaluate differences (incremental cash flows) between alternatives until all have been considered. a. If incremental cash flow between next alternative and current alternative is acceptable, choose the next. b. Repeat, and select as the preferred alternative the last one for which the incremental cash flow was acceptable
Example 6-4 A B C D E F Capital 900 1,500 2,500 4,000 5,000 7,000 Annual R – 150 276 400 925 1,125 1,425 Annual E IRR 10.6% 13% 9.6% 19.1% 18.3% 15.6% The alternative is arranged based on increasing capital investment. IRR for alternative (C) 9.6<MARR(10%) Then C is rejected
A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)ΔCapital 900Δ(Annual R 150– Annual E)IRRΔ 10.6%Is Yesincrementjustified A is the base alternative then we will compare A with B
A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)ΔCapital 900 600Δ(Annual R 150 126– Annual E)IRRΔ 10.6% 16.4%Is Yes Yesincrementjustified B is better then A because the additional capital investment gives IRR >MARR(16.4%>10%) Then we will compare B with D since C is rejected from the beginning
A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)ΔCapital 900 600 2,500Δ(Annual R 150 126 649– Annual E)IRRΔ 10.6% 16.4% 22.6%Is Yes Yes Yesincrementjustified D is better then B because the additional capital investment gives IRR >MARR(22.6%>10%) Then we will compare D with E
A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)ΔCapital 900 600 2,500 1,000Δ(Annual R 150 126 649 200– Annual E)IRRΔ 10.6% 16.4% 22.6% 15.1%Is Yes Yes Yes Yesincrementjustified E is better then D because the additional capital investment gives IRR >MARR(15.1%>10%) Then we will compare E with F
A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)ΔCapital 900 600 2,500 1,000 2,000Δ(Annual R 150 126 649 200 300– Annual E)IRRΔ 10.6% 16.4% 22.6% 15.1% 8.1%Is Yes Yes Yes Yes Noincrementjustified F is NOT better then E because the additional capital investment gives IRR < MARR(8.1%>10%) Then E is the best alternative
Three Errors Common To Incremental Investment Analysis Procedure Applied To IRRChoosing the feasible Alternative with:1. the highest overall IRR on total cash flow (choose D) A B C D E FCapital 900 1,500 2,500 4,000 5,000 7,000Annual R – 150 276 400 925 1,125 1,425Annual EIRR 10.6% 13% 9.6% 19.1% 18.3% 15.6%
Three Errors Common To Incremental InvestmentAnalysis Procedure Applied To IRRChoosing the feasible Alternative with:2.the highest IRR on an incremental capital investment(choose D-B instead of E-D) A Δ(B-A) Δ(D-B) Δ(E-D) Δ(F-E)ΔCapital 900 600 2,500 1,000 2,000Δ(Annual R 150 126 649 200 300– Annual E)IRRΔ 10.6% 16.4% 22.6% 15.1% 8.1%Is Yes Yes Yes Yes Noincrementjustified
Three Errors Common To Incremental Investment Analysis Procedure Applied To IRR3.the largest capital investment that has an IRR greater than or equal to the MARR (choose F) A B C D E FCapital 900 1,500 2,500 4,000 5,000 7,000Annual R – 150 276 400 925 1,125 1,425Annual EIRR 10.6% 13% 9.6% 19.1% 18.3% 15.6%
Incremental analysis must be used with rate of return methods to ensure the best alternative is selected
Comparing MEAs with Unequal lives.• Repeatability assumption• Coterminated assumption
Rate of Return with unequal lifes A BCapital investment $3,500 $5,00Annual cash flow 1,255 1,480Useful life 4 6 When the repeatability applies 1.we need to find the AW of each alternative over it’s own useful life 2.Find the interest rate that make them equal AW A(i*%) = AW B(i*%) -3,500(A/i*%,4) +1,255 = -5,000(A/i*%,6) +1,480 Trail and error i= 26% >MARR (10%) the increment is justified and B is preferred
Useful life > Study period The imputed Market Value Technique i. e find the estimated market value at any year that is before the end of the useful life MVT = PW at EOY T of remaining CR amounts + PW at EOY T of original market value at end of useful life
Example If the capital investment is 47,600 useful life = 9 years market value at the end of 9 years = $5,000 find the market value et the end of 5 years if MARR = 20%Step 1 : CR = 47,000 (A/P,20%,9) – 5,000(A/F,20%,9)Step 2 : PW at EOY 5 of the remaining CR PW(20%) CR = CR (P/A, 20%,4) = $29,949Step 3 : PW at EOY 5 of MV 9 PW(20%) MV = 5,000 (P/F, 20%,4) = $2,412The estimated market value at EOY 5 =MV 5 = PW(20%) CR + PW(20%) MV = 29,949 + 2,412 = 32,361