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Here the process design engineer must rely on intuitive judgment or rules of thumb. As a result, his design may not always be the most economic, and a...
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Given the anticipated growth in product. demnd, how much plant capacig should be installed now?

he design of chemical plants has been predicated

Ton a known and constant production rate over the life of the plant. The demand, however, for many chemical products such as urethane foams, trialkylaluminum catalysts, and Freons exhibits strong growth. Here the process design engineer must rely on intuitive judgment or rules of thumb. As a result, his design may not always be the most economic, and a central plant-design question remains to be answered: Given an anticipated growth in demand, how much plant capacity should be installed now? Despite its importance, this problem has received little attention. The authors of a recent article (3) have followed an empirical approach based on a six-year demand forecast with four possible courses of action. They offer a nomographical method for calculating total costs for various decisions and outcomes. Another article (4) presents a considerably more quantitative treatment, particularly with respect to the uncertainty of equipment performance. The model, however, is somewhat limited because it does not consider explicitly future expansions in capacity. I n this article, a mathematical model based on the criterion of present value (7, Z ) , is formulated and solved for some average assumptions. Market refers to total consumption of the product, and demand is that share of the market which isheld or expected to be held by a single producer. For a monopolist, the terms are synonymous.

Market Growth and Plant Expansion

Experience with growing markets shows that their growth pattern is characterized by four distinct phases which can be described by the classical S-shaped Gomperz curve :


incubation period where the market increases gradually growth period where the market increases ra@idly at an exponential rate -A leveling-out period where the market stabilizes -A period of decline


The incubation period is relatively short-about two to five years. T h e growth period extends five to 20 years, depending on the rate of customer acceptance and substitution. T h e leveling-off period is five to 30 years and, depending on the degree to which the product is vulnerable to substitution by other, new products. The period of decline may run its course rather quickly, say in two years. Or, it may never occur at all. Basic raw materials such as sulfur or ammonia which have widely diverse uses will probably never reach the decline period. Rather they may grow with the over-all economy. This growth pattern bears a definite relationship to the evolution of the chemical plant. For a plant which is to produce a new product, the incubation period is when the chemical is being produced in pilot-plant quantities for customer testing and evaluation. Based on these results, some estimate of demand can be made for design and construction of a full-scale. Then the growth period begins as the product finds increasing applications and a plant expansion follows. As the market matures and/or competitors enter the field with new or similar processes, the demand level, as viewed by the company, begins to stabilize. Coincident with this leveling-off period plant expansion comes to a halt, and process equipment is replaced as it fails. This period is often characterized by falling prices and, as the old process becomes increasingly unprofitable, the plant is finally scrapped when a major piece of equipment requires replacement. For plant design, we are primarily concerned with the demand pattern during the growth and leveling-off periods. We shall describe all such demand patterns by means of two parameters: a n estimate of the rate of growth, and an estimate of the ultimate demand level. These parameters yield a growth curve which is identical to the classical S-shaped curve, except in the transition region between growth and leveling-off (Figure 1). Where the classical curve is smooth, the approximation is discontinuous. The duration of the growth phase, from a n individual company’s standpoint, may be as long as the complete market growth period or much shorter if, for instance, the rate of entry by other producers is high. O n the average, we may expect it to be less than the average economic life of most types of chemical process equipment which is about 15 years. This implies that plants are more likely to be expanded by parallel additions of equipment rather than by replacement of old equipment with a larger-sized unit.

Consequently, after the growth period, most equipment is replaced because of failure. And, as previously mentioned, all essential equipment will continue to be replaced until continued operation can no longer justify the required replacement expenditure. I n practice, this situation is likely to occur eventually, but for the model described here, the contrary is assumed-i.e., that replacement continues forever. This assumption may be justified for the following reason. The combined duration of the growth and stability periods can be 10 to 50 years, or longer if the product is a basic raw material or intermediate. An average production life, perhaps, is approximately 30 years. For a cost of capital of lo%, which is typical, $100 received 30 years hence has a present value of only $5.70. Or, alternatively, although we would be willing to pay up to $942.46 now for a n investment that would yield $100 per year for the next 30 years, we would be willing to pay only 6% more, or $1000, if the investment were to yield $100 per year forever. I n short, so long as we are dealing with a relatively long production life, say greater than 20 years, assumptions about termination of the plant and equipment are not very important. Thus, the “replacement forever” assumption yields as accurate results as any other arbitrary assumption of plant life, and it avoids the difficulties of making a n explicit prediction of the cash flows occurring at the time when the plant is scrapped or sold. Process Equipment and Economies of Scale

That the installed costs of process equipment exhibit economies of scale is certainly one of the most basic principles of plant design. Over a wide range of equipment capacities, the following relation gives a reasonable fit to the cost data (6): Installed cost = C’k”. T h e scale-up factor is, of course, specific to the type of equipment; it ranges between 0.2 and 1.0 with a n average value of about 0.7. This factor provides a measure of the economies of scale for each plant component-i.e., the lower the factor, the greater the economies of scale. Herein lies the basic problem of designing a plant for a growing demand: If each plant component is designed now to meet the ultimate demand level, it is possible to take complete advantage of the inherent economies of scale. O n the other hand, the capital invested in this over-capacity is idle ; it will not return profits until some time in the future. Just how much overcapacity should be installed, then, depends on the value of the scale-up factor as opposed to the alternative use for the capital. Clearly, if the scale-up factor were zero-i.e., the equipment cost is independent of the capacity, it would always pay to install the ultimate capacity now. At the other extreme (a = l.O), where any capacity level is attained by some multiple of a basic unit, such as a n electrolytic chlorine cell, then there is no incentive for building any overcapacity. We simply install more units as they are required. Since the scale-up factor for most types of process equipment falls between these limits, some compromise and, hence, a n optimum exists. VOL. 5 6

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Formulation and Solution of the Growth Model

We are given a type of process equipment with installed cost C'k" and with annual fixed cash expenses equal to some fraction /? of the installed cost. The annual fixed cash expenses are those out-of-pocket costs which are independent of the rate of output-e.g., maintenance (to a large extent), taxes, and insurance. The forecast of demand is : D = D,e@for 0 2 0 2 L and D = D L for 8 > L Given these costs and demand parameters, the objective is to find the initial capacity and the size of the future capacity additions so that the present value of the costs of meeting this demand are minimum. We assume continuous discounting, economic life of the equipment equals the depreciable life, the salvage value equals the cost of removal, and replacement forever. If we assume that the plant capacity is always greater or equal to the product demand, and that variable costs are strictly variable-Le., the equipment is equally efficient over all capacities and all rates of outputthe quantity (revenue-variable costs) depends only on demand and is independent of the choice of capacities. The only economic variables which will depend upon the capacity decision, are the installed cost and the fixed annual cash expenses. That is, regardless of whether the problem is approached from a profit-maximizing or cost-minimizing view, the solution will be that choice of initial capacity and future capacity increments which minimizes the present value of the installed costs and fixed annual cash expenses associated with the equipment. The present value of the total costs ( P V T C ) incurred by the installation of k units of equipment capacity, which possesses an economic life of m years, may be computed from the estimated costs as follows:

tion made at time Tiin units/time; and ko size of initial installed capacity in units/time.

K,, =

k, i


If demand is always satisfied, then additional capacity must be added when demand is equal to total capacity. The time when the ith addition to capacity must be made is given by : 1 Ki-l T t -- -1n(3) g


Substitution of this relation in Equation 2 and imposition of the constraint that, at the end of the growth period the total capacity must equal the ultimate demand level, yields upon integration :

To minimize P V T C , we set the partial derivatives of Equation 4 with respect to each of the ki's equal to zero and solve the resulting equations simultaneously. The solution, for n additions to capacity, is : (5) Fori= 1 , 2 , , . . , n - 2 :

Finally, for i = n - 1:

or the present value of $1.OO received continuously over each year for m years, discounted at r% interest. r is cost of capital or minimum acceptable rate of return, fraction per time. The present value of the total costs incurred by the installation of n additions to the initial capacity, where k , is the capacity of the equipment installed at time Ti, and where each addition is replaced forever, may be expressed as :

where y = growth ratio = r / g . The solution to these equations-i.e., the kt's which satisfy Equations 5 through 7-does not yield the optimum solution directly, but rather the lowest cost obtainable for a given number of capacity additions. The global minimum, or the optimum, is determined by calculating (PVTC),i, in sequence for n = 0, 1, 2, . . . , using the minimum cost ki's for each n. This procedure is continued until n 1 additions yield a greater cost than n additions. The design of a n evaporation unit will illustrate this method : Suppose that when the plant goes on stream, the initial product demand rate is 5 million pounds per year which is expected to grow at a rate of 17.9% on a continuous basis and, ultimately, to level out at a rate of 30 million pounds per year. Design calculations show that this initial production rate requires 1000 square feet of evaporator heating surface, and that the required heating surface is proportional to the production rate. Or, at the ultimate de-

where T , is time of ith addition to the initial capacity installation, i = 1, 2, , , 1 : ; ki, size of capacity addi-

AUTHORS John R. Coleman, Jr., is a Research Engineer with the Marathon Oil Co., Denver Research Center, Littleton, Colo. Robert York i s a Professor in the School of Chemical Engineering at Cornell University, Ithaca, N . Y .

P V T C = C'k" [l - td;,


+ p(1 - t)A+]




where C = C'(1 - tdq7 p(1 - t ) A g r ); t , income tax rate (decimal fraction) ; d+, present value of the depreciation charges resulting from $1.OO of assets depreciated over m years at ry0interest.





mum solution represents a 20Oj, saving over the worst possible case considered-Le., install 6000 square feet now. I n practice, of course, the expected savings may be more or less than 20%, depending upon what would have been done had no calculations of this kind been made. I n an actual design problem, primary concern is with the optimum capacity to install now; because of the uncertainty of the demand forecast, the question of whether to make three or four additions later borders on the trivial. Because the optimum initial capacity depends upon only three variables, we can determine once and for all the optimum initial capacity installation for a range of demand cases. A chart or table of these capacities would then offer a convenient design reference. Optimum lnifial Capacity Chart


. _"-__ ~

X r lX lIl"


^ . p I y I x p I I I I I ^ y I I _




TIME Figure 7. The model is concerned primarily with the demand pattern during the growth and leveling-off periods

mand level of 30 million pounds per year, 6000 square feet of evaporator heating surface will be required. The installed cost of the evaporator, which is of the long tube vertical type with a cast iron body and copper tubes, can be estimated by ( 5 ) : $4600(k)0.sswhere k is heating surface in hundreds of square feet. Other costs, including maintenance, cleaning, taxes, and insurance, which are independent of the rate at which the evaporator is operated, are estimated at 5'% of the installed cost on a n annual basis. Other data include an estimated 20-year economic and depreciable life of the evaporator, a salvage value equal to expected costs of removal, a 52y0 tax rate, and a 10% cost of capital to the company. Given the above data, the problem is to determine size of evaporator to be installed now and size of future capacity additions. Equation 4 may be written for the above conditions as follows : i-1

(60 - K,- 1)0~~~(10/K,1)OJS9 (8) = 0.547 by the sum-of-the-years-digits

where dml method. Since three additions yield the lowest total cost, the optimum solution is to install 1900 square feet now, followed by additions of 1000, 1500, and 1600 square feet a t 3.6, 6.0, and 8.3 years, respectively. The opti-

For simplicity, the product demand rate when the plant goes on-stream is assumed as 1 0 0 ~ ocapacity and the optimum design capacity is computed as a percentage of this base. T h a t is, the ultimate demand level ( D L )is estimated as a percentage of the on-stream rate and the continuous rate of growth (9) as it applies to this base. Because this is a four-dimensional problem, we must illustrate by taking orthogonal "slices" of the optimum capacity surface. Two of the many possible charts are shown in Figures 2 and 3. Although these charts hold for only three values of the scale-up factor, these values are representative of large, average, and small economies of scale and they serve to outline the pattern. I n Figure 2, where the optimum initial capacity is given as a function of D Lfor a constant 7, the optimum capacity is identical to DL for low values of DL. As DL increases beyond a certain critical value which depends on the scale-up factor, the optimum capacity drops substantially and then increases gradually with further increases in DL. A similar discontinuity is exhibited by Figure 3, in which the optimum initial capacity is shown as a function of the reciprocal growth ratio (7-l) at constant DL. As expected, the optimum capacity is identical to DL for high values of 7-l; that is, high rates of growth, and for large economies of scale. As 7-l decreases, a critical value is encountered, after which the optimum capacity falls abruptly. And, as 7-l approaches zero, the optimum capacity approaches the on-stream demand rate. Although the optimum initial capacity curve exhibits this discontinuous property, the total cost curve (PVTC) is continuous. T h a t is, a t the critical value of D L or 7-l, we are indifferent between installing the higher or lower capacity given by the end points of the dashed lines in Figures 2 and 3. As the value of D L or y-' moves away from the critical value, we prefer the indicated optimum capacity roughly in proportion to the displacement. This discontinuous property of the optimum has its greatest importance in situations involving risk. Given any set of demand parameters, these charts VOL. 5 6

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The procedure can be modified when special problems are encountered can be used to size optimally each component of the plant according to its characteristic scale-up factor. The resulting plant is then an optimum; if the capacity of any component is altered, the present value of the cash outflows is increased. Although this is true when the underlying assumptions are valid, it is perhaps worthwhile to anticipate a few of the difficulties with this method. Here the largest question centers in accuracy of the demand parameter estimates. Modifications of tho R o c d u n

The problem of design dependency may or may not be bothersome. Design dependency means that the design of one piece of equipment directly affects the design of another. For example, in the design of a compressor house, we are required to lay at least as many foundations as the gumber of compressors, even though the optimum calculations may dictate otherwise. A more difficult problem of this kind is the optimum sizing of a distillation column. Although the column, reboiler, and condenser have peculiar values of the scale-up factor, these units cannot be optimally sized independently of the others. That is, for each rate of output, there exists an optimum reflux ratio which is a function of the cost of the column, the steam, and the cooling water costs. It is best to treat the collection of interdependent equipment items as one unit, which is characterized by a single, overall value of the scale-up factor. A second problem may be that the capacity is not a linear function of the demand rate; for example, sizing of a pump for a liquid product pipeline. Here, the capacity of the pump required not only depends on the throughput but also upon the pressure drop, which is a function of the pipe diameter and the throughput. Hence, for a given pump and pipe system operating at some rate of output, to double the output by the addition of a second pump, we must provide more than the original capacity. This type of problem is best handled by first translating the product demand curve into terms of the required equipment capacity. Although there is no guarantee that the translation will result in an exponential growth curve as we have assumed, some form of this curve is likely to be a close approximation. The other alternative is to solve the mathematical model with the translated demand curve. For certain types of equipment, efficiency may vary with capacity. Often this variation results in a relatively s m a l l error in the calculations, but occasionally the difference in efficiency is too large to neglect. This cost can be easily accounted for by taking the most efficient capacity as a base. We then add the present value of the incremental variable costs incurred by installing other capacities which are less efficient to the installed cost of these capacities. The subsequent evalu32


Ftgwe 2. Optimum inilinl ca&cz?y and d6imafrc h n d as a p” cent of on-~lruamdemand fm (I reciprocal of growth ratio of 7.5


F i p e 3. OprimMI imfiol cnpacify whnc dfimafu demand is 409% of on-strcnm demand












D LEstimate Evaporator Capaciw Addition, Sq. A. Initial 1 6000 2700 2100 1900 1800


3300 1450 1000 600



2450 1500 900

I 3rd I

1600 1200




PvTCmin.. $

180,000 151,000 146,000 144,000 147,000

ation of the optimum using the scale-up factor calculated on this basis will then include the effect of the efficiency variations. Also, the scale-up factor itself may vary with equipment capacity. This is particularly true in the upper range of equipment capacities, where the scale-up factor tends to approach unity. Here, it is most convenient to assume some average value to establish the approximate range of capacities under consideration. We may &hen re-calculate the optimum using the value of the scale-up factor which holds for this range. Implicit in the preceding development is the assumption that no increase in equipment capacity is achieved through technological advances or accumulated knowhow. Where these factors are believed to contribute to increases in equipment capacity without an associated capital outlay, the growth ratio should be computed as: y = r / ( g - p) where p is the estimated rate of technological advance, fraction per time. Finally, if inflation of equipment costs is expected to occur in the future, then this factor may be taken into account by subtracting the anticipated rate of inflation from the cost of capital before calculating the inverse growth ratio. T h a t is, if equipment costs are expected to grow uniformly at a rate b fraction per time (continuous basis), then y is computed by y = ( r - b ) / g . Uncertainty in the Demand Forecast

The problem of risk in relation to design arises because the demand curve parameters cannot be estimated exactly. Thus, the a priori optimum plant will turn out more or less profitable, depending upon the accuracy of these estimates. This is a n exceedingly complex problem; therefore, we shall attempt only to show how the optimization method might be used to outline the magnitude and character of the gamble involved, rather than try to state explicitly a n optimum. Perhaps the easiest way to approach this problem is by some simplified examples. We shall assume first that the proposed plant is to consist of only one process unit with installed cost of $100,000(k)0.6 where k is a capacity of 1 million pounds per year. Further, we assume that there are no annual fixed cash expenses, a process unit life of 25 years, a cost of capital of lo%, and a n on-stream product demand level of 4.0 million pounds per year. With respect to the demand parameters, we assume that the rate of growth is known fairly well, but that the ultimate size of the market is in question. This



DL,million lb./yr.


Optimum initial capacity, million Ib./yr.



12.0 252,000 285,000

PVTCzs,$ PVTC,, $




28.0 394,000 426,000


14.8 287,000 313,000

might be true, for instance, if another producer had recently begun development work on a competing product, but the outcome of this work was uncertain. I n particular, we suppose a firm estimate of the rate of growth is given as 15%, and the expected, or most probable, ultimate demand level is 28.0 million pounds per year. If the development effort on the competing product is unsuccessful, the ultimate demand level might be as high as 36.0 million pounds per year. If, however, the development is a n overwhelming "s*uccess, then the ultimate demand level could be as low as 12.0 million pounds per year. For the expected case, where y-l = 1.5 and DL = 28.0 million pounds per year, the optimum initial capacity, from Figure 2, is 28 million pounds. T h e present value of the total cash flows over the 25-year life is:


$100,000(28)0~6 [l - (0.52)dfiloI


= $

81,300(28)O.5 = $394,000

And, assuming replacement of the unit forever :




d0 = $426,000

For the optimistic case, where y-l = 1.5 and DL = 36 million pounds per year, the corresponding optimum initial capacity is 14.8 million. And, for the pessimistic case, where y-' = 1.5 and DL = 1 2 million, the optimum capacity is 12 million pounds per year. These capacities and their associated costs are given in Table 11. Assuming first that we build the optimum capacity for the expected D, estimate and the pessimistic estimate turns out in fact to be correct, a present value loss of $142,000 is incurred. This represents the difference between the cost incurred of $394,000 and the cost which would have been incurred had we designed for the pessimistic estimate, or $252,000. This loss assumes that a t the end of 25 years the oversized unit is replaced with a smaller one. O n the other hand, if the optimistic estimate of DL is correct, a loss is incurred also because our incremental capacity addition is not of the optimum size. O n a replacement-forever basis, this loss is computed as $25,000 : Install 28 million pounds per year now and 8 million later : PVTC, = $81,300[28°*6 8°*5(4/28)0*667]



$494,000 (Continued on next page)

VOL. 5 6

NO. 1




Install 14.8 pounds per year now and 21.2 million later:


= $

+ 21 .2°*5(4/14.8)O*667]

81,300 [14.80.5

Correct Estimate of

= $469,000


I n a similar manner the possible losses incurred by designing for the other estimates of DLcan be computed (Table 111). Following the minimax principle-Le., choose that course of action which minimizes the maximum losswe would design for the pessimistic estimate. Here the maximum possible loss, which would be incurred if the expected estimate proved to be correct, is $12,000. Curiously, the next best choice is to design for the optimistic estimate which has a maximum loss of only $35,000. Only if we were willing to bet a t least $130,000 that the pessimistic estimate would not turn out to be correct, and 320,000 that the optimistic estimate would be false, would we design for the expected D, estimate. T h e above situation, of course, is a consequence of the discontinuity in the optimum design capacity curve (Figure 2). Thus, the optimum capacity for the pessimistic estimate does not differ greatly from the optimum capacity for the optimistic estimate. I n other words, by sacrificing some of the economies of scale, we are able to hedge against an unfavorable demand outcome and, at the same time, assume a reasonably good position with respect to the most favorable outcome. However, large economies of scale are essential for this hedging. If, for instance, the process unit scale-up factor were 0.9, then the optimum capacity is relatively insensitive to varying estimates of DL. Hence, the lower the scale-up factor, the more we should be concerned with the possibility of hedging. As a second example, consider essentially the same problem, but with the rate of growth as the main unknown. I n particular, we suppose that the process unit scale-up factor is 0.7 and that the ultimate demand level has been established at 16 million pounds per year. TABLE 111. POSSIBLE LOSSES I F O P T I M U M CAPACITY I S INSTALLED FOR EACH ESTIMATE O F DL

Correct Estimate of Pessimistic

Pessimistic Expected Optimistic

0 12,000 5,000





Estimate of g, 70 Optimum initial capacity, million lb./yr. Later capacity addition, million lb./yr.



142,000 0 25,000

Estimate of g

PVTC,, $


35,000 10,000 0






Pessimistic Expected Optimistic

Loss, 0

Pessimislic 0 11,000 47,000





16,000 0 35,000

Obtimistic 133,000 8,000 0

The major question is now the rate of product acceptance. T h e three estimates of g, the optimum initial and later capacity additions, and the total costs are shown by Table IV. The corresponding possible losses are given in Table V. By the minimax principle we should choose to design for the expected estimate of g. If, however, the chances were remote that the pessimistic forecast would prove correct, then we might be tempted to design for the optimistic estimate and risk an $8000 loss in order to avoid the $35,000 loss incurred if the optimistic estimate were to prove correct. In a n actual design, to be sure, the problem is much more complicated. Ordinarily, there exists some degree of uncertainty in both the rate of growth and the ultimate demand level. And, too, there are many more possible outcomes than pessimistic, expected, and optimistic. I n addition, the plant will consist of many components, each with a particular value of the scale-up factor as well as a share of the difficulties mentioned previously. I n spite of these shortcomings, the method does offer a means of achieving a more optimum design than is generally possible by intuition alone. Its knowledgeable and responsible use should at least give the design engineer more time for those problems which can be solved only by intuition, experience, and good judgment. Nomenclature A?;;;, = continuous annuity discount factor b = estimated rate of inflation, fraction/time = installed cost per unit capacity, S/units/time C’ = present value of installed cost and fixed cash expenses per unit capacity C $/units/time


= demand, units of product per time = ultimate demand level, units/time

DL = on-stream demand level, units/time DO &,, = present value depreciation factor

Loss, I


’ ’




g K


= continuous rate of growth, fractionjtime = total capacity, units/time = equipment capacity, units/time

L m

= duration of growth period, time = economic and depreciable life of equipment, time p = rate of technological advance, fraction/time I = cost of capital, fraction/time T = time of addition to capacity, time t = income tax rate, fraction a = scale-up factor p = fixed annual cash expenses, fraction of installed cost y = growth ratio e = time P V T C = present value of total costs, 8

5 literature Cited 7.6

8.4 436,000



6 0 561,000


(1) Bierman, H., Jr., Smidt, S., “The Capital Budgeting Decision,” MacMillan, New York, 1960. ( 2 ) Happel, J., “Chemical Process Economics,” Wiley, New York, 1958. (3) ~. Lawless. R. M.., Haas., P. R.., Jr.., Harvard Bur. Rev. 40. 97 (March 1962). (4) Saletan, D. I., Caselli, A. V., Chcm. Eng. Prog. 59, 69 (May 1963). ( 5 ) Vilbrandt, F. C., Dryden, C. E., “Chemical Engineering Plant Design,” (3rd ed.), McGraw-Hill, New York, 1959. (6) Williams, R., Jr., Cham. Eng. 54, 124 (December 1947). I

569,000 0