engineering economics lec5

7/21/2019 Engineering Economics Lec5

1/19

2/26/20

$

,

$


2/19

2/26/20

,

Principle:

How fast can I recover my initial investment?

Method:

Based on the cumulative cash flow (or

accounting profit)

Screening Guideline:

If the payback period is less than or equal to

some specified payback period, the projectwould be considered for further analysis.

Weakness:

Does not consider the time value of money


3/19

2/26/20

Pizza-in-a-Hurry operates a pizza delivery, they use,two eight year-old vehicles for delivery, both of whichare large, consume a great deal of gas and are startingto cost a lot to repair. The owner, Ray, is thinking ofreplacing one of the cars with a smaller, three-year-oldcar that his sister-in-law is selling for $8000. Rayfigures he cansave $3000, $2000, and $1500 per year for the nextthree years and $1000 per year for the following twoyears by purchasing the smaller car. What is the paybackperiod for this decision?

Example 1

yearyearyearyear AAAA BBBB

0 -$1000 -$2783

1 +200 +1200

2 +200 +1200

3 +1200 +1200

4 +1200 +1200

5 +1200 +1200

6 +1200 +1200

The cash flows for two alternatives are as follows:

Find the Pay Back Period


4/19

2/26/20

$1000

$200

$1200

0

1 2 3 4 5 6

YearsAnnual

cashflow

$200

$1200 $1200 $1200

-1000-500

0

500

1000

1500

0 1 2 3 4 5 6Years (n)

2.5 years

Payback period

Cumulativecashflow($)

How long does it take to recover the initialinvestment for Project B in Example 2?


5/19

2/26/20

A firm is trying to decide which of two weighingscales it should install to check a package-fillingoperation in the plant. If both scales have a 6-yearlife, which one should be selected? Assume an8% interest rate.

AlternativeAlternativeAlternativeAlternative Cost ($)Cost ($)Cost ($)Cost ($) Uniform AnnualUniform AnnualUniform AnnualUniform AnnualBenefitBenefitBenefitBenefit

($)($)($)($)

EndEndEndEnd ofofofof useful Lifeuseful Lifeuseful Lifeuseful Life(salvage Value) ($)(salvage Value) ($)(salvage Value) ($)(salvage Value) ($)

Atlas Scale 2000 450 100

Tom Thumb 3000 600 700


6/19

2/26/20

Discounted Payback Period

Principle:How fast can I recover my initial investment

plus interest?

Method:

Based on the cumulative discounted cash flow

Screening Guideline:

If the discounted payback period (DPP) is less

than or equal to some specified payback period,

the project would be considered for further

analysis.

Weakness:

Cash flows occurring after DPP are ignored

Autonumerics company has just bought a new

spindle machine at a cost of $105,000 to

replace one that had a salvage value of 20,000.

the projected annual after-tax savings via

improved efficiency, which will exceed the

investment cost, are as follows:

What is the Discounted Pay back Period ?


7/19

2/26/20

$85,000

$15,000

$25,000$35,000

$45,000 $45,000

$35,000

0

1 2 3 4 5 6

Years

Annualcashflow

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

**** Cost of funds = (Unrecovered beginning balance) X (interest rate)Cost of funds = (Unrecovered beginning balance) X (interest rate)Cost of funds = (Unrecovered beginning balance) X (interest rate)Cost of funds = (Unrecovered beginning balance) X (interest rate)


8/19

2/26/20

Two equivalent pieces of quality inspectionequipment are being considered for purchase bySquare D Electric. Machine 2 expected to beversatile and technologically advanced enoughto provide net income longer than machine 1.Assume i= 15%.

Machine 1 Machine 2

First cost($) 12,000 8,000

Annual NCF($) 3,000 1,000(years 1-5)3,000(years 6-14)

Maximum life (years) 7 14


9/19

2/26/20

Principle: Compute the equivalent net surplus at n = 0 for a given

interest rate of i.

Decision Rule for Single Project Evaluation: Accept the project if the

net surplus is positive.

Decision Rule for Comparing Multiple Alternatives: Select the

alternative with the largest net present worth.

2 3 4 5

0 1Inflow

Outflow

0

PW(i)inflow

PW(i)outflow

Net surplus

PW(i) > 0


10/19

2/26/20

$75,000

$24,400 $27,340$55,760

01 2 3

outflow

inflow

An electrical motor rated at 15HP needs to bepurchased for $1,000.

The service life of the motor is known to be 10years with negligible salvage value.

Its full load efficiency is 85%.

The cost of energy is $0.08 per kwh.

The intended use of the motor is 4,000 hoursper year.

Find the total cost of owning and operating themotor at 10% interest.


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2/26/20

0 1 2 3 4 5 6 7 8 9 10

$4,211

$1,000


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2/26/20

Given: Cash flows and MARR (i)

Find: The net equivalent worth at aspecified period other thanpresent, commonly the end ofproject life

Decision Rule: Accept the project ifthe equivalent worth is positive.

$75,000

$24,400 $27,340$55,760

01 2 3

Project life

NFW is based on the equivalent worth of all cash inflows and outflows at theend of the study period at an interest rate that is generally the MARR.

A $45,000 investment in a new conveyor

system is projected to improve throughput and

increasing revenue by $14,000 per year for five

years. The conveyor will have an estimated

market value of $4,000 at the end of five years.

Using FW and a MARR of 12%, is this a good

investment?


13/19

2/26/20

NFW = -$45,000(F/P, 12%, 5)+$14,000(F/A, 12%, 5)+$4,000

NFW = $13,635.70 >0.0

NFW = -$45,000(1.7623)+$14,000(6.3528)+$4,000

Higgins Corporation (HC) has developed a robot called Helpmate The firm would need a new plant for manufacturing. Plant could be built and would be ready for production in 2 years 12 hectares site is needed (could be purchased with a cost of 1.5

million in year 0 ) Building construction would begin early in year 1 and continue

throughout year 2 Building cost 10 million (4 million at the end of year 1 and 6 million

at the end of year 2) Manufacturing equipment 13 million at the end of year 2 After termination land after tax worth 2 million , building 3 million

and equipment 3 million The plant would begin operation at the beginning of year 3, first cash

flow at the end of year 3 and economic life is 6 years i = 15%


14/19

2/26/20

Barcewell, built a hydroelectric plant using his personal savings

of $800,000

Power generating capacity of 6 million kwhs

Estimated annual power sales after taxes - $120,000

Expected service life of 50 years

Was Bracewell's $800,000 investment a wise one?

How long does he have to wait to recover his initial investment,

and will he ever make a profit?


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2/26/20


16/19

2/26/20

Principle: PW for a project with an annual receipt ofA over

infinite service life

Equation:

CE(i) =A(P/A, i, ) =A/i

A

0

P = CE(i)

Capitalized worth is a special case of the present worth, it is the

present worth of all revenues or expenses over an infiniteproject life time.

Capitalized costExpenses only

10

$1,000

$2,000

P = CE (10%) = ?

0

Given: i= 10%, N=

Find: Por CE (10%)


17/19

2/26/20

10

$1,000

$2,000

P = CE (10%) = ?

0

Construction cost = $2,000,000

Annual Maintenance cost = $50,000

Renovation cost = $500,000 every 15 years

Planning horizon = infinite period

Interest rate = 5%


18/19

2/26/20

$500,000 $500,000 $500,000 $500,000

$2,000,000

$50,000

0 15 30 45 60

Years

Solution:

Construction CostP1 = $2,000,000

Maintenance Costs

P2 = $50,000/0.05 = $1,000,000

Renovation Costs

P3 = $500,000(P/F, 5%, 15)

+ $500,000(P/F, 5%, 30)

+ $500,000(P/F, 5%, 45)

+ $500,000(P/F, 5%, 60)

.

= {$500,000(A/F, 5%, 15)}/0.05= $463,423

Total Present Worth

P = P1 + P2 + P3 = $3,463,423


19/19

2/26/20

Concept: Find the effective interest rate per payment period

Effective interest rate for a 15-year cycle

i = (1 + 0.05)15 - 1 = 107.893%

Capitalized equivalent worth

P3 = $500,000/1.07893= $463,423

15 30 45 600

$500,000 $500,000 $500,000 $500,000

engineering economics lec5

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