1. RATE OF RETURN ANALYSIS CHAPTER 7

2. Rate of Return (ROR) Analysis
Methods for Finding ROR
Internal Rate of Return (IRR) Criterion
Incremental Analysis
Mutually Exclusive Alternatives

3. RATE OF RETURN
A relative percentage method which measures the annual rate of return as a percentage of investment over the life of a project.

4. Meaning of Rate of Return
In 1970, when Wal-Mart Stores, Inc. went public, an investment of 100 shares cost $1,500. That investment would have been worth $9,990,920 on year 2011.
What is the rate of return on that investment?

5. Given: P = $1,500 Find i = ? Rate of Return Solution: $9,990,920 4141
$1,500
Given: P = $1, Find i = ?
F = $9,990,920
N = 41
$9,990,920 = $1,500 (1 + i )41
i = 23.95%
Rate of Return

6. What is the meaning of this 6% interest here?
Suppose that you invested the amount $1,500 in a savings account at 6% per year. Then, you could have only $16, on year 2011.
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!

7. 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 [MARR] (or required rate of return).
So, you can apply the following decision rule, to see if the proposed investment is a good one.
ROR (23.95%) > MARR(6%)

8. 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.

9. LOAN BALANCE CALCULATION
2.4869
$10,000 = $4,021(P/A, i , 3)
Unpaid Return on Unpaid
balance unpaid balance
at beg. balance Payment at the end
Year of year (10%) received of year 1 2 3
-$10,000
-$6,979
-$3,656
-$10,000
-$6,979
-$3,656
-$1,000
-$698
-$366
+$4,021

10. 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:

11. RECONSIDER THE LOAN TRANSACTION
$4,021
$4,021
$4,021 1 2 3
Rate of Return = 10%
$10,000

12. RETURN ON INVESTED CAPITAL
Definition 3:
The internal rate of return (IRR) is the interest rate charged on the unrecovered project balance of the investment such that, when the project terminates, the unrecovered project balance is zero.
Example:
A company invests $10,000 in a computer with a three year useful life 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.

13. PROJECT BALANCE
We see that 10% is earned on $10,000 during year one, 10% is earned on $6,979 during year two, and 10% is earned during year three. Or, the firm earns a 10% rate of return on funds that remain internally invested in the project.

14. 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 to determine its rate of return.
Computational Methods
Direct Solution Method
Trial-and-Error Method

15. 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, $ (-, +, +)
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, $300, -$ (-, +, -)
ROR: A possibility of multiple RORs

16. Figure 7-1 Classification of investments

17. Example 7.2 Finding i * by Direct Solution:
Two Flows and two periods compute the rate of return N
Project 1
Project 2
-$3,000
-$2,000 1
$1,300 2
$1,500 3 4
$4,500

18. Example 7.2 Finding i * by Direct Solution:
Project 1:
FW( i ) = -$3,000(F/P, i , 4) + $4,500 = 0
Setting FW( i ) = 0, we obtain
$4,500 = $3,000 (F/P, i , 4) = $3,000 (1+ i )4
1.5 = (1+ i )4 solving for i* =
= , or 10.67% Or
PW( i ) = -$3,000+ $4,500 (P/F, i , 4) = 0
i* = = , or 10.67% N
Project 1
-$3,000 1 2 3 4
$4,500
$3,000 = $4,500 (1+ i )-4
= = (1+ i )-4

19. Example 7.2 Finding i * by Direct Solution:
Project 2: N
Project 2
-$2,000 1
$1,300 2
$1,500 3 4

20. Example 7.3 Finding i* by Trial and Error Method
ACME Corporation distributes agricultural equipment. The board of directors is considering a proposal to establish a facility to manufacture an electronically controlled "intelligent" crop sprayer invented by a professor at a local university. This crop-sprayer project would require an investment of $10 million in assets and would produce an annual after-tax net benefit of $1.8 million over a service life of eight years. All costs and benefits are included in these figures. When the project terminates, the net proceeds from the sale of the assets would be $1 million (Figure 7.2). Compute the rate of return of this project.
Figure 7-2 Cash flow diagram for a simple investment

21. Solution 7.3 Finding i * by Trial and Error Method
Given: I = $10M, A = $1.8M, S = $1M, N = 8 years
Step 1: Guess an interest rate, say, i = 8%
Step 2: Compute PW ( i ) at the guessed i value.
PW (8%) = -$10 +$1.8(P/A, 8%, 8) +$1(P/F, 8%, 8) = $0.88
Step 3: If PW ( i ) > 0, then increase i If PW ( i ) < 0, then decrease i.
Since NPW is positive, raise i, in order to bring this value toward zero.
Use i = 12%. Therefore,
PW (12%) = -$10 +$1.8(P/A, 12%, 8) +$1(P/F, 12%, 8) = -$0.65
Using straight-line interpolation, we approximate that
i*= 8% +(12% - 8%) [0.88 / ] = 8 + 4(0.5752) = 10.30%

22. Solution 7.3 Finding i* by Trial and Error Method
PW (10.30%) = -$10 +$1.8(P/A, 10.3%, 8) +$1(P/F, 10.3%, 8) = -$0.045
Since the result is not zero, we may re-compute i* at a lower interest rate
PW (10%) = -$10 +$1.8(P/A, 10%, 8) +$1(P/F, 10%, 8) = $0.069
With another straight-line interpolation, we approximate that
i*= 10% +(10.30% - 10%)[0.069/ ] = (0.6053) = 10.18%
PW (10.18%) = -$10 +$1.8(P/A, 10.18%, 8) +$1(P/F, 10.18%, 8) = $0.0007
Which is practically zero, so we may stop here.
Therefore, Rough estimate through computer gives us i = %.

23. Finding i * by Trial and Error
$55,760
$27,340 n
Cash Flow
-$75,000 1
24,400 2
27,340 3
55,760
$24,400 1 2 3
$75,000

24. i Solution 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: Use a linear interpolation to approximate the solution.
3,553
-749
15% i
18%

25. Internal-Rate-of-Return Criterion

26. Incremental Analysis for Comparing Mutually Exclusive Alternatives Based on IRR
Issue: Can we rank the mutually exclusive projects by the magnitude of its IRR? n
Project A1
Project A2 1
-$1,000
$2,000
-$5,000
$7,000
IRR
1000/1000
100% >
2000/5000
40%

27. Incremental Analysis for Comparing Mutually Exclusive Alternatives Based on IRR
Issue: Can we rank the mutually exclusive projects by the magnitude of its IRR? n
Project A1
Project A2 1
-$1,000
$2,000
-$5,000
$7,000
IRR
PW(10%)
1000/1000
100% >
$ <
2000/5000
40%
$1,364

28. Incremental Investment
Project A1
Project A2 1
-$1,000
$2,000
-$5,000
$7,000
ROR
PW(10%)
100%
$ 818
40%
$1,364
Project A1, need $1,000 from investment pool. Remaining $4,000 will continue to earn
10% interest. One year later, $2,000 from the project investment and $4,400 from investment pool. At the end of one year for $6,400 (28% return on $5,000).
The equivalent PW of this will be; PW(10%)= -$5,000 + $6,400(P/F, 10%, 1)=$818
Project A2, need $5,000 from investment pool, will generate $7,000 after one year.
The equivalent PW of this will be PW(10%)= -$5,000 + $7,000(P/F, 10%, 1)=$1,364

29. Incremental Investment
Project A1
Project A2
Incremental Investment (A2 – A1) 1
-$1,000
$2,000
-$5,000
$7,000
-$4,000
$5,000
ROR
PW(10%)
100%
$ 818
40%
$1,364
25%
$ 546
Project A1, The equivalent PW of this will be;
PW(10%)= -$5,000 + $6,400(P/F, 10%, 1)=$818
Project A2, The equivalent PW of this will be
PW(10%)= -$5,000 + $7,000(P/F, 10%, 1)=$1,364
PW(10%)= -$4,000 + $5,000(P/F, 10%, 1)=$546

30. Incremental Analysis (Procedure)
For a pair of mutually exclusive projects (A and B, with B defined as a more costly option), we may rewrite B as: B =A+ (B - A).
In other words, B has two cash flow components:
(1) the same cash flow as A and
(2) the incremental component ( B - A).
Therefore, the only situation in which B is preferred to A is when the rate of return on the incremental component ( B - A) exceeds the MARR

31. 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.

32. n B1 B2 1 2 3
-$3,000
1,350
1,800
1,500
-$12,000
4,200
6,225
6,330
IRR
25%
17.43%
Example 7.7
Abdulqadir, a KFUPM student, wants to start a small-scale painting business during his off-school hours. To economize the start-up business, he decides to purchase some used painting equipment. He has two mutually exclusive options. Do most of the painting by himself by limiting his business to only residential painting jobs (B1) or purchase more painting equipment and hire some helpers to do both residential and commercial painting jobs (B2). He expects option B2 will have a higher equipment cost, but provide higher revenues as well (B2). In either case, he expects to fold the business in three years when he graduates from college. The cash flows for the two mutually exclusive alternatives are given above as follows:

33. n B1 B2 1 2 3
-$3,000
1,350
1,800
1,500
-$12,000
4,200
6,225
6,330
IRR
25%
17.43%
Solution 7.7 a)
With the knowledge that both alternatives are revenue projects, which project would Abdulqadir select at MARR = 10%? (Note that both projects are profitable at 10%) a)
PW(10%)B1 = -$ $1350 (P/F, 10%, 1) +$1800 (P/F, 10%, 2) + $1500 (P/F, 10%, 3)
= $842
PW(10%)B2 = -$12,000 + $4200 (P/F, 10%, 1) +$6225 (P/F, 10%, 2) + $6330 (P/F, 10%, 3)
= $1718.5

34. b)
Find The IRR on the increment and which alternative is preferable. To choose the best project, we compute the incremental cash flow for B2 - B1. n B1 B2
B2 - B1 1 2 3
-$3,000
1,350
1,800
1,500
-$12,000
4,200
6,225
6,330
-$9,000
2,850
4,425
4,830
IRR
25%
17.43%
15 %
b) We compute the IRR on this increment of investment by solving PW( i %)B2-B1 = -$ $2850 (P/F, i, 1) + $4425 (P/F, i, 2) + $4830 (P/F, i, 3) = 0 PW(10 %)B2-B1 = -$ $2850 (P/F, 10%, 1) + $4425 (P/F, 10%, 2) + $4830 (P/F, 10%, 3) = $ PW(20 %)B2-B1 = -$ $2850 (P/F, 20%, 1) + $4425 (P/F, 20%, 2) + $4830 (P/F, 20%, 3) = -$757 IRRB2-B1 = i* B2-B1 = (876 / ) = 15% Since IRRB2-B1 = 15% > 10%, and also IRRB2 > 10%, select B2.

35. .
4,755 .
1,650 . . .
17.43

38. When you have more than two mutually exclusive alternatives, D1 D2 D3
-$2,000
-$1,000
-$3,000 1
1,500
800 2
1,000
500
2,000 3
IRR
34.37%
40.76%
24.81%
When you have more than two
mutually exclusive alternatives,
they can be compared in pairs
by successive examination
Step 1:
Examine the IRR for each project to eliminate any project that fails to meet MARR (15%).
Step 2:
Compare lowest investment cost D2 and the second lowest investment cost D1 in
pairs. Compute the incremental analysis IRRD1-D2. If IRRD1-D2 > MARR select D1
otherwise choose D2.
Step 3:
Compare the result from step 2 with the highest investment cost D3. Compute the
incremental analysis IRRD3-D1 and chose D3 if IRRD3-D1 > MARR otherwise choose D1.

39. IRR on Increment Investment: Three AlternativesD1 D2
D1 - D2
-$2,000
-$1,000 1
1,500
800
700 2
1,000
500 3
300
IRR
34.37%
40.76% ?
Step 2: Compare D1 and D2 in pairs.
PW( i ) = -$1,000+$700(P/F, i, 1)+ $500(P/F, i, 2)+ $300(P/F, i, 3) = 0
PW( 25% ) = -$1,000+$700(0.8)+ $500(0.64)+ $300(0.512) = $33.6
PW( 30% ) = -$1,000+$700(0.7692)+ $500(0.5917)+ $300(0.4552) = -$29.15
i* D1-D2= 25% +(30% - 25%) [33.6 / ] = (0.535) = = %
IRRD1-D2=27.67% > 15%, so select D1. D1 becomes the current best.

40. IRR on Increment Investment: Three AlternativesD1 D3
D3 – D1
-$2,000
-$3,000
-$1,000 1
1,500 2
1,000
2,000 3
800
200
IRR
34.37%
24.81% ?
Step 3: Compare D1 and D3.
PW( i ) = -$1,000+$0(P/F, i, 1)+ $1,000(P/F, i, 2)+ $200(P/F, i, 3) = 0
PW( 6% ) = -$1,000+$0(0.9434)+ $500(0.89)+ $300(0.8396) = $57.92
PW( 10% ) = -$1,000+$0(0.9091)+ $500(0.8264)+ $300(0.7513) = $-23.34
i* D3-D1 = 6% +(10% - 6%) [57.92 / ] = 6 + 4(0.7127) = = 8.85 %
IRRD3-D1= 8.8% < 15%, so select D1 again. We conclude that D1 is the best Alternative.

41. PRACTICE PROBLEMS

42. 7.12) Consider an investment project with the following net cash flow:
What would be the value of X if the project’s IRR is 10%?
Year
Net Cash Flow
-$1,500 1 X 2
$650 3

43. SOLUTION

44. 7.13) Consider an investment project with the following net cash flow:     What value of X if the project’s IRR is 25%?
Year
Net Cash Flow
-$12,000 1
$2,500 2
$5,500 3 X 4

45. SOLUTION

46. 7.21
You are considering a luxury apartment building project that requires an investment of $12,500,000. The building has 50 units. The firm expects that the maintenance cost for the apartment building will be $250,000 in the first year, and to rise to $300,000 in the second year, and to continue to increase by $50,000 in subsequent years. The cost to hire a manager for the building is estimated to be $80,000 per year. After five years of operation, the apartment building can be sold for $14,000,000. What annual rent per apartment unit will provide a return on investment of 15%? Assume that the building will remain fully occupied during the five years.

47. SOLUTION 7.21
PW(15%) = - $12,500,000 - $250,000 (P/A, 15%, 5) - $50,000 (P/G, 15%, 5)
+ $14,000,000 (P/F, 15%, 5) - $80,000 (P/A, 15%, 5) + A(P/A, 15%, 5) = 0
PW(15%) = - $12,500,000 - $250,000 (3.3522) - $50,000 (5.7751)
+ $14,000,000 (0.4972) - $80,000 (3.3522) + A (3.3522) = 0
PW(15%) = - $12,500,000 - $838,050 - $288,755 + $6,960,800 - $268,176 + A(3.3522) = 0
PW(15%) = -$13,894,981 + $6,960,800 + A(3.3522) = 0
PW(15%) = -$6,934,181 + A(3.3522) = 0
A = $2,068,546 income per year for the building
A = $2,068, / 50 = $41,371 per unit

48. 7.23
You are considering a CNC machine. This machine will have an estimated service life of 10 years, with a salvage value of 10% of the investment cost. Its expected savings from annual operating and maintenance costs are estimated to be $60,000. To expect a 15% rate of return on investment, what would be the maximum amount that you are willing to pay for the machine?

49. SOLUTION

50. 7.36) Consider the following two mutually exclusive alternatives:
Determine the IRR on the incremental investment
in the amount of $4,000. (Assume that MARR is 10%)
b) If the firm’s MARR is 10%, which alternative is the
better choice? A1 A2
-$16,000
-$20,000 1
$7,500
$5,000 2
$15,000 3
$8,000

51. a) b) Select project A2 A1 A2 A2 – A1 -$16,000 -$20,000 -$4,000 1
SOLUTION a)
b) Select project A2 A1 A2
A2 – A1
-$16,000
-$20,000
-$4,000 1
$7,500
$5,000
-$2,500 2
$15,000 3
$8,000
$500