The document discusses various methods for evaluating the economic profitability of capital investment projects, including present worth, future worth, annual worth, internal rate of return, and external rate of return. It provides examples of how to use these methods to calculate metrics like net present value, internal rate of return, payback period, and compare them to required rates of return to determine if projects are economically justified. The goal is to evaluate if a project's revenues over time can recover its costs and provide an adequate return given the risks.

1. Evaluating a Single project
1

2. Introduction
In this chapter we will answer the following
question :
Whether a proposed capital and its expenditures
can recovered by revenue over time in addition
to a return on the capital sufficiently attractive in
view of the risks .
2

3. Methods of Evaluating the
Economic Profitability of a
Problem Solution
 Present Worth ( PW )
 Future Worth ( FW )
 Annual Worth ( AW )
 Internal Rate of Return ( IRR )
 External Rate of Return ( ERR )
3

4. Assumptions
1. We know the future with certainty
2. We can borrow and lend with the same i%.
4

5. The most-used method is the
present worth method.
The present worth (PW) is found by
discounting all cash inflows and
outflows to the present time at an
interest rate that is generally the
MARR.
A positive PW for an investment
project means that the project is
acceptable (it satisfies the MARR).
5

6. Present Worth ( PW )
Eq..(5-1)
 =F0(1+i)0+F1(1+i)-1+F2(1+i)-2+………..+FN(1+i)-N
 i = effective interest rate or MARR
 K = index of each compounding period
 Fk = future cash flow at the end of period k
 N = # of compounded periods . 6

7.  PW Decision Rule:
If PW (i=MARR) ≥ 0 , the project is economically
justified
7

8. Note
 The higher the interest rate and the further into the
future a cash flow occurs , the lower its PW is
 See figure 5-2
 PW of 1,000 10 years from now i =5% is $613.90
 PW of 1,000 10years from now i=10% is $385.5
8

9. Figure 5-2
9

10. Example(PW)
Consider a project that has an initial
investment of $50,000 and that returns
$18,000 per year for the next four years.
If the MARR is 12%, is this a good
investment?
PW = -50,000 + 18,000 (P/A, 12%, 4)
PW = -50,000 + 18,000 (3.0373)
PW = $4,671.40  This is a good investment!
10

11.  Example 5.1:
 A piece of new equipment has been proposed by engineers to
increase the productivity of a certain manual welding
operation. The investment cost $25,000 and the equipment
will have a market value of $5,000 at the end of a study
period of five years. Increased productivity attributable to the
equipment will amount to $8,000 per year after extra
operating costs have been subtracted from the revenue
generated by the additional production. A cash flow diagram
for this investment opportunity is given below. If the firm’s
MARR is 20% per year, is this proposal a sound one? Use the
PW method.
11

12.  Example. 5-1
 PW = PW(inflows) – PW (outflows)
 PW= $8,000(P/A,20%,5) +$5,000(P/F,20%,5)-$25,000
 PW= $934,29
12

13. Bond Value
 The value of the bond at any time is the PW of
future cash receipts
 Two types of payments
1. Series of periodic interest payments (rZ)
2. A single payment = C ,When the bond is sold or
retired .
13

14. VN = C(P/F, i%, N) + rZ (P/A, i%,N) Eq.5-2
 Z= Face ,or parValue,
 C = redemption or disposal price (usually = Z).
 r = bond rate (nominal interest) per interest period
 N= # of periods before redemption
 i = bond yield rate per period
 VN = PW
14

15. 15

16. Example5-4
16

17. The Capitalized-Worth Method
 A special variation of the PW
 To find PW for infinite length of time
 N ∞
CW = PWN  ∞ = A ( P / A, i%, ∞ )
( 1+i )N - 1
= A lim ------------------ =A ( 1 / i )=A/i
N
∞ i ( 1 + i )N
17

18. Ex. 5-5 page 219
18

19. (a)As we discussed before when N= ∞
(p/A,i%,N)=1/i, then
(p/A,8%,N)=1/0.08 = 12.5
go to App.C then N= 100 this assumed
as (∞ )
19

20. Ex. 5-5 page 219
(b)
30,000 20,000 A / F ,8%,4)
(
CW  (100000 A / P,8%, )) / .08
, (
0.08
 30,000  20,000 ( A / F ,8%,4) 
CW  100,000     $530,475
 0.08 
20

21. Future Worth ( FW )
 Equivalent worth of all cash flows (In& Out) at the end of the
planning horizon at MARR
Eq.(5-3)
 i = effective interest rate or MARR
 K = index of each compounding period
 Fk = future cash flow at the end of period k
 N = # of compounded periods .
21

22.  FW Decision Rule:
If FW (i=MARR) ≥ 0 , the project is economically
justified
22

23.  Ex 5-6 page 221
23

24. Example5-7
24

25. The Annual Worth Method
 Annual worth is an equal periodic series of dollar amounts
that is equivalent to the cash inflows and outflows, at an
interest rate that is generally the MARR.
 The AW of a project is annual equivalent revenue or savings
minus annual equivalent expenses, less its annual capital
recovery (CR) amount.
 AW(i%) = R – E – CR(i%)
 R = Annual equivalent Revenue or saving
 E = Annual equivalent Expenses
 CR = Annual equivalent Capital Recovery amount
25

26.  AW Decision Rule:
If AW (i=MARR) ≥ 0 , the project is economically
justified
26

27. CR (Capital Recovery)
 CR is the equivalent uniform annual cost of the capital
invested that covers
1. Loss of value of the asset.
2. Interest on invested capital (at MARR)
27

28. CR
 CR (i%) = I(A/P, i%, N) – S(A/F, i%,N) Eq. 5-5
 I = initial investment for project
 S = salvage (market) value at the end of the study
period
 N = project study period
28

29.  CR (i%) = (I – S)(A/F, i%, N) +I(i%)
 CR (i%) = (I –S)(A/P, i%, N) + S(i%)
29

30. Example 5-8 page 224
30

31. Example 5-9
31

32. Example 5-10
32

33. Example 5-11
33

34. A project requires an initial investment of
$45,000, has a salvage value of $12,000 after
six years, incurs annual expenses of $6,000,
and provides an annual revenue of $18,000.
Using a MARR of 10%, determine the AW of
this project.
Since the AW is positive, it’s a good
investment.
34

35. Internal Rate Of Return Method ( IRR )
 IRR solves for the interest rate that equates the equivalent
worth of an alternative’s cash inflows (receipts or savings) to
the equivalent worth of cash outflows (expenditures)
 IRR is positive for a single alternative only if:
 both receipts and expenses are present in the cash flow
diagram
 the sum of inflows exceeds the sum of outflows
35

36. Internal Rate of Return
 It is also called the investor’s method, the discounted
cash flow method, and the profitability index.
 If the IRR for a project is greater than the MARR, then
the project is acceptable.
36

37. INTERNAL RATE OF RETURN METHOD ( IRR )
 IRR is i’ %, using the following PW formula:
R k = net revenues or savings for the kth year
E k = net expenditures including investment costs for the
kth year
N = project life ( or study period )
Note: FW or AW can be used instead of PW
37

38. Solving for the IRR is a bit more
complicated than PW, FW, or AW
 The method of solving for the i'% that equates
revenues and expenses normally involves trial-and-
error calculations, or solving numerically using
mathematical software.
38

39. 39

40. 40

41. Installment Financing
 An application of IRR method
 Ex.5-14 will be discussed
41

42. 42

43. 43

44. 44

45. 45

46. 46

47. 47

48. Challenges in Applying the IRR
Method.
 It is computationally difficult without proper tools.
 In rare instances multiple rates of return can be
found(take a look to page 255 example 5-A-1).
 The IRR method must be carefully applied and
interpreted when comparing two more mutually
exclusive alternatives (e.g., do not directly compare
internal rates of return).
48

49. The External Rate Of Return Method ( ERR )
 ERR directly takes into account the interest rate
(  ) external to a project at which net cash flows
generated over the project life can be reinvested
(or borrowed ).
 If the external reinvestment rate, usually the
firm’s MARR, equals the IRR, then ERR method
produces same results as IRR method
49

50. Calculating External Rate of Return ( ERR )
1. All net cash outflows are discounted to the present (time 0) at ε%
per compounding period.
2. All net cash inflows are discounted to period N at ε %.
3. Solve for the ERR, the interest rate that establishes equivalence
between the two quantities.
The absolute value of the present equivalent worth of the net cash
outflows at ε % is used in step 3.
 A project is acceptable when i ‘ % of the ERR method is greater
than or equal to the firm’s MARR
50

51. ERR is the i'% at which
where
Rk = excess of receipts over expenses in period k,
Ek = excess of expenses over receipts in period k,
N = project life or number of periods, and
ε = external reinvestment rate per period.
51

52. Calculating External Rate of Return ( ERR )
2
N
 R ( F / P,  %, N - k )
k = 0k
0
1 Time N
N
E ( P / F,  %, k )
k=0 k 52

53. Calculating External Rate of Return ( ERR )
2
3 N
 R ( F / P,  %, N - k )
k = 0k
0 i ‘ %= ?
1 Time N
N
 Ek ( P / F,  %, k ) ( F / P, i ‘ %, N )
k=0 53

54. Example 5-17
 $25,000 (F/P,i’%,5)=$8,000(F/A,20%,5)+$5,000
 (F/P,i’%,5)= $64,532.80  2.5813  (1  i ' )5
$25,000
 i’ =20.88%
54

55. Example 5-18
55

56. Example
For the cash flows given below, find the ERR when the
external reinvestment rate is ε = 12% (equal to the MARR).
Year 0 1 2 3 4
Cash Flow -$15,000 -$7,000 $10,000 $10,000 $10,000
Expenses
Revenue
Solving, we find
56

57. The payback period method
 The simple payback period is the number of years
required for cash inflows to just equal cash outflows.
 It is a measure of liquidity rather than a measure of
profitability.
57

58. Payback is simple to calculate.
The payback period is the smallest value of θ (θ ≤ N) for
which the relationship below is satisfied.
For discounted payback future cash flows are discounted
back to the present, so the relationship to satisfy becomes
58

59.  A low- valued pay back period is considered desirable
59

60. Finding the simple End
End of
Net Cumulat Present Cumulati
of Cash ive PW worth at ve PW at
and discounted Year
Year
Flow at 0% i = 6% 6%
payback period for a
set of cash flows. 0 0 -$42,000 -$42,000 -$42,000 -$42,000
The cumulative cash
1 1 $12,000 -$30,000 11,320.8 -$30,679
flows in the table were
calculated using the
formulas for simple 2 2 $11,000 -$19,000 9,790 -$20,889
and discounted
payback. 3 3 $10,000 -$9,000 8,396 -$12,493
From the calculations 4 4 $10,000 $1,000 7,921 -$4,572
θ = 4 years and θ' = 5
years. 5 5 $9,000 6,725.7 $2,153
60

61. Col.3:
EOF1= -42,000-12,000=-30,000 End of Net Cash Cumulative Cumulative
End Net Cumulat Present Cumulati
Year Cash
of Flow ive PW at 0% at PW at 6%
PW worth ve PW at
EOF2 : 11,000-30,000=-19,000 Year Flow at 0% i = 6% 6%
EOF3 : 10,000-19,000=-9,000
00 -$42,000 -$42,000 -$42,000 -$42,000
-$42,000 -$42,000 -$42,000
EOF4 : 10,000-9,000=1,000>0
θ=4
Col.4: 11 $12,000 -$30,000 11,320.8 -$30,679
$12,000 -$30,000 -$30,679
12,000 (p/F,6%,1) =
12,000* 0.9434= 113,208 22 $11,000 -$19,000
$11,000 -$19,000 9,790 -$20,889
-$20,889
11,000 (p/F,6%,2) = 11,000*0.8900=
9,790
33 $10,000
$10,000 -$9,000-$9,000
8,396 -$12,493
-$12,493
10,000*(P/F,6%,3) =
10,000*0.8396= 8,396
44 $10,000$1,000 $1,000
$10,000 7,921 -$4,572
-$4,572
10,000*(P/F,6%,4) =
10,000*0.7921 =7,921
9,000*(P/F,6%,5) = 55 $9,000
$9,000 6,725.7 $2,153
$2,153
9,000*0.7473 = 6,725.7
61

62. End Net Cumulat Present Cumulati
Col.5: of Cash ive PW worth at ve PW at
Year Flow at 0% i = 6% 6%
EOF1:
-42,000 + 11,320.8 = -30,679.2 0 -$42,000 -$42,000 -$42,000 -$42,000
EOF2 :
1 $12,000 -$30,000 11,320.8 -$30,679
9,790 -30,679.2 = -20,889.2
EOF3 :
2 $11,000 -$19,000 9,790 -$20,889
8,396 -20,889.2 = -12,493.2
EOF4 : 3 $10,000 -$9,000 8,396 -$12,493
7,921-12,493.2 = -4,572.2 <0
EOF5 : 4 $10,000 $1,000 7,921 -$4,572
6,725.7 -4,572.2 = 2,153.5 >0
5 $9,000 6,725.7 $2,153
62

63. Problems with the payback period method.
 It doesn’t reflect any cash flows occurring after θ, or θ'.
 It doesn’t indicate anything about project desirability except
the speed with which the initial investment is recovered.
 Recommendation: use the payback period only as
supplemental information in conjunction with one or more
of the other methods in this chapter.
63

64. Suggested Problems
 4, 10, 16, 21, 25, 36, 46, 49, 55, 60
64