CHAPTER 5 Engineering Economic

CHAPTER 5 EVALUATING SINGLE PROJECT

Introduction • • 1. 2. 3. 4. 5.

Capital projects should consider return on the capital that is sufficiently attractive in view of the risks involved and potential alternative uses. We will discuss 5 methods for evaluating the economic profitability of a project: Present worth (PW) Future worth (FW) Annual worth (AW) Internal Rate of Return (IRR) External Rate of Return (ERR)

Convert cash flows into equivalent worth at some point (s) in time using an interest rate known as M i n i m u m A t t r a c t i v e r at at e o f R e t u r n    (MARR)  (MARR)

Introduction • • 1. 2. 3. 4. 5.

Capital projects should consider return on the capital that is sufficiently attractive in view of the risks involved and potential alternative uses. We will discuss 5 methods for evaluating the economic profitability of a project: Present worth (PW) Future worth (FW) Annual worth (AW) Internal Rate of Return (IRR) External Rate of Return (ERR)

Convert cash flows into equivalent worth at some point (s) in time using an interest rate known as M i n i m u m A t t r a c t i v e r at at e o f R e t u r n    (MARR)  (MARR)

The basic question to be addressed: is the proposed c a p i t al   and its associated a l in i n v e s t m e n t  and e x p e n d i t u r e s  (cash out flow)

can be recovered by   over time r e v en e n u e (s (s a v i n g s )  over

(cash in flow)

in addition to a r e tu t u r n o n t h e c a p i t al a l  (rate of return ~ MARR) that is sufficiently attractive in view of the risk involved.

MINIMUM ATTRACTIVE RATE OF RETURN ( MARR )

•  An interest rate used to convert cash flows flows into equivalent worth at some point(s) in time • Usually a policy issue based on: - amount, source and cost of money available for investment - number and purpose of good projects available for investment - amount of perceived risk of investment opportunities and estimated cost of administering projects over over short and long run - type of organization involved

MARR is sometimes referred to as hurdle rate

CAPITAL RATIONING • Establishing MARR involves opportunity cost viewpoint results from phenomena of CAPITAL RATIONING

• Exists when management decides to restrict the total amount of capital invested, by desire or limit of available capital • Select only those projects which provide annual rate of return in excess of MARR •  As amount of investment capital and opportunities

available change over time, a firm’s MARR will also change •

See slide

PRESENT WORTH METHOD ( PW ) • Based on concept of equivalent worth of all cash flows relative to the present as a base •  All cash inflows and outflows discounted to present at interest -- generally MARR • PW is a measure of how much money can be afforded for investment in excess of cost • PW is positive if dollar amount received for investment exceeds minimum required by investors

FINDING PRESENT WORTH

• Discount future amounts to the present by using the interest rate over the appropriate study period N

PW = k = 0Fk ( 1 + i ) - k

 – i  = effective interest rate, or MARR per compounding period  – k = index for each compounding period  – Fk = future cash flow at the end of period k  – N = number of compounding periods in study period • interest rate is assumed constant through project • The higher the interest rate and further into future a cash flow occurs, the lower its PW

FUTURE WORTH METHOD (FW ) • FW is based on the equivalent worth of all cash inflows and outflows at the end of the planning horizon at an interest rate that is generally MARR • The FW of a project is equivalent to FW = PW ( F / P, i %, N ) • If FW > 0, it is economically justified  N

N-k i  FW ( i % ) =k  F  ( 1 +   ) =0 k

 –i  = effective interest rate  –k = index for each compounding period  –Fk = future cash flow at the end of period k  –N = number of compounding periods in study period

ANNUAL WORTH METHOD ( AW ) •  AW is an equal annual series of dollar amounts, over •

• • • •

a stated period ( N ), equivalent to the cash inflows and outflows at interest rate that is generally MARR  AW is annual equivalent revenues ( R ) minus annual equivalent expenses ( E ), less the annual equivalent capital recovery (CR)  AW ( i  % ) = R - E - CR ( i  % )  AW = PW ( A / P, i  %, N )  AW = FW ( A / F, i  %, N ) If AW > 0, project is economically attractive  AW = 0 : annual return = MARR earned

 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 salvage value $5,000 at the end of a five year study period. Increased productivity attributable to this equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. The firm MARR is 20% per year. a) Draw the cash flow diagram b) Determine the equivalent present worth (PW)

CAPITAL RECOVERY ( CR ) • CR is the equivalent uniform annual cost  of the capital invested • CR is an annual amount that covers:  – Loss in value of the asset  – Interest on invested capital ( i.e., at the MARR ) CR ( i  % ) = I ( A / P, i  %, N ) - S ( A / F, i  %, N ) I = initial investment for the project S = salvage ( market ) value at the end of the study period N = project study period

Example of capital recovery calculation Consider a machine that cost $10,000, last for five years, and have a salvage (market) value $2,000. Thus, the loss in value of this asset over five years is $8,000. Additionally, the MARR is 10% per year. Thus CR (i%)=I ( A/P ,i%,N ) – S( A/F,i%,N ) Solution: CR (10%) = $10,000( A/P,10%,5 ) - $2,000( A/F,10%,5 ) = $10,000 (0.2638) - $2,000 (0.1638) = $2,310.40

Example of Annual Worth Method:  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 $5,000 at the end of a study period of five years. Increased productivity attributable to this equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the firm MARR is 20% per year, is this proposal a sound one? Use the AW method. Solution:  AW (i%) = R - E – CR (i%)  AW (20%) = $8,000 – [$25,000(A/P,20%,5) - $5,000(A/F,20%,5)] = $8,000 – [$8360 - $672] = $312 Conclusion: because AW (20%) is positive, the equipment more than pays for itself over the planning horizon.

Problem 4.19a  A certain service can be performed satisfactorily by process X, which has a capital investment cost of $8,000, an estimated life of 10 years, no salvage value, an annual net receipts (revenue  – expenses) of $2,400. Assuming a MARR of 18% before income taxes, find the AW of this process and specify whether you would recommend it. Solution:  AW (i%) = R - E – CR (i%)  AW (18%) = $2,400 – [$8,000(A/P,18%,10) - $0(A/F,18%,10)] = $2,400 – [$1,780 - $0] = $620 Conclusion: because AW (18%) is positive, the process X is recommended.

CAPITAL RECOVERY ( CR) • CR is also calculated by adding sinking fund amount (i.e., deposit) to interest on original investment CR ( i  % ) = ( I - S ) ( A / F, i  %, N ) + I ( i  % ) • CR is also calculated by adding the equivalent annual cost of the uniform loss in value of the investment to the interest on the salvage value CR ( i  % ) = ( I - S ) ( A / P, i  %, N ) + S ( i  % )

• 5 methods for evaluating the economic profitability of a project: 1. Present worth (PW) √ 2. Future worth (FW) √ 3. Annual worth (AW) √ 4. Internal Rate of Return (IRR) 5. External Rate of Return (ERR)

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) •  Also referred to as:  – investor’s method  – discounted cash flow method  – profitability index • IRR is positive for a single alternative only if:  – both receipts and expenses are present in the cash flow pattern the sum of receipts exceeds sum of cash outflows

INTERNAL RATE OF RETURN METHOD ( IRR ) • IRR is i  %, using the following PW formula: ’

 N

 R k ( k=0

 N

P / F, i  %, k ) =  E k ( P / F, i  %, k ) ’

’

k=0

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 ) • If i  > MARR, the alternative is acceptable ’

• To compute IRR for alternative, set net PW = 0  N  N PW =  R k ( P / F, i  %, k ) -  E k ( P / F, i  %, k ) = 0 k=0 k=0 • i   is calculated on the beginning-of-year unrecovered investment through the life of a project ’

’

’

DISADVANTAGES OF IRR • The IRR method assumes recovered funds, if not consumed each time period, are reinvested at i   %, rather than at MARR • The computation of IRR may be unmanageable ‘

• Multiple IRR’s may be calculated for the same problem • The IRR method must be carefully applied and interpreted in the analysis of two or more alternatives, where only one is acceptable • Basic IRR method cannot rank mutually exclusive projects, the project with higher IRR potentially having lower net present value . . . inconsistent ranking.

 ADVANTAGES OF IRR • IRR can be calculated without having to estimate cost of capital or MARR • IRR, in the form of rate of return is more appealing to evaluate investment • IRR is more appealing to communicate profitability

• When unique (not multiple IRR’s), it provides valuable information about the return on the investment and is often viewed as a measure of efficiency

Using Multiple Internal Rates of Return: The IRR Parity Technique PW vs rate of return 50 rate of return (%)

0        0

8        6        6        6        1        2        3

       5        4        4        5

       4        4        4        6        7        8

-50    )    $    (    W-100    P

-150

-200

-250

       4        4        9        0        1

Multiple IRRs @ 30% and 60% EOY

Cash flows ($)

0

-1000

1

2900

2

-2080

       4        4        1        2        1        1

       4        4        3        4        1        1

       4        5        1

Using Multiple Internal Rates of Return: The IRR Parity Technique  According to IRR parity technique proposed by Zhang (2005)

If the number of real IRRs which is greater than the MARR is: Even (including zero), reject the project; Odd, accept the project. •

•

Referring to the earlier chart of PW vs. rate of return • If 30% < MARR < 60%, there is one real IRR (60%) that is greater than MARR; therefore the project should be accepted. Then we identify the relevant IRR as 60%.

• If MARR < 30%, there are two real IRRs (both 30% and 60%) that are greater than MARR. Hence, the project should be rejected.

• If MARR > 60%, there is no real IRR that is greater than MARR; therefore the project should be rejected.

D Zhang (2005), Engineering Economist, Vol. 50, Issue 4, pp 237-335

Example of Internal Rate of Return method  A capital investment of $10,000 can be made in a project that will produce a uniform annual income revenue of $5,310 for five years and then have a salvage value of $2,000. Annual expenses will be $ 3,000. The company is willing to accept any project that will earn at least 10% per year on all invested capital. Determine whether it is acceptable by using IRR method. Solution: Set net PW = 0 PW = 0 = -$10,000 + ($5,310  – $3,000)(P/A,i’%,5 ) + $2,000(P/F,i’%,5 ) ; i’ ’% = ? By trial-and-error process; at i = 5%; PW = $1,568 at i = 15%; PW = -$1,262 By linear interpolation; i’ = 10.5%. Therefore, the project is minimally acceptable ( i’% > MARR)

Problem 4.4 b

Determine the IRR of the following engineering project when the MARR is 15% per year. Investment cost Expected life Salvage value  Annual receipts  Annual expenses

$10,000 5 years -$ 1,000 $ 8,000 $ 4,000

Solution: PW = 0 = -$10,000  – $1,000(P/F,i’%,5 ) + (8,000-$4,000)( P/A,i’%,5 ) By trial and error process, at i’ = 20%, PW = $1560.50 at i’ = 25%, PW = $ 429.50 B linear extrapolation, i’ = 27%, hence the project is acceptable; > MARR

Example of IRR Method:

 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 $5,000 at the end of a study period of five years. Increased productivity attributable to this equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the firm MARR is 20% per year, is this proposal a sound one? Use the IRR method. Solution: PW = 0 = -$25,000 + $8,000( P/A,i’%,5 ) + $5,000(P/F,i’%,5 ) By trial and error process, at i’ = 20%, PW = $ 934.30 at i’ = 25%, PW = -$1847.10 By linear interpolation, i’ = 22 %. Therefore, the project is acceptable ( i’% > MARR)

Example of IRR method Barron Chemical uses a thermoplastic polymer to enhance the appearance of certain RV panels. The first cost of one process was $126,000 with annual costs of $49,000 and revenues of $88,000. A salvage value of $33,000 was realized when the process was discontinued after 8 years. What rate of return did the company make on the process ?

Solution: Set FW = 0 FW = 0 = -$126,000 (F/P,i’%,8 ) + ($88,000 - 49,000) (F/A,i’%,8 ) + $33,000 By trial and error process, at i’ = 25%, FW = -$55,811 at i’ = 30%, FW = $64,370 By linear interpolation, i’ = 27.3%

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

CALCULATING EXTERNAL RATE OF RETURN ( ERR ) 1. All net cash outflows are discounted to the present (time 0) at  % per compounding period. 2. All net cash inflows are discounted to period N at  %. 3. ERR -- the equivalence between the discounted cash inflows and cash outflows -- is determined. The absolute value of the present equivalent worth of the net cash outflows at  % 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

CALCULATING EXTERNAL RATE OF RETURN ( ERR )  N  Ek (

k=0

P / F,  %, k )( F / P, i   %, N ) ‘

=

 N



Rk ( F / P,  %, N - k )

k= 0 receipts of

Rk = excess over expenses in period k Ek = excess of expenses over receipts in period k N = project life or period of study  =

external reinvestment rate per period  N

 R

i   %= ?

0  N  E

‘

Time  ( P / F,  %, k )( F / P, i   %, N ) ‘

 ( F / P,  %, N - k )

k = 0k

N

ERR ADVANTAGES

• ERR has two advantages over IRR: 1. It can usually be solved for directly, rather than by trial and error. 2. It is not subject to multiple rates of return. (see slide Problem 4.38)

Example of ERR Method:

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 $5,000 at the end of a study period of five years. Increased productivity attributable to this equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the external investment rate = MARR = 20% per year, what is the alternative’s ERR, and is this proposal a sound one?

Solution: $25,000(F/P,i’%,5) = $8,000(F/A,20%,5)   + $5,000 (F/P,i’%,5) = $64,532.80/$25,000 = 2.5813 i’  = 20.88% Since i’ > MARR , the alternative is minimally justified.

Problem 4.32 Summary of the projected costs and annual receipts for a new product line is presented as follows: End of Year Net Cash Flow 0 - $450,000 1 - 42,500 2 92,800 3 386,000 4 614,600 5 - 202,200 The company’s external reinvestment rate per year = MARR = 10% per year

Solution: [$450,000 + $42,500(P/F,10%,1 ) + $202,200(P/F,10%,5 )](F/P,i’%,5 ) = $92,800(F/P,10%,3 ) + $386,000(F/P,10%,2 ) + $614,600(F/P,10%,1 ) $614,182.73(F/P,i’%,5 ) = $1,265,544 (F/P,i’%,5 ) = 2.0622 By interpolation, i’% , ERR = 15.6%

Problem 4.34 Given a cash flow as stated below: End of Year Net Cash Flow 0 - $10,000 1 – 7 1,400 7 10,000 The external rate of reinvestment for the company = 8% per year. Solution: $10,000(F/P.i’%,7 ) = $1,400(F/A,8%,7 ) + $10,000 (F/P,i’%,7 ) = $22,491.92 / 10,000 = 2.2492 By interpolation, ERR = i’% = 12.3%

Problem 4.48: A certain project has a net receipts equaling $1,000 now, has costs of $5,000 at the end of the first year, and earns $6,000 at the end of the second year. If the external reinvestment rate of 10% is available, what is the rate of return for this project using the ERR  method ? Solution: $5,000(P/F,10%,1 )(F/P,i’%,2 ) = $1,000(F/P,10%,2 ) + $6,000 (F/P,i’%,2 ) = $7,210 / $4,545.50 = 1.5862 By interpolation, ERR = i’% = 25.9%

PAYBACK PERIOD METHOD

• Sometimes referred to as simple payout method • Indicates liquidity (riskiness) rather than profitability • Calculates smallest number of years (  ) needed for cash inflows to equal cash outflows -- break-even life •  ignores the time value of money and all cash flows which occur after  



( Rk -Ek) - I > 0

k=1

• If  is calculated to include some fraction of a year, it is rounded to the next highest year

PAYBACK PERIOD METHOD • The payback period can produce misleading results, and should only be used with one of the other methods of determining profitability •  A discounted payback period   ( where   < N ) may be calculated so that the time value of money is considered ‘

‘

’

 k=1

( Rk - Ek) ( P / F, i %, k ) - I > 0

i   is the MARR ‘

I is the capital investment made at the present time ( k = 0 ) is the present time   is the smallest value that satisfies the equation ‘

Simple Payback Period vs. Discounted Payback Period

EOY

Net c / flow

0 1 2 3 4 5

- $25,000 8,000 8,000 8,000 8,000 13,000

Cum PW @ i=0% - $25,0000 - 17,000 9,000 1,000 7,000 20,000

The payback period is at 4th years, because the cumulative balance turns positive at EOY 4 [ t i m e v al u e o f m o n e y i s n o t considered]

PW @ i=20% - $25,000 6,667 5,556 4,630 3,858 5,223

Cum PW @ MARR 20% - $25,000 - 18,333 - 12,777 - 8,147 - 4,289 934

The payback period is at 5th year, because the cumulative discounted balance turns positive at EOY 5 [ t i m e v al u e o f m o n e y i s considered]

INVESTMENT-BALANCE DIAGRAM

Describes how much money is tied up in a project and how the recovery of funds behaves over its estimated life.

INTERPRETING IRR USING INVESTMENT-BALANCE DIAGRAM P (1 + i ) ‘

[ P (1 + i ) - (R 1 - E1) ] (1 +i ) Unrecovered 1 + i 1 + i Investment 1 + i Balance, $ (R1 - E1) ‘

‘

‘

‘

‘

(R2 - E2)

(R3 - E3)

Initial investment =P 0

1

2

3

(RN-1 - EN-1)

1 + i

‘

(RN - EN)

$0

N

• downward arrows represent annual returns (R k - Ek) : 1 < k < N • dashed lines represent opportunity cost of interest, or interest on BOY investment balance • IRR is value i  that causes unrecovered investment balance to equal 0 at the end of the investment period. ‘

INVESTMENT-BALANCE DIAGRAM EXAMPLE

• Capital Investment ( I ) = $10,000 • Uniform annual revenue = $5,310 •  Annual expenses = $3,000 • Salvage value = $2,000 • MARR = 5% per year

   t   n   $   e  ,   e 5,000   m   c    t   s   n   a   e   l   v   a   n   B    I

MARR = 5% $2,001 ( = FW )

’ + $4,310

Years 0 1

- 5,000

2

3

4

5

- $2,199 Area of Negative - $2,310 Investment - $2,310 - $4,294 Balance - $6,290 - $8,190 - $2,310

- 10,000 -$10,500

- $2,310 - $2,310 - $6,604 - $8,600

- $4,509

WHAT INVESTMENT-BALANCE DIAGRAM PROVIDES • Discounted payback period (  ) is 5 years • FW is $2,001 • Investment has negative investment balance ‘

until the fifth year • Investment-balance diagram provides additional insight into worthiness of proposed capital investment opportunity and helps communicate important economic information

INVESTMENT-BALANCE DIAGRAM EXAMPLE 4-17 Construct an investment balance diagram for the following cash flow table. The MARR is 10% per year. End of Year

Net Cash Flow

Three sign Changes

0 1 2 3

- $5,000 6,000 - 1,000 4,000

--negative to positive positive to negative negative to positive