Lecture 11

ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Chapter 7 Present Worth Analysis † Describing Project Cash Flows † Initial Project Screening Method † Present Worth Analysis

Bank Loan vs. Investment Project Bank Loan Loan Customer

Bank Repayment

Investment Project Investment Project

Company Return

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Describing Project Cash Flows Year (n)

Cash Inflows (Benefits)

0

0

1

Cash Outflows (Costs)

Net Cash Flows

$650,000

-$650,000

215,500

53,000

162,500

2

215,500

53,000

162,500

…

…

…

…

8

215,500

53,000

162,500

Example Payback Period N 0 1 2 3 4 5 6

Cash Flow -$105,000+$20,000 $35,000 $45,000 $50,000 $50,000 $45,000 $35,000

Cum. Flow -$85,000 -$50,000 -$5,000 $45,000 $95,000 $140,000 $175,000

Payback period should occurs somewhere between N = 2 and N = 3.

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

$45,000

$45,000

$35,000

$35,000

Annual cash flow

$25,000 $15,000 0 1

2 Years

3

4

5

4

5

6

Cumulative cash flow ($)

$85,000 150,000

3.2 years Payback period

100,000 50,000 0 -50,000 -100,000 0

1

2

3

6

Years (n)

Payback Period Principle: How fast can I recover my initial investment? Method: Based on cumulative cash flow (or accounting profit) Screening Guideline: If the payback period is less than or equal to some specified payback period, the project would be considered for further analysis. Weakness: Does not consider the time value of money

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Discounted Payback Period Calculation Period

Cash Flow

Cost of Funds (15%)*

Cumulative Cash Flow

0

-$85,000

0

-$85,000

1

15,000

-$85,000(0.15)= -$12,750

-82,750

2

25,000

-$82,750(0.15)= -12,413

-70,163

3

35,000

-$70,163(0.15)= -10,524

-45,687

4

45,000

-$45,687(0.15)=-6,853

-7,540

5

45,000

-$7,540(0.15)= -1,131

36,329

6

35,000

$36,329(0.15)= 5,449

76,778

Payback period has been increased by a year !

Net Present Worth Measure Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i. Decision Rule: Accept the project if net surplus > 0 Inflow 0

1 2

3

Outflow

4

5 Net surplus

PW(i)inflow

0

PW(i) > 0 PW(i)outflow

Gives a measure of profitability of the project

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Example - Tiger Machine Tool Company inflow $24,400

$27,340

1

2

0

outflow

$55,760

3

i = 15%

$75,000

P W (15% ) inflow = $24 ,400 ( P / F ,15% ,1) + $27 , 340 ( P / F ,15% ,2 ) + $55,760 ( P / F ,15% ,3 ) = $78,553 P W (15% ) outflow = $75,000 P W (15% ) = $78,553 − $75, 000 = $3,553 > 0 , A ccept what if i = 0%?

what if i = 20%?

Present Worth Amounts at Varying Interest Rates i (%)

PW(i)

PW(i)

i(%)

0

$32,500

20

-$3,412

2

27,743

22

-5,924

4

23,309

24

-8,296

6

19,169

26

-10,539

8

15,296

28

-12,662

10

11,670

30

-14,673

12

8,270

32

-16,580

14

5,077

34

-18,360

16

2,076

36

-20,110

0

38

-21,745

-751

40

-23,302

17.45* 18 *Break even interest rate

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Present Worth Profile 40

Reject

Accept

30

PW (i) ($ thousands)

20

Break even interest rate (or rate of return)

10 $3553

17.45%

0 -10 -20 -30 0

5

10 15

20

25

30

35

40

i

i = Minimum Attractive Rate of Return (MARR)

Present Worth Analysis Net Present Worth of initial and future cash flows can be used to select among alternative projects. It is important to understand what Net Present Worth means, especially when the cash flows include both revenue and expenses.

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Terminology Salvage Value – the amount of money you can expect to receive by selling an asset when you are done with it. What value does it have when you are done with it? MARR – Minimum Attractive Rate of Return – I expect or need this return in order to be willing to invest my money.

Example Problem Project A costs $10,000 and will last for 10 years. Annual, end of the year revenues will be $3,000, and expenses will be $1,000. There is no salvage value. Project B costs $10,000 and will also last for 10 years. Annual revenues will be $3,000 with annual expenses of $1,500. Salvage value is $5,000. Conduct an economic analysis to select the preferred project using a MARR of 10% per year, compounded annually.

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Example Problem Project A costs $10,000 and will last for 10 years. Annual, end of the year revenues will be $3,000, and expenses will be $1,000. There is no salvage value. DIAGRAM: GIVEN: LIFETIME = 10 YRS NPWA ? MARR = 10%/YR, CPD ANNUALLY FIRST COST = $10,000 ANNUAL REVENUES = $3 000/YR 0 1 2 3 ANNUAL COSTS = $1,000/YR SALVAGE VALUE = $0 $10,000 FIND NPWA: NET ANNUAL = ANNUAL REVENUES – ANNUAL COSTS = $3,000/YR – $1,000/YR = $2,000/YR NPWA = A(P|A,i,N) – 1ST COST = $2,000(P|A,10%,10) – $10,000 = $2,000(6.1446) – $10,000 = $2,289

$3,000 4

10 $1,000

Example Problem Project B costs $10,000 and will also last for 10 years. Annual revenues will be $3,000 with annual expenses of $1,500. Salvage value is $5,000. DIAGRAM: GIVEN: LIFETIME = 10 YRS NPWB ? $5,000 MARR = 10%/YR, CPD ANNUALLY $3,000 FIRST COST = $10,000 ANNUAL REVENUES = $3,000/YR 10 0 1 2 3 4 ANNUAL COSTS = $1,500/YR $1,500 SALVAGE VALUE = $5,000 $10,000 FIND NPWB: NET ANNUAL = ANNUAL REVENUES – ANNUAL COSTS = $3,000/YR – $1,500/YR = $1,500/YR NPWB = A(P|A,i,N) + SALVAGE(P|F,i,N) – 1ST COST = $1,500(P|A,10%,10) + $5,000(P|F,10%,10) – $10,000 = $1,500(6.1446) + $5,000(0.3855) – $10,000 = $1,144 ►PREFER A

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

What does this mean? NPWA = $2,289 NPWB = $1,144 We prefer project A over project B. Does NOT mean $2289 profit! Concept: We favor Project A by $2,289 over taking $10,000 and putting it in an account earning 10%.

In other words… With expenses and revenues known, select the largest NPW > 0 ∴ Select Project A What does this mean? At i = 10%, $2,000 at the end of each of the next 10 years is worth, today, $2,289 more than the initial cost of $10,000.

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Further… You would be willing to pay as much as $10,000 + $2,289 = $12,289 for the project. At that price and at i = 10%, you are indifferent between: 1. Investing $12,289 for 10 years at i = 10%. 2. Obtaining $2000 at the end of each of the next 10 years (and reinvesting each receipt at 10%).

Illustrating… F10 = ($10,000 + $2,289) (F | P, 10%, 10) = ($10,000 + $2,289) (2.594) = $31,875 F10 = $2000 (F | A, 10%, 10) = $2000 (15.937) = $31,875

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Thus… The project only costs $10 000, but at i = 10% it is equivalent to investing $10 000 + 2 289 for 10 years. Since PW > 0, you are actually earning more than 10% on investment.

Problem 1 A company is considering the purchase of a new piece of testing equipment that is expected to produce $8,000 additional income during the first year of operation. This amount will decrease by $500 per year for each subsequent year of ownership. The equipment costs $20,000 and will have an estimated salvage value of $3,000 after 8 years of use. For a MARR of 15% compounded annually, determine the net present worth of this investment.

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Problem 1 GIVEN: 1ST COST = $20,000 ANNUAL COST: $500/YR (LINEAR GRADIENT) STARTING AT YR 2 ANNUAL REVENUE: $8,000 SALVAGE VALUE: $3,000 LIFE TIME: 8 YEARS MARR = 15%/YR, CPD ANNUALLY FIND: NPW DIAGRAM:

NPW = - $20,000 - $500(P|G,15%,8)

$3,000

+ $8,000(P|A,15%,8) + $3,000(P|F,15%,8) $8,000 0

= - $20,000 - $500(12.4807) 1

2

$500 $1,000

3

+ $8,000(4.4873) + $3,000(0.3269) 8

= - $20,000 - $6,240 + $35,898 + $981 NPW = $10,639

$3,500

Worth investing in !

$20,000

(Net) Present Worth Analysis When comparing projects, it is necessary to compare alternatives with the same project life (i.e., over the same period of time).

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ENGR 390 Lecture 11: Present Worth Analysis

Winter 2007

Present Worth Analysis When applied correctly, NPW can be used to select among various alternative projects. • The larger the NPW the better. • Requires establishing MARR. • MARR is used as the (i) in the equations.

Plotting NPW vs. i 10 000

NPW ($) IRR 2 289

0

10%

15.1% i (%)

Why does NPW decrease as i increases?

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