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CE 231 – ENGINEERING ECONOMY

REPLACEMENT AND ECONOMIC LIFE LECTURE NOTES

REPLACEMENT

The problem of replacement is different from the problem of simple alternative comparison

because it involves the use of special concepts such as sunk cost, economic life, and unused

value.

Usually, the duration and magnitude of cash flows for existing assets and new assets are quite

different. New assets characteristically have high capital costs and low operating costs.

Whereas, assets which are being considered for replacement have small capital costs and high

(and increasing) operating costs. In addition to these, it has a short life left and its future can

easily and accurately be estimated.

There are two basic reasons for replacement:

1. Physical impairment, and

2. Obsolescence

Physical impairment refers only to changes in the physical condition of the asset itself. This

may lead to a decrease in the value of the service rendered (increased operating cost, increased

maintenance cost, declining efficiency).

Obsolescence takes place because of the effects of changes in the environment external to the

asset. This happens either because the asset is not adequate any more for the type of service

one requires or as a result of the continuous improvement of the tools of production.

Normally, it is quite difficult for ordinary people to replace an asset when it is most economical

rather than when it is worn out. On top of that logic, there is also some sentimental link between

the machine and its user. It is important however to be able to replace a piece of machinery

which has reached its optimum economic life.

Another difficulty in replacement is the question of the general resistance to novelty. It is

always easy to continue with the existing system, but difficult to make a change. It is therefore

of paramount importance to create a need and then an enthusiasm for the novelty before

introducing it.

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Replacement Analysis for Equal Lives

Three methods of analysis will be considered in this section.

1. The “Outsider Viewpoint” Method:

In this method, only the costs and incomes incurred in the future for both the replacement and

the existing asset are considered. In other words, sunk costs are not considered. To put it more

clearly, it is assumed that there is an outsider who wants to select between the replacement and

the existing asset. Since he is an outsider, he is not involved with what happened to the existing

asset up to now. All he knows is that there is a replacement with an initial cost, a salvage value,

operating and maintenance costs; and there is an existing asset which has an initial cost (which

is the salvage value for that year), a salvage value, operating and maintenance costs. Analysis

is carried out as usual by using present worth, annual equivalent, or rate of return methods.

Example:

Existing Machine Replacement

Initial Cost 2.200 TL, 4 years ago 2.400 TL

Useful Life 10 years 6 years

Salvage Value 600 TL after 4 years

200 TL at the end of its economic life 300 TL

Operating Costs 700 TL/yr 400 TL/yr

MARR 15%

Cash flows from the point of view of an “outsider”, look as follows:

Existing Machine

Replacement

P2 = 2.400 TL

A2 = 400 TL/yr

F2 = 300 TL

0 1 2 3 4 5 6 years

0 1 2 3 4 5 6 years

P1 = 600 TL Initial Cost = 2.200 TL

A1 = 700 TL/yr

F1 = 200 TL

0 4 5 6 7 8 9 10 years

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AE1(15%) = – 700 – 600 (A/P,15%,6) + 200 (A/F,15%,6)

AE1(15%) = – 700 – 600 (0,26424) + 200 (0,11424)

AE1(15%) = – 700 – 158,544 + 22,848

AE1(15%) = – 835,69 TL/yr

AE2(15%) = – 400 – 2.400 (A/P,15%,6) + 300 (A/F,15%,6)

AE2(15%) = – 400 – 2.400 (0,26424) + 300 (0,11424)

AE2(15%) = – 400 – 634,176 + 34,272

AE2(15%) = – 999,90 TL/yr

∴ Prefer existing machine.

Then, carry out economic analysis by present worth, annual equivalent, or rate of return

methods.

2. Comparative Use Value Method:

In this method, the value of the machine to be replaced is calculated, so that this results in an

annual cost equal to the annual cost of operation with the replacement.

Example: Same as before:

Let us denote the initial cost of the existing machine by “X”.

Existing Machine

Replacement

P2 = 2.400 TL

A2 = 400 TL/yr

F2 = 300 TL

0 1 2 3 4 5 6 years

0 1 2 3 4 5 6 years

X Initial Cost = 2.200 TL

A1 = 700 TL/yr

F1 = 200 TL

0 4 5 6 7 8 9 10 years

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Let us calculate the annual equivalents of each alternative and equate them to each other:

AEE.M. = AEREP.

– X (A/P,15%,6) – 700 + 200 (A/F,15%,6) = – 2.400 (A/P,15%,6) – 400 + 300 (A/F,15%,6)

Let P1 to be the current salvage value of the existing machine as in the previous example;

If X comes out to be larger than P1 = 600.000 TL, then it is seen that the existing machine is a

good bargain, so we shall have to keep it.

If X comes out to be smaller than P1 = 600.000 TL, then it is understood that the existing

machine worths less than its selling price, therefore we prefer, this time, the replacement.

– X (A/P,15%,6) – 700 + 200 (A/F,15%,6) = – 2.400 (A/P,15%,6) – 400 + 300 (A/F,15%,6)

– 0,26424 X – 700 + 200 (0,11424) = – 2.400 (0,26424) – 400 + 300 (0,11424)

– 0,26424 X – 700 + 22,848 = – 634,176 – 400 + 34,272

0,26424 X = 322,752

X = 322,752 / 0,26424

X = 1.221,435 > P1 = 600 TL

∴ We select existing machine.

Solution with present worth approach:

PWE.M. = PWREP.

– X – 700 (P/A,15%,6) + 200 (P/F,15%,6) = – 2.400 – 400 (P/A,15%,6) + 300 (P/F,15%,6)

– X – 700 (3,7845) + 200 (0,4323) = – 2.400 – 400 (3,7845) + 300 (0,4323)

– X – 2.649,15 + 86,46 = – 2.400 – 1.513,8 + 129,69

– X – 2.562,69 = – 3.784,11

X = 1.221,42 > P1 = 600 TL

∴ We select existing machine.

3. Receipts and Disbursements Method:

In this method it is accepted that, one either uses the existing machine with its operating cost;

or he sells the old machine and buys a new one. The income from the sale of the old machine

is therefore considered in the cash flow of the replacement.

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Example: Same as before:

Existing Machine

Replacement

AE1(15%) = – 700 + 200 (A/F,15%,6)

AE1(15%) = – 700 + 200 (0,11424)

AE1(15%) = – 700 + 22,848

AE1(15%) = – 677,15 TL/yr

AE2(15%) = – 400 – (2.400 – 600) (A/P,15%,6) + 300 (A/F,15%,6)

AE2(15%) = – 400 – 1.800 (0,26424) + 300 (0,11424)

AE2(15%) = – 400 – 475,632 + 34,272

AE2(15%) = – 841,36 TL/yr

∴ Select existing machine.

The “Outsider Viewpoint” method is generally preferred over this method, because:

1. It is cumbersome to use this method when there are many mutually exclusive

alternatives.

2. This method may lead to errors when lives are not equal.

0 1 2 3 4 5 6 years

P1 = 600 TL Initial Cost = 2.200 TL

A1 = 700 TL/yr

F1 = 200 TL

0 4 5 6 7 8 9 10 years

P2 = 2.400 TL

A2 = 400 TL/yr

F2 = 300 TL SV = 600 TL

0 1 2 3 4 5 6 years

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Replacement Analysis for Unequal Lives

Since most replacement decisions are considered with the replacement of old assets by new

ones, the economic alternatives being examined are seldom of equal duration. However, study

periods for all alternatives in a given comparison should be equal.

In order to do this sort of comparison, use the Common Multiple of Lives approach. This

approach is used when we do not know exactly what will replace the machine with shorter life.

Therefore we assume that it will repeat itself.

Example:

Existing Machine Replacement

Initial Cost 6.000 TL, 4 years ago 6.000 TL

Useful Life 8 years 6 years

Salvage Value

3.000 TL after 4 years

No salvage value at the end of its economic

life

1.000 TL

Operating Costs 800 TL/yr 300 TL/yr

Maintenance Cost 500 TL at the 6th year of its economic life -

MARR 10%

a. Using the “Outsider Viewpoint” Method, decide whether the company should replace

the existing equipment or not. Use Annual Equivalent Approach.

b. Using the Comparative Use Value Method, decide whether the company should replace

the existing equipment or not.

i. Use Annual Equivalent Approach.

ii. Use Present Worth Approach.

c. Using the Receipts and Disbursements Method, decide whether the company should

replace the existing equipment or not. Use Present Worth Approach.

a. “Outsider Viewpoint” Method and Annual Equivalent Approach

Existing Machine

0 1 2 3 4 years

P1 = 3.000 Initial Cost = 6.000 TL

A1 = 800 TL/yr

MC = 500 TL

0 4 5 6 7 8 years

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Replacement

AE1(10%) = – 800 – 3.000 (A/P,10%,4) – 500 (F/P,10%,2) (A/F,10%,4)

AE1(10%) = – 800 – 3.000 (0,31547) – 500 (1,2100) (0,21547)

AE1(10%) = – 800 – 946,41 – 130,36

AE1(10%) = – 1.876,77 TL/yr

AE2(10%) = – 300 – 6.000 (A/P,10%,6) + 1.000 (A/F,10%,6)

AE2(10%) = – 300 – 6.000 (0,22961) + 1.000 (0,12961)

AE2(10%) = – 300 – 1.377,66 + 129,61

AE2(10%) = – 1.548,05 TL/yr

∴ Replace the existing machine.

b. Comparative Use Value Method

Existing Machine

P2 = 6.000 TL

A2 = 300 TL/yr

F2 = 1.000 TL

0 1 2 3 4 5 6 years

0 1 2 3 4 years

X Initial Cost = 6.000 TL

A1 = 800 TL/yr

MC = 500 TL

0 4 5 6 7 8 years

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Replacement

i. Annual Equivalent Approach

AE1(10%) = – 800 – X (A/P,10%,4) – 500 (F/P,10%,2) (A/F,10%,4)

AE1(10%) = – 800 – X (0,31547) – 500 (1,2100) (0,21547)

AE1(10%) = – 800 – X (0,31547) – 130,36

AE1(10%) = – 930,36 – 0,31547 X

AE2(10%) = – 300 – 6.000 (A/P,10%,6) + 1.000 (A/F,10%,6)

AE2(10%) = – 300 – 6.000 (0,22961) + 1.000 (0,12961)

AE2(10%) = – 300 – 1.377,66 + 129,61

AE2(10%) = – 1.548,05 TL/yr

PW1= PW2

– 930,36 – 0,31547 X = – 1.548,05

X = 1.958 TL < 3.000 TL

∴ Replace the existing machine.

ii. Present Worth Approach

AE1(10%) = – 930,36 – 0,31547 X

PW1(10%) = (– 930,36 – 0,31547 X) (P/A,10%,12)

PW1(10%) = (– 930,36 – 0,31547 X) (6,8137)

PW1(10%) = – 6.339,19 – 2,15 X

AE2(10%) = – 1.548,05 TL/yr

PW2(10%) = – 1.548,05 (P/A,10%,12)

PW2(10%) = – 1.548,05 (6,8137)

P2 = 6.000 TL

A2 = 300 TL/yr

F2 = 1.000 TL

0 1 2 3 4 5 6 years

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PW2(10%) = – 10.547,94 TL

PW1= PW2

– 6.339,19 – 2,15 X = – 10.547,94

X = 1.957,56 TL < 3.000 TL

∴ Replace the existing machine.

c. Receipts and Disbursements Method and Present Worth Approach

Existing Machine

Replacement

AE1(10%) = – 800 – 500 (F/P,10%,2) (A/F,10%,4)

AE1(10%) = – 800 – 500 (1,2100) (0,21547)

AE1(10%) = – 800 – 130,36

AE1(10%) = – 930,36 TL/yr

PW1(10%) = – 930,36 (P/A,10%,12)

PW1(10%) = – 930,36 (6,8137)

PW1(10%) = – 6.339,19 TL

AE2(10%) = – 300 – (6.000 – 3.000) (A/P,10%,6) + 1.000 (A/F,10%,6)

P2 = 6.000 TL

A2 = 300 TL/yr

F2 = 1.000 TL SV = 3.000 TL

0 1 2 3 4 5 6 years

0 1 2 3 4 years

P1 = 3.000 Initial Cost = 6.000 TL

A1 = 800 TL/yr

MC = 500 TL

0 4 5 6 7 8 years

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AE2(10%) = – 300 – 3.000 (0,22961) + 1.000 (0,12961)

AE2(10%) = – 300 – 688,83 + 129,61

AE2(10%) = – 859,22 TL/yr

PW2(10%) = – 859,22 (P/A,10%,12)

PW2(10%) = – 859,22 (6,8137)

PW2(10%) = – 5.854,46 TL

∴ Replace the existing machine.

Replacement Analysis with More than One Alternative

In case of existence of more than one replacement alternative:

1. Select the best option between the replacement alternatives by using Annual Equivalent

or Present Worth Comparison.

2. Proceed with the replacement analysis between the selected replacement alternative and

the existing machine by using the stated replacement analysis method.

Example:

Existing Machine Replacement 1 Replacement 2

Initial Cost 10.000 TL, 4 years ago 8.000 TL 5.000 TL

Useful Life 10 years 6 years 4 years

Salvage Value

7.000 TL after 4 years

2.000 TL at the end of its

economic life

1.000 TL -

Operating Costs 800 TL/yr 500 TL/yr 600 TL/yr

Maintenance

Cost

500 TL at the 5th and 6th years

of its economic life - 100 TL/yr

MARR 10%

Using the “Outsider Viewpoint” Method, decide whether the company should replace the

present equipment or not. Use Present Worth Approach.

Selection of the Best Replacement Alternative:

Replacement 1

8.000 TL 500 TL/yr

1.000 TL

0 1 2 3 4 5 6 years

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Replacement 2

AE1(10%) = – 500 – 8.000 (A/P,10%,6) + 1.000 (A/F,10%,6)

AE1(10%) = – 500 – 8.000 (0,22961) + 1.000 (0,12961)

AE1(10%) = – 500 – 1.336,88 + 129,61

AE1(10%) = – 2.207,27 TL/yr

PW1(10%) = – 2.207,27 (P/A,10%,12)

PW1(10%) = – 2.207,27 (6,8137)

PW1(10%) = – 15.039,67 TL

AE2(10%) = – 700 – 5.000 (A/P,10%,4)

AE2(10%) = – 700 – 5.000 (0,31547)

AE2(10%) = – 700 – 1.577,35

AE2(10%) = – 2.277,35 TL/yr

PW2(10%) = – 2.277,35 (P/A,10%,12)

PW2(10%) = – 2.277,35 (6,8137)

PW2(10%) = – 15.517,18 TL

∴ Select Replacement 1 for comparison with the existing machine.

Comparison of Existing Machine and Selected Replacement:

Existing Machine

0 1 2 3 4 5 6 years

P1 = 7.000 TL

Initial Cost = 10.000

A1 = 800 TL/yr

F1 = 2.000 TL

MC = 500 TL/yr

0 4 5 6 7 8 9 10 years

5.000 TL

700 TL/yr

0 1 2 3 4 years

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Replacement 1

PWEM(10%) = – 7.000 – 800 (P/A,10%,6) – 500 (P/A,10%,2) + 2.000 (P/F,10%,6)

PWEM(10%) = – 7.000 – 800 (4,3553) – 500 (1,7355) + 2.000 (0,5645)

PWEM(10%) = – 7.000 – 3.484,24 – 867,75 + 1.129

PWEM(10%) = – 10.222,99 TL

PWR1(10%) = – 8.000 – 500 (P/A,10%,6) + 1.000 (P/F,10%,6)

PWR1(10%) = – 8.000 – 500 (4,3553) + 1.000 (0,5645)

PWR1(10%) = – 8.000 – 2.177,65 + 564,5

PWR1(10%) = – 9.613,15 TL

∴ Replace existing machine with Replacement 1.

ECONOMIC LIFE

Before we carry on with comparison methods for existing and replacement assets, it will be

beneficial to introduce the concept of economic life, called also the minimum cost life or the

optimum replacement interval.

The economic life of an asset is the time interval that minimizes the asset’s total annual

equivalent costs or maximizes its total annual equivalent incomes.

P2 = 8.000 TL

A2 = 500 TL/yr

F2 = 1.000 TL

0 1 2 3 4 5 6 years

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To calculate the economic life of an asset, it is sufficient to calculate the annual equivalents

of the costs (or income) for different service lives and compare them. The life that minimizes

costs is the economic life.

Economic Life Problem

A small compressor can be bought for 10.000 TL. The salvage value is assumed to be negligible

regardless of the replacement interval. Annual operating and maintenance costs are expected

to increase by 750 TL/yr with the first year’s cost anticipated to be 1.500 TL. Using a MARR

of 8%, determine the year at which replacement should take place.

Let n = 1

AE(8%) = – 10.000 (A/P,8%,1) – 1.500

= – 10.000 (1,08) – 1.500

= – 10.800 – 1.500

= – 12.300 TL/yr

Let n = 2

AE(8%) = – 10.000 (A/P,8%,2) – 1.500 – 750 (A/G,8%,2)

= – 10.000 (0,56077) – 1.500 – 750 (0,4808)

= – 5.607,70 – 1.500 – 360,6

= – 7.467,76 TL/yr

Let n = 3

AE(8%) = – 10.000 (A/P,8%,3) – 1.500 – 750 (A/G,8%,3)

= – 10.000 (0,38803) – 1.500 – 750 (0,9487)

= – 3.880,30 – 1.500 – 711,525

= – 6.091,825 TL/yr

Let n = 4

AE(8%) = – 10.000 (A/P,8%,4) – 1.500 – 750 (A/G,8%,4)

= – 10.000 (0,30192) – 1.500 – 750 (1,4040)

= – 3.019,20 – 1.500 – 1.053

1.500 TL

10.000 TL MARR = 8%

2.250 TL

3.000 TL 3.750 TL

0 1 2 3 4 n years

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= – 5.572,20 TL/yr

Let n = 5

AE(8%) = – 10.000 (A/P,8%,5) – 1.500 – 750 (A/G,8%,5)

= – 10.000 (0,25046) – 1.500 – 750 (1,8465)

= – 2.504,60 – 1.500 – 1.384,875

= – 5.389,475 TL/yr

Let n = 6

AE(8%) = – 10.000 (A/P,8%,6) – 1.500 – 750 (A/G,8%,6)

= – 10.000 (0,21632) – 1.500 – 750 (2,2763)

= – 2.163,20 – 1.500 – 1.707,225

= – 5.370,425 TL/yr

Let n = 7

AE(8%) = – 10.000 (A/P,8%,7) – 1.500 – 750 (A/G,8%,7)

= – 10.000 (0,19207) – 1.500 – 750 (2,6937)

= – 1.920,70 – 1.500 – 2.020,275

= – 5.440,975 TL/yr

We see that the annual equivalent of costs reaches a minimum when n = 6 years, and that it

starts climbing again starting on the 7th year. It is best therefore to replace the compressor at

the end of the 6th year.