Electric utility expansion planning in the presence of existing capacity: A nondifferentiable, convex programming approach

Comput. Opns Res. Vol. 14, No. 1, pp. 19-31, 1987

Printed in Great Britain. All rights reserved

Copyright0

0305-0548/87 $3.00 + 0.00 1987 Pergamon Journals Ltd

ELECTRIC UTILITY EXPANSION PLANNING IN THE PRESENCE EXISTING CAPACITY: A NONDIFFERENTIABLE, CONVEX PROGRAMMING APPROACH F. H. Temple

University,

OF

MURPHY*

Philadelphia,

PA 19122, U.S.A.

S. SENT University of Arizona, Tucson, AZ 85721, U.S.A.

and A. L. SOYSTER$ College of Engineering, The Pennsylvania State University, University Park, PA 16802, U.S.A (Received August 1984; in revised form January 1986)

Abstract-This paper is about mathematical modeling ofcapacity expansion planning in the electric utility industry. In the absence of any existing capacity, an optimal expansion plan for a single period can be determined with an ordinary breakeven analysis applied to the load-duration curve. However, if some existing capacity already exists, which is the usual case, then the problem is considerably more complex. This paper shows how to formulate this problem as a convex program, one in which the load duration curve is directly incorporated into the.model as a monotone decreasing function. The algorithm is shown to be finitely convergent and several examples are provided which demonstrate the efficiency of the approach.

I. INTRODUCTION

This paper is about mathematical models of capacity expansion in the electric utility industry. The objective of these models is to specify a mix of equipment that minimizes the sum of the capital and operating costs. The notion of an “optimal” expansion plan used here is in a very specific context. The model and algorithm optimize the single period capacity expansion plan. There are algorithms for multiperiod models, e.g. Juseret [l] and Phillips et al. [2], however, a single period model is sufficient in many circumstances. Moreover, it is sometimes possible to decompose the multiperiod problem into a sequence of single period models, e.g. see Murphy et al. [3]. Rolling horizon models are commonly used because they allow one to add details, e.g. demand and price uncertainty, which would make a multiperiod problem too large. Also a simpler model allows one to expand, say, adding a financial submodel. Here one can address a larger set of issues associated with generation planning than just equipment selection. (See Sen et al. [4] for an example of how the generation expansion algorithm in this paper can be used as a repeatedly solved subroutine in addressing the effects of generation plans on the cost of capital.) Another example of where this type of model is appropriate is in energy system models that cover all energy sectors, where no one component can be too large. A version of the model described here is contained in the Intermediate Future Forecasting System, a new energy market model developed at the Energy Information Administration [5]. *Frederic H. Murphy is an Associate Professor in the School of Business of Temple University. He holds a B.A. in Mathematics and a Ph.D. in Administrative Sciences from Yale University. Dr Murphy’s research interests include energy policy analysis, electric utility planning and the computation of large-scale economic equilibria. He has published in such journals as Operations Research, Management Science, Mathematical Programming, the Energy Journal and AIIE Transactions.

tsuvrajeet Sen is an Assistant Professor in the SIE Department of the University of Arizona, Tucson. He holds a B.E. (M.E.) from Birla Institute of Technology and Science, Pilani (India), an M.E. (I.E.) from the University of Louisville and a Ph.D. in Industrial Engineering and Operations Research from VP1 and SU. He has published in journals such as Mathematical Programming, Operations Research and Management Science. His current research interests include nonconvex programming and nondifferentiable optimization with applications to large scale optimization. SAllen L. Soyster is Professor and Head of the Industrial and Management Systems Engineering Department at Penn State. He received his B.S. degree in Industrial Engineering from Penn State, his M.S. in I.E. from Cornell, and his Ph.D. in I.E. from Carnegie Mellon. His research interests are in mathematical programming and production planning with applications in the natural resources and energy. 19

20

F. H. MURPHY et al.

Using the single period formulation allows for the design of a simplified solution algorithm. The approach taken here fits into the class of feasible directions algorithms. The special structure of the problem drastically reduces the number of directions which must be considered. This results in computation times of just one or two seconds on the VAX 1l/780 for problems with as many as 15 alternative equipment types. The main elements of the problem are quite conveniently divided into two categories, demand and supply. The demand (kW) for electricity during a given time period, say a year, varies with time. This time varying demand is utilized in deriving a load-duration curve (or load curve) [6]. Such a curve is simply a reordering of the chronological occurrence of demand into one which is monotone nonincreasing. As such, the load curve gives the number of hours (horizontal axis) during which the demand either equals or exceeds some specified load level L (vertical axis). On account of the time varying demand, an electric utility typically chooses a supply-side mix of generating equipment that includes both capital and fuel intensive units. Roughly speaking, each equipment type is characterized by a unit (annualized) capital cost ($/kW) and a unit operating cost ($/kWh), say Ki and gi, respectively, for equipment type i. In the absence of any existing capacity, the determination of an optimal mix of equipment can be obtained with the use of ordinary breakeven chart methods [6]. The objective in this paper is to determine an optimal expansion plan for a significant generalization in which there is some existing capacity (bi 2 0) associated with each equipment type. Hence the unit capital cost for equipment i is zero if the amount used, xi, is such that xi6 b,; and the capital cost is Ki for units in excess of bi. Such a generalization makes the determination of optimal expansion plans considerably more difficult. The remainder of this paper is divided into four sections and two appendices. In the next section a mathematical program is formulated to determine the optimal capacity expansion plan when some or all equipment types have existing capacity. This formulation is somewhat unique in that the presence of existing capacities is incorporated via a nondifferentiable objective. Furthermore, no discretization of the load curve is necessary. Indeed, we will simply require the load curve to be montone decreasing and differentiable. Section II also provides some key results characterizing the mathematical program (a convex program) and some properties of optimal solutions. In Section III we derive the optimality conditions which are associated with certain directional derivatives of the nondifferentiable objective. The next section formalizes the algorithm and provides conditions under which the method is finite. (The proof of this result is given in Appendix B.) Section V illustrates the nature of the algorithm with a simple example. This section also summarizes some computational experience. In a subsequent paper we plan to present some comprehensive studies concerning the validation of the model presented in this paper. These studies, similar to Singleton et al. [7] will compare the model results obtained from simulating a previous historical period. II. MATHEMATICAL

MODEL

FORMULATION

The demand-side specification of the model is fully characterized by a load duration curve f(t). It is convenient to let the domain of this function be the unit interval [0, 11. The horizontal axis then becomes the capacity factor as opposed to time. On the supply-side, the electric utility can choose any mix of n different equipment types. Each equipment type is characterized by a triple (Ki, gi, bi) where Ki is the (annualized) unit capital cost, ($/kW), gi is the unit operating (fuel) cost ($/kW) and bi 2 0 is the amount of existing equipment of type i. The decision vector x = (x,, . . ., x,) represents the amount of each equipment type used to satisfy the demand. Hence, the capital cost associated with the use of Xi kW of equipment type i is defined by Ci(Xi) =

0

Xi< bi

Ki(xi - bi)

Xi >

bi.

(1) (2)

The only formal constraint imposed on the decision process is that (3)

Electric utility expansion planning

21

where P = f(0) is the peak load. The mathematical formulation of the problem will use a property commonly known as merit order dispatching [6]. This property states that for a given set of capacities, it is optimal to dispatch equipment in such a manner that as one traverses the load axis (vertical axis of the load curve), the fuel cost of equipment is nondecreasing. Thus, ordering the equipment in increasing fuel costs, (gl < g2 < . . .6 g,,), the mathematical program is min i

;’

Ci(Xi)+g,

i= 1

f-

l(z)

dz

s

s

x,+...+y,

+Si

:,+.

_il,_,

f-‘(z)dz

s

‘i,+...+Y,,

+

&

I,+..

+‘i,,_,

f-‘(z)dz

(4)

subject to

i= 1 Xi

3 0,

where f- ‘(* ) is the inverse load curve. We assume that f(s) is monotone decreasing and differentiable on (0,l). Throughout, we will also assume that Ki and gi are nonnegative for all i. The mathematical program described by (4) is of course a highly simplified characterization of the actual capacity expansion problem. Issues pertaining to restrictions on unit size, availability and downtime considerations on the supply-side, as well as demand-side issues such as seasonality and diurnal demand patterns have been suppressed to simplify the model analysis. However, considerations for unit availability and forced outages could be directly integrated into (4) by simply using “derated” capacities. It is important also to make an observation about the mathematical structure of (4). The objective function is not differentiable. Hence, certain solution algorithms which depend upon differentiability would not be appropriate. However, a differentiable formulation is possible if one is willing to add the “sparse” constraints of the form -zi + Xi6 bi, where zi is the amount of new equipment type i. Consider the following basic property of (4). Theorem 1

The mathematical

program described by (4) is convex.

Proof See Phillips et al. [S].

A model similar to the one presented here, has been solved through the explicit use of KuhnTucker conditions in Ref. [2]. The main observation in that paper is that the presence of existing capacities shifts the lines in a breakeven chart in such a manner as to reflect the marginal value of old plants. A trial-anderror method is then adopted for the solution of a nonlinear system of equations. The solution methodology described in our paper may be classified as a feasible directions algorithm and as such, it is somewhat more structured. To begin the analysis, we first provide an important characterization of optimal solutions to (4). Theorem 2

In any optimal solution to (4) there exists at most one equipment type i for which 0 < XT< bi. Furthermore, in this case xj* = 0 for j = i + 1, . , . , n.

F. H. MURPHY et al.

22

Proof

Suppose for i, < i, one has 0 < xc < b,, k = 1, 2 in some feasible solution to (4). Observe that slightly increasing the use of type i, and slightly decreasing the use of type i2 by the same amount keeps the current solution feasible. Such an interchange does not involve any change in capital cost, but since gi, < gi2, the fuel cost decreases and the interchanges would result in a feasible solution with less total cost. The second part of the theorem follows in a similar manner. If for some equipment type i, 0 < xi < bi, then slightly increasing the use of equipment type i is always less costly than using some equipment type j > i since gi < gj (which surely means gi < Kj + gj). A more subtle result about optimal solutions to (4) is given in the following corollary. Corollary 2.1

Let {XT}be an optimal solution to (4) and suppose equipment type Tis the last type loaded, i.e. loaded at the peak. If x: > bi; then bj = 0 for j = ?+ 1, . , ., n, i.e. there is no existing capacity for unit types numbered i+ 1, . . . , n. Proof

To ease the notation, rewritten as

define F(y) = JyOf- ’ (2) dz. Then the objective function of (4) may be

Now to prove the result, suppose x? > br but bj > 0 for some j > i Consider the cost of satisfying the load between f(0) - E and f(0) for some small E > 0. For the given solution this cost is K$ + g&F(P) - F(P - &)I.

(5)

On the other hand, if this part of the load was satisfied by typej equipment then no capital cost component is applicable and the total cost is gj[F(p)

-

F(P - &I].

(6)

The cost savings imparted by replacing E kW type rwith E kW of type j is thus K;& + (SF- gj)[F(p) - F(P -

Dividing by E> 0 yields

&)I.

(7)

13

F(P) - F(P - E)

K~+ (Cl;- Sj)

E

[

which for small enough E becomes arbitrarily close to (9)

Where D-{ } represents the left-hand derivative (which is assumed to exist). But since F(o) is a primitive of f-‘(s), then D-{F(P)} =f-‘(P) = 0 which implies (8), and thus (7), is positive for small enough E.Hence, x: > brand bj > 0, j > < cannot be optimal since a lower cost feasible solution exists and the corollary is proven. III.

OPTIMALITY

CONDITIONS

VIA DIRECTIONAL

DERIVATIVES

An interesting feature of program (4) is that a certain class of directional derivatives are easily determined. This class of directional derivatives will be used to determine improving directions for a direct search procedure to solve (4).

Electric utility expansion planning

23

Recall that the directional derivative of a function h: R” + R at a point x in the direction d, denoted D,P(x)l, is D,[h(%)] = lim

/I(: + cd) - h(x) E

e-0+

Define the direction d, E R” as the vector with + 1 in the ithcomponent, - 1 in thejth component and zeroes elsewhere. Neglecting nonnegativity, observe that if X is feasible for (4), then X + Edij is also feasible for all E> 0. The next lemma shows how a certain subset of vectors {dij} can be chosen to express any d E R” with the property e* d = 0. Lemma I

Let X be a feasible solution d= (d,, &, . . ., d,) and define

to (4) and suppose

for E> 0, X + Ed is also feasible. Let

and N- ={kNIdi
where N = {1,2,. . ., n). Then, one can choose iij > 0 so that di=

1

2,

ieN’

(loa)

je N-.

(lob)

jcN_

and -dj

= c

A,

IEN+

Proof

Since d is a feasible direction for (4) it follows that e-d = 0, i.e.

(11)

Now observe that equations (1Oa) and (lob) define an ordinary transportation problem with supply rows given by i E N + and demand rows given by j E N- . Furthermore, (11) states that the rim conditions balance, i.e. total supply equal total demand. Such a transportation problem always has a non-negative solution and, hence, the lemma is proven. Equations (1Oa) and (lob) also show how to write a vector d (with e-d = 0) as a nonnegative combination oft he {d,) . Since the ith component Ofd,, say d$), is 1 and the jt h component of d,, say d$f, is - 1, it follows from (1Oa) and (lob) that di = 1

&,d’,j,

ieN+

,sN-

and dj =

1 iijd$’ !GN+

jEN_.

F. H. MURPHY et al.

24

Furthermore,

since d$) = 0 for iE N+ and k # [, and d@’= 0 for jE N- and k Pj, it follows that d= 1

1

ieN+

I,d,.

(12)

jsN-

The interesting characteristic oft his representation of d in (12) is that if di > 0 then l,i = 0 for all k E N. This characterization is important in the following theorem. Also, for this next theorem we use the following notation: if Xi2 bi ifXi
-Ki 0

C’i=

if Xi> bi if Xi6 bi.

The usefulness of this notation follows from the fact that if C(x) = xipN ci(Xi)is the total capital cost of a solution x, then D,,j[C(X)] = C~+ C’_j. Theorem

3

Let the objective function of (4) be written as h(x) = C(x) + G(x), where C(x) denotes the capital cost and G(x) represents the operating cost. Let Xbe a feasible solution such that D,,,[h($] 2 Ofor all (i, j), i #j. Then X is an optimal solution for (4). Proof we focus individually on each of the two terms. Firstly,

Since D,[h(!i)] = DJC(x)] + D,[G($], consider D&C(X)] and note that

Dd[C(X)] = ,e:+ Cidi + 1 C’_j( -dj).

(13)

jsN_

Substituting for (di} in (13) from (1Oa) and (lob) yield

c

~&WI =

C:(j~_i,)+j~_C-j(iZ+li_i)

isN+

C

=

ieNC

C ~ij(C:+C'_j) jcN_

(*) Next we examine D,[G(x)]. Since G(x) is differentiable it follows that

(**I Combining (*) and (**) it follows that for any feasible direction d, and 2, satisfying (lOa), (lob) we have

= i;+ j;-

ii#‘,i,[h(x)l.

Hence, if for all pairs D,i,[h(X)] 2 0, it follows that D,[h($] theorem is proven.

2 0 for all feasible directions d, and the

Electric utility expansion planning

25

The general framework suggested for solving (4) is to start with some feasible solution, X,and move along some improving direction d,, i.e. a direction for which D,,j[h(X)] < 0. If an jl is found such that &,[&)I

2 0

for all (i, j), i #j

(14)

then, according to Theorem 3, an optimal solution has been found. However, it will be shown presently, that if D,Jh(X)] > 0 for only a “selected few pairs” of (i, j), then (14) will automatically follow. This simplification will enhance the efficiency of the algorithm. The following lemma provides a useful decomposition of D,,,[h(X)]. Lemma 2

Assume i -c k b,. Then

4i;jrwl = 4l&[W] + D,Jh(i,].

(15)

Proof

To simplify matters assume that X, > 0 for m = i +‘l, . . ., k - 1, i.e. some of each equipment type strictly between i and k is used in the current solution X. Let G,, I represent the breakpoint (capacity factor associated with equipment types i and i + 1). Then, it is easily seen that j-l

D,ij[h(X)] = d&i -

sjKj

1 bm-

+

9, + 1 h+ 1)

(16)

m=i

where ai equals zero or one depending upon whether Xi< bi or Xi> bi. Similarly, Sj equals zero or one depending upon whether Xj < bj or Xi > bj. Observe that the summation term in (16) represents the instantaneous change in the operational cost caused by substituting type i for type k capacity, On the other hand, k-l

(17)

and

Combining (17) and (18) clearly yields (16) and hence (15) follows. We now present an optimality condition that considerably reduces the number of directional derivatives that need to be evaluated. Theorem 4

. . . $ gn and let K, 2 0 for all j. Let X be a feasible solution such that for j < i, Xi > bj, LetO bj. For j # p, and j = 1,2, . . . , i+ 1, let j; satisfy the following:

&,pcw)l= 0

for j such that Xj = b,

(1%

for j such that 2, # b,.

(19b)

Then X is an optimal solution for problem (4). (For ease of presentation, we have assumed that b;, 1 # 0. However, if br+ i = 0, then in addition to (19) we require Ddr+,,Jh(X)] = 0.)

26

F. H. MURPHYet al.

Proof

For notational simplicity we will simply write D, for the directional derivative D,Jh(x)] and for any pair i, j, i 0, SOthat D, > 0. Similar reasoning also shows that Dji 3 0. (b) p < i
(20)

But from (19a), Di, 3 0, which implies that Ki 2 Kp + Spi.

(21)

Substituting (21) in (20) we have D, 2 0. Next suppose that dj = 0. If Xj < bj, then the same argument as above holds. Hence let Xj = bj. We now have to show that Ki + S, 2 0. From (19b), we have DP,i+ 1 = K, + S,i + S, + Sj,;+ 1 = 0.

(22)

By merit ordering Sj,;+ r < 0, and further Dip 2 0 implies (21). Hence it follows that Ki + S, > 0. In an analogous fashion, one can show that Dji~ 0. (c) i
From (19a), D, 3 0, or that K i - Kp + Sij + Sjp 2 0.

D, = Ki + S, ~ K, - Sj, > 0

(23)

where the last inequality is due to the assumption that K, > 0 and merit ordering implies S, 6 0. The reverse case Dji3 0 also follows analogously. Having considered all the above cases, the proof is complete. IV.

THE

ALGORITHM

AND

ITS

FINITENESS

By virtue of Theorem 4, at any iteration we are able to check optimality by evaluating O(n) directional derivatives instead of O(n’). Furthermore, if we find an index that violates (19), an improving direction is automatically at hand. There are several other implications that we can draw from the results of the previous sections. From Theorem 2 we know that the minimum capacity level of each plant type used in the solution is the existing capacity unless it is the last plant type used in the merit order. This property will be maintained throughout the algorithm. Next consider the case where there are two adjacent plant types in the merit order operating above existing capacity. From Lemma 2, the capacity factor at the boundary between the capacity types can be determined from the breakeven chart without existing capacities, since we are finding the capacity factor where the capital plus operating costs for the planttypes are equal.

Electric utility expansion

planning

27

The solution algorithm essentially positions groups of plant types that are at existing capacity levels between plant types above existing capacity. Groups may split or join, depending on the steps of the solution process. Next, we give a formal statement of the algorithm. Some further discussion is provided in Appendix A. The algorithm

(1) Initialize. Set t = 1 and find a feasible solution denoted x’. (One effective scheme, which provides a solution that satisfies the necessary conditions of Theorem 2 is described in Ref.

Pm

(2) Optimality Test. Given x’, determine the smallest indexj such that xi > bj. Denote this index by pr. (Here pt plays the role of p in Theorem 4.) Determine whether x’ satisfies (19). If so, x’ is

optimal and stop. Otherwise go to 3. Iterative step. Find the smallest index j, that violates (19). Let rt denote an index such that (3) ifj, < ptr then r, = pt and ifjj > pt, then r1 is the largest j bj. Obtain x’+ ’ so as to satisfy the necessary conditions of Theorem 2 and (24).

4,,,Mx’+ ‘,I 3 0

(244

Dd,,,,[h(x’+ ‘,] > 0.

(24b)

With respect to the load curve, $+I is obtained by translating a group of plant types either up or down on the load curve. The determination of x1+’ 1s . carried out according to various cases that arise, and these steps are given in Appendix A. (4) Having obtained x’+ ‘, increment t by 1 and go to 2. While theoretically the exact line search in the above algorithm will help us prove the finiteness of the procedure, in practice we conduct only a quadratic approximation. These details are not provided in this paper but are available in Ref. [8]. It is worth noting however that for the numerous test problems and the working version used daily by the Department of Energy, the quadratic approximation has not posed any difficulties with respect to convergence. Theorem 5

Let Kj > 0 for all j = 1,2, . . ., n and let 0 < gi < . . . < gn. Further let f(s) be continuous on (0,l). Then the algorithm presented above is finitely convergent. Proof (See Appendix B.) V. AN

EXAMPLE

PROBLEM

AND

COMPUTATIONAL

RESULTS

In this section we illustrate the use of the algorithm via an example problem and report limited computational experience with the procedure. Consider the equipment data given in Table 1 and let the load curve be the following 5th degree polynomial. f(t) = 100 - 235.16t + 1112.39t’ - 2665.93t3 + 2771.33t4 - 1043.74t2.

The above load curve is quite representative of actual data. Observe that f(1) = 38.88 # 0. This is

F. H. MURPHY et al.

28

Table 1.

Equipment

data for example

Equipment type

Annualized capital cost

Fuel cost (WW

1 2 3 4 5

3.4 3.3 4.0 2.8 1.0

0.60 0.73 0.95 3.30 16.50

Existing capacity WV) 10 10 10 10 10

Table 2

(24

(2c)

(2b)

Equipment

x’

cI1

*+’

x’f’

1 2 3 4 5

53.05 26.95 10.00 IO.00 0.00

1.00 0.77 0.21 0.06 _

53.05 19.09 10.00 17.86 0.00

1.00 0.77 0.43 0.15 _

Objective value (?.a--Solution

$245.50

X*

(L*

53.05 12.05 10.00 10.00 14.90

1.00 0.77 0.54 0.37 0.10

8216.50

$243.77

at iteration t; (2b)--Solution

at iteration

t + 1; (2c)-Optimal

solution.

termed as the base load and must be served “round-theclock.” Now, the analysis presented earlier requires the load curve to be differentiable at every Ui, In the presence of a base load however, we approximate the above curve by a load curve f such that f(1) = 0 and _? is differentiable almost everywhere on the interval. In order to illustrate the algorithmic step, suppose that the feasible solution x’ in Table 2(a) is given. Let us conduct the optimality test. Note that for this solution, pt = 1. Next, we compute the directional derivatives to test whether (19) holds. We have D,, = K, - K, - (gl -g&

= (3.3) - (3.4) - (0.6 - 0.73) x 0.77 2: 0.

D,, = K, - K, - (gl - g&t2 - (g2 - g3)t13 = 0.7463 > 0. D,, = D,, = K, - K, - (g2 - g&t, - (g3 - g&t4 = - 0.3128 < 0.

Thus, we have j, = 4 and I, = 2. Hence we will increase xq and decrease x2. We now compute dt’i in such a manner as to satisfy (24). Utilizing the procedure of Appendix A we have x’+ ’ as given in Table 2(b). This completes one iteration of the algorithm. Continuing in this fashion the optimal solution shown in Table 2(c) is obtained. The above algorithm was coded in FORTRAN and has been implemented as an electric utility submodel in the IFFS system at the Department of Energy (DOE). In the actual implementation of the model, the static problem (4) is utilized sequentially as discussed earlier. While computational

Table 3. Computational

results

Problem

No. of equipment types

CPU s on VAX

2 3 4 5 6 7 8 9 10 11 12 13 14 15

6 6 6 8 8 8 10 10 10 12 12 12 15 15 15

0.410 1.09 0.75 0.75 0.80 0.949 1.27 1.410 0.93 1.383 1.352 1.301 0.789 2.961 0.973

29

Electric utility expansion planning

problem outside of the DOE market model (IFFS) are limited, our preliminary are quite encouraging. We summarize, in Table 3, the CPU times on a VAX 1l/780 system (VMS Operating System, version 3.0). The problems are “semi” real cases in that most of the data are obtained from a few DOE regions. The main observation that we would like to make in Table 3 is that execution times for problems with 15 plant types are not very different from those for smaller problems. Further experimentation with the algorithm is currently underway. The algorithm is used almost daily at DOE and it accounts for a very small portion of the IFFS solution time.

results

for this static

results

REFERENCES 1. R. Juseret, Long term optimization of electrical system generation by convex programming. Mathematical Programming Study 9. North-Holland, Amsterdam (1978). 2. D. Phillips, F. P. Jenkin, J. A. T. Pritchard and K. Rybicki, A mathematical model for determining generating plant mix. Third Power Systems Computation Congress, Rome, Italy (1969). 3. F. H. Murphy, S. Saraf and A. L. Soyster, The replication of multiyear solutions using single period models of electric utility capacity expansion planning. IIE Trans. In press. 4. S. Sen, S. K. Saraf, A. L. Soyster and F. H. Murphy, The capital supply curve in capacity expansion models: some economic and algorithmic aspects. Nav. Res. Logist. Q. 31, 199-212 (1984). 5. F. H. Murphy, Design strategies for energy market models. Energy Models and Studies (Edited by Ben Lev). NorthHolland, Amsterdam (1983). 6. R. Turvey, Optimal Pricing and Investment in EIectriciry Supply. MIT Press, Cambridge, Mass. (1968). I. F. D. Singleton, M. Sankar and R. G. Thompson, Verification/validation of an economic process model for electric power generation in the United States. Comput. Opns Res. 8, 311-340 (1981). 8. F. H. Murphy, A. L. Soyster and S. Sen, A convex programming approach for determining capacity expansion plans in the electric utility industry. Working Paper 81-110, Department of Industrial and Management Systems Engineering, The Pennsylvania State University, University Park, Pa.

APPENDIX

A

In this Appendix we illustrate the procedure to compute xl+‘. The basic idea is to translate the loading of a group of equipment types up or down the load duration curve. Since the number of cases which can arise is quite large, a formal statement of the procedure is cumbersome and requires much detail. As such, we present a small example that illustrates the basic idea. (Th: actual flowchart to handle all possible cases is available from the authors.) Consider x’ as shown in Fig. Al(a). The dotted equipment type denotes a plant whose existing capacity has not been fully utilized in a given solution, cross hatched equipment denote plant types that have not been expanded but all existing capacity is utilized, and the white regions denote plant types that have been expanded in the given solution. Thus, in Fig. Al(a), plant types 1 and 2 have been expanded while plant type 3 has not utilized all the available existing capacity. For the sake of illustration, let the current switch decrease xi and increase x& Then, X; will be first decreased to the point where either Ya = b,

t

(a)

t

(b)

kW

kW

1 Time

Time

kW

kW

Fig. Al. Illustration of step length computations.

30

F. H. MURPHYet al.

or xi = bz. In this illustration, suppose that the former occurs. This is shown in Fig. Al(b). At this solution, say z’ , one checks whether D,,(z’) i 0. Suppose (for the sake of illustration) that D,,(z’) < 0. Then, one decreases x2 so that x2 = b,, and xs is adjusted accordingly. This solution is shown in Fig. Al(c) and denote this point by z2. Next one evaluates D,,(z*). Suppose that it so happens that D,,(z2) 2 0. Then, one evaluates D,,(z2). Suppose that D,‘(z’) < 0. Hence one decreases x, and increases xs accordingly. Next, decrease x, to the point at which x’ = b,, Denote this point by z3. One now evaluates D’ 3(z3). Suppose that D,,(z3) < 0. Hence, there is a convex combination of z2 and z3 such that D13 = 0. One finds this point by searching for I E (0,l). In this example, such a solution is shown in Fig. Al(d). The point thus obtained is x1+‘. The essence of all the other cases in the procedure is similar. APPENDIX

B

The proof of Theorem 5 is structured as follows. We show the lexicographic increases of a complexity vector C’ = (c’,, whose components are defined (in iteration t) as follows:

c; =

0

ifj = p,

0

if xi = bj and Ddjr[h(x’)] 2 0

0

if xi # bj and D,,,,[h(x’)] = 0

-1 Because of the nondifferentiability

, cf).

otherwise.

of the objective function several subcases have to be considered. They are as follows:

I. j, < p,. Here we increase xj, and p,+, = j,. II. j, > p,. that is increase xj, (a) xi,+’ > x;, the capacity of plant type r, is reduced by the increase in xj, but remains above existing capacity (1) $+’ >b, (2) x’,:‘=b is reduced to existing capacity by the increase in xj, decrease xj,. which does not improve the complexity vector immediately, but improvement is (b) S,+ ’ < x:; guaranteed within a finite number of steps. For notational simplicity we denote D,,[h(x’)] by D:, and for any pair i < j, S:, = ILL: increases lexicographically in a finite number of steps.

(g,,, - gm+ ,)c&,+, We show that C’

Proof (Theorem 5) First let us consider jj < p,. Then r’ = p, and 4z = bj,. Hence if (19) is violated for j = j,, then Dj,P,< 0, or c,!,= - 1. Next note that p,+, =J, and therefore cj,‘+’ = 0 Thus if we show that ci+’ 2 c) forj i’. First observe that (25) and, by (24) D’+.‘=K P,It

0.6)

PI

Hence combining (25) and (26) we have K.-K. J

JI

+S’.

JP,

-S!+‘>O

’

IfPI

or that Kj - K, + S:,, + S;,, - S:,;,’ a 0.

(27)

Note that (S>,, - S$‘) < 0 (since 9, < gm+, and rk > a:’ for m = j, + 1, . , p,). Hence (27) implies that D>i ’ = K, - K, + S;i ’ 2 0. Hence for j and the lexicographic increase for this case is proven. We next consider the case j, > pt. Here we will consider two subcases: ,$,+’ > x’. and x!+’ < x>,. Let us first consider the subcase .$’ ’ > x;!.In this case, either 4,” > b,, or xl,’ 1Lb,. Let :‘s examine these further subcases independently. Considering the former (i.e. .x:,” > b,,), it is clear that p,+ 1= p, and for j < r,, it is clear that c:+’ = c:. Let us therefore considerj such that r,
Y,

,I,

+S’

.-S’+.‘.

r,,

111

(28)

Butgm 0 and combining this with the fact that 4, > 0 [sincej satisfies (19a)J, we conclude from (28) that DjP, ‘+ ’ 2 0 for r, < j - 1 = c;,. To complete this subcase, we have to still consider the s’tuation in which xi,+’ = b,,. In this (sub-sub) case if r, > p, then the same arguments as the previous subcase apply and hence we will consider the situation in which r, = p,. Since this (sub-sub) case considers 2,: ’ = b,,, r, = p, implies that p ,+ , = j, and hence the increase of the vector C’ has to be studied by examining

31

Electric utility expansion planning D;,:‘.j
For 1
1, we have D’.t’=K -K +S’ +S+’ 111 =(I;:-~,+~:~)+~~,,-xj,+s~~~) = D;, + D;,;,’ .

(29)

‘+‘>O,l
(30)

D;, = K, - K,, - S;,, 3 0.

(31)

- S;, j) 0 G Kj - K, + S;i: ’ + (S::,’ - S;, j) = 0;; ‘ + (S;,; .'

(32)

Also, our choice of j, ensures

(30) and (31) imply that

By noting that (St:’ - S:,j) < 0, (32) implies that 0;;’ > 0, r’ + 1 i j < j,- 1. Hence for all j < j,, ci” = c: and as before, c;,’ ’ = 0 > - 1 = cl,. This completes the subcase x’j,+’ > xi,. We finally consider the last subcase that 4,’ ’ < X,,.Unlike the previous sub-cases, thecomplexity vector does not necessarily increase at the end of one iteration. However, we are able to show that in a finite number of iterations, the vector does increase lexicographically. We first observe that in this case, pt+ 1 = pt. Now forj = 1,2, . , r, there is no change in capacity factors and hence for such j, c:’ ’ = ci. Next let us examine j = r, + 1, . . ., j, - 1, (if any). For these j, we have D’.+‘=D! JPt

IP,

+(S’ .-S:;‘). ‘rl

(33)

I

In (33), Dip, 2 0, however (S:,j - S$‘) < 0. Thus in the process of realizing (24) it is possible to violate (19) for some j in r, + 1
. . . . j,}/{T,-q,}

andc>,+‘=O>

-l=ci,

Now q, may be dehed from 17;and the resulting set is called ‘I;+, If T+, = 4, then clearly we are done. Otherwise our rule for thechoiceofj, will choosej,+z as thesmallest index in ‘I+ i, and once again we are in the previous subcase. Thus with respect to C’+‘, the complexity vector must increase at the end of iteration t + 2. Continuing in this fashion, after n, steps, C’+“~*’ is lexicographically greater than C’. This completes our consideration of all the cases and hence the proof.