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IE 343 Engineering Economics

Lecture 14: Chapter 5 – Evaluating a Single Project

Instructor: Tian Ni Sep.23, 2011

IE 343 Fall 2011 1

New Plan: 5th Week Sep 19 Sep 21 Sep 23 (Quiz 3&Chap4, 5) (Hw4 due &Chap 5) 6th Week Sep 26 Sep 28 Sep 30 (Table&Excel&VBA (Quiz 4& (Exam 1 &Chap 4 Problems) Review 1) &HW5 out) 7th Week Oct 3 Oct 5 Oct 7 8th Week Oct 10 Oct 12 Oct 14 (Oct Break) (HW5 due

& HW6 Out) IE 343 Fall 2011

Exam 1 Date Change!

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No homework next week. Homework 5 will be assigned next Friday Sep. 30

after Exam 1. Homework 5 is due Oct. 14 Exam 1 will have two parts: (Chapter 1 – 4) Part 1: Old problems randomly picked from Homework,

Lecture Notes and Textbook. (Just like Quizzes) Part 2: New problems to test your understanding

Details will be announced next Monday You can start to review Homework, Lecture Notes

and Textbook now! IE 343 Fall 2011

Announcement

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Textbook Tables (Appendix C) Excel VBA Solve some Practice Problems in Chapter 4 Note: Monday’s class is part of review 1. I will

teach how to use textbook tables (appendix C), Excel, and VBA to simply the computations, which is very useful in the Exam 1.

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Next Class … Monday (Sep. 26)

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Question: What is the corresponding present equivalent value of the cash flow diagram below under a nominal interest rate r = 12% compounded annually, monthly and continuously? Is it a good investment?

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*Example 4.22*

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*Example 4.22*

0 1 13 14 15

A = 3,000

….. 22 23 24 ……..…... 12

$19,000

2 3 4 5

A A 1.06A A06.1 2

A06.1 9

A06.1 10

A06.1 11

EOY

6

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*Example 4.22* - Decomposition

0 1

$19,000

2 3 23 24

0 1

A = 3,000

………... 2 3 4 5 12

A

Single Cash Flow

Deferred Annuities Start at EOY 3

23 24

PS

PD

7

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*Example 4.22* - Decomposition

Geometric Gradient Series Start at the end of year 13

13 14 15 ….. 22 23 24

A 1.06A A06.1 2

A06.1 9

A06.1 10

A06.1 11

0 1 PG

12 ……..

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Basic Idea: Evaluate the present equivalent value of the three decomposed cash flow diagrams separately and then sum them up.

Notation: PS : Present Value for the single cash flow PD : Present Value for the Deferred Annuities PG : Present Value for the Geometric Gradient Series iA : Effective interest rate with annual compounding iM : Effective interest rate with monthly compounding iC : Effective interest rate with continuous compounding

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*Example 4.22* – Solution

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General Solution: PS = -19,000 PD = 3,000(P/A, i%, 10)(P/F, i%, 2) f = 6%, r = 12%, f ≠ r PG = (P/F, i%, 12)

P = PS + PD + PG

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06.0%)]12,06.0,/)(12%,,/(1[000,3

−−

iPFiFP

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Solution: Annual Compounding iA = r = 12% per year PS = -19,000 PD = 3,000(P/A, 12%, 10)(P/F, 12%, 2) = 13,513 f = 6%, i = 12%, f ≠ i PG = (P/F, 0.12, 12)

= 6,205 P = PS + PD + PG = 718 It is a good investment since P > 0!

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06.012.0)]12,06.0,/)(12,12.0,/(1[000,3

−− PFFP

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Solution: Monthly Compounding r = 12% per year iM = > iA = 0.12 PS = -19,000 PD = 3,000(P/A, 0.126825, 10)(P/F, 0.126825, 2) =

12985 f = 6%, i = 12.6825%, f ≠ i PG = (P/F,

0.126825, 12) = 5,569

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06.0126825.0)]12,06.0,/)(12,126825.0,/(1[000,3

−− PFFP

126825.011212.01

12

=−

+

12

Solution: Monthly Compounding r = 12% per year iM = > iA = 0.12

P = PS + PD + PG = -446

It is a bad investment since P < 0!

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126825.011212.01

12

=−

+

13

Solution: Continuous Compounding r = 12% per year iM = > iM = 0.126825 (> iA = 0.12) PS = -19,000 PD = 3,000(P/A, 0.1275, 10)(P/F, 0.1275, 2) = 12,934 f = 6%, i = 12.75%, f ≠ i PG = (P/F, 0.1275,

12) = 5,510

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06.01275.0)]12,06.0,/)(12,1275.0,/(1[000,3

−− PFFP

( ) 1275.0112.0 =−e

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Solution: Continuous Compounding r = 12% per year iM = > iM = 0.126825 (> iA = 0.12)

P = PS + PD + PG = -556

It is a bad investment since P < 0!

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( ) 1275.0112.0 =−e

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Chapter 5 - Evaluating a Single Project

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Time value of money - Application

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We will learn how to evaluate the profitability and liquidity of a single problem solution (or alternative). Minimum Attractive Rate of Return (MARR) is useful for this analysis. MARR ("hurdle rate") is usually organization-specific and determined based on the following: – Cost of money available for investment – Number of good projects available for investment – Risks involved in investment opportunities

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Time value of money - Applications

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How to use MARR? Use it as an interest rate to convert cash flows into

equivalent worth at some point in time. The proposed problem solution (project or alternative) is profitable if it generates sufficient cash flow to recover the initial investment and earn an interest rate that is at least as high as MARR.

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Time value of money - Applications

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Methods to evaluate profitability Present Worth (PW) Future Worth (FW) Annual Worth (AW) Internal Rate of Return (IRR) External Rate of Return (ERR)

Methods to evaluate liquidity Simple payback period Discounted payback period

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Methods

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Present Worth method (PW)

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Discount all cash flows to the present time by using the MARR as the interest rate.

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Present Worth (PW) method

NN iFiFiFiFiPW −−− ++++++++= )1(...)1()1()1(%)( 2

21

10

0

∑=

−+=N

k

kk iFiPW

0)1(%)(

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: represents future cash flow at the end of the compounding period k;

i: is the effective interest rate (MARR) per compounding period, which is assumed to be constant;

N represents the number of compounding periods in the planning horizon (i.e., study period).

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Present Worth (PW) method N

N iFiFiFiFiPW −−− ++++++++= )1(...)1()1()1(%)( 22

11

00

∑=

−+=N

k

kk iFiPW

0)1(%)(

kF

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Rule: If PW(MARR) ≧ 0, the project is profitable If PW(MARR)< 0, the project is not profitable.

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Present Worth (PW) method

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Consider a project that has the following cash flows over a study period of 5 years: Initial investment: $100,000 Annual revenues: $40,000 Annual expenses: $5,000 Salvage value: $20,000 MARR: 20%.

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Present Worth (PW) method - Example 5.1

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Example 5.1 - Cash Flow Diagram

0 1 2 3 4

$100,000

5

MARR = 2%

$40,000

$5,000

S=$20,000

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Solution: PW(20%)=-100,000+(40,000-

5,000)(P/A,20%,5)+20,000(P/F,20%,5)=$12,709. Is this a profitable project? Yes! Because: We recovered $100,000 (investment). We

earned an interest rate of 20% which was desired. We even made a profit that has a present equivalent value of $12,709, which is beyond what was required.

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Example 5.1 – Solution

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Suppose we had an interest rate higher than 20%. For example, 20% nominal interest rate

compounded continuously. Then, the effective interest rate is

PW(i=22.14% discrete compounding) = PW(r= 20% continuous compounding) = $7,285.7 Still profitable but PW is lower when the effective

interest rate increases.

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Example 5.1 – Cont’d

%.14.2212.0 =−= ei

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Note: If the first cash flow of a project is negative and its subsequent cash flows are positive, then PW is a decreasing function of the effective interest rate i.

For a general pattern of cash flows, PW may not be the decreasing function of i.

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Example 5.1 – Cont’d

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Future Worth method (FW)

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Find the equivalent worth of all cash flows at the end of the study period by using the MARR as the interest rate.

Note that FW and PW of a project are equivalent at the interest rate of i%, i.e., FW=PW(F/P,i%,N).

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Future Worth (FW) method

022

110 )1(...)1()1()1(%)( iFiFiFiFiFW N

NNN ++++++++= −−

∑=

−+=N

k

kNk iFiFW

0)1(%)(

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Rule: If FW(MARR) ≧ 0, the project is profitable If FW(MARR)< 0, the project is not profitable.

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Future Worth (FW) method

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Back to Example 5.1 with the following project: Initial investment: $100,000 Annual revenues: $40,000 Annual expenses: $5,000 Salvage value: $20,000 MARR: 20%. N: 5 years.

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Future Worth (FW) method - Example 5.1

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IE 343 Fall 2011

Example 5.1 - Cash Flow Diagram

0 1 2 3 4

$100,000

5

MARR = 2%

$40,000

$5,000

S=$20,000

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Solution: FW(20%)=-100,000(F/P,20%,5)+(40,000-

5,000)(F/A,20%,5)+20,000=$31,624. Since FW(20%) ≧ 0, the project is profitable.

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Future Worth (FW) method - Example 5.1

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Capitalized Worth method (CW)

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Definition: Capitalized–Worth method involves determining the PW of all revenues or expenses over an infinitely length of time.

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Capitalized – Worth (CW) Method

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CW of a perpetual series of end-of-period uniform payments A, with interest at i% per period is A(P/A, i%, ∞)

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0 ∞

A A A . . . . . . . . . . . . . . . . . . . . . . . . . . …

1 2 3 . . . . . . . . . . . . . . . . . . . . . . . . ….

To find P?

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CW of a perpetual series of end-of-period uniform payments A, with interest at i% per period is A(P/A, i%, ∞)

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=

+

−+=

∞==

∞→

∞→

iA

iiiAiCW

iAPAPWiCW

N

N

N

N

1)1(

1)1(lim%)(

)%,,/(%)(

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A practical approximation of “forever” (infinity) is dependent on the interest rate i%

E.g.1: i=8% (1/i = 12.5000), (P/A,8%,N) = 12.4943 when N = 100. So a good approximation for infinity when i=8% is N ≧ 100

E.g.2: i=20% (1/i =5.0000), (P/A,20%,N) = 4.9966 when N = 40. So a good approximation for infinity when i=20% is N ≧ 40

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Annual Worth method (AW)

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The AW of a project is an equal annual series of dollar amounts, for a stated study period, that is equivalent to the cash inflows and outflows at an interest rate that is generally the MARR. AW(i%)=R – E – CR(i%) R: Annual equivalent revenues E: Annual equivalent expenses CR: Annual equivalent Capital Recovery cost

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Annual Worth (AW) method

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CR includes the loss in value of asset and interest on invested capital.

The CR can be computed by: CR(i%)=I(A/P,i%,N)-S(A/F,i%,N) I: Initial investment S: Salvage value.

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The above formula of CR can be interpreted in the way that the annual equivalent of the initial capital investment minus the annual equivalent of the salvage value.

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Back to Example 5.1: Initial investment (I): $100,000 Annual revenues (R): $40,000 Annual expenses (E): $5,000 Salvage value (S): $20,000 MARR (i%): 20%. N: 5 years.

IE 343 Fall 2011

Annual Worth (AW) method - Example 5.1

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Back to Example 5.1: I = $100,000 R = $40,000 E = $5,000 S = $20,000 MARR = 20%. N: 5 years.

CR(20%)=100,000(A/P,20%,5)-20,000(A/F,20%,5) =$30,750.

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CR cost calculation - Example 5.1

Year

Value of investment

at BOY

Uniform loss in value of

year (A)

Interest on BOY (B) investment

(i=20%)(B)

CR cost for year

(A+B)

PW of CR amount at

i=20% 1 100,000 16,000 20,000 36,000 30000 2 84,000 16,000 16,800 32,800 22777.78 3 68,000 16,000 13,600 29,600 17129.63 4 52,000 16,000 10,400 26,400 12731.48 5 36,000 16,000 7,200 23,200 9323.56

Sum= 91962.45

Assume that the loss in value is uniform per year. So loss per year is (I – S)/N = (100,000 – 20,000)/5 = 16,000

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BOY stands for Beginning-of-Year. PW of CR cost= 36,000(P/F,20%,1)

+32,800(P/F,20%,2)+...+23,200(P/F,20%,5) =91,961.44

Uniform annual equivalent of CR cost =91,961.44(A/P,20%,5) = $30,750

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CR cost calculation - Example 5.1

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AW(i%)= R – E – CR(i%) AW(20%)=40,000-5,000-30,750=4,250 Since AW(20%) ≧ 0, the project is profitable.

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Note that AW, PW, and FW are equivalent measures of profitability of a project: AW=PW(A/P,i%,N), AW=FW(A/F,i%,N), PW=FW(P/F,i%,N), FW=PW(F/P,i%,N). (Check that these expressions hold for the project in

Example 5.1.)

Therefore if PW ≧ 0, then FW ≧ 0, and AW ≧ 0.

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