Analysis of Alternatives, Selections, and Replacements
A decision is made between alternatives or the selection is based in this chapter on applying the economic indicators presented before in Chapters 6 and 7.
8.1 INTRODUCTION
This technique of analysis of alternatives and selection can assume many different aspects.
Typical example are: the choice among alternative processes proposed for enhanced oil recovery in oil fields, the choice among alternative methods of cooling process streams in gas plants or the choice among alternative designs of heat exchangers, wasteheat boilers, pumps, or any piece of equipment.
A SELECTED CASE (EXAMPLE 8.1)
An oil company is offered a lease of a group of oil wells in which primary production is getting to an end, and the major condition of this offer is to undertake a secondary recovery project (water injection) by the end of the 5th year. The capital investment of this project is estimated to be $650,000. In return, the revenue in the form of cash flow realized from this lease is as follows:
 • $50,000/year for the 1st 4 years
 • $100,000/year for the 1st 4 years from the 6th to 20th years
A comparison has to be made between the two alternatives: To invest or not to invest? In other words, should the project be accepted or not?
SOLUTION
Such a situation could be handled by using the annual cost/present worth economic approach as will be explained later.
Economic alternatives can be classified into two main categories: ^{• [1]}
• To choose among alternative assets or equipment doing the same job. where mutually exclusive choices are considered and decision is made by technical people. Mutually exclusive projects imply that when two alternatives are compared, one project or the other is selected (but not both).
Many cases of alternative analysis can be handled with the "differential technique” or finding the “rate of return on the extra investment”, for the difference between two alternate investments. The following methods are presented:
 1. Differential approach (Д approach) or return on extra investment (R.O.E.I.)
 2. Total equivalent annual cost (T.E.A.C.)/present value method
 3. Total capitalized method
To identify the problem at hand is of prime importance to identify the problem at hand to be either income expansion, or cost reduction:
 • Profit or income expansion: where revenues (cash flows) are generated, and maximization of the profit is required, or
 • Cost reduction: where no cash flows are given; instead expenses are known and reduction in costs is the criterion.
Replacement analysis, on the other hand, can be considered some sort of alternative analysis for investment tied up with an old asset versus an additional or (a replacement) investment. This situation is encountered to replace worn, inadequate or obsolete equipment, and physical assets.
8.2 DIFFERENTIAL APPROACH (A APPROACH),
OR RETURN ON EXTRA INVESTMENT (R.O.E.I.)
The differential approach is a concept, which could be applied for selection among alternatives for a group of equipment, plants, processes, or oilrelated venture projects. The principle of minimum capital investment is applied in this method in the following sense: For a set of alternatives needed for a given job and doing the same function, choose the minimum investment as the base one on base plan.
The differential approach to be used as a criterion for selecting alternatives is summarized by the following procedure: ^{[2]}
3. Calculate the rate of R.O.E.I. as follows:
• Check to see that the preferred choice should have an R.O.E.I. greater than a minimum value prescribed by management.
For alternatives involving small increments in capital investment, the best (most economical) alternative is arrived at by either graphic or analytical solutions.
The following solved examples illustrates these principles.
Example 8.2
In the alkanolamine sweetening process of natural gas, two types of coolers have been suggested for the amine solvent: type A and type B. Using the data given next (Table 8.1) recommend which alternative should be used if both types are acceptable technically. The minimum rate of return on money invested is 15% and the economic lifetime is 10 years for the coolers.
SOLUTION
Use straightline depreciation of 10% of C.l.
The problem is a costreduction type.
TABLE 8.1
Data for Example 8.2
Type A 
Type В 

Capital Investment (C.l.) 
10,000 
15,000 

n, years 
10 
10 

A. dep. =C.I./n 
1,000 
1,500 

A. oper. cost 
3,000 
1,500 

Total annual cost = A. dep. + A. oper. cost 
4,000 
3,000 

A. rate of return (given) 
0.15 
0.15 

Diff. in C.l. 
5,000 

Diff. in annual cost (saving) 
1,000 

Annual percentage saving = diff. in annual cost (saving)/diff. in C.l. 
20% 

Rate of return on the extra capital is greater than 15%. 

Therefore, Type В is recommended. 

Assuming that n for type В changes between 5 and 15 years. 

Plot the annual % saving versus n, as shown next in Figure 8.1. 
FIGURE 8.1 Change of A.% saving versus lifetime of type B.
Example 8.3
Insulation thickness is important for heat exchangers in the oil industry. One situation was encountered in the sulfur recovery plant from hydrogen sulfide gas (H,S) (which has to be removed from natural gas). A heat exchanger was designed and recommendation was made for four possible thicknesses of insulation. The costs and savings related to these cases are as follows. Which one is recommended for 15% minimum R.O.I.?
ILLUSTRATION OF SOLUTION IS GIVEN IN TABLE 8.2
For 15% minimum R.O.I., calculations indicate that all four proposals are acceptable, since they generate R.O.I. > 15%, each. Now, we can apply the differential approach as indicated above. However, let us use the graphic analysis technique, since the problem involves smallinvestment increments. Referring to Figure 8.2, the annual savings/C.I. curve is drawn as shown using the above data. As can be seen, by increasing the C.I., the annual savings are increased until we hit the optimum point, M, which represents the maximum savings. Then, by drawing
FIGURE 8.2 Graphical presentation of differential solution (Example 8.3).
TABLE 8.2
Data for Example 8.3
Parameter 
1 inch Insulation 
2 inch Insulation 
3 inch Insulation 4 
inch Insulation 
Cost of insulation ($) 
1,200 
1,600 
1,800 
1,870 
Savings (Btu/hr) 
300,000 
350,000 
370,000 
380,000 
Value of savings ($/yr)^{a} 
648 
756 
799 
821 
Annual depreciation cost ($/yr)^{b} 
120 
160 
180 
187 
Annual profit ($) 
528 
596 
619 
634 
R.O.I. 
44.0% 
37.3% 
34.4% 
33.9% 
" Based on $0.3 per million Btu of (he heat recovered and 300 working days per year. ^{b} Based on 10year lifetime.
our tangent line at P, we can achieve an R.O.E.I. of about 17% when using C.l. of nearly $1,600, or an insulation of 2inch thickness.
The R.O.E.I. or differential method has one big drawback if applied to alternatives with different economic lifetimes. This puts a constraint on using it for these situations, which can be handled by other methods to be discussed next. Figure 8.2 illustrates this case.
1 2 
3 
4 

1st comparing 1 to 2 
acceptable as a basis 17.0% 
— 
— 
2nd comparing 2 to 3 
— basis 
11.5% 
— 
3rd comparing 2 to 4 
— basis 
— 
14.1% 
Conclusion: Design 2 is recommended; it gives more profit than design 1.
While, R.O.E.I. is 17%, which is >15% (minimum).
Figures 8.3 and 8.4 are graphical plots to illustrate the results obtained in solving this example.
FIGURE 8.3 Change of R.O.I. savings versus types.
FIGURE 8.4 Comparison between types of pumps.
8.3 TOTAL EQUIVALENT ANNUAL COST (T.E.A.C.)/PRESENT VALUE METHOD
In this method, all costs incurred in buying, installing, operating, and maintaining an asset are put on the some clcitum: that is, on annual basis. Generally, the annual equivalent costs are brought to the present value for all alternatives.
Specifically, the T.E.A.C. is the sum of the annual cost of capital recovery (initial capital plus interest on it) and other annual operating costs. (Remember that depreciation costs cannot be included with the annual operating costs. They are taken care of in the cost of capital recovery.)
Where:
Example 8.4
Recommend which arrangement to select out of the following two cases, where energy saving is required by using higher capital investment:
A pump with control discharge valve (1) 
A pump with a variable speed drive (II) 

C.l. ($) 
13,000 
17,000 
Annual cost of energy for pumping ($) 
6,000 
2,800 
Annual maintenance costs ($) 
1,500 
3,000 
Lifetime (yr) 
10 
10 
Assume i = 10% and the salvage value is negligible.
TABLE 8.3
Data for Example 8.4
Capital Investment (C.l.) 
A Pump with Control Discharge Valve (1) 13,000 
A Pump with a Variable Speed Drive (II) 17,000 
Lifetime (yr) 
10 
10 
Annual maintenance costs ($) 
1,500 
3,000 
Annual cost of energy for pumping ($) 
6,000 
2,800 
Ar 
2,116 
2,767 
T.E.A.C. 
9,616 
8,567 
Assume i = 10% and the salvage value is negligible. Which design is to be recommended?
System II is recommended, since T.E.A.C. is less than for system I. Figure 8.4 represents the results.
Figure 8.4 is a graphical plot to illustrate such results.
SOLUTION IS GIVEN IN TABLE 8.3
Example 8.5
Given: Assume the same two heat exchangers given in Example 7.2 with the same annual costs, economic lives, salvage values, and investments, and with the cost of capital once again 8%.
Wanted. Compare the two alternatives using the present worth values for each of the possibilities, as well as total equivalent capital at the "present" time of consideration of purchase of heat exchangers.
SOLUTION
Using the present value method, a series of known uniform annual costs are reduced to an equivalent present value. This allows one to estimate the dollar value at the present time that is equivalent to the amount of annual costs for some fixed years of service by two alternatives. But uniform annual costs must first be determined, and this is what the present value method does. (The annual cost method does not determine uniform annual costs.)
TABLE 8.4
Data for Example 8.6
Purchase Possibility A 
Purchase Possibility В 

1. Present value of original (initial) costs 
$15,000 
$40,000 
2. Present value of salvage value; formula "Find P, Given F", or factor 
$500x0,4632 =232 
$1,000x0.4632 = 463 
3. Present value of annual costs: (total costs) x factor of formula "Find P, Given A" 
$12,100 x 6.710 = $81.191 
$6.800x6.710 = 545.628 
4. (1)  (2) + (3) 
$95,959 
$85,165 
Now, for each of the possibilities, the present values of installations and the salvage values must be added and deducted, respectively, to current value of annual costs for 10 years in order to get total equivalent capital requirements.
The following calculations are carried out to find the the equivalent capital at 8% Table 8.4 illustrates the solution.
A comparison of the calculations for equivalent capital involved, $95,959 for possibility A and $85,165 for possibility В for the present time on an economic basis, indicates $10,794 less favoring possibility B. In other words, a savings of $1,610 in annual costs, as given by the annual cost method, favoring possibility В is reflected in a $10,794 reduction in equivalent present value of possibility В when annual costs are uniform and determined with the use Total "present" equivalent capital at 8%: of the present value method.
Under conditions of low interest rates, the present or current value of possibility B, which is $40,000, can still be less than the current worth of $15,000 for possibility A. Thus, the interest rate is important in order to determine the present value. Lower interest rates, such as say 5% instead of 8%, favor even more the use of higher initial investments, in this case the $40,000 stainless steel heat exchanger rather than the $15,000 steelcopper exchanger, because the relative cost for the use of money is lower. Example 8.7 confirms these results. A summary of the solution using excel is given next, followed by a graphical chart as shown in Figure 8.5.
Example 8.6: (summary)
FIGURE 8.5 Change of total annual costs for different types of H.E.
Example 8.7
Compare the relative annual costs and current present values of the two alternatives in Examples 8.5 and 8.6 for 10 years of service if money is worth 5% instead of 8%.
SOLUTION
a. For the annual cost method at 5%:
Purchase Possibility A 
Purchase Possibility В 

Annual costs: 

Capital recovery = $14,500 x 0.1295 ("Find A, Given P") + (0.0S)($500) = 
$1,903 
$39,000x0.1295 + (0.05)($ 1,000) = 5,101 
Capital recovery = $14,500 x 0.1295 ("Find A, Given P") + (0.0S)($500) = 
$1,903 
$39,000x0.1295 + (0.05)($ 1,000) = 5,101 
Labor, maintenance, etc. 
11,500 
4,000 
Other direct costs 
600 
2.800 
Total annual costs 
$13,903 
$11,901 
Compared to the costs when the interest rate is 8% (see Example 7.2, total annual costs for each possibility are lower when the interest rate is 5%. But the difference in annual costs is greater when the interest rate is lower. At 8%, the difference is $1,610 less in favor of possibility B, whereas at 5% it is $2,002 in favor of possibility B. Thus, lower costs of borrowing favor alternatives with large investment amounts more than alternatives with lower investment amounts.
b. For the present value (present worth) at 5%:
Purchase Possibility A 
Purchase Possibility В 

Present worth of original costs (initial) 
$15,000 
$40,000 
Present worth of salvage value ("Find P, Given F") 
$500x0.6139 = 307 
$1,000x0.6139 = 614 
Present worth of annual costs ("Find P, Given A") 
$12,100x7.722 = 93,436 
$6,800 x 7.722 = 52,509 
Total "present" equivalent capital at 5% 
$108,129 
$91,895 
At 5%, the equivalent capital for purchase possibility В is $16,234 ($108,129  $91,895) less than purchase possibility A. With lower interest rates, differences in equivalent capital are greater: $16,234 between alternatives at 5% and $10,794 between alternates at 8%. However, total present equivalent capital amounts are greater with lower interest rates: Totals at 5% are $108,129 and $91,895 for possibilities A and B, respectively, and at 8% are $95,959 and $85,165.
A comparison of these results and those obtained when money is worth 8% shows that (a) the timemoney series is equivalent to larger capital requirements, and (b) the difference in equivalent present value is greater in favor of purchase possibility В than it is for A when money is worth only 5%.
Example 8.8
The overhead condenser in a stabilization unit of a natural gasoline plant has to be made of corrosionresistant material. Two types are offered; both have the same capacity (surface areaj; however, the costs are different because of different alloying materials:
Condenser A 
Condenser В 

C.I. ($) 
23,000 
39,000 
n (years) 
4 
7 
If money can be invested at 8%, which condenser would you recommend based on the T.C.C.?
Therefore, condenser type A is selected (lower K).
Example 8.9
Solve Example 8.5 using the capitalized cost technique for 8% and 5% annual interest rates.
SOLUTION
Two methods are presented:
1. Using the relationship givenbefore, (direct application): a. For i = 8%:
Purchase Possibility A 
Purchase Possibility В 

n (year) 
10 
10 
C_{R}($) 
14,500 
39,000 
V_{s}($) 
500 
1,000 
Total operating cost ($/yr) 
12,100 
6,800 
К_{А}($) = 500 +14,500(1.8629) + 
K„($) = 1000 + 39,000(1.8629) + 
12,100 
6800 
0.08 
0.08 
= $178,762 
= $158,653 
b. For i = 5%: 

К_{л} 500 + 14,500(2.59)+ ^{12,100 }0.05 = $280,055 
К,, 100 + 39,000(2.59)+ ^{6800 }0.05 = $238,010 
2. Using stepbystep procedure (detailed):
a. Capital requirements through capitalization, with interest at 8%, are as follows:
Purchase Possibility A 
Purchase Possibility В 

Total annual costs: 

Net capital invested factor of formula ("Find A, Given F") 
$14,500 x 0.06903 = $1,001 
$39,000 x 0.06903 = $2,692 
Annual labor, maintenance, operational costs, etc. 
11,500 
4,000 
Other direct annual costs 
600 
2.800 
Total annual costs to be capitalized 
$13,101 
9,492 
Capitalization of annual costs 
$13,101/0.08 = 163,763 
$9,492/0.08 = 118,650 
Initial costs of annual costs 
15,000 
40,000 
Total capitalized cost when money is worth 8% 
$178,763 
$158,650 
b. Capital requirements through capitalization, with interest at 5%, are as follows:
Purchase Possibility A 
Purchase Possibility В 

Total annual costs: 

Net capital invested factor of formula ("Find A, Given F") 
$14,500x0.0795 = $1,153 
$39,000 x 0.0795 = $3,100 
Annual labor, maintenance, operational costs, etc. 
1,500 
4,000 
Other direct annual costs 
600 
2,800 
Total annual cost to be capitalized 
$13,253 
$9,900 
Capitalization of annual costs 
$13,253/0.05 = 265,060 
$9,900/0.05 = 198,000 
Initial costs of investment 
15,000 
40,000 
Total capitalized cost when money is worth 5% 
280,060 
$238,000 
It is clear that both the direct and detailed methods give the same final answer; however, one would be reluctant to use the latter approach.
At the lower interest rate of 5%, the capitalized cost is $42,060 less for possibility В ($280,060  $238,000]. The results illustrate the peculiar effect of the interest rate and emphasize the potential difficulties in comparing alternates on either a present value or a capitalized cost basis. When cost of capital is high, total capitalized costs become lower, but differences between capitalized costs of higher and lower investment amounts favor higher investments more when cost of capital (interest rate) is lower.
The interest rate is the determining factor, although the relative size of such individual items as initial costs, annual labor costs, annual material, repairs, maintenance, and other costs, when compared to capital recovery costs, can affect total equivalent capital involved.
The important point is that the interest based on the going value of money is always lower than the rate for a venture involving a risk. The engineer using the going rate for interest will bias his comparisons in favor of the alternative equivalent to oil capital requirements. Because of this, the annual cost method is preferred, but the service lives of the alternatives should be equal and annual costs of alternatives should be uniform. When different service lives are involved, or where nonuniform annual expenditures must be compared for alternatives, it is better to use the present value method and put all costs on a comparable basis in order to get accurate results and avoid "distortions" of costs.
8.4 REPLACEMENT ANALYSIS
In the oil industry, the usual experience is that assets are retired while they are still physically capable of continuing to render their service either in the oil field, in transportation systems or in the refining operations. The question is: how can we make the decision to replace an asset?
The decision to make such replacement should generally be made on the grounds of economy along with engineering fundamentals applicable to oil operations. That is, replacing a worn, obsolete, or inadequate asset can be translated into the language of economics.
Reasons behind a replacement can be defined either as “a must”, that is, we have to replace, otherwise the operation will come to a halt, or “optional” in w'hich case the asset is functioning, but there is a need for a more efficient or modern type. Such a classification is illustrated in Figure 8.6.
The principles governing replacement are best explained by using the word “defender” to stand for the old asset and the word “challenger” to identify the possible new candidate that will make the replacement. In order to utilize the challenger/ defender analogy for replacement comparison, the following factors must be considered (ValleRiestra, 1983):
All input/output of cash flows associated with the asset have to be known or estimated. This applies in particular to maintenance and operating costs of both defender and challenger.
Cost estimation of the value of the defender (market value/book value) must be made.
FIGURE 8.6 Replacement analysis.
Methods recommended earlier for the comparison of alternatives such as T.E. A.C., present worth or Д approach could be applied. In other words, no new' techniques are provided. Tax obligations or credits should be considered.
Example 8.10
A tank farm is receiving crude oil through a pipeline. Periodic measurements of the crude oil level are made. The annual labor cost for the manual operation is estimated to be $50,000. However, if an automated levelmeasuring system is installed, it will cost $150,000. Maintenance and operating expenses of the system are $15,000 and $5,000, respectively. The system will be operated for 5 years.
Should the automated levelmeasuring system be installed? Assume interest rate is 10%.
SOLUTION
Two alternatives must be compared:
Alternative 1: Manual operation
Alternative 2: Automated levelmeasuring system The manual operation, alternative 1, is less expensive.
Example 8.11
An oil company has an existing steamgeneration unit. Its cost when new is $30,000, its lifetime is 10 years and it has a salvage value of zero. The annual operating cost is $22,000. After it has been in use for 5 years, the estimated book value of the unit is found to be $6,000. The remaining lifetime now is only 3 years.
It has been proposed to replace this unit by another new one. Its cost is $40,000, lifetime 10 years, operating costs $15,000/yr and zero salvage value. $hould we continue using this unit or go for the replacement?
The company requires 10% R.O.I.
SOLUTION
Old Unit 
Replacement 
V„($): 30,000 
$40,000 
n: 10 years 
n: 10 years 
Operating: $22,000/yr After 5 years of use 
Operating: $15,000/yr 
V_{5} = $6,000 

3 years are left only 

V_{s} = o 
Now, take these 3 years for comparison:
d  ^{6000}  $_{2},000/yr 
_{d=}4°'^{000} _{=} 4,000 10 
Operating costs = $22,000/yr 
Operating costs = 15,000 
Total cost = $24,000/yr 
Total cost = $19,000 
Therefore, savings = 24,000  19,000 = $5,000/yr.
If replacement takes place, R.O.E.I. = (5,000/$34,000)100 = 14.7%.
Example 8.12
Consider a control valve that becomes obsolete 3 years before it has been fully depreciated. When fully depreciated, the valve will have a salvage value of $400, but at this time (3 years before], it has a tradein (or resale) value of $1,000. If the book value (original cost  total depreciation to date) is $760, there is a favorable "bonus" to management of $240 in tradein.
But the bonus of $240 is irrelevant as a sunk cost. If a minimum rate of return is assumed as 10% before taxes, the question is whether the obsolete control valve with 3 years to go before being fully depreciated should be replaced now by a new valve. Calculations are needed to compare the old valve with a new valve, which would cost $5,000 and have an eventual salvage value of $500 and a service life of 10 years.
SOLUTION
Annual cost of old valve: Capital recovery costs (760)(0.40211)10% for 3 years + 0.10 x $400 
= $346.00 
Operating and maintenance costs (estimated) 
= $1,870.00 
Total annual cost of old valve 
= $2,166.00 
Then, tentative annual cost of new valve: Capital recovery costs ($4,500)(0.16275) 10% for 10 years + 0.10 x $500 
782.00 
Operating and maintenance cost (estimated) 
= $1,000.00 
Total annual cost of new valve 
= $1,782.00 
By comparing the old control valve with the new valve, we can see that purchasing the new valve now would mean an annual savings of $384 or ($2,166  $1,782). If the old valve is depreciated out, only the salvage value of $400 could be allowed on capital recovery.
 [1] To choose among different ways to invest money not necessarily toaccomplish the same job, in which case the decision is influenced bymanagement rather than by technical people;
 [2] Select the minimum capital investment (C.I.) as our base plan, computeДС.1. (difference in capital investment) for the alternatives. 2. Compute Aprofit (difference in cash income) for the alternatives, for theincomeexpansion problem, and Asaving (difference in annual costs) forthe alternatives, for the costreduction problem.