Chapter 8: Rate of Return Analysis – Multiple Alternatives
For a single project, ROR or i* is determined by PW or AW method. We either solve for i* in
0 = - PWD + PWR
or
0 = -AWD + AWR
where D: disbursements (costs) and R: revenue (income).
i* can only be calculated if we have disbursements and revenue.
If i*  MARR, the project is economically acceptable. If i* < MARR, it is not.
Comparing Alternatives Using ROR Method
We can again use PW or AW methods. We have two types of selections:
i) If projects are independent – Here we select all projects that have i* equal or greater than
MARR. We, therefore, determine for each project whether or not i*  MARR. All those with
i*  MARR are selected.
These choices can only be among projects with revenue and cost. If there is no revenue, it
means we cannot calculate i* individually and hence no independent selection.
To determine whether or not i*  MARR, substitute i* = MARR in the PW or AW equation.
If positive, then i*  MARR.
ii) If projects are mutually exclusive – Here we must select only ONE best one. For these
comparisons, we must use incremental analysis.
If alternatives have revenue, then the problem will state whether they are mutually exclusive
(select only ONE) or not. If alternatives have only costs, then, it automatically indicates that
they are mutually exclusive and, therefore, only ONE shall be selected.
Incremental analysis must be carried out over the same period. Therefore, if project lives are
different, the cash flow for each project must be extended to LCM years.
Example. Consider two alternatives A and B. Cash flows are as follows. Determine which
should be selected if MARR is 15% per year.
Initial cost
Annual cost
SV
Life, years
A
B
-8000
-3500
0
10
-13000
-1600
2000
5
Note: Only costs are involved. Therefore, we automatically select the one that is best.
Here lives are 10 and 5 years. Analysis must be over LCM years which is 10 years. We must
now write down the incremental cash flow. We always order projects with lower initial cost
first. (In multiple projects, lowest to highest ordering is done).
1
Year
Cash flow A
0
1–5
5
-8000
-3500
-
6 – 10
10
-3500
-
Cash flow B
Incremental cash flow (B – A)
-13000
-1600
{+2000
-13000}
-1600
+2000
-5000
+1900
-11000
+1900
+2000
Using PW method:
0 = -5000 + 1900.(P/A,i*,10) – 11000.(P/F, i*,5) + 2000. (P/F,i*,10)
[i* is used for the cash flow (B – A) to distinguish it from iA* or iB* ]
If we substitute i* = 15%, in the above
= -5000 + 1900.(5.0188) – 11000.(0.4972) + 2000.(0.2472) = -439.08
It is negative, therefore i* < MARR (=15%)
Therefore, extra investment for B is not justified. Select A.
We can also solve using AW method. Here, we have two options.
1. Use incremental cash flow. Then, we must use cash flow up to LCM. For the cash flow
for 10 years,
0 = -5000.(A/P,i*,10) + 1900 – 11000.(P/F,i*,5).(A/P,i*,10) + 2000.(A/F,i*,10)
Substituting i* = MARR (=15%), we get -87.488 which is negative.
Therefore, i* < MARR.
B is not justified; we select A.
2. Use AW of each project for one cycle only.
We write,
AWA for one cycle, which is for 10 years,
and
Then,
AWB for one cycle, which is for 5 years.
0 = AWB – AWA
Hence,
0 = -13000.(A/P,i*,5) – 1600 + 2000.(A/F,i*,5) – [-8000.(A/P,i*,10) – 3500]
For i* = 15%, the value of above = -87.3
Therefore, i* < 15% (MARR)
B is not justified; we select A.
2
ROR Analysis: Multiple Alternatives
The procedure will be demonstrated by means of solved problems.
Problem 1.
We are evaluating five alternatives for which the cash flows are shown below. The MARR is
18% per year and all the alternatives are expected to have a life of 8 years.
(a) If the proposals are independent, which should be selected?
(b) If the proposals are mutually exclusive, determine which one should be selected using the
incremental-rate-of-return method.
A
B
C
D
E
-10000
-12000
-18000
-24000
-33000
AOC
-5000
-5500
-7000
-11000
-16000
SV
2000
2500
3000
3500
4500
Annual income
9000
10000
10500
12500
14500
Initial investment
a) If proposals are independent, we select every alternative that has i*  MARR or i*  18%.
Using PW,
For A: 0 = -10000 – 5000.(P/A,i*,8) + 2000.(P/F,i*,8) + 9000.(P/A,i*,8)
To see if i*  18%, we substitute i* = 18% in this equation:
= -10000 – 5000x4.0776 + 2000x0.266 + 9000x4.0776 = 6842.4
Since it is positive, it means i* > MARR or i* > 18%.
Therefore, we select A.
We repeat the above for B, C, D, and E:
We find that, for B: i* > 18%, therefore, we select B also.
For C:
i* < 18%, therefore, we do not select C.
For D:
i* < 18%, therefore, we do not select D.
For E:
i* < 18%, therefore, we do not select E.
For independent alternatives we select A and B and reject C, D, and E.
b) For mutually exclusive selection, since we have already shown that C, D, and E have ROR,
i.e. i*, less than MARR, we do not consider them for further calculations. To find which is
best, we need to compare A and B only (Note that we do not have to consider DO Nothing,
i.e. ‘DN’, here since i* greater than MARR for A and B already tells us that A and B are
better than DN). For A and B, we do an incremental analysis to see which is better. We put
the lower cost alternative first.
A
B
-10000
-12000
AOC
-5000
-5500
SV
2000
2500
Annual income
9000
10000
Initial investment
3
Alternatives compared: B to A
Incremental investment
-2000
Incremental AOC
-500
Incremental SV
500
Incremental income
1000
PW method: 0 = -2000 –500.(P/A, i*,8) + 500.(P/F, i*,8) + 1000.(P/A, i*,8)
For i* = 18%,
= -2000 – 500x4.0776 + 500x0.266 + 1000x4.0776 = 171.8
Therefore, i* > 18% or MARR
We select B.
Note that in this problem the alternatives have revenue (income, receipts, etc.). It is because of
this that we can calculate i* for each alternative and it is because of this fact that we can
evaluate them as independent alternatives.
Problem 2.
Determine which one of the following projects should be selected using the incremental-rateof-return method. All the projects are expected to have a 10-year life and MARR is 15% per
year.
First cost
Annual income
A
B
C
D
E
-15000
-18000
-35000
-25000
-52000
5000
6000
9000
7000
12000
Problem indicates only ONE, best one, to be selected, i.e. mutually exclusive. Here, the
alternatives have revenue or income as well. We can do the selection by either one of the
following two methods:
i) Determine, individually, which alternatives have i*  MARR. Then, select those with
i*  MARR for incremental analysis to select the best one (as in Problem 1), or
ii) Start with the first alternative as challenger to “Do Nothing” (DN). Depending on the
outcome (if i*  MARR, select first alternative as defender to be challenged by the second
one. If i* < MARR, reject first alternative and go to second alternative as challenger against
DN, and so on).
We’ll follow (ii) as i* values are not available as in Problem 1.
To start the process, we line up the alternatives with increasing first costs.
A
First cost
Annual income
B
D
C
E
-15000
-18000
-25000
-35000
-52000
5000
6000
7000
9000
12000
Alternative compared: A to DN
0 = -15000 + 5000.(P/A,i*,10)
or (P/A,i*,10) = 3
(From the Table for i = 15%, if (P/A,i*,10)  5.018, then i*  MARR or  15%)
4
Therefore, i* > MARR
We, therefore, eliminate DN, select A as defender and B as challenger.
B – A: 0 = -3000 + 1000.(P/A,i*,10)
or (P/A,i*,10) = 3
Again, based on the Table for i=15%, i*>MARR; select B as defender, and eliminate A.
D – B: 0 = -7000 + 1000.(P/A,i*,10)
or (P/A,i*,10) = 7
Therefore, i*<MARR; eliminate D, B is still defender.
C – B: 0 = -17000 + 3000.(P/A,i*,10)
or (P/A,i*,10) = 17/3 = 5.667
Therefore, i*<MARR; eliminate C, B is still defender.
E – B: 0 = -34000 + 6000.(P/A,i*,10)
or (P/A,i*,10) = 5.667
Therefore, i*<MARR; eliminate E also.
Select B.
Problem 3.
Determine which of the following projects should be selected using the incremental-rate-ofreturn method. All the projects are expected to have a 10-year life and MARR is 18% per
year.
A
B
C
D
E
First cost
-28000
-33000
-22000
-51000
-46000
AOC
-20000
-18000
-25000
-12000
-14000
Here, we do not have any revenue (income). We must, therefore, choose ONE, best one, using
incremental method. (There cannot be comparison to DN or calculation for i* for each
alternative).
We start by ordering alternatives with increasing costs.
C
A
B
E
D
First cost
-22000
-28000
-33000
-46000
-51000
AOC
-25000
-20000
-18000
-14000
-12000
We start with the first one (C) as defender and the second one (A) as the challenger.
A – C: 0 = -6000 + 5000.(P/A,i*,10) or (P/A,i*,10) = 1.2
(From the Table for i = 18%, if (P/A,i*,10)  4.494, then i*  MARR or  18%)
Therefore, i* > MARR, eliminate C.
B – A: 0 = -5000 + 2000.(P/A,i*,10)
or (P/A,i*,10) = 2.5
Therefore, i* > MARR, eliminate A.
E – B: 0 = -13000 + 4000.(P/A,i*,10)
or (P/A,i*,10) = 3.25
Therefore, i* > MARR, eliminate B.
D – E: 0 = -5000 + 2000.(P/A,i*,10)
or (P/A,i*,10) = 2.5
Therefore, i* > MARR, eliminate E.
Select D.
5
Problem 4.
A company is considering the projects shown below, all of which can be considered to last
forever. If the company MARR is 18% per year, use rate-of-return analysis to determine
which should be selected
a) If the alternatives are independent, and
b) If the alternatives are mutually exclusive.
A
First cost
Annual income
B
C
D
E
-10000
-20000
-15000
-70000
-50000
2000
4000
2950
10000
6000
a) For infinite lives,
P = A/i
Then ROR for A:
2000 / i* = 10000; or i* = 0.2, i.e. 20%
Similarly for B, i* = 20%; for C, i* = 19.6%; for D, i* = 14.3%; and E, i* = 12%.
Since MARR is 18%, we select A, B, and C.
b) Since D and E have i* < MARR, we do not have to consider them for further calculations.
Ordering A, B, and C in increasing costs:
A
First cost
Annual income
C
B
-10000
-15000
-20000
2000
2950
4000
We have already determined that i* for A is greater than MARR. This means it is better than
Do Nothing. After stating this fact, proceed to compare A and C incrementally.
A vs. DN:
Select A as already shown that i* > MARR
C – A:
0 = -5000 + 950/i* or i* = 0.19 (19% which is greater than MARR)
Therefore, eliminate A and select C.
B – C:
0 = -5000 + 1150/i* or i* = 0.23 (23% which is greater than MARR)
Therefore, eliminate C and select B.
6
Problem 5.
A metal plating company is considering four different methods of recovering by-product
heavy metals from manufacturing site’s liquid waste. The investment costs and incomes
associated with each method have been estimated. All methods have a 10-year life. The
MARR is 12% per year.
a) If the methods are independent, because they can be implemented at different plants, which
ones are acceptable?
b) If the methods are mutually exclusive, determine which one method can be selected using
ROR evaluation.
Method
First Cost
SV
Annual Income
A
-15000
1000
4000
B
-18000
2000
5000
C
-25000
-500
6000
D
-35000
-700
8000
a) Using PW method:
Method A:
0 = -15000 + 4000.(P/A,i*,10) + 1000.(P/F,i*,10)
Substituting MARR(=12%) for i*, the RHS is positive (=7922.8), hence i*>MARR
Therefore, select A.
Method B:
0 = -18000 + 5000.(P/A,i*,10) + 2000.(P/F,i*,10)
Substituting MARR(=12%) for i*, the RHS is positive (=10895), hence i*>MARR
Therefore, select B also.
Method C:
0 = -25000 + 6000.(P/A,i*,10) - 500.(P/F,i*,10)
Substituting MARR(=12%) for i*, the RHS is positive (=8740.2), hence i*>MARR
Therefore, select C as well.
Method D:
0 = -35000 + 8000.(P/A,i*,10) - 700.(P/F,i*,10)
Substituting MARR(=12%) for i*, the RHS is positive (=9976.2), hence i*>MARR
Therefore, select D too
Hence, we select A, B, C and D.
Alternatively, we could have used AW method. Then,
Method A:
0 = -15000.(A/P,i*,10) + 1000.(A/F,i*,10) + 4000
At 12%, the value to right hand side (RHS) = 1402.28. Hence, i* > MARR
We select A.
Similar analysis for B, C, and D also show that i* > MARR for all of these methods.
Hence, we select A, B, C and D.
7
b) A – DN: It is already shown in (a) that i* for Method A > MARR. Therefore, A is selected
over DN. Alternatively, we can write (using PW method):
0 = -15000 + 4000.(P/A,i*,10) + 1000.(P/F,i*,10)
i* > MARR, select A
B – A:
0 = -3000 + 1000.(P/A,i*,10) + 1000.(P/F,i*,10)
It can be shown that i* > 12% (RHS is positive). Therefore, select B
C – B:
0 = -7000 + 1000.(P/A,i*,10) - 2500.(P/F,i*,10)
With i* = 12%, RHS is negative; Hence, i* < MARR (=12%). Therefore, select B again.
D – B:
0 = -17000 + 3000.(P/A,i*,10) - 2700.(P/F,i*,10)
Here, again i* < MARR.
Therefore, B is still the selection. Select B.
Problem 6.
Company X is designing a processing facility. Two options are being considered with the
following cash flows:
Capital investment
Annual expenses:
Operation
Maintenance
Salvage value
Useful life, years
A
$66400
B
$95200
$4330
$2200 in year 1,
and increasing
$1000/yr thereafter
0
5
$3440
$1000 in year 4,
and increasing
$200/yr thereafter
$10000
9
The MARR is 20% per year. Using the rate of return method, determine which option
should be selected?
In this example if we wish to use either of the equations:
0 = -AWD – AWR
or
0 = - PWD - PWR
we have to determine the incremental cash flow for LCM of years, i.e. 45 years in this case.
Incremental cash flow will be obtained by considering 9 cycles of A and 5 cycles of B. This
makes it a laborious exercise. It is, therefore, better to use AW formula,
0 = AWB – AWA
where AW’s of A and B for one cycle only is used. (Note that if we wish to use PWA and
PWB values, then we have to calculate the values over LCM of years).
0 = -95200(A/P,i*,9) + 10000(A/F,i*,9) – 3440
- [1000(P/A,i*,6) + 200(P/G,i*,6)](P/F,i*,3)(A/P,i*,9)
- {-66400(A/P,i*,5) – 4330 – [2200 + 1000.(A/G,i*,5)]}
Substituting i* = 20%,
RHS = 3130. Hence i* > 20%.
Select B.
8
Problem 7.
The four alternatives described below are being evaluated.
a) If the proposals are independent, which should be selected when MARR = 18% per year?
b) If the proposals are mutually exclusive, which one should be selected when MARR=9.5%
per year?
Incremental ROR,%, When compared
with alternative
Alternative
Init. Investment
ROR,%
A
B
A
-40000
29
B
-75000
15
1
C
-100000
16
7
20
D
-200000
14
10
13
a) Only alternative A has a ROR > 18%.
Select A
b) A to DN:
i* = 29% > 9.5%, eliminate DN.
B to A:
i* = 1% < 9.5%, eliminate B.
C to A:
i* = 7% < 9.5%, eliminate C.
D to A:
i* = 10% > 9.5%, eliminate A.
Therefore, select alternative D.
9
C
12