1
Chapter Outline
 Incremental Analysis
 Graphical Technique in Solving problems with
Mutually Exclusive Alternatives
 Using Spreadsheets in Incremental Analysis
2
Learning Objectives
 Define Incremental Analysis
 Apply Graphical Technique in Solving Problems with
Mutually Exclusive Alternatives
 Use Spreadsheets in Incremental Analysis
3
Internal Rate of Return (IRR)
By definition in Chapter 7:
 Given a cash flow stream, IRR is the interest rate i at which
the benefits are equivalent to the costs. This can be
expressed mathematically in five different ways.





NPW=0
PW of benefits - PW of costs =0
PW of benefits = PW of costs
PW of benefits / PW of costs=1
EUAB-EUAC=0
4
Example
Cash flows for an investment are shown in the following
figure. What is the IRR to obtain these cash flows?
YEAR
CASH FLOW
0
($500)
1
$100
2
$150
3
$200
4
$250
5
YEAR
CASH
FLOW
0
($500)
1
$100
2
$150
3
$200
4
$250
QUESTION CONTINUES
EUAW  EUAB  EUAC  0
100  50( A / G , i %, 4)  500( A / P, i %, 4)  0
Try i  5%
100  50( A / G ,5%, 4)  500( A / P,5%, 4)
100  50(1.439)  500(0.2820)  30.95
Try i  15%
100  50( A / G ,15%, 4)  500( A / P,15%, 4)
100  50(1.326)  500(0.3503)  -8.85
10.20
6
YEAR
CASH
FLOW
0
($500)
1
$100
2
$150
3
$200
4
$250
QUESTION CONTINUES
EUAW  EUAB  EUAC  0
100  50( A / G , i %, 4)  500( A / P, i %, 4)  0
Try i  5%
100  50( A / G ,5%, 4)  500( A / P,5%, 4)
100  50(1.439)  500(0.2820)  30.95
Try i  15%
100  50( A / G ,15%, 4)  500( A / P,15%, 4)
100  50(1.326)  500(0.3503)  -8.85
10.20
7
INTERPOLATION
5%
30.95
30.95
5-X
10
X%
0
15%
-8.85
39.80
5 X
30.95  0
30.95


5  15 30.95  (8.85) 39.80
X  12.78  13%
8
INTERPOLATION
5%
30.95
30.95
5-X
10
X%
0
15%
-8.85
39.80
5 X
30.95  0
30.95


5  15 30.95  (8.85) 39.80
X  12.78  13%
9
EXCEL solution
IRR = irr(a1:a5) = 12.83%
10
Mutually Exclusive Alternatives
 Only one alternative may be implemented
 All alternatives serve the same purpose
 Objective of incremental analysis is to select the best
of these mutually exclusive alternatives
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Incremental Analysis
 When there are two alternatives, rate of return
analysis is performed by computing the
incremental rate of return, ΔIRR, on the difference
between the two alternatives, as discussed in
Chapter 7.
12
Incremental Analysis
 The cash flow for the difference between alternatives is
calculated by taking the higher initial-cost alternative
minus the lower initial-cost alternative.
 The below decision path is made for incremental rate of
return (ΔIRR) on difference between alternatives:
Two -Alternative
Situations
Decision
ΔIRR≥MARR
Choose the higher-cost alternative
ΔIRR<MARR
Choose the lower-cost alternative
13
Example
The cash flows for four different alternatives are given in
table below. If MARR 10%, which is the best alternative?
Using the incremental analysis, we need to repeat 3 times,
by comparing 2 alternatives at a time.
14
MARR = 10%
EUAB =EUAC (Increment)
ΔIRR≥MARR
ΔIRR<MARR
Choose the higher-cost alternative
15
Choose the lower-cost alternative
Incremental Analysis
 Could be applied to rate of return (IRR), present worth (PW),
equivalent uniform annual cost (EUAC), or equivalent uniform
annual worth (EUAW) approaches.
 [Higher-cost alternative] = [Lower-cost alternative]
+ [Increment between them]
 The “defender” is the best alternative identified so far in the
process, and “challenger” is the next higher-cost alternative to
be evaluated.
 For a set of N mutually exclusive alternatives, (N - 1)
“challenger/defender” comparisons must be made.
Copyright Oxford University Press 2009
16
Example
Given the alternatives below: Select the one best
alternative if MARR = 8%. Use incremental rate of return
analysis.
17
MARR = 8%
Since the MARR is 8%, Alt. D may be eliminated, as the ROR is less than 8%
Among the remaining alternatives A, B, and C, the two lower cost
alternatives are A and B.
(A - B) increment:
PW of benefit = PW of cost
(623 - 531)(P/A, i, 10) = (4,000 - 3,000)
(P/A, i, 10) = 1,000/92 = 10.86
(C - B) increment:
PW of benefit = PW of cost
(1,020 - 531)(P/A, i, 10) = (6,000 - 3,000) (P/A, i, 10) = 3,000/489 = 6.13
∆ROR is greater than 8%. Therefore, choose the higher-cost
alternative, Alt. C
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Example 8-1
High Capacity
$13,400
$4000/year
5 years
Cost
Benefit
Life
Low Capacity
$10,310
$3300/year
5 years
Increment
$3090
$700/year
5 years
PW LOW= -$10,310 + $3300(P/A,i,5)
PW HIGH= -$13,400 + $4000(P/A,i,5)
$8,000.00
IRRIncrement= 4.3%
$6,000.00
IRRHigh
PW
$4,000.00
IRRLow
$2,000.00
$0.00
0%
5%
10%
15%
20%
25%
($2,000.00)
($4,000.00)
i
19
Example 8-1 – Continued
In column B, input formula:
PW LOW = –$10,300 + $3300(P/A,i,5) = –$10300 + pv(A3, 5, –3300)
In column C, input formula:
PW HIGH = –$13,400 + $4000(P/A,i,5) = –$13400 + pv(A3, 5, –4000)
From a3 to a24, input interest from 0 to 0.21
In column D, input formula for incremental cost PW HIGH–LOW:
= C3 – B3
Then draw a line chart!
Or use EXCEL function npv(i, value range)
-10300+npv(A3, $A$28:$A$32)
-13400+npv(A3, $A$36:$A$40)
Example 8-1 – Continued
Interest Rate
PW LOW= -$10,300 + $3300(P/A,i,5)
0% ≤ i ≤ 4.3%
4.3 % ≤ i ≤ 18%
18% ≤ i
PW HIGH= -$13,400 + $40000(P/A,i,5)
PWhigh-low = -$3090 + $700(P/A, i, 5)
Best Choice
High Capacity
Low Capacity
Do Nothing
IRRIncrement= %4.3
$8,000
$6,000
IRRIncrement= %4.3
IRR
$40 High
$4,000
$20
$2,000
($4,000)
Diff
IRRLow
5.0%
10.0%
($20)3.5%
15.0%
($40)
High
Low
$0
$0
0.0%
($2,000)
Diff
Diff
20.0%
4.5%
25.0%
5.5%
30.0%
($60)
($80)
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Net Present Worth
Example 8-2
Rate
MARR = 10%
Machine X
$200
Machine Y
$700
Uniform Annual
Benefit
$95
$120
End-of-Useful-Life
Salvage Value
$50
$150
Initial Cost
Useful Life, in Years
6
In column C,
–700 + pv(A3, 12, –120, –150)
12
Machine X
Machine Y
0%
$840.00
$890.00
1.322
752.24
752.24
2
710.89
687.31
4
604.26
519.90
6
515.57
380.61
8
441.26
263.90
10
378.56
165.44
12
325.30
81.83
14
279.77
10.37
16
240.58
-51.08
18
206.65
-104.23
20
177.10
-150.47
In column B,
–200 + pv(A2, 6, -95, -50) + pv(A2,6, 0, 200 – pv(A2, 6, -95, -50))
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Net Present Worth
Rate
Example 8-2
$1,000.00
NPW
$600.00
IRRY
$400.00
$200.00
$840.00
$890.00
1.322
752.24
752.24
2
710.89
687.31
4
604.26
519.90
6
515.57
380.61
8
441.26
263.90
10
378.56
165.44
12
325.30
81.83
14
279.77
10.37
16
240.58
-51.08
18
206.65
-104.23
20
177.10
-150.47
Machine X
Machine Y
$0.00
0%
5%
10%
15%
($200.00)
($400.00)
i
For MARR ≤ 1.3%, Machine Y is the right choice
20%
Machine Y
0%
∆ IRRIncrement =1.3%
$800.00
Machine X
25%
For MARR ≥1.3%, Machine X is the right choice
23
Example 8-3
Consider the three mutually exclusive alternatives:
Initial Cost
Uniform Annual
Benefit
A
$2000
B
$4000
C
$5000
410
639
700
Each alternative has a 20 year life and no salvage value. If the MARR is
6%, which alternative should be selected?
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Initial Cost
Uniform Annual
Benefit
A
$2000
B
$4000
C
$5000
410
639
700
$10,000.00
∆ IRRC-B= 2%
$8,000.00
If MARR ≥ 9.6%,
Choose Alt. A
NPW
$6,000.00
∆ IRRB-A=9.6%
$4,000.00
$2,000.00
Alt. B
If 9.6% ≥ MARR≥2%,
Choose Alt. B
Alt. A
$0.00
($2,000.00)
0%
5%
10%
15%
Alt. C
($4,000.00)
20%
25%
If 2% ≥ MARR ≥ 0%,
Choose Alt. C
i
Net Present Worth Graph of Alternatives A, B, and C.
25
$10,000.00
∆ IRRC-B= 2%
$8,000.00
If MARR ≥ 9.6%,
Choose Alt. A
NPW
$6,000.00
∆ IRRB-A=9.6%
$4,000.00
$2,000.00
Alt. B
If 9.6% ≥ MARR≥2%,
Choose Alt. B
Alt. A
$0.00
($2,000.00)
0%
5%
10%
15%
Alt. C
20%
($4,000.00)
25%
If 2% ≥ MARR ≥ 0%,
Choose Alt. C
i
How to find the intersection points:
NPW(C-B) = -$5000+$4000+($700-$639)(P/A,i,20) = 0
∆IRR(C-B) = 2%
NPW(B-A) = -$4000+$2000+($639- $410)(P/A,i,20) = 0
∆IRR(B-A) = 9.6%
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Example 8-4
Brass
$100,000
4
Cost
Life
Stainless
$175,000
10
Titanium
$300,000
25
$80,000
$70,000
If 6.3% ≥ MARR ≥ 0%,
Choose Titanium
EUAC
$60,000
IRRTitanium - Stainless= 6.3%
$50,000
$40,000
If 15.3% ≥ MARR≥ 6.3%,
Choose Stainless
$30,000
$20,000
IRRStainless - Brass=15.3%
$10,000
If MARR ≥ 15.3%,
Choose Brass
$0
0%
5%
10%
15%
20%
25%
i
27
Example 8-5 Incremental Analysis
(with Do-Nothing option)
Machine X
$200
65
6
Initial Cost
Uniform Annual Benefit
Useful Life, in years
$60
IRRY-Z=3.5%
$40
IRRZ-X=11%
If MARR≥23%,
IRRX=23% Choose “Do-Nothing”
$0
($20) 0%
Machine Z
$425
100
8
X
$20
EUAW
Machine Y
$700
110
12
5%
10%
15%
20%
25%
If 23%≥MARR≥11%,
Choose X
Z
($40)
Y
($60)
If 11%≥MARR≥3.5%,
Choose Z
($80)
($100)
i
If 3.5%≥MARR≥0%,
Choose Y
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Example 8-5 Incremental Analysis
(without Do-Nothing option)
Machine X
$200
65
6
Initial Cost
Uniform Annual Benefit
Useful Life, in years
$60
IRRY-Z=3.5%
$40
IRRZ-X=11%
X
EUAW
$20
Machine Y
$700
110
12
If MARR≥11%,
IRRX=23%
Choose X
$0
($20) 0%
5%
10%
15%
20%
25%
Z
($40)
Y
($60)
Machine Z
$425
100
8
If 11%≥MARR≥3.5%,
Choose Z
If 3.5%≥MARR≥0%,
Choose Y
($80)
($100)
i
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Example 8-6 Incremental Analysis
using Graphical Comparison
A
$4000
639
Initial Cost
Uniform Annual Benefit
B
$2000
410
C
$6000
761
D
$1000
117
E
$9000
785
10,000
8,000
IRRC-A=2%
6,000
IRRA-B=9.6%
4,000
NPW
Life = 20 yrs
IRRB=20%
2,000
0
(2,000) 0%
5%
10%
15%
20%
25%
(4,000)
(6,000)
(8,000)
i
30
Example 8-6 Incremental Analysis
using Graphical Comparison
Initial Cost
Uniform Annual Benefit
A
$4000
639
B
$2000
410
C
$6000
761
D
$1000
117
E
$9000
785
Calculating Incremental Interest
∆IRR(C-A) = $6000-$4000 = ($761 - $639)(P/A, i, 20) = 2%
∆IRR(A-B) = $4000-$2000 = ($ 639 - $410)(P/A, i, 20) = 9.6%
And to find where the NPW of B crosses the 0 axis
IRR (B) = $2000 = $410(P/A, i, 20) = 20%
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Example 8-6 Incremental Analysis
using Graphical Comparison
A
$4000
639
Initial Cost
Uniform Annual Benefit
B
$2000
410
C
$6000
761
10,000
8,000
IRRA-B=9.6%
NPW
4,000
IRRB=20%
If 20%≥MARR≥9.6%,
Choose B
2,000
0
(2,000) 0%
E
$9000
785
If MARR≥20%,
Choose Do-Nothing
IRRC-A=2%
6,000
D
$1000
117
5%
10%
15%
(4,000)
20%
25%
If 9.6%≥MARR≥2%,
Choose A
(6,000)
(8,000)
i
If 2%≥MARR≥0%,
Choose C
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Spreadsheet and Incremental Analysis
Excel Functions
Purpose
Rate (n, A, -P, [F], [Type], [guess]) To find rate of return or incremental
rate of return given n, P, and A
IRR (range, [guess])
To find internal rate of return (or
incremental rate of return) of a series
of cash flow (or incremental cash flow)
Excel Tools
Purpose
Goal Seek
It varies the value in one specific cell until a formula that's
dependent on that cell returns the wanted result.
Solver adjusts the values in the changing cells to produce
the result from the target cell formula. Constraints are
applied to restrict the values Solver can use in the model.
Solver
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End of Chapter 8
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