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Chapter 7
Rate of Return Analysis
 Rate of Return
 Methods for Finding ROR
 Internal Rate of Return
(IRR) Criterion
 Incremental Analysis
 Mutually Exclusive
Alternatives
Rate of Return
Definition
A relative percentage method which
measures the annual rate of return as a
percentage of investment over the life of a
project.
In 1970, when Wal-Mart Stores, Inc. went
public, an investment of 100 shares cost
$1,650. That investment would have
been worth $12,283,904 on January 31,
2002.
What is the rate of return on that
investment?
Example: Meaning of Rate
of Return
Solution:
0
32
$12,283,904
$1,650
Given: P = $1,650
F = $12,283,904
N = 32
Find i:
$12,283,904 = $1,650 (1 + i )32
i = 32.13%
N
i
P
F )
1
( 

Rate of Return
Suppose that you invested that amount ($1,650) in a
savings account at 6% per year. Then, you could have
only $10,648 on January, 2002.
What is the meaning of this 6% interest here?
This is your opportunity cost if putting money in
savings account was the best you can do at that time!
So, in 1970, as long as you earn more than 6%
interest in another investment, you will take that
investment.
Therefore, that 6% is viewed as a minimum
attractive rate of return (or required rate of return).
So, you can apply the following decision rule, to see
if the proposed investment is a good one.
ROR (32.13%) > MARR(6%)
Return on Investment
Definition 1: Rate of return (ROR) is defined as the
interest rate earned on the unpaid balance of an
installment loan.
Example: A bank lends $10,000 and receives annual
payment of $4,021 over 3 years. The bank is said to
earn a return of 10% on its loan of $10,000.
Loan Balance Calculation:
A = $10,000 (A/P, 10%, 3)
= $4,021
Unpaid Return on Unpaid
balance unpaid balance
at beg. balance Payment at the end
Year of year (10%) received of year
0
1
2
3
-$10,000
-$10,000
-$6,979
-$3,656
-$1,000
-$698
-$366
+$4,021
+$4,021
+$4,021
-$10,000
-$6,979
-$3,656
0
A return of 10% on the amount still outstanding at the
beginning of each year
Rate of Return:
Definition 2: Rate of return (ROR) is the break-even
interest rate, i*, which equates the present worth of a
project’s cash outflows to the present worth of its cash
inflows.
Mathematical Relation:
PW i PW i PW i
( ) ( ) ( )
* * *
 

cash inflows cash outflows
0
Return on Invested Capital
Definition 3: Return on invested capital is defined as the
interest rate earned on the unrecovered project balance of
an investment project. It is commonly known as internal
rate of return (IRR).
Example: A company invests $10,000 in a computer and
results in equivalent annual labor savings of $4,021 over
3 years. The company is said to earn a return of 10% on
its investment of $10,000.
Project Balance Calculation:
0 1 2 3
Beginning
project balance
Return on
invested capital
Payment
received
Ending project
balance
-$10,000 -$6,979 -$3,656
-$1,000 -$697 -$365
-$10,000 +$4,021 +$4,021 +$4,021
-$10,000 -$6,979 -$3,656 0
The firm earns a 10% rate of return on funds that remain internally
invested in the project. Since the return is internal to the project, we
call it internal rate of return.
Methods for Finding Rate of Return
 Types of Investment (cash flow) Classification
 Simple Investment
 Non-simple Investment
 Once we identified the type of investment cash flow,
there are several ways available to determine its rate
of return.
 Computational Methods
 Direct Solution Method
 Trial-and-Error Method
 Computer Solution Method
13
Investment Classification
Simple Investment
 Definition: Initial cash
flows are negative,
and only one sign
change occurs in the
net cash flows series.
 Example: -$100, 250,
$300 (-, +, +)
 ROR: A unique ROR
 If the initial flows are positive and
one sign change occurs referred to
simple-borrowing.
Non-simple Investment
 Definition: Initial cash
flows are negative,
but more than one
sign changes in the
remaining cash flow
series.
 Example: -$100, 300,
-$120 (-, +, -)
 ROR: A possibility of
multiple RORs
Period
(N)
Project
A
Project
B
Project
C
0 -$1,000 -$1,000 +$1,000
1 -500 3,900 -450
2 800 -5,030 -450
3 1,500 2,145 -450
4 2,000
Project A is a simple investment.
Project B is a non-simple investment.
Project C is a simple borrowing.
Computational Methods
Direct
Solution
Direct
Solution
Trial &
Error
Method
Computer
Solution
Method
Log Quadratic
n Project A Project B Project C Project D
0 -$1,000 -$2,000 -$75,000 -$10,000
1 0 1,300 24,400 20,000
2 0 1,500 27,340 20,000
3 0 55,760 25,000
4 1,500
Example 7.2 Direct Solution Methods
Project A Project B
PW i
i i
x
i
PW i x x
x
x
i
i
i
i
i
i i
( ) $2,
$1,
( )
$1,
( )
,
( ) , , ,
:
.
. .
, *
  






   



   

  
    
000
300
1
500
1
0
1
1
2 000 1300 1500
08
08
1
1
25%, 1667
1
1
160%
100% 25%.
2
2
Let then
Solve for
or -1.667
Solving for yields
Since the project's
$1, $1, ( / , , )
$1, $1, ( )
. ( )
ln .
ln( )
. ln( )
.
.
000 500 4
000 500 1
0 6667 1
0 6667
4
1
0101365 1
1
1
4
4
0 101365
0 101365

 
 

 
 
 
 



P F i
i
i
i
i
e i
i e
10.67%
Trial and Error Method – Project C
 Step 1: Guess an interest
rate, say, i = 15%
 Step 2: Compute PW(i)
at the guessed i value.
PW (15%) = $3,553
 Step 3: If PW(i) > 0, then
increase i. If PW(i)< 0,
then decrease i.
PW(18%) = -$749
 Step 4: If you bracket the
solution, you use a linear
interpolation to approximate
the solution
3,553
0
-749
15% i 18%









749
553
,
3
553
,
3
%
3
%
15
i
%
45
.
17

Basic Decision Rule:
If ROR > MARR, Accept
This rule does not work for a situation where
an investment has multiple rates of return
Comparing Mutually Exclusive
Alternatives Based on IRR
Issue: Can we rank the mutually exclusive
projects by the magnitude of its IRR?
n A1 A2
0
1
IRR
-$1,000 -$5,000
$2,000 $7,000
100% > 40%
$818 < $1,364
PW (10%)
Who Got More Pay Raise?
Bill Hillary
10% 5%
Can’t Compare without Knowing Their
Base Salaries
Bill Hillary
Base Salary $50,000 $200,000
Pay Raise (%) 10% 5%
Pay Raise ($) $5,000 $10,000
For the same reason, we can’t compare mutually exclusive projects based on
the magnitude of its IRR. We need to know the size of investment and its timing
of when to occur.
Incremental Investment
 Assuming a MARR of 10%, you can always earn that rate from
other investment source, i.e., $4,400 at the end of one year for $4,000
investment.
 By investing the additional $4,000 in A2, you would make
additional $5,000, which is equivalent to earning at the rate of 25%.
Therefore, the incremental investment in A2 is justified.
n Project A1 Project A2
Incremental
Investment
(A2 – A1)
0
1
-$1,000
$2,000
-$5,000
$7,000
-$4,000
$5,000
ROR
PW(10%)
100%
$818
40%
$1,364
25%
$546
Incremental Analysis (Procedure)
Step 1: Compute the cash flow for the difference
between the projects (A,B) by subtracting
the cash flow of the lower investment
cost project (A) from that of the higher
investment cost project (B).
Step 2: Compute the IRR on this incremental
investment (IRR ).
Step 3: Accept the investment B if and only if
IRR B-A > MARR
B-A
NOTE: Make sure that both IRRA and IRRB are greater than MARR.
Example 7.7 - Incremental Rate of Return
n B1 B2 B2 - B1
0
1
2
3
-$3,000
1,350
1,800
1,500
-$12,000
4,200
6,225
6,330
-$9,000
2,850
4,425
4,830
IRR 25% 17.43% 15%
Given MARR = 10%, which project is a better choice?
Since IRRB2-B1=15% > 10%, and also IRRB2 > 10%, select B2.
IRR on Increment Investment:
Three Alternatives
n D1 D2 D3
0 -$2,000 -$1,000 -$3,000
1 1,500 800 1,500
2 1,000 500 2,000
3 800 500 1,000
IRR 34.37% 40.76% 24.81%
Step 1: Examine the IRR for each
project to eliminate any project
that fails to meet the MARR.
Step 2: Compare D1 and D2 in pairs.
IRRD1-D2=27.61% > 15%,
so select D1. D1 becomes the
current best.
Step 3: Compare D1 and D3.
IRRD3-D1= 8.8% < 15%,
so select D1 again.
Here, we conclude that D1 is the best
Alternative.
Items CMS Option FMS Option
Investment $ 4,500,000 $ 12,500,000
Total annual operating costs $ 7,412,920 $ 5,504,100
Net salvage value $ 500,000 $ 1,000,000
Example 7.9 Incremental Analysis for Cost-Only Projects
 The firm’s MARR is 15%. Which alternative would be a better choice, based
on the IRR criterion?
 Discussion: Since we can assume that both manufacturing systems would
provide the same level of revenues over the analysis period, we can compare
these alternatives based on cost only. (these systems are service projects).
 Although we can not compute the IRR for each option without knowing the
revenue figures, we can still calculate the IRR on incremental cash flows.
 Since the FMS option requires a higher initial investment than that of the
CMS, the incremental cash flow is the difference (FMS – CMS)
Example 7.9 Incremental Analysis for Cost-Only Projects
(cost are itemized)
Items CMS Option FMS Option
Annual O&M costs:
Annual labor cost $1,169,600 $707,200
Annual material cost 832,320 598,400
Annual overhead cost 3,150,000 1,950,000
Annual tooling cost 470,000 300,000
Annual inventory cost 141,000 31,500
Annual income taxes 1,650,000 1,917,000
Total annual operating costs $7,412,920 $5,504,100
Investment $4,500,000 $12,500,000
Net salvage value $500,000 $1,000,000
Example 7.9 Incremental Cash Flow (FMS – CMS)
n CMS Option FMS Option
Incremental
(FMS-CMS)
0 -$4,500,000 -$12,500,000 -$8,000,000
1 -7,412,920 -5,504,100 1,908,820
2 -7,412,920 -5,504,100 1,908,820
3 -7,412,920 -5,504,100 1,908,820
4 -7,412,920 -5,504,100 1,908,820
5 -7,412,920 -5,504,100 1,908,820
6 -7,412,920 -5,504,100
$2,408,820
Salvage + $500,000 + $1,000,000
Solution:
PW i
P A i
P F i
IRR
FMS CMS
FMS CMS
( ) $8, ,
$1,908, ( / , , )
$2, , ( / , , )
.43%


 



 
000 000
820 5
408 820 6
0
12 15%,
select CMS.
 Although the FMS would provide an incremental annual savings of $1,908,820
in operating costs, the savings do not justify the incremental investment of
$8,000,000.
COMMENTS:
 Note that the CMS option was marginally preferred to the FMS option.
 However, there are dangers in relying solely on the easily quantified savings in
input factors – such as labor, energy, and materials – from FMS and in not
considering gains from improved manufacturing performance that are more
difficult and subjective to quantify.
 Factors such as improved product quality, increased manufacturing flexibility
(rapid response to customer demand), reduced inventory levels, and increased
capacity for product innovation are frequently ignored in financial analysis
because we have inadequate means for quantifying benefits.
 If these intangible benefits were considered, as they ought to be, however, the
FMS option could come out better than the CMS option.

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SE 307-CHAPTER_7_RATE_OF_RETURN_ANALYSIS_new.pptx

  • 1. Chapter 7 Rate of Return Analysis  Rate of Return  Methods for Finding ROR  Internal Rate of Return (IRR) Criterion  Incremental Analysis  Mutually Exclusive Alternatives
  • 2. Rate of Return Definition A relative percentage method which measures the annual rate of return as a percentage of investment over the life of a project.
  • 3. In 1970, when Wal-Mart Stores, Inc. went public, an investment of 100 shares cost $1,650. That investment would have been worth $12,283,904 on January 31, 2002. What is the rate of return on that investment? Example: Meaning of Rate of Return
  • 4. Solution: 0 32 $12,283,904 $1,650 Given: P = $1,650 F = $12,283,904 N = 32 Find i: $12,283,904 = $1,650 (1 + i )32 i = 32.13% N i P F ) 1 (   Rate of Return
  • 5. Suppose that you invested that amount ($1,650) in a savings account at 6% per year. Then, you could have only $10,648 on January, 2002. What is the meaning of this 6% interest here? This is your opportunity cost if putting money in savings account was the best you can do at that time!
  • 6. So, in 1970, as long as you earn more than 6% interest in another investment, you will take that investment. Therefore, that 6% is viewed as a minimum attractive rate of return (or required rate of return). So, you can apply the following decision rule, to see if the proposed investment is a good one. ROR (32.13%) > MARR(6%)
  • 7. Return on Investment Definition 1: Rate of return (ROR) is defined as the interest rate earned on the unpaid balance of an installment loan. Example: A bank lends $10,000 and receives annual payment of $4,021 over 3 years. The bank is said to earn a return of 10% on its loan of $10,000.
  • 8. Loan Balance Calculation: A = $10,000 (A/P, 10%, 3) = $4,021 Unpaid Return on Unpaid balance unpaid balance at beg. balance Payment at the end Year of year (10%) received of year 0 1 2 3 -$10,000 -$10,000 -$6,979 -$3,656 -$1,000 -$698 -$366 +$4,021 +$4,021 +$4,021 -$10,000 -$6,979 -$3,656 0 A return of 10% on the amount still outstanding at the beginning of each year
  • 9. Rate of Return: Definition 2: Rate of return (ROR) is the break-even interest rate, i*, which equates the present worth of a project’s cash outflows to the present worth of its cash inflows. Mathematical Relation: PW i PW i PW i ( ) ( ) ( ) * * *    cash inflows cash outflows 0
  • 10. Return on Invested Capital Definition 3: Return on invested capital is defined as the interest rate earned on the unrecovered project balance of an investment project. It is commonly known as internal rate of return (IRR). Example: A company invests $10,000 in a computer and results in equivalent annual labor savings of $4,021 over 3 years. The company is said to earn a return of 10% on its investment of $10,000.
  • 11. Project Balance Calculation: 0 1 2 3 Beginning project balance Return on invested capital Payment received Ending project balance -$10,000 -$6,979 -$3,656 -$1,000 -$697 -$365 -$10,000 +$4,021 +$4,021 +$4,021 -$10,000 -$6,979 -$3,656 0 The firm earns a 10% rate of return on funds that remain internally invested in the project. Since the return is internal to the project, we call it internal rate of return.
  • 12. Methods for Finding Rate of Return  Types of Investment (cash flow) Classification  Simple Investment  Non-simple Investment  Once we identified the type of investment cash flow, there are several ways available to determine its rate of return.  Computational Methods  Direct Solution Method  Trial-and-Error Method  Computer Solution Method
  • 13. 13 Investment Classification Simple Investment  Definition: Initial cash flows are negative, and only one sign change occurs in the net cash flows series.  Example: -$100, 250, $300 (-, +, +)  ROR: A unique ROR  If the initial flows are positive and one sign change occurs referred to simple-borrowing. Non-simple Investment  Definition: Initial cash flows are negative, but more than one sign changes in the remaining cash flow series.  Example: -$100, 300, -$120 (-, +, -)  ROR: A possibility of multiple RORs
  • 14. Period (N) Project A Project B Project C 0 -$1,000 -$1,000 +$1,000 1 -500 3,900 -450 2 800 -5,030 -450 3 1,500 2,145 -450 4 2,000 Project A is a simple investment. Project B is a non-simple investment. Project C is a simple borrowing.
  • 15. Computational Methods Direct Solution Direct Solution Trial & Error Method Computer Solution Method Log Quadratic n Project A Project B Project C Project D 0 -$1,000 -$2,000 -$75,000 -$10,000 1 0 1,300 24,400 20,000 2 0 1,500 27,340 20,000 3 0 55,760 25,000 4 1,500
  • 16. Example 7.2 Direct Solution Methods Project A Project B PW i i i x i PW i x x x x i i i i i i i ( ) $2, $1, ( ) $1, ( ) , ( ) , , , : . . . , *                              000 300 1 500 1 0 1 1 2 000 1300 1500 08 08 1 1 25%, 1667 1 1 160% 100% 25%. 2 2 Let then Solve for or -1.667 Solving for yields Since the project's $1, $1, ( / , , ) $1, $1, ( ) . ( ) ln . ln( ) . ln( ) . . 000 500 4 000 500 1 0 6667 1 0 6667 4 1 0101365 1 1 1 4 4 0 101365 0 101365                  P F i i i i i e i i e 10.67%
  • 17. Trial and Error Method – Project C  Step 1: Guess an interest rate, say, i = 15%  Step 2: Compute PW(i) at the guessed i value. PW (15%) = $3,553  Step 3: If PW(i) > 0, then increase i. If PW(i)< 0, then decrease i. PW(18%) = -$749  Step 4: If you bracket the solution, you use a linear interpolation to approximate the solution 3,553 0 -749 15% i 18%          749 553 , 3 553 , 3 % 3 % 15 i % 45 . 17 
  • 18. Basic Decision Rule: If ROR > MARR, Accept This rule does not work for a situation where an investment has multiple rates of return
  • 19. Comparing Mutually Exclusive Alternatives Based on IRR Issue: Can we rank the mutually exclusive projects by the magnitude of its IRR? n A1 A2 0 1 IRR -$1,000 -$5,000 $2,000 $7,000 100% > 40% $818 < $1,364 PW (10%)
  • 20. Who Got More Pay Raise? Bill Hillary 10% 5%
  • 21. Can’t Compare without Knowing Their Base Salaries Bill Hillary Base Salary $50,000 $200,000 Pay Raise (%) 10% 5% Pay Raise ($) $5,000 $10,000 For the same reason, we can’t compare mutually exclusive projects based on the magnitude of its IRR. We need to know the size of investment and its timing of when to occur.
  • 22. Incremental Investment  Assuming a MARR of 10%, you can always earn that rate from other investment source, i.e., $4,400 at the end of one year for $4,000 investment.  By investing the additional $4,000 in A2, you would make additional $5,000, which is equivalent to earning at the rate of 25%. Therefore, the incremental investment in A2 is justified. n Project A1 Project A2 Incremental Investment (A2 – A1) 0 1 -$1,000 $2,000 -$5,000 $7,000 -$4,000 $5,000 ROR PW(10%) 100% $818 40% $1,364 25% $546
  • 23. Incremental Analysis (Procedure) Step 1: Compute the cash flow for the difference between the projects (A,B) by subtracting the cash flow of the lower investment cost project (A) from that of the higher investment cost project (B). Step 2: Compute the IRR on this incremental investment (IRR ). Step 3: Accept the investment B if and only if IRR B-A > MARR B-A NOTE: Make sure that both IRRA and IRRB are greater than MARR.
  • 24. Example 7.7 - Incremental Rate of Return n B1 B2 B2 - B1 0 1 2 3 -$3,000 1,350 1,800 1,500 -$12,000 4,200 6,225 6,330 -$9,000 2,850 4,425 4,830 IRR 25% 17.43% 15% Given MARR = 10%, which project is a better choice? Since IRRB2-B1=15% > 10%, and also IRRB2 > 10%, select B2.
  • 25. IRR on Increment Investment: Three Alternatives n D1 D2 D3 0 -$2,000 -$1,000 -$3,000 1 1,500 800 1,500 2 1,000 500 2,000 3 800 500 1,000 IRR 34.37% 40.76% 24.81% Step 1: Examine the IRR for each project to eliminate any project that fails to meet the MARR. Step 2: Compare D1 and D2 in pairs. IRRD1-D2=27.61% > 15%, so select D1. D1 becomes the current best. Step 3: Compare D1 and D3. IRRD3-D1= 8.8% < 15%, so select D1 again. Here, we conclude that D1 is the best Alternative.
  • 26. Items CMS Option FMS Option Investment $ 4,500,000 $ 12,500,000 Total annual operating costs $ 7,412,920 $ 5,504,100 Net salvage value $ 500,000 $ 1,000,000 Example 7.9 Incremental Analysis for Cost-Only Projects  The firm’s MARR is 15%. Which alternative would be a better choice, based on the IRR criterion?  Discussion: Since we can assume that both manufacturing systems would provide the same level of revenues over the analysis period, we can compare these alternatives based on cost only. (these systems are service projects).  Although we can not compute the IRR for each option without knowing the revenue figures, we can still calculate the IRR on incremental cash flows.  Since the FMS option requires a higher initial investment than that of the CMS, the incremental cash flow is the difference (FMS – CMS)
  • 27. Example 7.9 Incremental Analysis for Cost-Only Projects (cost are itemized) Items CMS Option FMS Option Annual O&M costs: Annual labor cost $1,169,600 $707,200 Annual material cost 832,320 598,400 Annual overhead cost 3,150,000 1,950,000 Annual tooling cost 470,000 300,000 Annual inventory cost 141,000 31,500 Annual income taxes 1,650,000 1,917,000 Total annual operating costs $7,412,920 $5,504,100 Investment $4,500,000 $12,500,000 Net salvage value $500,000 $1,000,000
  • 28. Example 7.9 Incremental Cash Flow (FMS – CMS) n CMS Option FMS Option Incremental (FMS-CMS) 0 -$4,500,000 -$12,500,000 -$8,000,000 1 -7,412,920 -5,504,100 1,908,820 2 -7,412,920 -5,504,100 1,908,820 3 -7,412,920 -5,504,100 1,908,820 4 -7,412,920 -5,504,100 1,908,820 5 -7,412,920 -5,504,100 1,908,820 6 -7,412,920 -5,504,100 $2,408,820 Salvage + $500,000 + $1,000,000
  • 29. Solution: PW i P A i P F i IRR FMS CMS FMS CMS ( ) $8, , $1,908, ( / , , ) $2, , ( / , , ) .43%          000 000 820 5 408 820 6 0 12 15%, select CMS.
  • 30.  Although the FMS would provide an incremental annual savings of $1,908,820 in operating costs, the savings do not justify the incremental investment of $8,000,000. COMMENTS:  Note that the CMS option was marginally preferred to the FMS option.  However, there are dangers in relying solely on the easily quantified savings in input factors – such as labor, energy, and materials – from FMS and in not considering gains from improved manufacturing performance that are more difficult and subjective to quantify.  Factors such as improved product quality, increased manufacturing flexibility (rapid response to customer demand), reduced inventory levels, and increased capacity for product innovation are frequently ignored in financial analysis because we have inadequate means for quantifying benefits.  If these intangible benefits were considered, as they ought to be, however, the FMS option could come out better than the CMS option.