1

Evaluating a Single project

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 .

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 )

4

Assumptions 1. We know the future with certainty

2. We can borrow and lend with the same i%.

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).

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 .

7

PW Decision Rule:

If PW (i=MARR) ≥ 0 , the project is economically justified

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

9

Figure 5-2

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!

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.

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

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 .

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

15

16

Example5-4

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

18

Ex. 5-5 page 219

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 (∞ )

20

Ex. 5-5 page 219

(b)

08.0

)4%,8,/(000,20000,3008./))%,8,/(000,100(

FAPACW

475,530$08.0

)4%,8,/(000,20000,30000,100

FACW

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 .

22

FW Decision Rule:

If FW (i=MARR) ≥ 0 , the project is economically justified

23

Ex 5-6 page 221

24

Example5-7

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

26

AW Decision Rule:

If AW (i=MARR) ≥ 0 , the project is economically justified

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)

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

29

CR (i%) = (I – S)(A/F, i%, N) +I(i%)

CR (i%) = (I –S)(A/P, i%, N) + S(i%)

30

Example 5-8 page 224

31

Example 5-9

32

Example 5-10

33

Example 5-11

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.

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

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.

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

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.

39

40

41

Installment Financing

An application of IRR method

Ex.5-14 will be discussed

42

43

44

45

46

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).

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

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

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.

52

Calculating External Rate of Return ( ERR )

Time N

0

Rk ( F / P, %, N - k ) N

k = 0

Ek ( P / F, %, k ) N

k = 0

1

2

53

Calculating External Rate of Return ( ERR )

i ‘ %= ?

Time N

0

Rk ( F / P, %, N - k ) N

k = 0

Ek ( P / F, %, k ) ( F / P, i ‘ %, N ) N

k = 0

1

2 3

54

Example 5-17

$25,000 (F/P,i’%,5)=$8,000(F/A,20%,5)+$5,000

(F/P,i’%,5)=

i’ =20.88%

5)'1(5813.2$25,000

$64,532.80i

55

Example 5-18

56

Example

Year 0 1 2 3 4

Cash Flow -$15,000 -$7,000 $10,000 $10,000 $10,000

For the cash flows given below, find the ERR when the

external reinvestment rate is ε = 12% (equal to the MARR).

Expenses

Revenue

Solving, we find

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.

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

59

A low- valued pay back period is considered desirable

60

Finding the simple and discounted payback period for a set of cash flows.

End

of

Year

End of

Year

Net

Cash

Flow

Cumulat

ive PW

at 0%

Present

worth at

i = 6%

Cumulati

ve PW at

6%

0 0 -$42,000 -$42,000 -$42,000 -$42,000

1 1 $12,000 -$30,000 11,320.8 -$30,679

2 2 $11,000 -$19,000 9,790 -$20,889

3 3 $10,000 -$9,000 8,396 -$12,493

4 4 $10,000 $1,000 7,921 -$4,572

5 5 $9,000 6,725.7 $2,153

The cumulative cash flows in the table were calculated using the formulas for simple and discounted payback.

From the calculations θ = 4 years and θ' = 5 years.

61

End of

Year

Net Cash

Flow

Cumulative

PW at 0%

Cumulative

PW at 6%

0 -$42,000 -$42,000 -$42,000

1 $12,000 -$30,000 -$30,679

2 $11,000 -$19,000 -$20,889

3 $10,000 -$9,000 -$12,493

4 $10,000 $1,000 -$4,572

5 $9,000 $2,153

Col.3:

EOF1= -42,000-12,000=-30,000

EOF2 : 11,000-30,000=-19,000

EOF3 : 10,000-19,000=-9,000

EOF4 : 10,000-9,000=1,000>0

θ = 4

Col.4:

12,000 (p/F,6%,1) =

12,000* 0.9434= 113,208

11,000 (p/F,6%,2) = 11,000*0.8900= 9,790

10,000*(P/F,6%,3) =

10,000*0.8396= 8,396

10,000*(P/F,6%,4) =

10,000*0.7921 =7,921

9,000*(P/F,6%,5) =

9,000*0.7473 = 6,725.7

End

of

Year

Net

Cash

Flow

Cumulat

ive PW

at 0%

Present

worth at

i = 6%

Cumulati

ve PW at

6%

0 -$42,000 -$42,000 -$42,000 -$42,000

1 $12,000 -$30,000 11,320.8 -$30,679

2 $11,000 -$19,000 9,790 -$20,889

3 $10,000 -$9,000 8,396 -$12,493

4 $10,000 $1,000 7,921 -$4,572

5 $9,000 6,725.7 $2,153

62

End

of

Year

Net

Cash

Flow

Cumulat

ive PW

at 0%

Present

worth at

i = 6%

Cumulati

ve PW at

6%

0 -$42,000 -$42,000 -$42,000 -$42,000

1 $12,000 -$30,000 11,320.8 -$30,679

2 $11,000 -$19,000 9,790 -$20,889

3 $10,000 -$9,000 8,396 -$12,493

4 $10,000 $1,000 7,921 -$4,572

5 $9,000 6,725.7 $2,153

Col.5:

EOF1:

-42,000 + 11,320.8 = -30,679.2

EOF2 :

9,790 -30,679.2 = -20,889.2

EOF3 :

8,396 -20,889.2 = -12,493.2

EOF4 :

7,921-12,493.2 = -4,572.2 <0

EOF5 :

6,725.7 -4,572.2 = 2,153.5 >0

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.

64

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