by assoc. prof. dr. ahmet ÖztaŞ

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By

Assoc. Prof. Dr. Ahmet ÖZTAŞ

Gaziantep UniversityDepartment of Civil Engineering

CHP 6. Rate of Return (ROR) Analysis

CE 533 - ECONOMIC DECISION ANALYSIS IN CONSTRUCTION

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CHP 6. Rate of Return (ROR) Analysis

1. Interpretation of ROR Values

2. ROR Calculations

3. Cautions When Using ROR Method

4. Understanding Incremental ROR Analysis

5. ROR Evaluation

6. Multiple ROR Values

7. Removing Multiple ROR Values

8. Using Excel for ROR Analysis

Topics

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Introduction

• This chapter presents the methods to evaluate alternatives using Rate of Return (ROR) procedure.

• Other names of ROR: IRR, ROI, PI

• Interest rate--- when discussing borrowed money; ROR when dealing with investments.

•More than one ROR may exist.

•How to recognize multiple ROR, if exists?

•A unique ROR value can be obtained by using an external investment rate.

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DEFINITION

Rate of return (ROR) is;

• either the interest rate paid on the unpaid balance of barrowed money (a loan),

• or the interest rate earned on the unrecovered balance of an investment

so that the final payment or receipt brings the balance to exactly “0” with interest considered.

6.1 Interpretation of ROR Values

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•ROR is expressed as a % per period …i=10%

• Mathematically, i rate must be:*100% i

• If an i <= -100% entire amount is lost..

• One can have a negative i value (feasible), but not less than -100%!

•Note: Definition does not state that ROR is on initial amount of investment; rather it is on the unrecovered balance, which changes each time period.

6.1 Interpretation of ROR Values

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6.1 Unrecovered Investment Balance

Example 6.1: Consider the following loan

You borrow $1,000 at 10% per year for 4 years

You are to make 4 equal end-of-year payments to pay off this loan

Your payments are:

A=$1,000(A/P,10%,4) = $315.47

a)Compute ROR on unrecovered balance

b)Compute ROR on initial investment

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6.1 The Loan Schedule

Year Beginning Unrecovered Balance

Payment

InterestAmount

Recovered

Amount

Ending Unrecovered Balance

0 $1,000 0 -- --- $1,000

1 1,000 315.47

100.00 215.47 784.53

2 784.53 315.47

78.45 237.02 547.51

3 547.51 315.47

54.75 260.72 286.79

4 286.79 315.47

28.68 286.79 0

a) Compute ROR on unrecovered balance

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6.1 Unrecovered Investment Balance

For this loan the unpaid loan balances at the end of each year are:

$1,000

784.53

547.51

286.79

0

0

1

2

3

4

Unpaid loan balance is now

“0” at the end of the life of the

loan

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6.1 Reconsider the Following

Assume you invest $1,000 over 4 years

The investment generates $315.47/year

Draw the cash-flow diagram0 1 2 3 4

P=$-1,000

A = +315.47

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Example 6.1 (b) Table 6.2

Year BOY Bal Cash Flow

InterestAmount

Prin. Red.

Amount

UnPaid Balanc

e

0 -1000 -- --- ---

1 -1,000 315.47

100.00 215.47 $784.53

2 -784.53

315.47

100.00 215.47 569.06

3 -569.06

315.47

100.00 215.47 353.59

4 -353.59

315.47

100.00 215.47 138.12400 861.88

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Table 6.2 shows unrecovered balance if 10% return is always figured on the initial $1000. Unrecovered amount: $138.12, because only $861.88 is recovered in 4 years.

Example 6.1 (b) The Loan Schedule

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6.1 Investment Problem

What interest rate equates the future positive cash flows to the initial investment?

We can state:

-$1,000= 315.47(P/A, i*,4)

Where i* is the unknown interest rate that makes the PW(+) = PW(-)

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6.1 Investment Problem

$1,000= 315.47(P/A, i*,4)

Solve the above for the i* rate

(P/A,i*,4) = 1,000/315.47 = 3.16987

Given n = 4, what value of i* yields a P/A factor value = 3.16987?

Interest Table search yields i*=10%

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6.1 ROR – Explained

ROR is the interest rate earned on the unrecovered investment balances throughout the life of the investment.

ROR is not the interest rate earned on the original investment

ROR (i*) rate will also cause the NPV(i*) of the cash flow to = “0”.

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Objective is interest rate represented as i*

Calculations are reverse of previous methods (PW, AW), where i was known.

In rate of return problems you seek an unknown interest rate (i*) that satisfies the following:

0= - PWD + PWR

PWD = present worth of costs

PWR = present worth of incomes

6.2 ROR Calculation

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6.2 ROR using Present Worth

•Finding the ROR for most cash flows is a trial-and-error effort.

•The interest rate, i*, is the unknown

•Solution is generally an approximation effort

•May require numerical analysis approaches

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•See Figure 6.1

0 1 2 3 4 5

-$1,000

+$500

+$1,500

•Assume you invest $1,000 at t = 0: Receive $500 @ t=3 and $1,500 at t = 5.

• What is the ROR of this project?

• Example:

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•Write a present worth expression, set equal to “0” and solve for the interest rate that satisfies the formulation.

1,000 = 500(P/F, i*,3) +1,500(P/F, i*,5)

•Can you solve this directly for the value of i*?

•NO!

•Resort to trial-and-error approaches

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1,000 = 500(P/F, i*,3) +1,500(P/F, i*,5)

• Guess at a rate and try it

• Adjust accordingly

• Interpolate

• i* approximately 16.9% per year on the unrecovered investment balances

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6.2 Trial-and Error-Approach

General procedure for PW-based equation is;

•1- Draw a cash flow diagram

•2- Set up the rate of return equation in the form of

0= PWD + PWR

•3- Select values of i by trial and error until the equation is balanced.

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6.2 Trial-and-Error Approach

• If the NPV is not = 0, assume another i*.

• If NPV > 0 increase “i* guess value”

• If NPV < 0 reduce “i* guess value”

•The objective is to obtain a negative PW and a positive PW value.

• Then, interpolate between the two i* values

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6.2 i* Spreatsheet Approach

• Fastest way is to apply RATE function.

•Format of function is:

RATE(n,A,P,F)

If CF vary over the years, IRR function is used to determine i*.

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6.2 ROR Criteria

• Determine the i* rate

• If i* => MARR, accept the project

• If i* < MARR, reject the project

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6.3 Cautions when using the ROR Method

• Important Cautions to remember when using the ROR method……

•There are some assumptions and difficulties with ROR analysis that must be considered when calculating i* and in interpreting its real world meaning for a particular project.

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6.3 Cautions when using the ROR

1-Computational Difficulties

• ROR method is computationally more difficult than PW/AW

•Can become a numerical analysis problem and the result is an approximation

•Conceptually more difficult to understand

2- Special Procedure for Multiple AlternativesFor ROR analysis of multiple alternatives, apply an incremental analysis approach.

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3- Multiple i* values

•Many real-world cash flows may possess multiple i* values

• More than one i* value that will satisfy the definitions of ROR

• If multiple i*’s exist, which one, if any, is the correct i*???

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4- Reinvestment Assumptions

• Reinvestment assumption for the ROR method is not the same as the reinvestment assumption for PW and AW

• PW and AW assume reinvestment at the MARR rate

• ROR assumes reinvestment at the i* rate

• Can get conflicting rankings with ROR vs. PW and AW

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6.4 Understanding Incremental ROR Analysis

• From previous chapters, we know that using PW, AW or FW one mutually exclusive alternative can be identified. In this chapter We also learned that ROR can be used to identify the best alternative; However, it is not always as simple as selecting the highest rate of return (ROR) alternative.

•Give an example:

• Investment amount:90,000 MARR= 16%

•Alt A: P=50,000, ROR=35%;

•Alt B: P=85,000, ROR=29%;

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•Major dilemma of ROR method when comparing alternatives: Under some circumstances, alternative ROR (i* ) values do not provide same ranking of alternatives as do PW and AW analyses.

•To resolve this dilemma… conduct an incremental analysis between two alternatives at a time and base the alternative selection on the ROR of the incremental cash flow series.

•A standardized format (A table, see Table 6.3) simplifies the incremental analysis.

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Year Cash Flow Incremental

Alternative A Alternative B Cash Flow

(1) (2) (3) = (2) - (1)

0

1

2

.

.

.

•Table shows a “format for incremental Cash Flow Tabulation”

•If Equal lives: “Year” column will go from 0 to n.

•If Equal lives: “Year” column will go from 0 to LCM of two lives.

• Incremental ROR analysis requires equal-servis comparison between alternatives.

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•When LCM of lives is used, salvage value and reinvestment in each alternative are shown at their respective times.

• If study period is used, the cash flow tabulation is for the specified period.

• Simplification: Use convention that between 2 alternatives, the one with the larger initial investment will be regarded as Alternative B. Then, for each year in table:

•Incremental CF = Cash Flow (B) – Cash Flow (A)

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•There are 2 types of alternatives:

•Revenue alternative: There are both – and + CFs

•Cost alternative: All CF estimates are negative (-).

•In either case, use Equation Inc. CF = CF(B) – CF(A) to determine the incremental CF series with the sign of each cash flow carefully. See Examples 6.4 and 6.5.

• Then, determine incremental rate of return (Δi*) on extra amount required by the larger investment alternative.

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•This rate, termed Δi*, represents the return over n years expected on optional extra investment in year 0.

•Selection criteria:

•If Δi* ≥ MARR, select larger investment alt. (Alt. B)

Othervise, select lover investment alt. (Alt. A).

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•If analysis is between multiple revenue alternatives:

•Determine each alternative’s i* value and remove those alternatives with i* ≤ MARR since, their return is too low.

•Then, complete the incremental analysis for the remaining alternatives.

•If all alternatives i* ≤ MARR choose “DN” alternative.

•Note: this can not be done for cost alternatives since they have no positive CF.

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•When independent alternatives compared:

•No incremental analysis is necessary.

•All alternatives i* ≥ MARR are acceptable.

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6.5 ROR Evaluation Two or More Mutually Exclusive Alternatives

• When selecting from two or more mutually exclusive alternatives on the basis of ROR, equal-service comparison is required, and an incremental ROR analysis must be used.

•Incremental ROR value between two alternatives (B and A) is identified as Δi*B-A, (or shortly Δi*).

Select the alternative that:

•1- Requires the largest initial investmet.

•2- has a Δi*≥ MARR, indicating that extra initial investment is economically justified.

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• Before conducting incremental evaluation, classify alternatives as cost or revenue alternatives.

• Cost: Evaluate alternatives only against each other.

• Revenue: First evaluate against do-nothing (DN), against each other.

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• Procedure: (To compare multiple, mutually exclusive alternatives using PW-based equivalence relation)

•1- Order alternatives by increasing initial investment For revenue alternatives add DN as the first one.

•2- Determine incremental CF between the first two ordered alternatives (B-A).

•3- Set up a PW-based relation of this incr. CF series and determine Δi*, the incremental ROR.

•4- If Δi*≥ MARR, eliminate A; B is the survivor. Otherwise A is survivor.

•5- Compare survivor to the next alternative.

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Example 6.6

… Alternative machines from manufacturers in Asia, America, Europe, and Africa are available with the cost estimates in Table 6.6.Annual cost estimates are expected to be high to ensure readiness at any time. The company representatives have agreed to use the average of the corporate MARR values, which results in MARR = 13.5%. Use incremental ROR analysis to determine which manufacturer offers the best economic choice.

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Follow the procedure for incremental ROR analysis.1. These are cost alternatives and are arranged by increasing first cost.2. The lives are all the same at n = 8 years. The B - A incremental cash flows are indicated in Table 6.7. The estimated salvage values are shown separately in year 8.

Example 6.6 - Solution

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3. Following PW relation for (B-A) results in Δi*= 14.57%.0= -1500 + 300(P/A, Δi*,8) + 400(P/F, Δi*,8)4. Since this return exceeds MARR = 13.5%, A is eliminated and B is the survivor.5. The comparison of C-to-B results in the eliminatian of C based on Δi* = -18.77% from the incremental relatian0= -3500 + 200(P/A, Δi*,8) - 200(P/F, Δi*,8)The D-to-B incremental cash flow PW relation for the final evaIuatian is O = -8500 + 1800(P/A, Δi*,8) + lOO(P/F, Δi*,8)With Δi* = 13.60%, machine D is the overall, though marginal, survivor of the evaluation; it should be purchased and located in the event of oil spill accidents.

Example 6.6 - Solution

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6.6 Multiple ROR Values

• A class of ROR problems exist that will possess multiple i* values

• Capability to predict the potential for multiple i* values

• Two tests can be applied:

• 1. Cash Flow Rule of Signs (Descartes’ rule)

• 2. Cumulative Cash Flow Test (Norstom’s criterion)

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1) Cash Flow Rule of Signs Test (Descartes’ rule):

•“The maximum number of i* values is equal to the number of sign changes in the cash flow series.

• Follows from the analysis of a n-th degree polynomial.

• A “0” value does not count as a sign change


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2) Cumulative Cash Flow Test (Norstrom Criterion):

•“There is one nonnegative i* value if the cumulative cash flow series, S0, S1,… Sn, changes sign only once and Sn # 0.

• To perform the test, cont the number of sign changes in Sn seires.

•Sn= C.C.F through period n.


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6.6 C.F. Rule of Signs Example

• Consider Example 6.8 (Page 140)Year Cash Flow

0 $2,0001 -$5002 -$8,1003 $6,800

+--+

Result: 2 sign changes in the Cash Flow

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6.6 Results: CF Rule of Signs Test

• Two sign changes in this example

• Means we can have a maximum of 2 real potential i* values for this problem

• Beware: This test is fairly weak and the second test must also be performed.

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6.6 Cumulative CF Sign Test (Norstrom Criterion)

• For the problem form the cumulative cash flow from the original cash flow.

Year Cash Flow Accum.C.F

0 $2,000 $2,0001 -$500 $1,5002 -$8,100 -$6,6003 $6,800 $200

Count sign changes here

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6.6 Cumulative CF Signs – Example


0 $2,000 $2,0001 -$500 $1,5002 -$8,100 -$6,6003 $6,800 $200

++-+

2 sign changes in the CCF.

C.C.F

• This indicates that there is not just 1 nonnegative root.

•So, as many as two i* values can be found.

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6.6 CCF Test – Rules

•If the value of the CCF for year “n” is “0”, then an i* of 0% exists

•If the value of CCF for year “n” is > 0, this suggests an i* > 0

•If CCF for year n is < 0, there may exist one or more negative i* values – but not always.

•If the number of sign changes in the CCF is 2 or greater, this strongly suggests that multiple rates of return exist.

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6.6 Example 6.8 – CCF Sign Test


0 $2,000 $2,0001 -$500 $1,5002 -$8,100 -$6,6003 $6,800 $200

2 Sign Changes here

•Strong evidence that we have multiple i* values

•CCF (n=3) = $200 > 0 suggests positive i* (s)

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6.6 Example 6.8 - Continued

•b) PW relation is;

PW = 2000-500(P/F, i*,1)- 8100(P/F, i*,2)- 6800(P/F, i*,3)

• Select values of i to find two i* values, and plot PW versus i.

• i1* = 8%, i2* = 41% approximately.

•Or i1* = 7.47%, i2* = 41.35% exactly (see next figure).

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6.6 PW Plot of 6.8

-150.000

-100.000

-50.000

0.000

50.000

100.000

150.000

200.000

250.000

0.00 0.20 0.40 0.60

i* = 7.47%i*=41.35%

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6.6 PW Plot – Continued

-150.000

-100.000

-50.000

0.000

50.000

100.000

150.000

200.000

250.000

0.00 0.20 0.40 0.60

PW < 0If the

MARR is between the two i*

values, this

investment would be rejected!

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6.6 Comments on ROR

• Multiple i* values lead to interpretation problems

• If multiple i*’s – which one, if any is the “correct” one to use in an analysis?

• Serves to illustrate the computational difficulties associated with ROR analysis

• Section 6.7 provides an alternative ROR approach – Composite ROR (C-ROR)

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6.7 Removing Multiple ROR Values By Using Reinvestment Rate

•All ROR values calculated so far can be termed as IRR. IRR guaranties that last receipt or payment is exactly 0 with the interest considered. No excess funds are generated in any year, so all funds are kept internal to the project.

• A project can generate excess funds prior to the end of project when net CF in any year is positive (NCFt >0).

•This can result in a nonconventional series.

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• ROR method assumes these excess funds can earn at any one of the multiple i* values. This generates ambiguity.

•Example: assume an alternative’s nonconventional CF have two ROR roots at -2% and 40%. Also, assume that neither is a realistic reinvestment assumption. Instead excess funds will likely earn at the MARR of 15%.

•“Is the project economically justified?”

•ROR analysis can not answer to this question. So what to do?

•Answer: Set a reinvestment rate (External Rate of Return) and determine the unique composite ROR value.


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• Composite ROR:

•This method works when excess funds (+ CF) earn at a rate that is stated and is determinedexternally to the project. This rate is commonly different from any of the multiple ROR values.

•In the preceeding example, if excess funds can actually earn 40%, then the one correct, unique i* is 40%.

• However, if excess funds can be expected to earn at the MARR of 15%, this technique will find the correct andunique (composite) rate for the project. So, dilemma of multiple ROR values is eliminated.


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6.7 Composite ROR Approach

•Consider the following investment

-$10,000

0 1 2 3 4 5

+$8,000

+$9,000

Determine the ROR as….

Example:

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• Analysis reveals:

•PW equation to determine i* is

0 = -10,000 + 8,000(P/F,i,2) + 9,000(P/F,i,5)

i* = 16.815%

i* = 16.82%/year on the unrecovered investment balances over 5 years

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• i* = 16.82%

-$10,000

0 1 2 3 4 5

+$8,000

+$9,000

Question: Is it reasonable to assume that the +$8,000 can be invested forward at 16.82%?

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• Remember …….

• ROR assumes reinvestment at the calculated i* rate

• What if it is not practical for the +$8,000 to be reinvested forward one year at 16.82%?

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6.7 Reinvestment Rates

• Most firms can reinvest surplus funds at some conservative market rate of interest in effect at the time the surplus funds become available.

• Often, the current market rate is less than a calculated ROR value

• What then is the firm to do with the +$8,000 when it comes in to the firm?

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6.7 Reinvesting

• Surplus funds must be put to good use by the firm

• These funds belong to the owners – not to the firm!

• Owners expect such funds to be put to work for benefit of future wealth of the owners

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• Consider the following representation:

TheFirm

Project

Invested Funds

Returns back

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• Or, put in another context….

TheFirm

Project

Project borrows from the firm

Project Lends back to the firm

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• Or, put in another context….

TheFirm

Project

Project borrows from the firm

At the i* rate (16.82%)

Project Lends back to the firm

But can the firm reinvest these funds at 16.82%? Probably not!

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• So, we may have to consider a reinvestment rate that is closer to the current market rate for reinvestment of the $8,000 for the next time period(s)

• Assume a reasonable market rate is, say, c = 8% per year.

•This is called the Reinvestment rate (or External Rate of Return) - c

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6.7 The external rate – c

• The one interest rate that now satisfies the rate of return equation is called composite rate of return (CRR) and symbolized by i’.

• The external interest rate – c, is a rate that the firm can reinvest surplus funds for at least one time period at a time.

• c is often set to equal the firm’s current MARR rate

•Thus, a procedure (Net-invest procedure) has been developed that will determine the following:

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•Net-investment procedure:

• Find i* given c – if multiple ROR’s exist.

• For multiple i*’s in a problem, the analysis determines a single i* given c.

• Denoted i*/c or, i’

• i’ is called the composite rate = i*/c

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• Net-investment procedure:

•Involves finding the future worth of the net investment amount 1 year in the future.

•Find project’s net-investment value Ft in year t from Ft-1 by using F/P factor for 1 year at the reinvestment rate c if the previous net investment ispositive or

•At CRR rate i’ if net investment value is negative.

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• Net-investment procedure:

•Mathematically, for each year t set up the relation

Ft = Ft-1 (1+i) + Ct

Where t= 1,2,…,n and Ct = net CF in year t

i = c if Ft-1 > 0 ; i = i’ if Ft-1 < 0

Set net-investment relation for year n equal to zero (Fn= 0) and solve for i’.

i’ value is unique for a stated reinvestment rate c

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6.7 Composite ROR Example

• CF is 50 at time 0, -200 at time1, 50 at time 2 and 100 at time 3. Reinvestment rate c=MARR= 15%.

•Find compsite rate i’?

• F0=50 > 0 so i = c = 15%

• F1 = F0 (1+0.15) – 200 = -142.50 < 0 i=i’

• F2 = -142.5 (1+i’) +50 = ..< 0 i=i’

• F3 = F2 (1+i’) +100 = 0 solve i’

•i’ = 3.13% and -168%.

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6.7 Example 6.10

• Compute the composite ROR of below cash flow if reinvestment rate is a) 7.47% b) MARR of 20%.

Year Cash Flow

0 $2,0001 -$5002 -$8,1003 $6,800

• a) Use procedure to determine i’ for c=i*= 7.47%

• F0=2000>0 i=c

• F1=2000(1.0747)-500=1649.40 >0 i=c• F2= 1649.40(1.0747)-8100=-6327.39 <0 use i=i’

• F3= -6327(1+i’)+6800= 0 then solve i’= 7.47% which is same as c, the reinvestment rate and one of the multiple i* values.

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• b) c = 20%

• Positive IB’s are reinvested at 20% -

• Not at the computed i* rate

• Now, what is the modified ROR – i’?

• Must perform a recursive analysis

6.7 Example 6.10

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• b) c = 20%

• F0 = +2,000 (> 0; invest at 20%)

• F1 = 2,000(1.20) – 500 = +1,900

•+1,900 > 0; invest at 20%

• F2 = 1,900(1.20) – 8100 = -5,820

• -5,820 < 0; invest at i’ rate

Under-recovered

6.7 Example 6.10

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F3 =-5,820 (1+i’) + 6,800 Since N = 3, F3 = 0 Solve; -5,820(1+i’) + 6,800 = 0

• -5,820(1+i’) + 6,800 = 0

• i’ = [6,800/5,820] –1

• i’ = 16.84% given c = 20%

• If MARR = 20% and i’ = 16.84%, this investment would be rejected

• Reduced the problem to a single conditional ROR value.

6.7 Example 6.10

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6.8 Using Excel for ROR Analysis

• 2 functions are used:

• 1) RATE (n,A,P,F)

• 2) IRR(first_cell:last_cell,guess)

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Chapter Summary

• ROR is a popular analysis method

• Understood in part by practitioners

• Presents computational difficulties

• No ROR may exist

• Multiple ROR’s may exist

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Chapter Summary

• For cash flows where multiple i* values exist, one must decide to:

• Go with one of the i* rates or,

• Calculate the composite rate, i’

• If an exact ROR is not required, it is suggested to go with PW or AW

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End of Chapter 6

by assoc. prof. dr. ahmet ÖztaŞ

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