U i

The three most commonly used methods in life cycle costing are the annual cost, present worth and rate-of-return analysis.

In the present worth method a minimum rate of return (i) is stipulated. Ail future expenditures are converted to present values using the interest factors. The alternative with lowest effective first cost is the most desirable.

A similar procedure is implemented in the annual cost method. The difference is that the first cost is converted to an annual expenditure. The alternative with lowest effective annual cost is the most desirable.

In the rate-of-return method, a trial-and-error procedure is usually required. Interpolation from the interest tables can determine what rate of return (i) will give an interest factor which will make the overall cash flow balance. The rate-of-return analysis gives a good indication of the overall ranking of independent alternates.

The effect of escalation in fuel costs can influence greatly the final decision. When an annual cost grows at asteady rate it may be treated as a gradient and the gradient present worth factor can be used.

Special thanks are given to Rudolph R. Yanuck and Dr. Robert Brown for the use of their specially designed interest and escalation tables used in this text.

When life cycle costing is used to compare several alternatives the differences between costs are important. For example, if one alternate forces additional maintenance or an operating expense to occur, then these factors as well as energy costs need to be included. Remember, what was previously spent for the item to be replaced is irrelevant. The only factor to be considered is whether the new cost can be justified based on projected savings over its useful life.

Figure 9-7. Gradient present worth (GPW).

Figure 9-7. Gradient present worth (GPW).


Throughout the text you will experience job situations and problems. Each simulation experience is denoted by SIM. The answer will be given below the problem. Cover the answers, then you can "play the game."

Problem 9-1

An evaluation needs to be made to replace all 40-watt fluorescent lamps with a new lamp that saves 12 percent or 4.8 watts and gives the same output. The cost of each lamp is $2.80.

Assuming a rate of return before taxes of 25 percent is required, can the immediate replacement be justified? Hours of operation are 5800 and the lamp life is two years. Electricity costs 7.5^/KWH.



From Table 9-5 a rate of return of 25 percent is obtained. When analyzing energy conservation measures, never look at what was previously spent or the life remaining. Just determine if the new expenditure will pay for itself.

Problem 9-2

An electrical study indicates electrical motor consumption is 4 X 106 KWH per year. By upgrading the motor spares with high efficiency motors a 10% savings can be realized. The additional cost for these motors is estimated at $80,000. Assuming an per KWH energy charge and 20-year life, is the expenditure justified based on a minimum rate of return of 20% before taxes? Solve the problem using the present worth, annual cost, and rate-of-return methods.


Present Worth Method

Alternate 1 Present Method

Alternate 2

Use High Efficiency Motor Spares

USPW (Table 9-4) (2) R X USPW = Present Worth (D + <2>



$1,402,560 ,$1,482,560 -Choose Alternate with

Lowest First Cost

Annual Cost Method Alternate 1


Alternate 2 $80.000 $288,000 .2



Alternate with Lowest First Cost

Rate of Return Method

80,000 USPW=l2^0b-2-5

What value of i will make USPW = 2.57 i = 40% (Table 7, App. I).

Problem 9-3

Show the effect of 10 percent escalation on the rate of return analysis given the Energy equipment investment = $20,000

After tax savings Equipment life(n)


Without escalation:

= 15 years


P 20.000

From Table 9-1, the rate of return is 10 percent, With 10 percent escalation assumed:

From Table 9-10, the rate of return is 15.6%.

Thus we see that taking into account a modest escalation rate can dramatically affect the justification of the project.

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Project Management Made Easy

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