Lecture 7B Comparison Methods PW Analysis PDF

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Lecture 7B Comparison Methods PW Analysis...

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ENGR301 Engineering Management Principles and Economics Comparison Methods: Present Worth Analysis

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Learning Objectives In this lecture learners will: •  Review the present worth analysis method of evaluating and comparing alternatives.

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Why Comparison Methods? •  Engineers play a major role in making decisions about investment opportunities. –  Engineers estimate the expected cost of and returns from an investment. –  Engineers must decide whether the expected returns outweigh the costs to see if the opportunity is potentially acceptable. –  Engineers may have to examine competing opportunities to see which one is the best. 3

Types of Comparison Methods •  Methods of evaluating and comparing projects are called comparison methods: –  Present worth method –  Annual worth method –  Internal rate of return –  Future worth method –  Payback period –  Benefit-cost ratio analysis 4

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General Assumptions 1.  Costs and benefits are always measureable in terms of money. –  In reality, costs and benefits may not be measureable in terms of money (e.g. providing safe working conditions has many benefits, including improving working morale!)

2.  Future cash flows are known with certainty –  In reality, future cash flows can only be estimated. –  The further into to future we try to forecast, the less certain our estimates become. 5

General Assumptions 3.  Cash flows are unaffected by inflation or deflation – 

In reality, the purchasing power of money typically declines over time.

4.  Sufficient funds are available to implement all projects – 

In reality, cash constraints on investments may be very important, especially for new enterprises with limited ability to raise capital.

5.  Taxes are not applicable – 

In reality, taxes are pervasive.

6.  All investments have a cash outflow at the start called first costs – 

In reality, some projects have cash inflows at the start, but involve a commitment of cash outflows at a later period. (e.g. a consulting engineering may receive a cash advance from the client). 6

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Relations Among Projects •  There are three types of connections among projects that cover all the possibilities. •  Projects may be: 1.  Independent 2.  Mutually exclusive, or 3.  Related but not mutually exclusive

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Relations Among Projects •  Independent Projects –  The simplest relation between projects occurs when they are independent. –  Two projects are independent if the expected costs and expected benefits of each project do not depend on whether the other one is chosen. –  If there are more than two projects under consideration, they are said to be independent if all possible pairs of projects in the set are independent. –  Evaluation is done by considering each opportunity one at a time, and accept or reject it on its own merit. 8

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Relations Among Projects •  Mutually Exclusive Projects –  In the process of choosing one, all other alternatives are excluded. –  Two projects are mutually exclusive if it is impossible to do both or it clearly would not make sense to do both. •  E.g. A student about to invest in a computer printer can get an inkjet printer or a laser jet printer. It would not make sense to get both.

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Relations Among Projects •  Projects that are related but not mutually exclusive –  The expected costs and benefits of one project depend on whether the other one is chosen. –  E.g. Shell may be considering a service station at Rue Mackay and Rue Du Fort. •  The cost and benefits of either service station will depend on whether the other is built. •  It may be possible, and make sense, to have both built 10

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Relations Among Projects •  Evaluation of related but not mutually exclusive projects can be simplified by combining them into exhaustive, mutually exclusive sets. –  E.g. the two projects being considered by shell can be put into four mutually exclusive sets. 1.  Neither station (the do nothing option) 2.  Just the station at Rue Mackay 3.  Just the station at Du Fort 4.  Both stations 11

Relations Among Projects •  In general, n related projects can be put into 2n sets, including the “do nothing option”. •  Once the related projects are put into mutually exclusive sets, the analyst treats these projects as alternatives.

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Minimum Acceptable Rate of Return (MARR) •  MARR is a lower limit for investment acceptability. •  MARR is an interest rate that must be earned for any project to be accepted. •  Projects that earn at least the MARR are desirable. –  This means that the money is earning at least as much that can be earned elsewhere. 13

Minimum Acceptable Rate of Return (MARR) •  The MARR can be viewed as the rate of return required to get investors to invest in a business. –  If a company accepts projects that earn less than the MARR, investors will not be willing to invest in the company.

•  The minimum return required to induce investors to invest in a company is the company’s cost of capital. 14

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Present Worth Method •  Project A and project B are compared by computing the net present worth of the two projects at the MARR. –  The preferred project is the one with the greater net present worth.

•  The value of a company can be considered to be the present worth of all its projects. –  Choosing projects with the greatest net present worth maximizes the value of the company. 15

Present Worth Method •  Net present worth (NPW) = PW of Benefits – PW of Costs •  A project is economically feasible if its: •  NPW ≥ 1 •  PW of costs is lower than PW of costs of alternative projects •  PW of benefits is greater than PW of benefits of alternative projects 16

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Present Worth Method: Rules of Economic Efficiency Input or Output

Situation

Rule

Fixed input

Amount of money or other input resources in fixed

Maximize present worth of benefits or other outputs

Fixed Output

There is a fixed task, Minimize present worth of benefit, or other output costs or other inputs to be accomplished

Neither input nor output is fixed

Typical general situation

Maximize (present worth of benefits minus present worth of costs), that is, maximize net present worth

Present worth analysis typically follows these rules 17

Present Worth Comparisons for Independent Projects •  The alternative to investing money on an independent project is to “do nothing”. –  Doing nothing means the money is used for some other project, earning interest at a rate at least equal to the MARR.

•  If an independent project has a net present worth greater than zero, it is acceptable. •  If an independent project has a net present worth less than zero, it is unacceptable. •  If an independent project has a net present worth of exactly zero, it is marginally acceptable. 18

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Present Worth Comparisons for Independent Projects •  A graduate student sees an opportunity in setting up an arcade business at the mall next to his university. The first cost for equipment, furniture, and software is $70,000. Net annual cash flow, after paying for labour, supplies, and other costs, is expected to be $30,000 a year for five years. When the student finishes his program in five years, he plans to close the business and expects the old equipment and computers to have zero value. If investors in this type of enterprise demand a return of 20% per year, is this a good investment? 19

Present Worth Comparisons for Independent Projects •  •  •  • 

The net present worth of the project is: NPW = -70,000 + 30,000 (P/A,20%,5) NPW = -70,000 + 30,000 (2.9906) NPW = 19,718

•  The project is acceptable, since the net present worth of about $20,000 is greater than zero. 20

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Present Worth Comparisons for Mutually Exclusive Projects •  Compute the present worth of each project using the MARR. –  The project with the greatest net present worth is the preferred project because it is the one with the greatest profit.

•  The service lives of the projects being assessed should be the same.

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Present Worth Comparisons for Mutually Exclusive Projects •  A manufacturing company is considering purchasing a new lathe. It is considering four lathes, each of which has a life of 10years with no scrap value. Given a MARR of 15%, which alternative should be taken? Lathe

1

2

3

4

First cost

$100,000

$150,000

$200,000

$255,000

Annual savings

$25,000

$34,000

$46,000

$55,000

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Present Worth Comparisons for Mutually Exclusive Projects •  NPW1 = -100,000 + 25,000(P/A,15%,10) = $25,468 •  NPW2 = -150,000 + 25,000(P/A,15%,10) = $20,636 •  NPW3 = -200,000 + 46,000(P/A,15%,10) = $30,860 •  NPW4 = -255,000 + 55,000(P/A,15%,10) = $21,029 23

Comparison of Alternatives with Unequal Lives •  When using PW comparisons, we must always use the same time period –  Take into account the full benefits and costs of each alternative.

•  We can transform into equal lives using one of the following methods 1.  Repeat the service life of each alternative to arrive at a common time period for all alternatives. Assume that each alternative can be repeated with the same costs and benefits in the future. Use the least common multiple of lives. 24

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Comparison of Alternatives with Unequal Lives 2.  Adopt a specified study period. To set an appropriate study period, a company will take into account the time of required service or the length of time it can be relatively certain of its forecasts. • 

An additional assumption about the salvage value has to be made.

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Comparison of Alternatives with Unequal Lives •  A mechanical engineer has decided to introduce automated materials-handling equipment for a production line. She must choose between two alternatives: building the equipment or buying the equipment off the shelf. Each alternative has a different service life and a different set of costs. •  Alternative 1:build custom materials-handling equipment –  –  –  –  –  – 

First cost Labour Power Maintenance Taxes and Insurance Service life

$15,000 $3,300 per year $400 per year $2,400 per year $300 per year 10 years 26

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Comparison of Alternatives with Unequal Lives •  Alternative 2: buy off-the-shelf standard automated materials-handling equipment –  First cost –  Labour –  Power –  Maintenance –  Taxes and Insurance –  Service life

$25,000 $1450 per year $600 per year $3075 per year $500 per year 15 years

•  If the MARR is 9%, which alternative is better? 27

Comparison of Alternatives with Unequal Lives •  Using the repeated lives method, each alternative is repeated enough times until there is a point in time when their service lives are simultaneously completed.

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Comparison of Alternatives with Unequal Lives Alternative 1 •  NPW (1)= -15,000 -15,000(P/F,9%,10) – 15,000(P/ F,9%,20) – (3300 + 400 + 2400 +300)(P/A,9%,30) •  NPW (1) = -$89,760 Alternative 2 •  NPW(2)= -25,000 – 25,000(P/F,9%,15) – (1450+600+3075+500)(P/A,9%,30) •  NPW(2) = -$89,649

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Comparison of Alternatives with Unequal Lives •  E.g., 10 year project life versus a 5 year project life •  Least common multiple is 10 years •  Reformulate 5 year life to 10 year life: –  including the cash flows in years 0 to 5 and also in years 5 to 10 –  the alternatives can now be compared

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Comparison of Alternatives with Unequal Lives •  A purchasing agent is considering the purchase of some new equipment for the mailroom. The following are the costs and useful lives of the equipment. Manufacturer

Cost

Useful Life (Years)

End-of-usefullife Salvage value

Speedy

$1500

5

$200

Allied

$1600

10

$325

•  Which manufacturer’s equipment should be selected? Assume 7% interest and equal maintenance costs. 31

Comparison of Alternatives with Unequal Lives

PW of cost = 1500 + (1500-200)(P/F, 7%,5) – 200(P/F, 7%, 10) = 1500 + 1300(0.7130) – 200(0.5083) = 1500 + 927-102 = $2,325 32

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Selecting an Arbitrary Analysis Period

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Selecting an Arbitrary Analysis Period • 

The LCM of lives would be 91 years. An analysis period of 91 years for office equipment is hardly realistic.

• 

Instead, a suitable analysis period should be based on how the equipment is likely to be needed.

• 

This may require that terminal values be estimated for the alternatives at some point before the end of their useful lives. –  Terminal value is the market value of the equipment at some point in time. It is the residual value of the equipment, similar to the salvage value

• 

Here we use an analysis period of 10 years. Alternative 1 is replaced at year 7 by an identical piece of equipment. At year 10, the terminal values of both alternatives are determined.

• 

The present worth of their costs are then determined by applying the appropriate single payment present worth factor

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Infinite Analysis Period: Capitalized Cost •  The need for large-scale infrastructure projects (bridges, pipelines, etc.) is considered to be permanent. •  These types of projects are considered to have an infinite analysis period. •  We call this particular analysis capitalized cost •  Capitalized cost is the present sum that is required to provide the service indefinitely. 35

Infinite Analysis Period: Capitalized Cost •  Capitalized cost is the present sum of money that would need to be set aside now, at some interest rate, to yield the funds required for the service or whatever indefinitely. •  To achieve this, the money set aside for future expenditures must not decline. –  The interest received on the money set aside can be spent, but not the principal.

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Capitalized Cost, cont’d. •  There can be an end-of-period withdrawal of A which is equal to P(i): –  These withdrawals will never decrease the original principal. –  A = Pi for n = infinity –  Therefore: –  Capitalized Cost = P = A/i •  The money set-aside that can provide the funds for the project forever. 37

Capitalized Cost (Example) •  How much should one set aside to pay $50 per year for maintenance on a gravesite if interest is assumed to be 4%? For perpetual maintenance, the principal sum must remain undiminished after the annual disbursement. Capitalized Cost P= Annual Disbursement A/ Interest Rate i P = 50/0.04 = $1,250

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