Lecture # 14 investment alternatives ii

Lecture # 14
Investment Alternatives
Incremental ROI/IRR Method
1
Dr. A. Alim

1. Incremental ROI Analysis of Multiple
Mutually Exclusive Alternatives
The concept of incremental investment
Suppose a company has $90,000 to invest in a project. Two
mutually exclusive alternatives are proposed; MARR is 16%:
Alternative A requires an investment of $50,000 and has a ROI of 35%
Alternative B requires an investment of $85,000 and has a ROI of 29%
The remaining funds after the selection would naturally be invested at MARR.
One would intuitively think that option A is preferred since it has a higher ROI,
but this would be incorrect .
Material sourced from "Plant Design and Economics for Chem.
Engineers", 5th ed. by Peters et al. and also from " Engineering
Economy", 6th edition,2005, by Blank and Tarquin. 2
© 2003 and 2005 by McGraw-Hill,
New York, N.Y All Rights Reserved

Since the company can always invest the unused funds at
MARR, the overall ROI is:
With option A, ROI = (50000(0.35) + 40000(0.16))/90000
= 26.6 %
With option B, ROI = (85000(0.29) + 5000(0.16))/90000
= 28.3 %
Option B is recommended.
Economy", 6th edition,2005, by Blank and Tarquin. 3

COMPARISON OF ALTERNATIVES
The concept of incremental investment
Definition: incremental investment means every
additional dollar invested must result in an
incremental ROI at least equal to MARR.
If investment B is more than investment A, then:
Investment B is recommended ONLY IF :
(Profit ”B” – Profit “A”) / (Investment “B” – Investment “A”) ≥ MARR
i.e. incremental ROI ≥ MARR
Material sourced from "Plant Design and Economics for Chem. Engineers",
5th ed. by Peters et al. and also from " Engineering Economy", 6th
edition,2005, by Blank and Tarquin.
4© 2003 and 2005 by McGraw-Hill,

Material sourced from "Plant Design
and Economics for Chem.
Engineers", 5th ed. by Peters et al.
and also from " Engineering
Economy", 6th edition,2005, by
Blank and Tarquin.
5
Profit
$
Investment
$
B
A
Select B and reject A only if :
Incremental ROI =(Profit ”B” – Profit “A”) / (Investment “B” – Investment “A”) ≥ MARR

Profit
Or
Savings
InvestmentMaterial sourced from "Plant Design and Economics for Chem. Engineers",
6

Profit
Or
Savings
Investment
Incremental ROI = delta profit / delta investment
At the limit of delta investment approaches zero, the incremental ROI is the slope to the curve.
Economy", 6th edition,2005, by Blank and Tarquin. 7© 2003 and 2005 by McGraw-Hill,

Profit
Or
Savings
Investment
Incremental ROI = delta profit / delta investment
At the limit of delta investment approaches zero, the incremental ROI is the slope to the curve.
Material sourced from "Plant Design and Economics for
Chem. Engineers", 5th ed. by Peters et al. and also from
" Engineering Economy", 6th edition,2005, by Blank and
Tarquin.

Profit
Or
Savings
Investment
Line of slope = MARR
Recommended investment for
Incremental ROI = MARR
Material sourced from "Plant Design and Economics for Chem. Engineers", 5th ed.
by Peters et al. and also from " Engineering Economy", 6th edition,2005, by Blank
and Tarquin.
All investments here have
incremental ROI over “X” less than MARR
X

10
© 2003 and 2005 by McGraw-Hill, New York, N.Y All
Rights Reserved
Incremental investment analysisIncremental Investment Analysis

ROI (A) is larger than ROI(B)
Yet we reject A and accept B
Material sourced from "Plant Design and Economics for Chem. Engineers", 5th ed. by
Peters et al. and also from " Engineering Economy", 6th edition,2005, by Blank and
Tarquin.
11
Incremental Investment Analysis

Basic concept: We want the return to be higher than MARR for every
dollar invested.
Below the tangent point:
 ROI > MARR , hence, + ve incremental investment
Beyond the tangent point:
 ROI < MARR , hence, - ve incremental investment
Investments beyond the tangent point are therefore not advised when
Compared to alternative investment at the tangent point.
Calculate incremental ROI:
 ROI =  ( profit or savings) /  ( investment)
Accept if  ROI ≥ MARR
Material sourced from "Plant Design and Economics for Chem. Engineers", 5th
ed. by Peters et al. and also from " Engineering Economy", 6th edition,2005, by
Blank and Tarquin.

When comparing several investments using the ROI method,
we apply the incremental analysis and follow this principle:
Select the one alternative
• That requires the largest investment, and
• Indicates that the extra investment over another acceptable
alternative is justified, i.e. yielding an incremental ROI ≥
MARR.
13

Example 8-6, Page 344
Plant Design and Economics For chem. Engineers, 5th ed, 2003 by Peters et al.
14

and Tarquin.
15
© 2003 and 2005 by McGraw-Hill, New York, N.Y
All Rights Reserved

• We now apply incremental ROI analysis among alternatives 1,2, and 3:
• First step is to arrange the investments from left to right in ascending
order. In this case the order is 1, 2, then 3.
• We then determine Inc. ROI between pairs starting from the left and
moving right.
• Inc. ROI (2-1) = (3,000 – 2,000) / (16,000 – 10,000) = 16.7 %
which is > MARR.
Therefore: Reject alternative 1
• Inc. ROI (3-2) = (3,200 – 3,000) / (20,000 – 16,000) = 5 %
which is < MARR.
Therefore: Reject alternative 3
Conclusion: Accept alternative 2
Blank and Tarquin.
16

Modified example 8-3, page 331
• 3 investments. Need to evaluate profitability of each.
• Use ROI, PBP, NPV, and DCFRR.
• Assume straight line depreciation.
• Tax rate is 35%
• MARR is 15%
• Use MARR as interest rate for time value of money.
• Ignore land value.
Incremental ROI analysis
Blank and Tarquin.
17
Plant Design and Economics For chem. Engineers, 5th ed, 2003 by Peters et al.

Blank and Tarquin.

Inv. 1 Inv. 2 Inv. 3
Total inv. $ 110,000 180,000 225,000
NPAT, $/y 24,300 27,607 32,562
ROI (MARR) 22.1(15) 15.3(15) 14.5(15)
Accept? YES YES NO
Incremental ROI (2-1) = (27,607 – 24,300) / (180,000 -110,000)
= 4.7 %
Which is less than MARR. Therefore reject 2 and accept 1.
Blank and Tarquin. 19© 2003 and 2005 by McGraw-Hill,

2. Incremental DCFRR (IRR) Analysis of
Multiple Mutually Exclusive Alternatives
 Given N mutually exclusive alternatives, using
the incremental DCFRR method
 Select the one alternative that
 Requires the largest investment, and at the
same time
 Indicates that the extra investment over
another acceptable investment is justified.
 This means incremental IRR must be equal to
or higher than MARR.
Blank and Tarquin.

Blank and Tarquin.
21

and Tarquin.
22
IRR (B–A) is the incremental IRR or DCFRR

Which project to accept, A or B?
• Depends on the value of MARR !
• Accept B and reject A if:
IRR (B-A) ≥ MARR
• Accept A and reject B if:
IRR (B-A) < MARR

MARR
Reject both A and B

MARR
Reject B, accept A

MARR
Reject B, accept A
Since IRR (B-A) < MARR

MARR
Reject A, accept B
Since IRR (B-A) > MARR

Ranking Rules – Selection Process
Among Mutually Exclusive
Alternatives
1. Order the alternatives from smallest to
largest initial investment. For revenue
projects the DN alternative is the first on
the left (no investment!)
2. Compute the cash flows for each
alternative (DN has zero cash flows)
3. Ensure project lives are equal, apply LCM
if needed.
Material sourced from "Plant Design and
Economics for Chem. Engineers", 5th ed.
by Peters et al. and also from "
Engineering Economy", 6th edition,2005,
by Blank and Tarquin.

4. Compute the DCFRR value for all
alternatives in the considered set.
 If any alternative has an DCFRR < MARR
drop it from further consideration
 The candidate set will be those
alternatives with computed DCFRR values
> MARR.
 Call this the FEASIBLE set
Economics for Chem. Engineers", 5th ed. by
Peters et al. and also from " Engineering
Economy", 6th edition,2005, by Blank and
Tarquin.

 The first alternative is called the
DEFENDER
 The second (next higher investment cost)
alternative is called the CHALLENGER
 Compute the incremental cash flow as
(Challenger – Defender)
Economics for Chem. Engineers", 5th ed. by
Peters et al. and also from " Engineering
Economy", 6th edition,2005, by Blank and
Tarquin.

5. Compute DCFRR Challenger – Defender
 If DCFRR Challenger – Defender ≥ MARR drop
the defender and the challenger wins the
current round.
 If DCFRR Challenger – Defender < MARR, drop
the challenger and the defender moves on
to the next comparison round

 At each round, a winner is determined
 Either the current Defender or the
current Challenger
 The winner of a given round moves to the
next round and becomes the current
DEFENDER and is compared to the next
challenger

6. This process continues until there are
no more challengers remaining.
 The alternative that remains after all
alternatives have been evaluated is the
final winner.

Costs Only (Service) Problems –
DCFRR Approach
Remember
 Cost problems do not have computed
DCFRR’s since there are more cost amounts
that revenue amounts (salvage values may
exist)
 Thus there are no feasible DCFRR’s for
each alternative, but they do exist for the
delta between two alternatives.
Chem. Engineers", 5th ed. by Peters et al. and also from "
Engineering Economy", 6th edition,2005, by Blank and
Tarquin.
8-35 © 2003 and 2005 by McGraw-Hill, New York, N.Y All Rights
Reserved

Costs Only Problems - Rules
 Rank the alternatives according to their
investment requirements (low to high)
 For the first round compare:
 (Challenger – Defender) Cash Flow
 Compute DCFRR Challenger – Defender
 If DCFRR Challenger – Defender ≥ MARR,
Challenger wins; else Defender wins
Tarquin.
Reserved

Costs Only Problems -continued
 The current winner now becomes the defender
for the next round.
 Compare the current defender to the next
challenger and DCFRR Challenger – Defender
 The winner becomes the current champion and
moves to the next round as the defender
 Repeat until all alternatives have been
compared.
Tarquin.
Reserved

Equal or unequal service lives?
 Remember what we did when using PW
comparison?
 Projects can be compared only if they have
equal service lives.
 For projects with unequal service lives, we
should use the LCM concept.
 Alternatively we could use the AW approach to
find the breakeven IRR.
Tarquin.
Reserved

Tarquin.
© 2003 and 2005 by McGraw-Hill, New York, N.Y All Rights
Reserved41
The incremental IRR method resulted in alternative E being selected. Let’s try the PW method
to check this conclusion:
YEAR A B C D E F
0 -900 -1500 -2500 -4000 -5000 -7000
1 150 276 400 925 1125 1425
2 150 276 400 925 1125 1425
3 150 276 400 925 1125 1425
4 150 276 400 925 1125 1425
5 150 276 400 925 1125 1425
6 150 276 400 925 1125 1425
7 150 276 400 925 1125 1425
8 150 276 400 925 1125 1425
9 150 276 400 925 1125 1425
10 150 276 400 925 1125 1425
PW $21.69 $195.90 ($42.17) $1,683.72 $1,912.64 $1,756.01
Justified ? YES YES NO YES YES YES
WINNER !

• 10 year project
• New equipment is required
• Two vendors
• MARR = 15%
• Which vendor should be selected?
• Cost or Service Problem
• Lowest Common Multiplier (LCM) = 10 years
Example 8.3 Blank (7th ed.), p. 208
Example 8.3 Blank (6th ed.), p. 284
Unequal service lives
by Blank and Tarquin. 42© 2003 and 2005 by McGraw-Hill,

Incremental Cash Flow

PW analysis
• We could stop because the PW(15%) has
signaled that A is the winner!
• Lowest PW cost
• Proceed with a IRR analysis BUT….
• IRR must be performed on the
incremental investment

IRR (B-A) is less
than the MARR of
15%.
Therefore Reject B
and go with A
IRR (B-A) = 12.65 %
Inc. Cash Flow $
0 -5,000
1 1,900
2 1,900
3 1,900
4 1,900
5 -9,100
6 1,900
7 1,900
8 1,900
9 1,900
10 3,900
IRR 12.65%
Inc. Cash Flow Results

This is the
breakeven rate
of return

Determining incremental DCFRR by
using the AW method
• Approach is particularly useful for comparing projects of unequal
service lives.
• Determine AWA and AWB from one cycle only.
• For projects A and B, express AWA and AWB as a function of
interest rate, then set (AWB – AWA = 0)
• The interest rate satisfying (AWB – AWA = 0) is the breakeven
incremental IRR

using the AW method
Example: Re-do example 8.3
Determine the incremental IRR using the AW method:
•10 year project (merger)
•New equipment is required
•Two vendors
•MARR = 15%
•Which vendor should be selected
•Cost or Service Problem

using the AW method
AWA = -8,000 (A/P, i*, 10) – 3,500
AWB = -13,000(A/P, i*, 5) + 2,000 (A/F, i*, 5) – 1,600
Set AWB - AWA = 0
Solve for breakeven i* (inc. IRR) = 12.65 (using EXCEL Solver)
This is the incremental IRR; being less than MARR
We then choose project A

Home Work # 6
Thursday, March 6, 2014

CHEE 5369 / 6369
Homework # 6
The following problem from Peters et al., fifth
edition, 2003:
Problem 8.16 page 356
The following solved examples from Blank and
Tarquin, 7th edition, 2012:
Example 5.1 Page 132
Example 6.1 Page 151
Example 8.4 page 211

8-16

Lecture # 14 investment alternatives ii

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