Engineering Economy - Lecture - RateOfReturnAnalysis, Lecture notes of Economics

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Rate of Return Analysis

Rate of Return Analysis

Learning Objectives



Rate of return (ROR) is the rate paid on the unpaid

balance of borrowed money, or the rate earned on

the unrecovered balance of an investment, so that

the final payment or receipt brings the balance to

exactly zero with interest considered.



In ROR problems you are trying to find an unknown

interest rate (i*) that satisfies the following:

PWi(+ cash flows) – PWi( - cash flows) = 0



This means that the interest rate (i*) is an unknown

parameter.



Wells Fargo Bank lent an engineer $1000 at i = 10% per year for 4

years to buy home office equipment. From the bank’s perspective (the

lender), the investment is expected to produce an equivalent net cash

flow of $315.47 for the next 4 years.

A = $1000( A/P ,10%,4) = $315.



This represents a 10% per year rate of return on the bank’s

unrecovered balance.



Compute the amount of the unrecovered investment for each of the 4

years using:



( a ) the ROR on the unrecovered balance (the correct basis) and



( b ) the return on the initial $1000 investment.



( c ) Explain why all the initial $1000 amount is not recovered by the

final payment in part ( b ).



( b ) Table 7–2 shows the unrecovered balance if the 10% return is

always figured on the initial $1000. Column 6 in year 4 shows a

remaining unrecovered amount of $138.12, because only $861.88 is

recovered in the 4 years (column 5).



( c ) A total of $400 in interest must be earned if the 10% return each

year is based on the initial amount of $1000. However, only $261.

in interest must be earned if a 10% return on the unrecovered

balance is used.



There is more of the annual cash flow available to reduce the

remaining loan when the rate is applied to the unrecovered balance

as in part ( a ) and Table 7–1. Figure 7–1 (next page) illustrates the correct

interpretation of rate of return in Table 7–1.



Each year the receipt of $315.47 represents 10% interest on the

unrecovered balance in column 2 plus the recovered amount in

column 5.



Because rate of return is the interest rate on the unrecovered

balance, the computations in Table 7–1 for part (a) present a correct

interpretation of a 10% rate of return. Clearly, an interest rate applied

only to the principal represents a higher rate than is stated. In

practice, a so-called add-on interest rate is frequently based on

principal only, as in part ( b ). This is sometimes referred to as the

installment financing problem.



An engineer for a company constructing one of the

world’s tallest buildings (Shanghai Financial Center

in the Peoples’ Republic of China) has requested

that $500,000 be spent now during construction on

software and hardware to improve the efficiency of

the environmental control systems. This is expected

to save $10,000 per year for 10 years in energy

costs and $700,000 at the end of 10 years in

equipment refurbishment costs. Find the ROR.

Solution



Use the trial-and-error procedure based on a PW equation.

0 = - 500,000 + 10,000 ( P/A , i *,10) + 700,000 ( P/F , i *,10) [7.5]



Use the estimation procedure to determine i for the first trial. All income will be

regarded as a single F in year 10 so that the P/F factor can be used. The P/F factor is

selected because most of the cash flow ($700,000) already fits this factor and errors

created by neglecting the time value of the remaining money will be minimized. Only

for the first estimate of i , define

P = $500,000, n = 10, and F = 10(10,000) + 700,000 = $800,000.

PWi(+ cash flows) – PWi( - cash flows) = 0

Solution

Year Amount Trial i PW value

0 -$500,000 4.00% $54,

1 $10,000 4.20% $44,

2 $10,000 4.40% $34,

3 $10,000 4.60% $25,

4 $10,000 4.80% $15,

5 $10,000 5.00% $6,

6 $10,000 5.20% -$1,

7 $10,000 5.40% -$10,

8 $10,000 5.60% -$19,

9 $10,000 5.80% -$27,

10 $710,000 6.00% -$35,

ROR =

5.16%

Solution



Use AW computations to find the ROR for the cash flows in Example 7.2.



The AW relations for disbursements (AW

D

) and receipts (AW

R

) are formulated using

Equation [7.2].

AW

D

= - 500,000 ( A/P , i ,10)

AW

R

= 10,000 + 700,000 ( A/F , i ,10)

0 = - 500,000 ( A/P , i *,10) + 10,000 + 700,000 ( A/F , i *,10)



Trial-and-error solution yields these results:



At i = 5%, 0 < $



At i = 6%, 0 > - $



By interpolation, i * = 5.16%, as before.

Multiple Rate of Return Values



So far we have dealt with conventional (or simple) cash flow series. The algebraic signs on

the net cash flows changed only once, usually from minus in year 0 to plus at some time

during the series.



However, for many series the net cash flows switch between positive and negative from one

year to another, so there is more than one sign change. Such a series is called

nonconventional (nonsimple).



When there is more than one sign change in the net cash flows, it is possible that there will

be multiple i * values in the - 100% to plus infinity range.

Multiple RORs



There are two tests to perform in sequence on the nonconventional series to

determine if there is one unique or multiple i * values that are real numbers. The first

test is the (Descartes’) rule of signs states that the total number of real-number roots

is always less than or equal to the number of sign changes in the series. This rule is

derived from the fact that the relation set up by Equation [7.1] or [7.2] to find i * is an

n th-order polynomial.



(It is possible that imaginary values or infinity may also satisfy the equation.) The

second and more discriminating test determines if there is one, real number, positive

i * value. This is the cumulative cash flow sign test, also known as Norstrom’s

criterion_._ It states that only one sign change in the series of cumulative cash flows

which starts negatively, indicates that there is one positive root to the polynomial

relation. To perform this test, determine the series

S

t

= cumulative cash flows through period t



Observe the sign of S 0 and count the sign changes in the series S

0

, S

1

,... , S

n

. Only

if S 0

0 and signs change one time in the series is there a single, real number,

positive i *.



( a ) Table 7–4 shows the annual cash flows and cumulative cash

flows. Since there are two sign changes in the cash flow sequence,

the rule of signs indicates a maximum of two real-number i * values.

The cumulative cash flow sequence starts with a positive number



S

0

= +2000, indicating there is not just one positive root.



The conclusion is that as many as two i * values can be found.



(b) The 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 the two i * values, and plot PW vs. i. The PW values are shown

below and plotted in Figure 7–5 for i values of 0, 5, 10, 20, 30, 40, and 50%.



The characteristic parabolic shape for a second-degree polynomial is obtained, with PW

crossing the i axis at approximately i

1

• = 8 and

i

2